instructor mohammed fazle baki

73-331

OPERATIONS MANAGEMENT I

INSTRUCTOR MOHAMMED FAZLE BAKI

ODETTE SCHOOL OF BUSINESS UNIVERSITY OF WINDSOR

FALL 2004

73-331 Operations Management I Fall 2004 Term

Table of Content

Chapter Lesson Topic Slide No.

Course Outline pp. 1-4

1 1 Introduction 1-4

1 2 Learning Curve 1-26

1 3 Capacity 27-47

2 4 Forecasting 28-71

2 5 Stationary Series 72-95

2 6 Trend 96-113

2 7 Trend with Seasonality 114-149

3 8 Aggregate Planning 150-178

3 9 Chase and Level Strategies 179-208

1 10a Lab 1a: Use of Excel for Learning Curve pp. 1-4

3 10b Lab 1b: Use of Excel for Aggregate Planning pp. 1-7

4 11 Deterministic Inventory – Economic Order Quantity Model 1-32

4 12 Finite Production Rate Model 33-47

4 13 Quantity Discount Models 48-66

4 14 Resource Constrained Models 67-92

4 15 EOQ Models for Production Planning 93-114

5 16 Stochastic Inventory – Single Period Models 115-151

5 17 Q, R Policy 152-175

5 18 Q, R Policy Optimization Without Service Constraints 176-185

5 19 Q, R Policy Optimization With Service Constraints 186-205

5 20 Lab 2: Use of Excel for Q, R Policy Optimization pp. 1-4

7 21 Material Requirements planning 1-40

7 22 Lot Sizing 41-80

7 23 Capacity Constraints 81-107

7 24 Just-In-Time (not included, available from website) ---

1

UNIVERSITY OF WINDSOR ODETTE SCHOOL OF BUSINESS

73-331 Operations Management I Fall 2004 Term

___________________________________________________________________________________ Instructor: Dr. M.F. Baki Office: 410 Odette Building Office Hours: T 1:30 pm-4:30 pm, TR 7:00 pm-8:00 pm, W 10:00 pm-1:00 pm or by appointment Extension: 3118 Email: [email protected] (preferred mode of communication) ___________________________________________________________________________________ Text: Production and Operations Analysis (5th ed.) by Steven Nahmias Available from the Bookstore Lecture Notes: Available from the Document Imaging Centre CHT 01 / course website Bring lecture notes to every class Lectures: Section 1: TR 10:00 am – 11:20 am Odette B04

Section 2: TR 11:30 am – 12:50 pm Odette B04 Computer Lab: Section 1: Oct 19 / Nov 23 10:00 am – 11:20 am OB 210

Section 2: Oct 19 / Nov 23 11:30 am – 12:50 am OB 210 Course Website: Follow the links from the University of Windsor website www.uwindsor.ca à Class

Notes à Business Administration. Visit the course website at least twice a week for lecture notes, Excel files, assignments, solutions, computer lab documents, midterm and exam information, and other announcements and information.

Course Objectives and Learning Outcomes:

After completing this course, the students will be able to • explain the role of operations management in manufacturing and service

organizations • identify the connections between operations management and the other parts

of the organization • recognize problems such as capacity planning, learning curves, forecasting,

aggregate planning, inventory control and material requirements planning • apply solution procedures developed in the area of operations management

and • use computer software (i.e., Excel) to solve operations management problems

Main Course Themes:

Two courses Operations Management I and II cover some important functions of operations management. Part I gives an overview of various functions of operations management and discusses

• Capacity Growth Planning • Learning and Experience Curves • Forecasting • Aggregate Production Planning • Inventory Control - Deterministic Demand • Inventory Control – Stochastic Demand • Material Requirements Planning

Part II covers operations and project Scheduling, supply chain management, facility layout and location, quality control and reliability of systems

2

Assignments: There are 5 equally weighted individual assignments. For each assignment, the lessons covered and resources available to complete the assignments are summarized in the Assignments page of the course website. Each assignment is due in the beginning of the class in which it is due. Late assignments may not be accepted. Submit every assignment in the section in which you are registered. You may discuss the problems with others, but must hand in your own work. Every assignment must be stapled with the cover sheet that includes name of the student, name of the course, section number and student ID number.

Midterm I: Thursday, Oct 14, class time, room TBA Midterm II: Thursday, Nov 25, class time, room TBA Midterm I & II Aides: Calculator, straightedge and a one-sided one-page note Final Exam: Wednesday, December 15, 12:00 pm – 3:00 pm. The final exam is comprehensive. Final Exam Aides: Calculator, straightedge and 3 one-sided pages of note Marking Scheme: Assignments 20% Midterm I 20%

Midterm II 20% Final 40%

Numeric/Letter Grade Conversion: A+ A A- B+ B B- C+ C C- D+ D D- F F- 93 86 80 77 73 70 67 63 60 57 53 50 35 0

Note: The above conversion scale is the one adopted by the University of Windsor Senate. The Odette School of Business prescribes a class average of C/C+/B-/B in all its 3rd year courses. The total marks may be adjusted, if required, to maintain the prescribed class average.

Missed Exam/Late Assignments:

It is the student’s responsibility to inform the instructor as soon as possible of the circumstances regarding a missed exam or late assignment. Unless the reason is obvious, such as a broken arm, documentation such as a medical note (or death certificate in the case of bereavement) will be required. Otherwise, there will be a penalty assessed and no accommodation made. Note that the instructor may reject any medical note that does not clearly state that you were ill and unable to attend the exam and that you were seen by the physician at the time of the illness. A makeup exam is administered by the Dean’s office at the end of the term after the final exam. The student is to complete a request for make up exam form, attach a fee of $30, and submit to the receptionist in the Dean's suite. The approval will be done by one of the Associate Deans.

Plagiarism: Cheating on exams, plagiarism, and unauthorized collaboration wi th colleagues on assignments will not be tolerated. The penalty for cheating is zero for the exam or assignment. In addition, you may be charged with student misconduct and subject to additional penalties depending on the outcome of judicial proceedings.

Informal and Formal Grade Announcement:

Informal final grades will be posted on the course website when they are available. Students should be aware that the informal grades are not final and subject to change until they are approved by the Grades Review Committee and signed by the Dean. For this class the recommended class average is C+/B-/B. Grades may be adjusted to comply with that recommended average. Formal final grades will be available on the student SIS system when approved.

3

73-331 Operations Management I Fall 2004 Term Tentative Lecture Schedule, Reading, Exercises and Assignment:

Week, Lesson and

Dates of Lectures

Reading Section and

Page Numbers (5th Edition text)

Suggested Exercises (4th/5th Edition)

Topic Exam Dates and Announcement

and Due Dates of Assignments

1. Sep 9 None None Introduction Strategy and Competition

2. Sep 14 3. Sep 16

1.10, 29-36 1.11, 36-45

29, 30, 33 34, 37, 38

Learning Curve Capacity

4. Sep 21 5. Sep 23

2.1-2.6, 51-63 2.7, 63-73

12, 13, 15, pp. 62-63 17, 18, 24

Forecasting Stationary Series

Assignment 1 Sep 14 / Sep 23

6. Sep 28 7. Sep 30

2.8, 74-77 2.9, 78-83

28, 30 33, 34

Trend Trend with Seasonality

8. Oct 5 9. Oct 7

3.1-3.3, 108-117 3.4, 117-125

8 9, 13, 14

Aggregate Planning Chase and Level Strategies

Assignment 2 Sep 28 / Oct 7

Oct 12

TBA

Midterm I: Thu, Oct 14, class time, room TBA

10. Oct 19 (computer lab Room OB 210) 11. Oct 21

See website 3.5-3.6, 125-135 4.1-4.5, 183-200

17, 19, 20 10, 12

Use of Excel for Learning Curve and Aggregate Planning Deterministic Inventory - EOQ

12. Oct 26 13. Oct 28

4.6, 202-204 4.7, 205-208

17, 19 22, 24

Finite Production Rate Quantity Discount

14. Nov 2 15. Nov 4

4.8, 212-215 4.9, 215-220

26, 28 29, 30

Resource Constrained Models EOQ Models for Production Planning

Assignment 3 Oct 26 / Nov 4

16. Nov 9 17. Nov 11

5.1-5.3, 232-245 5.4, 250-254

8, 12a, 12b 13b, p. 261

Stochastic Inventory – Single Period Models Q, R Policy

18. Nov 16 19. Nov 18

5.4, 253-255 5.5, 255-262 5.7, 265-270 (skim)

13a, p. 261 16, 17 None

Q, R Policy without Service Q, R Policy with Service Multi-Product Systems

Assignment 4 Nov 9 / Nov 18

20. Nov 23 (computer lab Room OB 210)

See website None Use of Excel for Q,R policy Optimization

Midterm II: Thu, Nov 25, class time, room TBA

21. Nov 30 22. Dec 2

7.1, 346-358 7.2-7.3, 358-366

4, 9 17, 25

Material Requirements Planning Lot-sizing

Assignment 5 Nov 23 / Dec 2

23/24. Dec 7

7.4, 366-369 7.6, 377-384

28 37

Capacity Constraints JIT

Notes: 1. Lesson 1 is not included in any exam or assignment. 2. Lessons 10 & 20 (computer labs) are not included in any exam, but covered by assignments 3 & 5. 3. Download Excel files used in class from the lecture notes page of the course website. 4. Before each lab, browse lab handouts and download Excel file from the computer lab page of the course

website.

4

73-331 Operations Management I Fall 2004 Term Tentative Lecture Schedule, Reading, Exercises and Assignment:

Week, Lesson and

Dates of Lectures

Reading Section and

Page Numbers (4th Edition text)

Suggested Exercises (4th/5th Edition)

Topic Exam Dates and Announcement

and Due Dates of Assignments

1. Sep 9 None None Introduction Strategy and Competition

2. Sep 14 3. Sep 16

1.10, 32-38 1.11, 38-48

29, 30, 33 34, 37, 38

Learning Curve Capacity

4. Sep 21 5. Sep 23

2.1-2.6, 55-65 2.7, 66-77

12, 13, 15, pp. 65-66 17, 18, 24

Forecasting Stationary Series

Assignment 1 Sep 14 / Sep 23

6. Sep 28 7. Sep 30

2.8, 77-81 2.9, 81-87

28, 30 33, 34

Trend Trend with Seasonality

8. Oct 5 9. Oct 7

3.1-3.3, 113-120 3.4, 121-127

8 9, 13, 14

Aggregate Planning Chase and Level Strategies

Assignment 2 Sep 28 / Oct 7

Oct 12

TBA

Midterm I: Thu, Oct 14, class time, room TBA

10. Oct 19 (computer lab Room OB 210) 11. Oct 21

See website 3.5-3.6, 129-139 4.1-4.5, 194-209

17, 19, 20 10, 12

Use of Excel for Learning Curve and Aggregate Planning Deterministic Inventory - EOQ

12. Oct 26 13. Oct 28

4.6, 211-213 4.7, 214-217

17, 19 22, 24

Finite Production Rate Quantity Discount

14. Nov 2 15. Nov 4

4.8, 221-225 4.9, 226-231

26, 28 29, 30

Resource Constrained Models EOQ Models for Production Planning

Assignment 3 Oct 26 / Nov 4

16. Nov 9 17. Nov 11

5.1-5.3, 243-254 5.4, 259-262

8, 12a, 12b 13b, p. 271

Stochastic Inventory – Single Period Models Q, R Policy

18. Nov 16 19. Nov 18

5.4, 262-264 5.5, 264-271 5.7, 275-280 (skim)

13a, p. 271 16, 17 None

Q, R Policy without Service Q, R Policy with Service Multi-Product Systems

Assignment 4 Nov 9 / Nov 18

20. Nov 23 (computer lab Room OB 210)

See website None Use of Excel for Q,R policy Optimization

Midterm II: Thu, Nov 25, class time, room TBA

21. Nov 30 22. Dec 2

7.1, 355-366 7.2-7.3, 366-375

4, 9 17, 25

Material Requirements Planning Lot-sizing

Assignment 5 Nov 23 / Dec 2

23/24. Dec 7 7.4, 375-380 7.6, 387-395

28 37

Capacity Constraints JIT

Notes: 5. Lesson 1 is not included in any exam or assignment. 6. Lessons 10 & 20 (computer labs) are not included in any exam, but covered by assignments 3 & 5. 7. Download Excel files used in class from the lecture notes page of the course website. 8. Before each lab, browse lab handouts and download Excel file from the computer lab page of the course

website.

1

1

73-331 OPERATIONS MANAGEMENT I

Content

• Forecasting

• Aggregate planning

• Inventory control

2

• Text (Nahmias, 5th edition)

• Website: lecture notes, assignments, solutions, computer lab documents, midterm and exam information, and other announcements and information

• Lecture Notes (purchase or download)• Computer Lab

• Background material and Software: Excel, linear programming, probability distributions (mostly normal distribution) and calculus.

• Grading– 5 Assignments – 2 Midterms– Comprehensive Final

COURSE INTRODUCTION

2

3

A WORD ABOUT CONDUCT

• Basic principles

1. Every student has the right to learn as well as responsibility not to deprive others of their right to learn.

2. Every student is accountable for his or her own actions.

4

In order for you to get the most out of this class, please consider the following:

a. Attend all scheduled classes and arrive on time. Late arrivals and early departures are very disruptive and violate the first basic principle.

b. Please do not schedule other activities during this class time. I will try to make class as interesting and informative as possible, but I can’t learn the material for you.

c. Please let me know immediately if you have a problem that is preventing you from performing satisfactorily in this class.

1

1

LESSON 2: LEARNING AND EXPERIENCE CURVES

Outline

• Rate of Learning • Learning Curve • Estimating Parameter Values

2

Rate of Learning

• As workers gain more experience with the requirements of a particular process, or as the process is improved with time, the number of hours required to produce an additional unit declines.

• The learning curve models this relationship.• Rate of learning, is defined as follows:

1002 ×=

item th- produce to required Timeitem th- produce to required Time

uu

L

L

2

3

Rate of Learning

• Let– Y(u) = time required for the u-th unit

• Then, from the definition of rate of learning,

( )( ) 1002 ×=uYuY

L

4

Rate of Learning

• For example, using rate of learning,

• Using rate of learning,

• Using rate of learning,

,1=u

10012

×=item produce to required Timeitem produce to required Time

st

nd

L

,2=u

10024


nd

th

L

1001020


th

th

L

,10=u

3

5

Rate of Learning

• Rewriting the definition of rate of learning,

• Hence,

• If we know the time required for the first unit, Y(1), we can find the time required for the 2nd, 4th, 8th, ….. units using the above equation iteratively.

( )( ) 1002

×=uYuY

L

( ) ( )uYLuY100

2 =

6

Rate of Learning

• The time required for the 1st unit = Y(1) [notation]• The time required for the 2nd unit,

• The time required for the 4 th unit,

• The time required for the 8 th unit,

( ) ( )1100

2 YLY =

( ) ( )2100

4 YLY =

( ) ( )4100

8 YL

Y =

4

7

Rate of Learning

• An example of 80% learning rate: Suppose that it requires 100 hours to produce the first unit. Then, the 2nd unit requires (0.80)(100)=80 hours. The 4th units requires (0.80)(80)=64 hours, and so on.Unit Number of Hours Required1st unit 100 hours2nd unit (0.80)100=80 hours4th unit (0.80)(80)=64 hours8th unit (0.80)(64)=51.2 hours

8

• In general, for any unit u, not necessarily 1, 2, 4, 8, …, the time required can be obtained from the learning curve equation.

• The learning curve is of the formY(u) = au-b

Where, a and b are parameters.a = time required for the first unitb = - ln (L) / ln (2), where L is the

rate of learning, 0.80 for 80% learning, 0.90 for 90% learning, etc.

Learning Curve

Here, ln = natural log. A review on logarithms follows.

Pro

cess

ing

time

per

unit,

Y(u

)

Units produced, u

5

9

Learning Curve

Suppose that

– a = 18 hours

– Learning rate = 80%

– What is time for the 9th unit?

Y(u) = au-b

= auln(L)/ln(2)

Y(9) =

Here, ln = natural log. A review on logarithms follows.

10

Logarithms (Review)

• Recall that if then, . • Here, p is the base. • If the base is e, “ln” (natural log) replaces “log”. So,

• Here, e is a constant:

qp x = qx plog=

qq elogln =

....71828.2

...241

61

21

11

11

...!4

1!3

1!2

1!11

!01

11

=

+++++=

+++++=

+

∞→=

e

nnLim

en

6

11

Logarithms (Review)

• A scientific calculator usually contains 2 buttons:• log x provides log10 x , logarithmic value of some

number x with base 10• ln x provides loge x , logarithmic value of some

number x with base e =2.71828…• To get a logarithmic value with a base other than 10 or e,

use the following formula:

pq

pq

q

pq

pq

q

e

ep

p

lnln

loglog

log

loglog

loglog

log10

10

==

==

or,

12

• Recall, that learning curve is of the form

Y(u) = au-b

Where, a and b are parameters.• If we observe the time required to

produce various units, we can estimate parameters a and b along with the rate of learning L.

• The relationship between u and Y(u), as shown on the left, is not linear. But, the relationship between ln(u) and ln(Y(u)) is linear.

Pro

cess

ing

time

per

unit,

Y(u

)

Units produced, u

Estimating Parameter Values

7

13


Y(u) = au-b (Learning Curve)or, ln(Y(u)) = ln(au-b) (Take logarithm on

both sides) or, ln(Y(u)) = ln(a)+ln(u-b)or, ln(Y(u)) = ln(a) - bln(u)

This equation has the form of a straight line y = c + mx (straight line, with slope m and intercept c)

Thus, a plot of ln(u) vs ln(Y(u)) fits a straight line

14


ln(Y(u)) = ln(a) - bln(u) (Learning Curve)y = c +mx (straight line)

Notice thatIntercept = ln(a) Hence, a = e intercept

Slope = - b Hence, b = -slope

Finally, Since, b = - ln (L) / ln (2), we have L = eslope*ln(2)

8

15

• It’s an important fact that the relationship between ln(u) and ln(Y(u)) is linear. Because if the relationship between two variables is linear, we can fit a straight line that provides the relationship.

• The slope and intercept of the straight line are obtained by using linear regression on ln(u) and ln(Y(u)).

• The slope and intercept can then be used to get paratmeters a and band rate of learning L.


ln(Y

(u))

ln(u)

16

• We estimate parameters as follows:• Step 1: Given a set of u and Y(u)

values, compute the set of ln(u) and ln(Y(u)) values.

• Step 2: Using linear regression on ln(u) and ln(Y(u)), compute slope, mand intercept, c of the straight line that best fits the set of ln(u) and ln(Y(u)) values.

ln(Y

(u))

ln(u)


c

1m

• Step 3: Compute a, b and L using the following formula: – a = e intercept = ec

– b = -slope = -m– L = eslope*ln(2) = em*ln(2)

9

17

• An interpretation of the intercept, c:– ec is an estimate of the time

required for the first unit denoted by a or Y (1).

• An interpretation of the slope, m:– em*ln(2) is an estimate of the

rate of learning, L.– Learning is demonstrated by

the negative slope. – If the slope is less, then the

line is steeper, L is less and the learning is faster.

ln(Y

(u))

ln(u)


c

1m

18

Estimating Parameter Values: Example

Consider the text example:

Cumulative Number of Number of Hours RequiredUnits Produced For the Next Unit

u Y (u )10 9.2225 4.85

100 3.8250 2.44500 1.71000 1.035000 0.6

10000 0.5

10

19

Relationship Between u and Y(u)A plot of u vs Y(u) is not linear.

u vs Y(u)

0123456789

10

0 2000 4000 6000 8000 10000

u

Y(u

)

20

A plot of ln(u) vs ln(Y(u)) is linear. Hence, linear regression is used on ln(u) and ln(Y(u)).

Relationship Between ln(u) and ln(Y(u))

ln(u ) vs ln(Y (u )

-1

0

1

2

3

0 5 10ln(u )

ln(Y

(u))

11

21

Step 1

Step 1: Compute the logarithmic values.

The Excel function for computing natural logarithms is LN()e.g., if a value of u is in B6, formula for ln(u) is =LN(B6)

u Y (u ) ln(u ) ln(Y (u ))10 9.2225 4.85

100 3.8250 2.44500 1.71000 1.035000 0.6

10000 0.5

22

Step 2 by Hand

Step 2: Compute i x y xy x ^2

ln(u ) ln(y (u )1 2.302585093 2.221375042 3.218875825 1.57897873 4.605170186 1.335001074 5.521460918 0.891998045 6.214608098 0.530628256 6.907755279 0.02955887 8.517193191 -0.51082568 9.210340372 -0.6931472

SumAverage

∑ ∑ ∑∑ yxxxyyx and ,,,, 2

12

23

Step 3 by Hand

Step 3: Compute the slope and intercept:

=−=

=

−

−

=

∑∑

∑∑∑

==

===

)

2

11

2

111

xy

xxn

yxyxn

n

ii

n

ii

n

ii

n

ii

n

iii

slope(Intercept

Slope

24

Steps 2 and 3 by Excel

• If Excel is used, steps 2 and 3 can be replaced by a single step. Two built-in Excel functions provides slope and intercept as shown below:

• Suppose thatln(u) values are in column A rows 18-25ln(Y(u)) values are in column B rows 18-25

• Excel formulae forIntercept is INTERCEPT(B18:B25,A18:A25)Slope is SLOPE(B18:B25,A18:A25)

• Thus, intercept = 3.1301, and slope = -0.42276.

13

25

Step 4

Step 4: Compute the parameters a, b, L

Suppose that the values of intercept and slope are in cells B29 and B30 respectively

Parameter Formula Excel formula Valuea eintercept = EXP(B29)b -slope = -B30L eslope*ln(2) = EXP(B30*ln(2))

26

READING AND EXERCISES

Lesson 2

Reading: Section 1.10, pp. 32-38 (4th Ed.), pp. 29-36 (5th Ed.)

Exercises: 29, 30, 33, pp. 37-38 (4th Ed.), pp. 35-36 (5th Ed.)

14

27

Outline

• Capacity• Economies and Diseconomies of Scale• Break Even Quantity• A Dynamic Capacity Expansion Policy

LESSON 3: CAPACITY GROWTH PLANNING

28

Capacity

• Capacity – Capacity of a plant is the number of units that the plant

can produce in a given time. – The term capacity may refer to average capacity,

design capacity, and peak capacity.• Best operating level

– Best operating level is the design capacity– Design capacity provides minimum average cost

• Capacity utilization

Level Operating BestUsedCapacity

=

15

29

Economies and Diseconomies of Scale

• Economies and diseconomies of scale – A higher volume of production or production of multiple

products results in savings• Two or more products can be produced in the same

location.• Duplication of support functions can be eliminated.

These functions include information storage and retrieval systems and clerical and support staff.

– However, too many products or product lines produced at the same facility could cause an increase in costs because the various manufacturing operations may interfere with each other. This is diseconomies of scale.

30

Economies and Diseconomies of Scale

250 roomhotel

Ave

rage

cos

t per

uni

t

Best operating

level

500 roomhotel

1000 roomhotel

Best operating

levelBest

operating level

Economies of scale Diseconomies of scale

Number of Units Produced

16

31

Cost of Capacity and Economies of Scale

Let f(y) = the cost of adding capacity y.

Economies of scale can be represented as:

Where, k is a constant and a<1.

akyyf =)(

Economies/Diseconomies of Scale

0.0

50.0

100.0

150.0

200.0

250.0

300.0

0 5 10 15 20

y

f(y)

with

k=1

0a=0.75a=1a=1.25

32

Break-Even Quantity

If you invest in a new product, how high must sales be so that you break even? — i.e., when does your net profit ≥ 0?

Example: Suppose a company produces Q units of a product each year. The selling price is c1 per unit. Production of this product results in a fixed cost of K per year and a variable cost of c2 per unit. What is the breakeven quantity Qb per year?

17

33

0

0.5

1

1.5

2

2.5

3

3.5

0 0.2 0.4 0.6 0.8 1 1.2

Production Quantity Q

Cos

ts a

nd R

even

ues

Profit

Loss

Qb

Break-Even Quantity

34

A Dynamic Capacity Expansion Policy

• Frequency of capacity additions– Too frequent: installation, training, premium for up to

date technology, loss of production time– Too infrequent: cost of capital tied to build and maintain

the excess capacity

• Let D = Annual increase in demandx = Years between successive capacity addition

• Then, each capacity addition = xD

18

35

Units

CapacityxD

Time

DemandD per year

Units

Time

Units

Time

Units

Incrementalexpansion

Time

Lead strategy Lag strategy

Average addition Incremental vs. one-step

One-step expansion


36

• First, a review on the present worth analysis• See Appendix 1-A, p. 51 for details• Let

i = discount rate per periodF = future value P = present valuen = number of periods

Then,


( )niF

P+

=1

19

37

• If r is the annual discount rate, and the compounding is continuous,

• If r is the annual discount rate, the compounding is continuous, and the future value is discounted over tyears,


rn

n

Fe

nr

FP −

∞→

=

+

=

1lim

rtFeP −=

38

• Let r = Annual discount ratef(y) = Cost of opening a plant of capacity y

• What is the present cost of adding a capacity of xD if the capacity is added x years from now and the compounding is annual?

• What is the present cost of adding a capacity of xD if the capacity is added x years from now and the compounding is semi-annual?


20

39

• Let r = Annual discount ratef(y) = Cost of opening a plant of capacity y

• What is the present cost of adding a capacity of xD if the capacity is added x years from now and the compounding is continuous?

• What is the present cost of adding a capacity of xD if the capacity is added 2x years from now and the compounding is continuous?


40


rx

a

rx

rxrx

rxrx

exDkexDf

eexDf

xDfexdfexDf

xCxxD

−

−

−−

−−

−=

−=

+++=

+++=

1)(

1)(

]1)[(

)()()(

)(

2

2

L

L

years,every capacity adding of costs present The

21

41


141- Figure from (ii) error and trial (i)

by computed bemay known, is If

:following the satisfies if minimum is function cost The

rxae

rxa

xxC

rx 1

)(

−=

42

Figure 1-14

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

0 2 4 6

rx

Fu

nct

ion

= a


22

43

Problem 1-37: Based on past experience, a chemicals firm estimates that the cost of new capacity additions obeys the law where y is measured in tons per year and f(y) in millions of dollars. Demand is growing at the rate of 3,000 tons per year and the accounting department recommends a rate of 12 percent per year for discounting future costs.

a. Determine the optimal timing of plant additions and the optimal size of each addition.

58.00205.0)( yyf =


44

b. What is the cost of each addition?Size of each addition

Cost of each addition


23

45

c. What is the present value of the cost of the next four additions? Assume an addition has just been made for the purposes of your calculation.


46

Problem: From the past experience it is known that each doubling of the size results in an increase in cost of about 75 percent. Compute the value of a, if the cost function is of the form akyyf =)(

)(yf


24

47


Lesson 3


Exercises: 34, 37, 38 pp. 47-48 (4th Ed.), pp. 44-45 (5th Ed.)

48

Outline

• Forecasting Success Stories• Decisions Based on Forecasts• Characteristics of Forecasts• Components of demand• Evaluation of forecasts

LESSON 4: FORECASTING

25

49

Forecasting Success Stories

• Sharing forecasting information along the supply chain is not very common. The result is forecast error as much as 60 percent of actual demand. US economy alone can save $179 billion in inventory investment with a coordinated forecasting. Wal-Mart has initiated such a process with Warner-Lambert, a manufacturer of Listerine.

• Hewlett-Packard uses forecasting method for new product development.

50


• Taco Bell developed a forecasting method that gives customer demand for every 15-minute interval. The forecast is used to determine the number of employees required. Taco Bell achieved labor savings of more than $40 million from 1993 to 1996.

• Compaq delayed the announcement of several new Pentium-based models in 1994. The decision was based on a forecasting method and contrary to the company belief.

26

51


• Forecasting is necessary to determine the number of reservations an airline should accept for a particular flight - overbooking, traffic management, discount allocation, etc.

52

Decisions Based on Forecasts

• Production– Aggregate planning,

inventory control, scheduling

• Marketing– New product introduction,

sales-force allocation, promotions

• Finance– Plant/equipment

investment, budgetary planning

• Personnel– Workforce planning,

hiring, layoff

27

53

Decisions Based on ForecastsThe decisions should not be segregated by functional area, as they influence each other and are best best made jointly. For example, Coca-cola considers the demand forecast over the coming quarter and decides on the timing of various promotions. The promotion information is then used to update the demand forecast. Based on this forecast, Coca-Cola will decide on a production plan for the quarter.

This plan may require additional investment, hiring, or perhaps subcontracting of production. Coke will make these decisions based on the production plan and existing capacity, and it must make them all in advance of actual production.

54

Characteristics of Forecasts

• Forecasts are always wrong; so consider both expected value and a measure of forecast error.

• Long-term forecasts are less accurate than short-term forecasts. For example, 7-Eleven Japan

has a replenishment process that enables it to respond to an order within hours. If a store manager places an order by 10 am, the order is delivered by 7 pm the same day. The store manager thus has to forecast what will sell that night less than 12 hours before the actual sale.

28

55

Characteristics of Forecasts

• Some time series (called aggregate series) are obtained by summing up more than one time series (called disaggregate series) . For example, annual sales are obtained by adding 12 monthly sales. The annual sales is an aggregate series and monthly sales is a disaggregate series. Aggregate forecasts are more accurate than disaggregate forecasts. Variation in GDP of a country is much less than the annual earnings of a company. Consequently, it is easy to forecast the GDP of a country with less than 2% error. However, it is much more difficult to forecast annual earning of a company with less than a 2% error.

56

Components of Demand

• Average demand• Trend

– Gradual shift in average demand• Seasonal pattern

– Periodic oscillation in demand which repeats• Cycle

– Similar to seasonal patterns, length and magnitude of the cycle may vary

• Random movements• Auto-correlation

29

57

Qan

tity

Time

(a) Average: Data cluster about a horizontal line.


58

Qu

anti

ty

Time

(b) Linear trend: Data consistently increase or decrease.


30

59

Qu

anti

ty

| | | | | | | | | | | |J F M A M J J A S O N D

Months

(c) Seasonal influence: Data consistently show peaks and valleys.

Year 1


60

Qu

anti

ty

| | | | | | | | | | | |J F M A M J J A S O N D

Months

(c) Seasonal influence: Data consistently show peaks and valleys.

Year 1

Year 2


31

61


62

Qu

anti

ty

| | | | | |1 2 3 4 5 6

Years

(c) Cyclical movements: Gradual changes over extended periods of time.


32

63


Suppose that a company has institute a sales incentive system that provides a bonus for the employee with the best improvement in bookings from one month to the next. With such an incentive in place, a month of poor sales is often followed by a month of good sales. Similarly, a month of good sales would usually be followed by a lull.

64


This means that sales in consecutive months tend to be negatively correlated. This information can be used to improve sales forecasts. Autocorrelation is the correlation among values of observed data separated by a fixed number of periods. In the example above, we would say that the series has a negative autocorrelation of order one.

33

65

Dem

and

Time

Trend

Randommovement

Dem

and

Time

Trend with seasonal pattern


66

Snow SkiingSeasonalLong term growth trend

Demand for skiing products increased sharply after the Nagano Olympics

34

67

Evaluation of Forecast

There are many forecasting techniques and software. The performance of a forecasting technique can be measured by the error produced over time. We shall now discuss various measures used to evaluate forecasting techniques. Let,

Error Percentage AbsoluteMeanMAPEDeviation AbsoluteMeanMADError Squared MeanMSE

errors forecast of deviation Standard period in error Forecast

period in Forecast

period in data Actual

===

=σ−==

=

=

ttt

t

t

DFtEtF

tD

68

Σ |Et |nΣEt

2

n

MAD =MSE =

MAPE = σ = MSE

Σ[|Et | (100)]/Dtn

Measures of Forecast Error

Et = Ft - Dt


35

69

Absolute Error Absolute Percent

Month, Demand, Forecast, Error, Squared, Error, Error, t Dt Ft Et Et

2 |Et| (|Et|/Dt)(100)

1 200 2252 240 2203 300 2854 270 2905 230 2506 260 2407 210 2508 275 240

-

Total


70

MSE = =

Measures of Error

MAD = =

MAPE = =


36

71


Lesson 4

Reading: Section 2.1-2.6, pp. 55-66 (4th Ed.), pp. 51-63 (5th

Ed.)

Exercises: 12, 13, 15, pp. 65-66 (4th Ed.), pp. 62-63 (5th Ed.)

72

Outline

• Simple Moving Average• Weighted Moving Average• Exponential Smoothing• Comparison of Simple Moving Average and

Exponential Smoothing

LESSON 5: FORECASTINGSTATIONARY TIME SERIES METHODS

37

73

Time Series Methods

• In this lesson we shall discuss some time series forecasting methods. All methods discussed in this lesson are designed for stationary series. Recall from the previous lesson that a stationary series contains only the average and no trend, seasonality, cyclicity, etc.

• No method is superior to any other method in every context. In a particular context, various methods can be used and evaluated using a suitable measure (e.g., MAD, MSE, MAPE etc.) discussed in the previous lesson. Then, it is possible to use the method that works best in that context. See the Taco Bell example.

• A comparison among the methods is done near the end of the lesson.

74

Time Series Methods

• All these methods will be illustrated with the following example: Suppose that a hospital would like to forecast the number of patients arrival from the following historical data:Week Patients Arrival

1 4002 3803 4114 415

• Note: Although week 4 data is given, some methods require that forecast for period 4 is first computed before computing forecast for period 5.

38

75

Time Series MethodsSimple Moving Average

Week

450 —

430 —

410 —

390 —

370 —Pat

ien

t arr

ival

s

| | | | | |0 5 10 15 20 25 30

Actual patientarrivals

A moving average of order N is simply the arithmetic average of the most recent Nobservations. For 3-week moving averages N=3; for 6-week moving averages N=6; etc.

76

450 —

430 —

410 —

390 —

370 —Pat

ien

t arr

ival

s

Week

| | | | | |0 5 10 15 20 25 30

PatientWeek Arrivals

1 4002 3803 411


Given 3-week data, one-step-ahead forecast for week 4 or two-step-ahead forecast for week 5 is simply the arithmetic average of the first 3-week data

39

77

=4F4for week

forecast ahead-step-One

450 —

430 —

410 —

390 —

370 —Pat

ien

t arr

ival

s

Week

| | | | | |0 5 10 15 20 25 30


1 4002 3803 411


78

450 —

430 —

410 —

390 —

370 —Pat

ien

t arr

ival

s

| | | | | |0 5 10 15 20 25 30


Week


1 4002 3803 411

5for week

forecast ahead-step-Two

=5F

40

79

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430 —

410 —

390 —

370 —Pat

ien

t arr

ival

s

Week

| | | | | |0 5 10 15 20 25 30


2 3803 4114 415


5for week

forecast ahead-step-One

=5F

One-step-ahead forecast for week 5 is computed from the arithmetic average of weeks 2, 3 and 4 data

80

450 —

430 —

410 —

390 —

370 —Pat

ien

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ival

s

Week

| | | | | |0 5 10 15 20 25 30


3-week MAforecast


41

81

Week

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| | | | | |0 5 10 15 20 25 30


3-week MAforecast

6-week MAforecast


82

Taco Bell determined that the demand for each 15-minute interval can be estimated from a 6-week simple moving average of sales.

The forecast was used to determine the number of employees needed.

42

83

Time Series MethodsWeighted Moving Average

In the simple moving average method each of the Nperiods is equally important for the purpose of forecasting. Weighted moving average is more general than the simple moving average and assigns different weights to different periods. Let,

Then, the one-step ahead forecast for period t

NtNtttttt DwDwDwF −−−−−− +++= L2211

( )( )

Ni

itDitw

it

it

,,2,1 K=

−=−=

−

−

period for data actual period to assigned weight

84

450 —

430 —

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390 —

370 —Pat

ien

t arr

ival

s

Week

| | | | | |0 5 10 15 20 25 30

3-week MAforecast Weighted Moving Average

Assigned weightst-1 0.70t-2 0.20t-3 0.10


=4F

43

85

450 —

430 —

410 —

390 —

370 —Pat

ien

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ival

s

Week

| | | | | |0 5 10 15 20 25 30

3-week MAforecast Weighted Moving Average

Assigned weightst-1 0.70t-2 0.20t-3 0.10


=5F

86

• Exponential smoothing method computes a forecast value which is the weighted average of the most recent data and forecast values.

• The weight assigned to the most recent data is called the smoothing constant, α and the weight assigned to the most recent forecast is (1- α).

• The method requires an initial forecast value. The initial forecast value may be obtained by some other forecasting technique.

• If the smoothing constant, α is large, the forecast values fluctuate with the actual data. If α is small, the fluctuation is less.

Time Series MethodsExponential Smoothing

44

87

• The one-step-ahead forecast for period t

• Notice that therefore,

• With further expansion of the expression for forecast for period t it can be seen that the forecast for period t depends on all previous data!!


( ) 11 1 −− α−+α= ttt FDF

( ) ( )( )( ) ( )( ) ( ) ( )( )( ) ( ) ( )

K3

33

221

332

21

22

21

221

111

111

11

11

−−−−

−−−−

−−−

−−−

α−+α−α+α−α+α=

α−+αα−+α−α+α=

α−+α−α+α=

α−+αα−+α=

tttt

tttt

ttt

tttt

FDDD

FDDD

FDD

FDDF

88

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s

Week

| | | | | |0 5 10 15 20 25 30

Exponential Smoothingα = 0.10

Ft = α Dt-1 + (1 - α)Ft - 1


45

89

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Week

| | | | | |0 5 10 15 20 25 30


Ft = α Dt-1 + (1 - α)Ft - 1

Initial forecast valueF3 = (400 + 380)/2=390D3 = 411


90

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Week

| | | | | |0 5 10 15 20 25 30


Ft = α Dt-1 + (1 - α)Ft - 1


Initial forecast valueF3 = (400 + 380)/2=390D3 = 411

=4F

46

91

Week

450 —

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410 —

390 —

370 —Pat

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ival

s

| | | | | |0 5 10 15 20 25 30

F4 = 392.1D4 = 415


Ft = α Dt + (1 - α)Ft - 1


=5F

92

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47

93

Comparison of Exponential Smoothing and Simple Moving Average

• Both Methods – Are designed for stationary demand– Require a single parameter– Lag behind a trend, if one exists– Have the same distribution of forecast error if

)1/(2 +=α N

94

• Moving average uses only the last N periods data, exponential smoothing uses all data

• Exponential smoothing uses less memory and requires fewer steps of computation; store only the most recent forecast!

Comparison of Exponential Smoothing and Simple Moving Average

48

95


Lesson 5


Exercises: 17, 18, 24, pp. 69, 75-76 (4th Ed.), pp. 66, 72 (5th Ed.)

96

Outline

Linear RegressionHolt’s Method

LESSON 6: FORECASTINGTREND-BASED TIME SERIES METHODS

49

97

Linear Regression

• Linear regression is a useful tool that can provide relationship between two variables.

• When a scatter diagram shows that two variables are linearly or nearly linearly related, linear regression can provide a mathematical relationship between two variables. The relationship is expressed in terms of an equation.

• Using linear regression, we can infer how good the relationship is.

98

Linear Regression

• We discuss linear regression in three contexts:– Lesson 2: To estimate the learning curve parameters.

This is the only context when we compute logarithms before using the method of regression. The reason for this is that u and Y(u) are not linearly related. However, log(u) and log(y(u)) are nearly linearly related. So, linear regression is used on log(u) and log(y(u)).

– Lesson 6: To forecast demand from a known value of some other variable.

– Lesson 7: To extend the deseasonalized demand series.

50

99

Linear Regression

Dep

end

ent v

aria

ble

Independent variableX

Y

The scatter diagram on this slide shows a linear relationship

between two variables.

100

Dep

end

ent v

aria

ble


Y Regressionequation:Y = a + bX

Linear Regression

A straight line of the form Y = a+bx nicely fits the points. Linear regression can provide values of parameters a and b

51

101

Dep

end

ent v

aria

ble


Y

Actualvalueof Y

Estimate ofY fromregressionequation

Value of X usedto estimate Y

Deviation,or error

{

Regressionequation:Y = a + bX

Linear Regression

The least square method finds a and b such that the sum of the squares of errors is minimum

102

Sales AdvertisingMonth (000 units) (000 $)

1 264 2.52 116 1.33 165 1.44 101 1.05 209 2.0

Linear Regression

Future sales are unknown, but future advertising expenses may be known. A known value of advertising expense may provide a forecast of sales if we can find the relationship between these two variables.

Since sales depends on advertising, sales is the dependent variable and shown on the y-axis. Advertising is the independent variable and shown on the x-axis.

52

103

a = y – bx=

b =

=

Σxy - nxy

Σx 2 - nx 2

Sales, y Advertising, xMonth (000 units) (000 $) xy x 2

1 264 2.52 116 1.33 165 1.44 101 1.05 209 2.0

Totaly= x =

Linear Regression

104

300 —

250 —

200 —

150 —

100 —

50

Sal

es (0

00s)

| | | |1.0 1.5 2.0 2.5

Linear Regression

Y = Interpretation:

53

105

• Coefficient of correlation, r, provides information on how strong the relationship is.The formula for computing r is shown in the next slide.

• The value of r varies from -1 to +1.• A negative value of r indicates a negative relationship (the

regression line will have a slope downward to the right) and a positive value of r indicates a positive relationship (the regression line will have a slope upward to the right).

• The value of r2 varies from 0 to 1. A value near 0 indicates a very weak or no relationship. A value near 1 indicates a very strong relationship.

Linear Regression

106

Sales, y Advertising, xMonth (000 units) (000 $) xy x 2 y 2

1 264 2.5 660.0 6.252 116 1.3 150.8 1.693 165 1.4 231.0 1.964 101 1.0 101.0 1.005 209 2.0 418.0 4.00

Total 855 8.2 1560.8 14.90y = 171 x = 1.64

nΣxy - Σx Σy

[nΣx 2 -(Σx) 2][nΣy 2 - (Σy) 2]r =

Linear Regression

54

107

Linear Regression

Sales, y Advertising, xMonth (000 units) (000 $) xy x 2 y 2

1 264 2.5 660.0 6.25 69,6962 116 1.3 150.8 1.69 13,4563 165 1.4 231.0 1.96 27,2254 101 1.0 101.0 1.00 10,2015 209 2.0 418.0 4.00 43,681

Total 855 8.2 1560.8 14.90 164,259y= 171 x = 1.64

r = 0.98, r 2 = 0.96Since r = 0.98>0, relationship is positiveSince r 2 = 0.96≈1, relationship is very strong

108

Linear Regression

The regression equation can be used to forecast sales of Month 6 from a known value of advertising expenditure in Month 6.

Forecast for Month 6:

Let advertising expenditure = $1750

Y =

55

109

| | | | | | | | | | | | | | |0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

80 —

70 —

60 —

50 —

40 —

30 —

Pat

ien

t arr

ival

s

Week

Yn = a + bXn

where

Xn = Weekn

Linear Regression

110

• The method uses two smoothing constants α and β• The method iteratively computes St and Gt

respectively slope and gradient at period t• Initial slope and gradient S0 and G0 are needed to

start the computation• Finally, St and Gt can be used to obtain a τ-step

ahead forecast, F t,t+τ from period t

Double Exponential SmoothingHolt’s Method

tttt

tttt

tttt

GSFGSSG

GSDS

τ+=β−+−β=

+α−+α=

τ+

−−

−−

,

11

11

)1()(

))(1(

56

111

α 0.05 S 0 3920β 0.25 G 0 180

==

==

8,47,4

6,45,4

FF

FF

Demandt (Units) S 0 G 0 F t -1,t

1 42002 43003 40004 4400

Double Exponential SmoothingHolt’s Method

112

A Comparison of Methods

60

65

70

75

80

85

90

0 5 10 15

Months

Dem

and

Actual

3-Mo MA

3-Mo WMA

Exp Sm

Double Exp Sm

57

113


Lesson 6


Exercises: 28, 30, pp. 79, 81 (4th Ed.), pp. 75, 77 (5th Ed.)

114

Outline

• Multiplicative Seasonal Influences• Additive Seasonal Influences• Seasonal influence with trend

LESSON 7: FORECASTINGMETHODS FOR SEASONAL SERIES

58

115

Turkeys have a long-term trend for increasing demand with a seasonal pattern. Sales are highest during September to November and sales are lowest during December and January.

116

Quarter Year 1 Year 2 Year 3 Year 4

1 45 70 100 1002 335 370 585 7253 520 590 830 11604 100 170 285 215

Total 1000 1200 1800 2200Average 250 300 450 550

Seasonal Influences

Consider the demand data shown above. Is the data seasonal? Are the demands in Quarter 1 consistently less than the average? What is the relationship of Quarter 1 demand with the average? How about the other quarters?

59

117

Seasonal Influences

• Is it not true that Quarter 1 demand is less than the average? How much less do you expect the Quarter 1 demand from the average?

• One can take different approaches to answer this question.– Answer 1: Quarter 1 demand is approximately 20% of

the average. – Answer 2: Quarter 1 demand is approximately 200 units

less than the average. • Are these answers true? We shall verify the correctness of

the answers in this lesson. The first answer uses the concept of multiplicative seasonal influence and the second answer additive seasonal influence. We shall now define these two seasonal influences.

118

Seasonal Influences

• A seasonal influence is multiplicative ifthe quarterly demand forecast of a quarter

= projected average quarterly demand × average seasonal index of that quarter.

• A seasonal influence is additive ifThe quarterly demand forecast of a quarter

= projected average quarterly demand + average seasonal index of that quarter.

60

119


1 45/250 = 0.18 70/300 = 0.23 100/450 = 0.22 100/550 = 0.182 335/250 = 1.34 370/300 = 1.23 585/450 = 1.30 725/550 = 1.323 520/250 = 2.08 590/300 = 1.97 830/450 = 1.84 1160/550 = 2.114 100/250 = 0.40 170/300 = 0.57 285/450 = 0.63 215/550 = 0.39

Multiplicative Seasonal Influences

For example, seasonal Index, Year 1, Quarter 1=

Seasonal index = Actual Quarterly Demand

Average Quarterly Demand

Step 1: For each period, compute

Multiplicative influence:

120


1 45/250 = 0.18 70/300 = 0.23 100/450 = 0.22 100/550 = 0.182 335/250 = 1.34 370/300 = 1.23 585/450 = 1.30 725/550 = 1.323 520/250 = 2.08 590/300 = 1.97 830/450 = 1.84 1160/550 = 2.114 100/250 = 0.40 170/300 = 0.57 285/450 = 0.63 215/550 = 0.39

Quarter Average Seasonal Index

1234


Step 2: For each quarter compute the average seasonal index

61

121


The average seasonal indices can be used to get forecasts of the quarterly demands if the average quarterly demand is projected. The quarterly demand forecast of a quarter = projected average quarterly demand × average seasonal index of that quarter.

For example, suppose that the next year, in Year 5, The projected annual demand is 2600. So, the projected average quarterly demand is 2600/4=650.

Then, the demand forecast in Quarter 1 = 650(0.20)=130. The next slide shows the demand forecast for the other quarters.

122


Given projected average quarterly demand =650The quarterly demand forecasts are obtained as follows:

Quarter Average Seasonal Index Forecast1234

62

123

Additive Seasonal Influences

For example, seasonal Index, Year 1, Quarter 1= 45-250 = -205

= Actual Quarterly Demand - Average Quarterly Demand

Step 1: For each period, compute seasonal index

Additive influence:


1 45-250 = -205 70-300 = -230 100-450 = -350 100-550 = -4502 335-250 = 85 370-300 = 70 585-450 = 135 725-550 = 1753 520-250 = 270 590-300 = 290 830-450 = 380 1160-550 = 6104 100-250 = -150 170-300 = -130 285-450 = -165 215-550 = -335

124

Quarter Average Seasonal Index

1 (-205-230-350-450)/4 = -308.752 (85+70+135+175)/4 = 116.253 (270+290+380+610)/4 = 387.504 (-150-130-165-335)/4 = -195.00


Step 2: For each quarter compute the average seasonal index


1 45-250 = -205 70-300 = -230 100-450 = -350 100-550 = -4502 335-250 = 85 370-300 = 70 585-450 = 135 725-550 = 1753 520-250 = 270 590-300 = 290 830-450 = 380 1160-550 = 6104 100-250 = -150 170-300 = -130 285-450 = -165 215-550 = -335

63

125


Given projected average quarterly demand =650The quarterly demand forecasts are obtained as follows:

Quarter Average Seasonal Index Forecast1 -308.75 650-308.75=341.25 2 116.25 650+116.25=766.253 387.50 650+387.50=1037.504 -195.00 650-195.00=455.00

126Difference between the highest and lowest demand increases

Multiplicative Influence

0

200

400

600

800

1000

1200

1400

0 5 10 15 20 25

Quarter

Dem

and

64

127Difference between the highest and lowest demand is constant

Additive Influence

0

200

400

600

800

1000

1200

1400

0 5 10 15 20 25

Quarter

Dem

and

128

Step 1: Compute moving averages– Compute N-period moving averages– The value of N is set to the length of season. For

example, if the data are quarterly demand, the length of the season is 4 (there are 4 quarters in a year), so Nshould be 4. If the data are monthly demand, N should be set to 12, etc.

– Center the moving averages– Put the centered values back on periods– Compute the values for the periods in the beginning and

end

Seasonal Influences with Trend

65

129

Step 2: Determine seasonal factors– For each period, compute a factor by dividing demands

by moving average values– Average the factors that correspond to the same periods

of each season– N seasonal factors will result– If the seasonal factors do not sum up to N, scale up or

down each factor so that the factors sum up to N.– Note that the seasonal factors usually do not sum up to

N. So, the seasonal factors should always be scaled up or down so that the factors sum up to N.


130

Step 3: Deseasonalize the original data– Divide the original data by the seasonal factors.– This step removes the seasonality from the data.– Since we assume that the data contains seasonality and

trend, there will only be trend after we remove the seasonality. Plus, recall that we already know how to deal with trend! For example, we can use regression to understand the trend.

– After we remove the seasonality, the deseasonalized series is expected to be smooth. Then, we can use regression or other trend-based methods in Step 4.


66

131

Step 4: Make a forecast using deseasonalized data– Use any trend-based method

• Double exponential smoothing (not used in this lesson)

• Linear regression (used in this lesson)– a. Compute

» The series x denotes periods 1, 2, 3, … and ydenotes the deseasonalized series. Note carefully that the series y does not denote the actual demand values but the deseasonalized demand that we get at the end of Step 3.


∑ ∑ ∑ ∑ yxxxyyx and ,,,, 2

132

Step 4: Make a forecast using deseasonalized data– a. Compute

» Recall that the deseasonalized series has only the trend and no seasonality. In the entire Step 4, we consider the deseasonalized series and get a mathematical understanding of the trend. This mathematical understanding refers to slope and intercept that we find in Step 4b.

– b. Compute slope and intercept» The slope and intercept approximately define

the straight line that best fits the deseasonalized series.


∑ ∑ ∑ ∑ yxxxyyx and ,,,, 2

67

133

Step 4: Make a forecast using deseasonalized data– c. Forecast deseasonalized series

» The deseasonalized series is projected to the required future period using slope and intercept.

» For example, if we have data of 8 periods, we can project the deseasonalized series in the 9th, 10th or any other period in future using slope and intercept.

» Note carefully that our job does not end with this step. Because, we have to put the effect of seasonality back. This is done in Step 5.


134

Step 5: Reseasonalize forecast using seasonal factors– Multiply the forecasted values by the seasonal factors– This step is necessary to reverse the effect of

deseasonalization that was done in Step 3.– Recall that the actual data is seasonal and we make a

projection using the deseasonalized data in Step 4.– The forecast obtained in Step 4 does not contain

seasonality.– Step 5 puts the effect of seasonality back into the

forecast.


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135

Step 1 (Problem) Centered Moving Average

Centered (B/D)Period Demand MA(4) MA Ratio

A B C D E1 2052 1403 3754 5755 4756 2757 6858 965

136

Sample computation:MA(4), Period 4: (205+140+375+575)/4=323.75MA(4), Period 5: (140+375+575+475)/4=391.25Observe that MA(4), period 4 is obtained from periods 1,2,3,4.

So, MA(4), period 4 represents period (1+2+3+4)/4 or period 2.5. Similarly, MA(4), period 5 represents period 3.5. Centered MA, period 3 is the average of these two values

Centered MA, Period 3:(323.75+391.25)/2=357.5Similalry, Centered MA is computed for periods 4, 5, and 6.

Step 1 (Sample Computation)Centered Moving Average

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137

Sample computation:Centered MA cannot be obtained similarly for the first two and

last two periods.For periods 1 and 2 centered MA = average of centered MA of

periods 3 and 4 (this is a simplified approach) = (357.5 +408.125)/2 = 382.813

For periods 7 and 8 centered MA = average of centered MA of periods 5 and 6 (again, a simplified approach) = (463.75 +551.25)/2 = 507.5

B/D ratio, period 1 = 205/382.813 = 0.54

Step 1 (Sample Computation)Centered Moving Average

138

FinalSeasonal Seasonal

Period Factors Factors1234

Total

Step 2 Seasonal Factors By CMA

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139

Step 3 Deseasonalize

Seasonal Deseasonalized ReseasonalizedPeriod Demand Factors Demand Forecast

A B C D=B/C E1 205 0.76712 140 0.42523 375 1.17974 575 1.62805 475 0.76716 275 0.42527 685 1.17978 965 1.6280910

140

Step 4a Forecast using Deseasonalized Data

DeseasonalizedDemand

x y xy x^21 267.24308752 329.2573623 317.88304754 353.18766245 619.22178816 646.75553247 580.66636688 592.7410333

SumAverage

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141

Step 4b Forecast using Deseasonalized Data

)slope(Intercept

Slope 2

11

2

111

xy

xxn

yxyxn

n

ii

n

ii

n

ii

n

ii

n

iii

−=

−

−

=

∑∑

∑∑∑

==

===

142


A B C D E1 205 0.7671 267.24308752 140 0.4252 329.2573623 375 1.1797 317.88304754 575 1.6280 353.18766245 475 0.7671 619.22178816 275 0.4252 646.75553247 685 1.1797 580.66636688 965 1.6280 592.74103339 0.767110 0.4252

Step 4c Forecast using Deseasonalized Data

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143


A B C D E1 205 0.7671 267.24308752 140 0.4252 329.2573623 375 1.1797 317.88304754 575 1.6280 353.18766245 475 0.7671 619.22178816 275 0.4252 646.75553247 685 1.1797 580.66636688 965 1.6280 592.74103339 0.7671 719.879247110 0.4252 776.8814164

Step 5Reseasonalize

144

Original Data

0

200

400

600

800

1000

0 5 10

Period

Dem

and

Originaldata

73

145

Step 3: Deseasonalize

0

200

400

600

800

1000

0 5 10

Period

Dem

and

Originaldata

Desea-sonalized

146

Steps 4a, 4b: Regression

0

200

400

600

800

1000

0 5 10

Period

Dem

and

Originaldata

Desea-sonalized

Reg-ression

74

147

Step 4c: Regression - Projection Forecast Deseasonalized Demand

0

200

400

600

800

1000

0 5 10

Period

Dem

and

Originaldata

Desea-sonalized

Reg-ression

148

Step 5: Reseasonalize

0

200

400

600

800

1000

0 5 10

Period

Dem

and

Originaldata

Desea-sonalized

Reg-ression

Reseasonalize

75

149


Lesson 7


Exercises: 33, 34, p. 87 (4th Ed.), pp. 82-83 (5th Ed.)

150

Outline

• Aggregate Planning– Issues– Costs

• Two Strategies– Chase Strategy– Level Strategy

• Optimization

LESSON 8: AGGREGATE PLANNING

76

151

Aggregate Production Planning

• Chapter 2 discusses forecasting. If the demand for a product changes over time, we need to plan the resource capacity the firm will need over time.

• Aggregate production planning determines the resource capacity a firm may need to meet its demand.

• Aggregate production planning is a short term decision making process. The planning horizon is typically 6 to 12 months.

• Since the planning horizon is short, it is usually not feasible to build or shut down plants, purchase or sell equipment, etc.

152


• However, capacity may be adjusted in some other ways. It is usually possible to– hire or lay off workers– increase or decrease number of shifts – increase or decrease number of working days in a

week– increase or decrease working hours (overtime or

undertime)– subcontract work to other firms– build up or deplete inventory levels

• Aggregate production planning meets the requirement of resource capacity using the above.

77

153


• The term aggregate refers to the fact that the plan is developed for a line of products, not just one individual product. For example, suppose that a firm produces three types of chairs: ladder-back chair, kitchen chair and desk chair. If an aggregate production plan is required for the chairs, the plan may consider an aggregate unit of chair. It will specify how many aggregate units of chair will be produced in various planning periods.

• An aggregate unit is some kind of an “average” unit. Different products of the same product line may require different amount of worker/machine time, raw materials, etc. and their selling prices may also be different. It is not always very obvious what the aggregating scheme should be.

154


• An example of an aggregation scheme is given in Example 3.1, pp. 114-115 of the text. A firm produces 6 different models of washing machines

Model Working Individual/Total SalesNumber Hours Required Ratio A5532 4.2 0.32K4242 4.9 0.21L9898 5.1 0.17L3800 5.2 0.14M2624 5.4 0.10M3880 5.8 0.06

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155


• The manager decides to define an aggregate unit of production as a fictitious machine requiring some weighted total working hours where the weights are taken from the individual/total sales ratio given in the previous slide. 0.32(4.2)

+ 0.21(4.9)+ 0.17(5.1)+ 0.14(5.2)+ 0.10(5.4)+ 0.06(5.8)= 4.856 hours

156

• The next slide presents a schematic view of the aggregate production planning function and its place in the hierarchy of the production planning decisions.

• Forecasting: First, a firm must forecast demand for aggregate sales over the planning horizon.

• Aggregate planning: The forecasts provide inputs for determining aggregate production and workforce levels over the planning horizon.

• Master production schedule (MPS): Recall, that the aggregate production plan does not consider any “real” product but a “fictitious” aggregate product. The MPS translates the aggregate plan output in terms of specific production goals by product and time period. For example,

Hierarchy of Production Decisions

79

157


Aggregate Planning

Master Production Schedule

Inventory Control

Operations Scheduling

Vehicle Routing

Forecast of Demand

158

suppose that a firm produces three types of chairs: ladder-back chair, kitchen chair and desk chair. The aggregate production considers a fictitious aggregate unit of chair and find that the firm should produce 550 units of chairs in April. The MPS then translates this output in terms of three product types and four work-weeks in April. The MPS suggests that the firm produce 200 units of desk chairs in Week 1, 150 units of ladder-back chair in Week 2, and 200 units of kitchen chairs in Week 3.

• Material Requirements Planning (MRP): A product is manufactured from some components or subassemblies. For example a chair may require two back legs, two front legs, 4 leg supports, etc. While forecasting, aggregate plan


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159

Hierarchy of Production DecisionsMaster Production Schedule

Ladder-back chair

Kitchen chair

Desk chair

1 2

April May

790790

3 4 5 6 7 8

200200

150150

120120

200200

150150

200200

120120

Aggregate production plan for chair family

550550

200200

160

and MPS consider the volume of finished products, MRP plans for the components, and subassemblies. A firm may obtain the components by in-house production or purchasing. MRP prepares a plan of in-house production or purchasing requirements of components and subassemblies.

• Scheduling: Scheduling allocates resource over times in order to produce the products. The resources include workers, machines and tools.

• Vehicle Routing: After the products are produced, the firm may deliver the products to some other manufacturers, or warehouses. The vehicle routing allocates vehicles and prepares a route for each vehicle.


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161

Hierarchy of Production DecisionsMaterials Requirement Planning

Back slats

Leg supports

Seat cushion

Seat-frameboards

Frontlegs

Backlegs

162

Issues in Aggregate Planning

• Smoothing– Smoothing refers to costs that result from changing

workforce level– Two of the key components of smoothing costs are

costs associated with hiring and firing workers– Firing workers could have far-reaching consequences

and the costs may be difficult to evaluate• Bottleneck problems

– Bottleneck refers to the inability of the system to respond to sudden changes in demand

– For example, a bottleneck may arise if the demand in one month is suddenly so high that the plant does not have sufficient capacity to meet the demand

82

163


• Planning horizon– Inaccurate forecasts are associated with too large

planning horizon and planning may be ineffective with too small planning horizon

– To minimize the inventory holding cost during the planning horizon, the aggregate plan may recommend zero inventory at the end of the planning horizon. This is known as the end-of-horizon effect. This may be a poor strategy, specially if demand increases at that time.

– Often, rolling schedules are used. In a rolling schedule, a plan is prepared for several periods more than the planning horizon over which the plan will be

164


implemented. At the time of the next decision, a new forecast is appended and old forecasts may be revised. The new plan may recommend different production and workforce levels than were recommended earlier.

– Often, it desired that because of production lead times that the schedule be frozen for a certain number of planning periods. Then, production and workforce levels over the frozen horizons cannot be changed.

• Nature of Demand– Aggregate planning ignores the possibility of forecast

errors. Assuming deterministic demand, aggregate planning can incorporate seasonal fluctuations and business cycles into the planning function.

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165


• Input– Demand forecast– Level of resources available– Relevant cost information

• Output– Aggregate production quantities

• production, inventory, backorder– Level of resources needed

• Workforce, overtime, machine capacity level, subcontracting

166

Costs in Aggregate Planning

• Smoothing costs– Hiring costs

• Time and cost to advertise positions• Interview candidates• Train new recruits

– Firing costs• Severance pay• The costs of a decline in worker morale• The potential for decreasing the size of the labor

pool in the future

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167

Costs in Aggregate Planning Smoothing Costs

Slope

= un

it hirin

g cos

t

Slope = unit firing cost

Cos

t

Number of hiresNumber of fires

168


• Inventory holding costs– Opportunity cost of capital – Cost of storage space– Taxes and insurance against fire, theft, and other

losses – Breakage, spoilage, deterioration and obsolescence– Example of calculation of inventory holding costs

Cost of capital - 15%Taxes and insurance - 2%Storage - 5%Breakage/spoilage - 3%

Total - 25%

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169


• Shortage costs– Shortages occur when the demand exceeds the

production capacity. One of two types of costs is charged depending on whether a shortage results in loss of sales or not:

• Backorder - if the excess demand is backlogged and fulfilled in a future period, backorder cost is charged (bookkeeping and/or delay costs).

• Lost sales - if the excess demand is lost because the customer goes elsewhere, the lost sales is charged. The lost sales include goodwill and loss profit margin (=selling price - unit variable cost)

170

Costs in Aggregate Planning Inventory Holding and Shortage Costs

Slope

= un

it hold

ing co

st

Slope = unit backorder cost

Cos

t

Positive inventoryBackorder

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171


• Production costs– Straight time costs

• Labor costs, regular time ($/hour)– Overtime and subcontracting costs

• Labor costs, overtime costs ($/hour)• Cost of subcontracting ($/unit or $/hour)

– Idle time costs• In most contexts, the idle time cost is zero. • However, a non-zero cost is appropriate if idle time

has some consequences. For example, when aggregate units are inputs to another process, idle times may delay the process.

172

A Feasible Aggregate Plan

• The next slide shows a chart in which cumulative number of units are plotted over a planning horizon.

• There are two lines: – Production line: this line corresponds to the production

schedule and shows the cumulative number of units produced

– Demand line: this line corresponds to the demand and shows the cumulative net demand

• The production line is straight and its slope does not change. This means that a constant number of units is produced in every period.

87

173


Period

Cum

ulat

ive

Num

ber o

f Uni

ts

A feasible productionschedule

Inventory

CumulativeNetDemand

174


• The production planning strategy in which a constant number of units is produced in each period is called the level strategy. Hence, the chart shows a level strategy.

• The demand line is not straight. This means that the demand is different in different period.

• At any point of time, the difference between cumulative production and cumulative net demand is the inventory. Hence, the vertical difference shows the inventory.

• If shortages are not allowed, a production schedule is feasible if the production line is always above the demand line. Hence, the chart shows a feasible production schedule.

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175


• If the production line lies below the demand line, the inventory is negative and shortages take place. Such a production schedule can also be feasible if shortages are allowed.

• Both the lines meet in the end of the planning horizon. That means that the inventory is zero at the end of the planning horizon. This is the end-of -horizon effect discussed before. It is is not necessary to have a zero inventory at the end of the planning horizon and sometimes it is undesirable. If there is a positive inventory at the end of the planning horizon, there will be a gap between the production and demand lines at the end of the planning horizon.

176


• If shortages are allowed, the production line may lie above or below the demand line. When the production line is above the demand line, inventory is positive and when the production line is below the demand line, inventory is negative.

• The opposite of the level strategy is the chase strategy. In a chase strategy the production volume is the same as the demand. So, the inventory is always zero. In such a case, there is no gap between the production line and the demand line.

89

177

Constraints

• Sometimes decision making can be constrained by – limits on overtime– limits on layoffs– limits on capital available– limits on stockouts and backorders

178


Lesson 8

Reading: Section 3.1 - 3.3 , pp. 113-120 (4th Ed.), pp. 108-117 (5th Ed.)

Exercise: 8, p. 121 (4th Ed.), p. 117 (5th Ed.)

90

179

Outline

• Examples– Chase Strategy– Level Strategy– Optimization

LESSON 9: AGGREGATE PLANNINGEXAMPLES

180

• Chase strategy– Produce as much as needed– Zero inventory, no holding cost, no shortages– Zero inventory is difficult to achieve because work

hours may not be flexible– Low inventory costs, high smoothing costs

• Level strategy– Produce a constant amount each period– Stable workforce, no hiring/firing, no overtime,– no subcontract– Low smoothing costs, high inventory costs

Two Simple Strategies

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181

Chase Strategy

02000400060008000

100001200014000

Fall

Winte

rSp

ring

Summer

Period

Num

ber o

f Uni

ts

DemandProduction

182

Chase Strategy

05000

10000

15000

20000

25000

30000

35000

40000

Fall Winter Spring Summer

Periods

Cum

ulat

ive

Num

ber o

f Uni

ts

CumulativeDemandCumulativeProduction

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183

Level Strategy

02000400060008000

100001200014000

Fall

Winte

rSp

ring

Summer

Period

Num

ber o

f Uni

ts

DemandProduction

184

Level Strategy

05000

1000015000200002500030000350004000045000

Fall Winter Spring Summer

Periods

Cum

ulat

ive

Num

ber o

f Uni

ts

CumulativeDemandCumulativeProduction

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185

Optimization

• The chase and level strategies are two extreme strategies. Chase strategy minimizes inventory costs and level strategy minimizes smoothing costs. The goal of optimization is to identify a production plan that minimizes the total inventory and smoothing costs.This can be done using linear programming.

• Lesson 10 discusses the application of linear programming using Excel Solver.

186

• Develop a production plan and calculate the annual cost for a firm whose demand forecast is fall, 10, 000; winter, 8,000; spring 7,000; summer, 12,000. Inventory at the beginning of fall is 500 units. At the beginning of fall you currently have 30 workers, but you plan to hire temporary workers at the beginning of summer and lay them off at the end of summer. In addition, you have negotiated with the union an option to use the regular workforce on overtime during winter or spring if the overtime is necessary to prevent stock-outs at the end of those quarters. Overtime is not available during fall. (Continued...)

Example

94

187

Relevant costs are: hiring, $100 for each temp; layoff, $200 for each worker laid off; inventory holding, $5 per unit-quarter; backorder, $10 per unit; straight time, $5 per hour; overtime $8 per hour. Assume that the productivity is 0.5 units per worker hour, with eight hours per day and 60 days per season.

• Develop a production plan using(1) all the constraints as stated (2) chase strategy, no overtime, work hours not flexible(3) chase strategy, no overtime, flexible hours (self study)

Example

188

(4) Suppose that a level strategy will be used without any overtime. What is the minimum number of workers required to avoid shortages? Develop a production plan using the minimum number of workers required to avoid shortages.

(5) Assuming that the shortages are allowed and that 6 new workers will be hired in the beginning of the fall term develop a production plan using level strategy and no overtime (self study)

(6) Assuming that the overtime will be used in fall and winter to prevent shortages and that 7 new workers will be hired in the beginning of the fall term, develop a production plan using level strategy with overtime (self study)

Example

95

189

Problem 1: The original problem

Example

Forecast Beginning Production Production Production Inventory Required Hours Hours

Required AvailableFall 10000 500Winter 8000Spring 7000Summer 12000

Overtime Workers Workers Actual Ending Hours Hired Fired Production Inventory

FallWinterSpringSummer

190

• Problem 1 sample computation:Production required in fall = forecast in fall - beginning inventory in fall = 10,000 - 500 = 9,500Production hours required in fall = production required in fall / productivity in units per worker = 9,500 / 0.50 = 19,000 hours Production hours available in fall = 30 workers × 60 days per season × 8 hours per day = 14,400 hoursOvertime and temporary workers are not available in fallActual production in fall = production hours available in fall × productivity in units per worker = 14,400 × 0.50 = 7,200 units

Example

96

191

• Problem 1 sample computation (continued):Ending inventory in fall = actual production in fall - production required in fall = 7,200 - 9,500 = -2,300 unitsBeginning inventory in winter = ending inventory in fall = -2,300 unitsOvertime hours required in winter = production hours required - production hours available = 20,600 - 14,400 = 6,200 hoursActual production in winter = (production hours available in winter + overtime hours in winter) × productivity in units per worker = (14,400+6,200) × 0.50 = 10,300 units

Example

192

• Problem 1 sample computation (continued):Workers hired in summer = (production hours required in summer - production hours available in summer) / number of working hours per worker in summer [Note: the result should be rounded up, the number of workers is an integer and enough workers should be hired to avoid shortages]= (23,600-14,400)/(60 days per season × 8 hours per day)= 19.167 rounded up to 20Note: Actual production in summer is 11,800 units, as much as required. The assumption is that temporary workers will not work for full 480 hours, but only as much as needed. So, they can be stopped after producing 11,800 units.

Example

97

193

Problem 1: The original problem

Example

Backorder Overtime Hiring FiringCost Cost Cost Cost


Inventory Straighttime Total H. Cost Cost Cost


Total cost

194

• Problem 1 sample computation:Straighttime cost in summer = actual production hours ×$5 per hour = 23,600 hour × 5 per hour = $118,000Note: the actual production hour in summer is the same as production hours required in summer because sufficient number of temporary worker are hired and the temporary workers can be stopped after producing the required amount of products.

Example

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195

Problem 2: Chase, no overtime, work hours not flexible

Example (Chase Strategy)

Forecast Beginning Net Production WorkersInventory Production Hours Required

Fall 10000 500Winter 8000Spring 7000Summer 12000

Workers Workers Actual Ending Hired Fired Production Inventory


196

• Problem 2 sample computation:Workers required in fall = production hours required in fall / number of working hours per worker in fall [Note: the result should be rounded up, the number of workers is an integer and enough workers should be hired to avoid shortages]= 19,000/ (60 days per season × 8 hours per day)= 39.583 rounded up to 40Number of workers hired in fall = Number of workers required in fall - number of workers available in the beginning of fall = 40 - 30 = 10

Example

99

197

• Problem 2 sample computation (continued):Actual production in fall = Number of workers available in fall × 60 days per season × 8 hours per day × 0.5 units per worker per hour = 40 × 60 × 8 × 0.50 = 9,600 unitsEnding inventory in fall = actual production in fall -production required in fall = 9,600--9,500 = 100 unitsBeginning inventory in winter = ending inventory in fall = 10 unitsNumber of workers fired in winter = Number of workers available in the beginning of winter - number of workers required in winter = 40 - 33 = 7.

Example

198

Problem 2: Chase, no overtime, work hours not flexible


Hiring Firing Straight Inventory TotalCost Cost time Holding Cost

Cost CostFallWinterSpringSummerTotal

100

199

Problem 3: Chase, no overtime, flexible hoursNet Production Workers Workers Workers

Production Hours Required Hired FiredRequirement Required

Fall 9500 19000 40 10 0Winter 8000 16000 34 0 6Spring 7000 14000 30 0 4Summer 12000 24000 51 21 0

Hiring Firing Straight TotalCost Cost time Cost

CostFall 1000 0 95000 96000Winter 0 1200 80000 81200Spring 0 800 70000 70800Summer 2100 0 120000 122100Total 370100


200

Problem 4: Constant workforce, no overtime, no shortages

Example (Level Strategy)

Workers hired Initial hiring costWorkers fired Initial firing costTotal workers Straighttime cost

Computation of the workforce required for avoiding shortagesNet Cumulative Cumulative Workers

Production Net units RequiredRequirement Production produced

Requirement per workerFall 9500Winter 8000Spring 7000Summer 12000

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201

• Problem 4 computation of number of workers required:Step1:

For each period compute the cumulative net production requirement

Step2:For each period compute the cumulative units produced per worker

Step 3:For each period compute the number of workers required to meet the cumulative demand upto that period by dividing the cumulative net production by the cumulative units produced and rounding up.


202

• Problem 4 computation of number of workers required:

Number of workers required to meet the cumulative demand upto FallWinterSpring Summer

Step 4:The number of workers required is the maximum of all the numbers obtained in Step 3

Number of workers required = max ( ) =


40583.39240/9500 ===

102

203

Problem 4: Constant workforce, no overtime, no shortages


Forecast Beginning Actual Ending Inventory Production Inventory

Fall 10000 500Winter 8000Spring 7000Summer 12000

Inventory Backorder Total Cost Cost Cost


Total cost

204

Problem 5: Constant 36 workers, no overtime, shortages allowed

Workers hired 6 Initial hiring cost 600Workers fired 0 Initial firing cost 0Total workers 36 Initial recruitment cost 600

Straighttime cost 345600


103

205

Problem 5: Constant 36 workers, no overtime, shortages allowed

Forecast Beginning Actual Ending Inventory Production Inventory

Fall 10000 500 8640 -860Winter 8000 -860 8640 -220Spring 7000 -220 8640 1420Summer 12000 1420 8640 -1940

Inventory Backorder Total Holding Cost Cost

CostFall 0 8600 8600Winter 0 2200 2200Spring 7100 0 7100Summer 0 19400 19400

Total 383500


206

Problem 6: Constant 37 workers, overtime to prevent shortages

Workers hired 7 Initial hiring cost 700Workers fired 0 Initial firing cost 0Total workers 37 Initial recruitment cost 700

Straighttime cost 355200


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207

Problem 6: Constant 37 workers, overtime to prevent shortagesForecast Beginning Regular Units Units

Inventory Production Available OvertimeBefore OT

Fall 10000 500 8880 -620 620Winter 8000 0 8880 880 0Spring 7000 880 8880 2760 0Summer 12000 2760 8880 -360 360

Ending Inventory Overtime Total Inventory Holding Cost Cost

CostFall 0 0 9920 9920Winter 880 4400 0 4400Spring 2760 13800 0 13800Summer 0 0 5760 5760

Total 389780


208


Lesson 9Reading: Section 3.4, pp. 121-127 (4th Ed.), pp. 117-

125 (5th Ed.)Exercises: 9, 13 and 14, pp. 127-129 (4th Ed.), pp. 123-

124 (5th Ed.)

Lesson 10Reading: Section 3.5-3.6, pp. 129-138 (4th Ed.), pp.

125-135 (5th Ed.)Exercises: 17, 19 and 20, pp. 138-139 (4th Ed.), pp.

133-134 (5th Ed.)

1

Lesson 10a: Learning Curve Estimating Learning Curve Parameters Using Excel Solver

Including Drawing Scatter Diagram

This lesson illustrates the procedure of estimating learning curve parameters using Excel. Download the Excel file Lesson_10a_learning_w03_331.xls from the course website and follow the instruction given below. The above Excel spreadsheet contains the following in cells A3 to B13:

Step 1: Compute logarithmic values We shall compute natural logarithms. Note that the choice between natural logarithm (base e) and common logarithm (base 10) is related to the formula for parameter a. If common logarithm is used, the formula for a must be changed from what we use. The Excel function for natural logarithm is LN(). To get the natural logarithm of the number typed in cell A6, choose a cell, say A18 and enter the following formula: = LN(A6) Now, using copy and paste all the other natural logarithms may be obtained. The resulting values are shown below:

A B3 Cumulative Number of Number of Hours Required4 Units Produced For the Next Unit5 u Y (u )6 10 9.227 25 4.858 100 3.89 250 2.44

10 500 1.711 1000 1.0312 5000 0.613 10000 0.5

A B17 Ln(u ) Ln(Y (u ))18 2.302585093 2.22137503819 3.218875825 1.57897870520 4.605170186 1.33500106721 5.521460918 0.89199803922 6.214608098 0.53062825123 6.907755279 0.02955880224 8.517193191 -0.51082562425 9.210340372 -0.693147181

2

Step 2: Compute the slope and intercept To get the slope and intercept, enter the following formula in two different cells, say B29 and B30:

= INTERCEPT(B18:B25,A18:A25) = SLOPE(B18:B25,A18:A25)

The resulting values are shown below:

Step 3: Compute the parameters a, b, L The Excel formula for the parameters are shown below:

Parameter Math formula Excel formula A eintercept = EXP(B29) B -slope = -B30 L eslope*ln(2) = EXP(B30*ln(2))

Enter the Excel formula for parameters a, b and L in cells B35, B36 and B37 respectively. The resulting values are shown below:

To get the “%” for the value of L, choose cell B37, select “Format” from the main menu, and then “cells” from a menu that pops down. From the “Format Cells” dialog box, select the “Number” tab. Choose “Percentage” from “Category” window and “0” from the “Decimal places.” Click on OK.

28 A B29 Intercept 3.13014773630 Slope -0.422762688

A B34 Parameter Value35 a 22.877359136 b 0.42276268837 L 75%

3

Construction of Scatter Diagram Suppose you want to create a scatter diagram showing the hours against the unit numbers. You can use the following procedure: 1. Open file Lesson_10a_learning_w03_331.xls. Use worksheet Regression 2. Highlight cells A6 to B13 and click Chart Wizard shown below. 3. Chart Wizard dialog box Step 1 of 4: choose Chart Type XY (Scatter), the one in the

middle row and first column. C lick Next. 4. Chart Wizard dialog box Step 2 of 4: Click Next. 5. Chart Wizard dialog box Step 3 of 4: provide the following information:

Chart title: Unit Numbers vs. Hours Category X axis: Unit Number, u Value Y axis: Number of Hours, Y(u)

6. Chart Wizard dialog box Step 4 of 4: click the radio button As new sheet: and type a label e.g., u vs Y(u). Click Finish.

7. Next, you can do some minor modifications. Click on the box that says Series1 and delete it. Double click the Chart Area (the white portion) and change font to 24 pt. Double click the Plot Area (the gray portion). You will see the Format Plot Area dialog box. On the left side of the dialog box there are radio buttons for border and on the right side of the box there are radio buttons for area. Choose none on both sides (i.e., no border and no color on the plot area).

The graph obtained by using the above procedure is shown on right using copy and paste.

Unit Numbers vs Hours

0123456789

10

0 2000 4000 6000 8000 10000 12000

Unit Numbers, u

Hou

rs, Y

(u)

Chart Wizard

4

Similarly, you can create another scatter diagram showing ln(hours) against ln(unit numbers). The procedure is stated below. 1. Open file Lesson_10a_learning_w03_331.xls. Use worksheet Regression 2. Highlight cells A18 to B25 and click Chart Wizard. 3. Chart Wizard dialog box Step 1 of 4: choose Chart Type XY (Scatter), the one in the

middle row and first column. Click Next. 4. Chart Wizard dialog box Step 2 of 4: Click Next. 5. Chart Wizard dialog box Step 3 of 4: provide the following information:

Chart title: ln(Unit Numbers) vs. ln(Hours) Category X axis: ln(u), u Value Y axis: ln(Y(u))

6. Chart Wizard dialog box Step 4 of 4: click the radio button As new sheet: and type a label e.g., ln(u) vs ln(Y(u)). Click Finish.

7. Next, you can do some minor modifications. Click on the box that says Series1 and delete it. Double click the Chart Area (the white portion) and change font to 24 pt. Double click the Plot Area (the gray portion). You will see the Format Plot Area dialog box. On the left side of the dialog box there are radio buttons for border and on the right side of the box there are radio buttons for area. Choose none on both sides (i.e., no border and no color on the plot area).

The graph obtained by using the above procedure is shown on the next page using copy and paste.

ln(Unit Numbers) vs ln(Hours)

-1

-0.5

0

0.5

1

1.5

2

2.5

0 2 4 6 8 10

ln(u)

ln(Y

(u))

1

Lesson 10b: Aggregate Planning Finding An Optimal Production Plan Using Excel Solver

This lesson illustrates the procedure of finding an optimal production plan using Excel Solver. Download the Excel file Lesson_10b_Agg_P_w03_331.xls from the course website and follow the instruction under “Setting up the Spreadsheet” and “Setting up the Excel Solver”. Input: Demand forecast for each quarter, number of workers available in the beginning, hiring and firing costs per worker, straighttime and over time costs per hour, inventory holding and backorder costs per unit, productivity (units per hour), working hours (hours per day and days per quarter) and beginning inventory. Output (Decision Variables): Decide for each quarter: number of workers hired/ fired, overtime hours, number of units in the ending inventory and backorder. Objective: Minimize the total cost of straight-time, overtime, hiring, firing, inventory holding and backorder Relationships 1. Inventory at the beginning of the current quarter (except the first quarter) =

positive inventory of the previous quarter – units backordered in the previous quarter

2. Number of workers available in the current quarter = number of workers

employed in the previous quarter + number of workers hired in the current quarter – number of workers fired in the current quarter

3. Regular production hours in a quarter = (Number of workers available in the

current quarter) × (Number of working hours per day) × (Number of working days per quarter)

4. Actual number of units produced in a quarter = (Regular production hours +

overtime production hours) × (productivity in units per hour) 5. Inventory at the end of the current quarter = Actual number of units produced

in the current quarter + Inventory at the beginning of the current quarter – Demand forecast of the current quarter

6. Inventory at the end of the current quarter = positive inventory of the current

quarter – units backordered in the current quarter

2

7. Costs: • Straight-time cost=(regular production hours)×(straight-time cost per hour) • Overtime cost=(overtime production hours)×(overtime cost per hour) • Inventory holding cost=(ending inventory)×(holding cost per unit) • Backorder cost=(units backordered)×(backorder cost per unit) • Hiring cost=(number of workers hired)×(hiring cost per worker) • Firing cost=(number of workers fired)×(firing cost per worker)

The Problem Minimize total cost such that ending inventory computed in (5) equals ending inventory computed in (6).

Setting up the Spreadsheet A spreadsheet is set up to compute the total cost from the given values of inputs and decision variables. Input: Known input values are entered in each of the cells below: • B5: Initial number of workers • B8: Hiring cost for each temporary workers • B9: Firing cost for each worker laid off • B11: Straight-time cost per hour • B12: Overtime cost per hour • B14: Holding cost per unit-quarter • B15: Backorder cost per unit • B16: Number of units produced per worker-hour • B17: Number of hours worked per day • B18: Number of days worked per quarter • B23 to B26: Forecast of demand in fall, winter, spring and summer • C23: Initial inventory, in the beginning of fall Output (Decision Variables): Excel solver will output the values of the decision variables in the following cells: • D23 to D26: numbers of workers hired in fall, winter, spring and summer • E23 to E26: numbers of workers fired in fall, winter, spring and summer • G23 to G26: overtime hours in fall, winter, spring and summer • E31 to E34: units in (positive) inventory in fall, winter, spring and summer • F31 to F34: units backordered in fall, winter, spring and summer

3

A B C D E F G

1 Lab Exercise 2 3 Problem 7: Linear optimization 4 5 Number of Workers 30 6 7 Recruitment costs 8 Hiring cost 100 for each temporary worker 9 Firing cost 200 for each worker laid off 10 Wages 11 Straight time cost 5 per hour 12 Overtime cost 8 per hour 13 Inventory costs 14 Holding cost 5 per unit-quarter 15 Backorder cost 10 per unit 16 Productivity 0.5 units per worker hour 17 Working hours 8 hours per day 18 60 days per season 19 20 Forecast Beginning Workers Workers Total Overtime 21 Inventory Hired Fired Workers Hours 22 23 Fall 10000 500 5 0 ? 0 24 Winter 8000 ? 0 5 ? 0 25 Spring 7000 ? 0 5 ? 0 26 Summer 12000 ? 2 0 ? 0 27 28 Regular Actual Ending Positive Backorder Ending 29 Production Production Inventory Inventory Inventory 30 Hours Units Computed Chosen 31 Fall ? ? ? 0 0 ? 32 Winter ? ? ? 0 0 ? 33 Spring ? ? ? 0 0 ? 34 Summer ? ? ? 0 5000 ? 35 36 Bakorder Overtime Hiring Firing Inventory Straight- 37 Cost Cost Cost Cost Holding time 38 Cost Cost 39 Fall ? ? ? ? ? ? 40 Winter ? ? ? ? ? ? 41 Spring ? ? ? ? ? ? 42 Summer ? ? ? ? ? ? 43 44 Total cost ?

4

Relationships Formulae are entered for relationships explained below: 1. Inventory at the beginning of the current quarter (except the first quarter) =

positive inventory of the previous quarter – units backordered in the previous quarter

C24 = D31 (copied to C25 and C26) 2. Number of workers available in the current quarter = number of workers

employed in the previous quarter + number of workers hired in the current quarter – number of workers fired in the current quarter

F23 = $B$5+D23–E23 F24 = F23+D24–E24 (copied to C25 and C26)

3. Regular production hours in a quarter = (Number of workers available in the current quarter) × (Number of working hours per day) × (Number of working days per quarter)

B31 = F23*$B$17*$B$18 (copied to B32, B33 and B34) 4. Actual number of units produced in a quarter = (Regular production hours +

overtime production hours) × (productivity in units per hour) C31 = (B31+G23)*$B$16 (copied to C32, C33 and C34)

5. Inventory at the end of the current quarter = Actual number of units produced in the current quarter + Inventory at the beginning of the current quarter – Demand forecast of the current quarter

D31 = C31+C23–B23 (copied to D32, D33 and D34) 6. Inventory at the end of the current quarter = positive inventory of the current

quarter – units backordered in the current quarter G31 = E31–F31 (copied to G32, G33 and G34)

7. Costs: • Straight-time cost=(regular production hours)×(straight-time cost per hour)

G39 = B31*$B$11 (copied to G40, G41 and G42) • Overtime cost=(overtime production hours)×(overtime cost per hour)

C39 = G23*$B$12 (copied to C40, C41 and C42) • Inventory holding cost=(ending inventory)×(holding cost per unit)

F39 = E31*$B$14 (copied to F40, F41 and F42) • Backorder cost=(units backordered)×(backorder cost per unit)

B39 = F31*$B$15 (copied to B40, B41 and B42) • Hiring cost=(number of workers hired)×(hiring cost per worker)

D39 = D23*$B$8 (copied to D40, D41 and D42) • Firing cost=(number of workers fired)×(firing cost per worker)

E39 = E23*$B$9 (copied to E40, E41 and E42) • Total cost

B44 = SUM(B39:G42)

5

Setting up the Excel Solver 1. Add-in Solver, if needed: Click on the “Tools” menu. If the “Solver” is not one

of the items of the “Tools” menu, choose the “Add-Ins.” Click on the box adjacent to “Solver Add-In” (a check mark will appear) and click “OK.”

2. Define objective: From the “Tools” menu, choose “Solver.” The “Solver

Parameters” Window pops up. Click on the arrow adjacent to the box “Set Target Cell.” Click on the Cell B44, that stores the total cost and click again on the arrow. Click on the circle adjacent to “Min.”

3. Decision variables: The cell addresses for the decision variables are entered

in the box below “By Changing Cells:.” The decision variables are $D$23:$E$26, $G$23:$G$26, $E$31:$F$34

6

4. Constraints: The constraints are entered in the box below “Subject to the Constraints.” Click on “Add” and the “Add Constraint” window pops up. Provide the information on constraint as shown below and click on “OK”:

5. Linearity and Non-negativity: From the “Solver Parameters” window click on

“Options.” The “Solver Options” window pops up. Click on a box adjacent to “Assume Linear Model” and another adjacent to “Assume Non-Negativity.” Click on “OK.”

6. Solve: Finally, from the “Solver Parameters” window click on “Solve.” A

“Solver Results” window pops up. Click on “OK.”

7

7. The Solution: See the changes to your spreadsheet. The cells for the decision variables are expected to contain the optimal solution and its cost in the cell for the total cost.

8. A limitation: The solution does not give integer values for workers hired and

fired. To avoid the limitation, add the constraint as shown below and deselect the “Assume linear model” using Solver Options window. However, While this change will provide integer values for workers hired and fired, an optimal solution is not guaranteed.

1

1

LESSON 11: INVENTORY MODELS (DETERMINISTIC)

Outline

• Hierarchy of Production Decisions• Inventory Control Questions• Inventory Control Costs• The Economic Order Quantity (EOQ) model• The EOQ Model Costs• The EOQ• Some Important Characteristics of the EOQ Cost

Function

2


• In Chapter 2, Lessons 4-7, we discuss demand forecasting

• Forecasting provides the demand of products in the near future

• In Chapter 3, Lessons 8-10, we discuss aggregate planning.

• Aggregate production plan is translated into a detail product-wise production plan by Master Production Schedule (MPS).

• The next question in the hierarchy of decision making is inventory control.

2

3

Aggregate Planning


Inventory Control


Vehicle Routing

Forecast of Demand


4

Inventory Control Questions

• Inventory control questions:– Raw materials, components (subassemblies), and

finished goods may be purchased or produced in-house.

– Timing and amount of purchase and production must be carefully planned. There are two major questions:

1. How much (lot size)? How much to purchase or produce?

2. When?When to purchase or produce?

3

5


• There can be two extreme strategies:– Small lot sizes and too frequent inventory

replenishment• This strategy yields higher ordering costs (if the

items are purchased) or setup costs (if the items are produced).

• The demand forecast is for a short term and it is more accurate.

– Large lot sizes and too infrequent inventory replenishment

6


– Large lot sizes and too infrequent inventory replenishment

• This strategy requires a large investment in the inventory and yields higher inventory holding costs.

• The demand forecast is for a long term and it is less accurate.

4

7

Inventory Control Costs

• Inventory holding costs,– Opportunity cost of capital – Cost of storage space– Taxes and insurance against fire, theft, and other

losses – Breakage, spoilage, deterioration and obsolescence– Example of calculation of inventory holding costs

Cost of capital - 15%Taxes and insurance - 2%Storage - 5%Breakage/spoilage - 3%

Total - 25%

Ich =

8


• Inventory holding costs,– Notation:

= Annual interest rate= Dollar value of one unit of inventory= holding cost in terms of dollars per unit per year

– Then, we have the relationship

Ich =

Ich =

Ich

5

9


• Order costs or setup costs, – If the items are purchased, a fixed ordering cost may be

incurred each time an order is placed. Similarly, if the items are produced, a fixed setup cost may be incurred each time the production facility is set up to produce the item.

– Some examples are bookkeeping expense, order processing fees, transportation costs, receiving costs, handling costs, etc.

K

10


• Order costs or setup costs,– If there is a variable part of the order cost/setup that

depends on the number of units ordered/produced, the variable cost is usually not considered in the order cost. Instead, the variable cost may be included in the cost of the item.

– For example, the salary paid to the purchasing clerk does not depend on the number of times orders are placed. Such a cost is an overhead expense, and not a part of the ordering cost.

– Notation:= ordering/setup cost per order/setup.K

K

6

11


• Penalty costs, – Shortages occur when the demand exceeds the amount

of inventory on hand. One of two types of costs is charged depending on whether a shortage results in loss of sales or not:

• Backorder - if the excess demand is backlogged and fulfilled in a future period, a backorder cost is charged (bookkeeping and/or delay costs).

• Lost sales - if the excess demand is lost because the customer goes elsewhere, the lost sales is charged. The lost sales include goodwill and loss of profit margin. So, penalty cost = selling price - unit variable cost + goodwill, if there exists any.

p

12


• Penalty costs, – In Chapter 4, Lessons 11-15, we assume that the

demands are known and fixed and shortages will not take place. So, penalty costs are not considered.

– In Chapter 5, Lessons 16-20, we assume that the demands are uncertain and shortages may occur. So, penalty costs are considered in Chapter 5.

p

7

13

• Major assumptions 1. Demand is known and fixed (uniform). The rate of

demand is units per year. This assumption is relaxed in Chapter 5, Lessons 16-20, Stochastic Inventory Models.

2. The cost parameters, unit cost, holding cost, and ordering cost are known and fixed. Given that the inventory control decisions are made for a short term, it’s likely that the costs will not change during the planning period. Still, this assumption is a simplification. The costs parameters may change over time.

3. Shortages are not permitted. This assumption is relaxed in Chapter 5, Lessons 16-20, Stochastic Inventory Models.

The EOQ Model

λ

hK

c

14

• Major assumptions 4. The inventory level increases instantaneously at one

point of time when an order is received. This assumption is appropriate in the context of purchasing. The EPQ model, discussed in Lesson 12, considers a gradual increase in the inventory level. The EPQ model is appropriate in the context of production.

5. There is no price discount for large order sizes. This assumption is relaxed later. See Lesson 13, the EOQ with price break.

The EOQ Model

8

15

• Constant order size and inventory cycle– Since there is no uncertainty, it’s optimal to plan an

order receipt only when the inventory level reaches zero. – Suppose that there is no inventory in the beginning. So,

an order will be received in the beginning.– Since the demand and cost parameters do not change

over time• If it’s optimal to order units in the beginning, it’s

also optimal to order units next time. – The above observation provides two important concepts.

One is the constant order size and the other is the inventory cycle.

The EOQ Model

QQ

16

• Constant order size and inventory cycle– The order size, is chosen to minimize the total

inventory control costs. The formula for optimal order quantity is given later.

– At the start of each inventory cycle, the inventory level is . The inventory level decreases uniformly. In the end of the inventory cycle the inventory level is zero. So, the length of the cycle is the length of time over which the demand is . Thus, the length of the cycle in years is

The EOQ Model

Q

Q

Q

λ=

QT

9

17

The EOQ Model

Demand rate

0 TimeLeadtime

Leadtime

Order Placed

Order Placed

Order Received

Order Received

Inve

ntor

y Le

vel

Reorder point, R

Order qty, Q

18

• Lead time and reorder point– Sometimes, there may be a lead time associated with

the orders. The lead time, is the length of time between order placement and order receipt.

– The presence of lead time requires the order be placed some time before it is needed. For example, if there is a lead time of 2 days, the order must be placed when the inventory is sufficient to meet the demand for 2 days.

– Reorder point, is the inventory level at the time of order placement.

– Since, reorder point must cover the lead time demand,

The EOQ Model

τ

R

λτ=R

10

19

The EOQ Model Costs

QKhQ

Qc

QK

Q

Qh

QQ

λ+=

λ

λ=

λ=

=

=

2

2

2

TC cost, annual total the Hence,

.considered not is and on tindependen is items, the buying of cost annual The

cost ordering annual the So,

orders of number annual The

cost holdinginventory annual the So,

inventory average the So,

to 0 from variesuniformly levelInventory

20

Slope = 0 Total Cost

Ordering Cost =

Order Quantity, Q

Annualcost ($)

Minimumtotal cost

Optimal solution, Q*, Economic Order Quantity (EOQ)

Holding Cost =2

hQ

QKλ

The EOQ Model Costs

QKhQ λ+=

2 TC

11

21

The EOQ

As the annual demand and cost parameters are fixed (see major assumptions), it’s important to analyze the effect of order quantity on the annual ordering, holding and total costs. The previous slide shows such an effect.

Annual ordering cost, decreases as the order quantityincreasesAnnual holding cost, increases as the order quantityincreasesThe total cost, . The total cost curve is nearly U-shaped.The total cost is minimum for some order quantity that’s neither too small nor too large.

QKhQ λ+=

2 TC

2hQQKλ

Khc and ,λ

22

The EOQ

( )( )

TC. minimizes EOQ, Hence,

by denoted and

(EOQ)quantity OrderEconomic called isquantity order This

or, or, Set :Answer

minimizes that Find :Question

hK

Q

Q

hK

QQKh

QdTCd

QKhQ

TCQ

λ=

λ==

λ−=

λ+=

2

.

20

20

2

*

*

2

12

23

The EOQ

• Recall that there are two major inventory control questions, how much and when. The EOQ model answers these questions as follows.

• How much to order? – Order

• When to order? – When the inventory on hand reaches the reorder point

hKQ λ= 2* EOQ,

λτ=R

24

Example 1: R & B beverage company has a soft drink product that has a constant annual demand rate of 3600 cases. A case of the soft drink costs R & B $3. Ordering costs are $20 per order and holding costs are 25% of the value of the inventory. R & B has 250 working days per year, and the lead time is 5 days. Identify the following aspects of the inventory policy:

a. Economic order quantity

13

25

b. Reorder point

c. Cycle time

d. Total annual cost

26

• Some notes on Example 1:

– The total cost curve is flat near the EOQ value. So, the total cost does not change much because of a little change in the order quantity from the EOQ value. See the discussion under some important characteristics. Since the number of cases of soft drinks is a whole number, the EOQ value has been rounded to the nearest integer.

– It’s not a coincidence that the annual holding cost is nearly the same as the annual ordering cost. If the EOQ units are ordered the annual ordering cost is the same as the annual holding cost. See the total cost curve and the discussion under EOQ model costs and some important characteristics. The little difference between two costs is due to the rounding.

14

27

Some Important Characteristics of the EOQ Cost Function

• At EOQ, the annual holding cost is the same as annual ordering cost.

hKQKhQ

hK

hK

KQK

hKhKhhQ

λ=λ

+=

λ=λ

λ=λ=

λ=λ==

22

22

22

22

*

*

*

*

cost annual Total

cost ordering Annual

cost holding Annual

28

• If the order quantity is near the EOQ value, the total cost does not change from the optimal value.– See the discussion under the EOQ model costs and

the figure showing various costs against order quantity. The total cost curve is flat near EOQ. This supports the point.

– The EOQ policy usually provides a good decision even when the cost parameters are little off than the assumed values (and, therefore, the EOQ value is incorrect). So, the EOQ model is insensitive to errors. See text pp. 208-209 for a detail discussion. In this note, the insensitivity is shown with an example.


15

29

CostOpt with

λ I K Q* Cost Q =4383600 25% 20 438 329 3293400 25% 20 426 319 3203600 35% 20 370 389 3943600 25% 30 537 402 411


30

• If the order quantity is near the EOQ value, the total cost does not change from the optimal value.– Consider the Example 1 data. The assumed values of

the annual demand, holding cost and ordering costs are 3600 units, 25% and $20/order respectively. So, the EOQ=438 units.


16

31

• If the order quantity is near the EOQ value, the total cost does not change from the optimal value.– Suppose that the correct value of the annual demand is

3400 units. So, the correct EOQ=426 units and the optimal total annual cost is $319. If the decision maker, being unaware of the correct value of the annual demand, uses an order quantity of 438 units, the total annual cost will be $320 which is less than 0.30% off the optimal value of $319.

– Similarly, the order quantity of 438 units, does not produce a large error when holding cost changes to 35% or ordering cost to $30/order.


32


Lesson 11


Exercise: 10 and 12, p. 210 (4th Ed.), pp. 201-202 (5th Ed.)

17

33

Outline

• Finite Production Rate Model– The model– Cost curves– Characteristics– Example

LESSON 12INVENTORY MODELS (DETERMINISTIC)

FINITE PRODUCTION RATE MODEL

34

Finite Production Rate

• Finite production rate model is an extension of the EOQ model

• The following assumption of the EOQ model is not used in the finite production rate model– All the quantities ordered are received at the same time

(the assumption is suitable when the quantities are purchased)

• An assumption in the finite production rate model– The quantity ordered is produced at an uniform rate,

(the assumption is suitable when the quantities are produced)

• The finite production run model is also called the Economic Production Quantity (EPQ) model

P

18

35


• As in the EOQ model, assume that the inventory at the beginning is zero.

• The production facility is set up to produce units. If it’s optimal to produce units in the beginning, it’s also optimal to produce units next time when the inventory level reaches zero, because no cost parameter changes in between two production runs.

• Therefore, like the EOQ model, the finite production rate model also has a constant order size and an inventory cycle.

Q

QQ

36


• The length of the inventory cycle is the length of time over which the demand is . Thus, the length of cycle in years

• In order to meet the demand (feasibility), the production rate, . Thus, for some time after the start of a new production run, the inventory level increases. The rate of increase is . The length of each production run is called the uptime and is denoted by . The uptime is the time required to produced order quantity, . So,

λ≥P

Q

λ−P

λ=

QT

1T

PQ

T =1

Q

19

37

Finite Production Rate • If the production must be stopped before the end of

the cycle in order to avoid producing more than the order quantity, units. Thus, there is a idle time or downtime in each cycle. The downtime is denoted by .

• Unlike the EOQ model, the maximum inventory in the finite production rate model is not because some items are used to meet the demand before the receipt of all of the order quantity. The maximum inventory is the inventory accumulated over the uptime. Since the inventory level increases at the rate of , the maximum inventory,

Q

λ>P

2T

PQQ

TTT −λ

=−= 12

Q

λ−P

( ) ( )

λ

−=λ−=λ−=P

QPPQ

PTH 11

38


Slope = P - λH

Slope = - λ

Inve

nto

ry

TimeT1 T2

T

20

39


• Note that the downtime demand is met entirely from the maximum inventory. Hence, downtime can also be obtained as follows:

• The formula for the optimal order quantity is shown in the next slide.

PQQ

PQH

T −λ

=

λ

−λ

=λ

= 12

40


λ = annual demandP = the production rate in units/yearQ = size of each production runK = cost of setting up productionh = annual holding cost per unith´ = h(1- λ/P)

'2

*

hK

Q

λ=

(EPQ)Quantity ProductionEconomic , solution, Optimal

21

41


'22'

2'

'2

2'

'2

2'

2'

*

*

*

*

hKQKQh

hK

hK

KQK

hKhKhQh

λ=λ

+=

λ=

λλ

=λ

=

λ=

λ==

cost annual Total

cost setup Annual

cost holding Annual

EPQ At

42

The Finite Production Rate Model Cost Curves

Slope = 0Total Cost

Ordering Cost =

Order Quantity, Q

Annualcost ($)

Minimumtotal cost

Optimal solution, Q*, Economic Production Quantity (EPQ)

Holding Cost =

QKQh λ

+2'

2'Qh

QKλ

22

43

Some Important Characteristics of the EPQ Cost Function

• As in the EOQ model– The annual holding cost is the same as the annual

setup cost at the EPQ– The total cost curve is flat near the EPQ

• So, the total cost does not change much with a slight change in the order quantity

44

Example 2: Vision Optics makes microscope lens housings. Annual demand is 100,000 units per year. Assume that the product can be produced at the rate of 200,000 units per year. Each production run costs $5,000 to set up, and the variable production cost of each item is $10. The annual cost per dollar value of holding items of inventory is $0.20. Compute

a. Economic production quantity

23

45

b. Cycle time

c. The length of each production run (uptime per cycle)

d. The length of downtime per cycle

46

e. The maximum inventory

f. Total annual cost

24

47


Lesson 12

Reading: Section 4.6 , pp. 211-213 (4th Ed.) pp. 202-204 (5th

Ed.)

Exercise: 17 and 19, pp. 213-214 (4th Ed.), p. 204 (5th Ed.)

48

Outline

• EOQ Model with Price Break (All Units)– The model– The cost function and feasibility– The procedure– An example


EOQ MODEL WITH PRICE BREAKS

25

49

• So far, we have assumed that the unit cost is independent of the order quantity.

• Often, the supplier encourages larger order sizes by providing price discounts. Such quantity discounts are common for many consumer products.

• The text discusses two types of discounts. In each case, there are one or more breakpoints defining changes in the unit cost.– All units: discount is applied to all units.– Incremental units: discount is applied to the

additional units beyond the breakpoint.

EOQ Model with Price Breaks (All Units)

50

• The difference between the two types of discounting can be discussed with an example. Suppose that the charge for photocopying is $0.10 per copy for 0-9 copies and $0.08 per copy for 10-49 copies. If 12 copies are made the total charge is computed as follows:– All units: each of the 12 units is charged $0.08. So,

the total charge = $0.08 × 12 = $0.96– Incremental units: the first 9 copies is charged $0.10

each and the remaining 3 units are charged $0.08 each. So, the total charge = $0.10 × 9 + $0.08 × 3 = $1.14.


26

51

• With all units discount schedule, the total charge may be less if a larger quantity is ordered. See the previous example again. If 9 units are ordered, the total charge = $0.10 × 9 = $0.90. But, if 10 units are, the total charge = $0.08 × 10 = $0.80. So, even if you need 9 units, order 10 units, throw away the 10th unit and save $0.10. See the figure on the next slide.

• Despite the above drawback, all units discount is often used because it is much simpler than the incremental units schedule. This course covers only the all units discount schedule.


52


Total Charge (All Units)

00.5

11.5

22.5

33.5

44.5

0 10 20 30 40

Quantity Ordered

To

tal C

har

ge

$

$0.10/unit$0.08/unit

27

53

• Assumptions– Demand occurs at a constant rate of items per

year.– Ordering Cost is per order.– Purchase Cost is per item if the quantity ordered is

between 0 and , if the order quantity is between and , etc.

– Holding Cost is per item in inventory per year (note holding cost is based on the cost of the item, ).

– Delivery time (lead time) is constant.

λ

K

ii Ich =


1c1x 2c 1x

2x

ic

54

• Some Formulae– Optimal order quantity: the procedure for determining

optimal order quantity will be demonstrated

– Number of orders per year:

– Time between orders (cycle time): years

– Total annual cost: (holding + ordering + purchase)

*Q

*Qλ

λ

*Q

isquantity order whencost holding unit

is quantity order whencost unit

==

λ+λ

+*

*

*

*

2 QhQc

cQKhQ

EOQ Model with Price Breaks (All Units)The Cost Function and Feasibility

28

55

• The total annual cost function is different at different price level


example. an withshown benow willThis curve. broken a like looks function

cost annual total (feasible) the result, a As.infeasible be alsomay cost annual total resulting the therefore, and, infeasible bemay quantity order the level price For

TC cost annual total The

and cost holding Thequantity order and level price For

ii

ii

iii

ii

ii

Qc

cQKQh

IchQc

λ+λ

+=

=

2

56

• The total annual cost function function is computed and plotted in the next slide for the following example:


88.2$976.2$20.3$

2025.0

25.0%25092,1

3

2

1

=+==

===

===λ

ccc

KcIch

I

iii

900 899-400

399-0cost Unit size Order

29

57

Total Annual Cost (All units)

3000

3200

3400

3600

3800

4000

4200

100 400 700 1000 1300

Quantity Ordered

To

tal A

nn

ual

Co

st

$3.2/unit$2.976/unt$2.88/unitFeasible cost

58

Steps1. Determine the largest (cheapest) feasible EOQ value

and list all the candidates: The most efficient way to do this is to compute the EOQ for the lowest price first, and continue with the next higher price. Stop when the first EOQ value is feasible (that is, within the correct interval). At each price level choose the candidate for optimal order quantity as follows:

• if the EOQ is feasible, the EOQ (the largest feasible EOQ) is a candidate for the optimal order quantity.

• if the EOQ is not feasible, a candidate for the optimal order quantity is the minimum order quantity (a breakpoint) available at that price range.

EOQ Model with Price Breaks (All Units)The Procedure

30

59

Steps2. Find cost given by each candidate and choose the

best candidate: Compare the total annual cost given by the candidates. The optimal order quantity is the candidate for which the total annual cost is the minimum. Note carefully that this means comparing costs at the largest feasible EOQ and at all the price breakpoints that are greater than the largest feasible EOQ.

EOQ Model with Price Breaks (All Units)The Procedure

60

Example 3: Nick's Camera Shop carries Zodiac instant print film. The film normally costs Nick $3.20 per roll, and he sells it for $5.25. Nick's average sales are 21 rolls per week. His annual inventory holding cost rate is 25% and it costs Nick $20 to place an order with Zodiac. If Zodiac offers a 7% discount on orders of 400 rolls or more and a 10% discount for 900 rolls or more, determine Nick's optimal order quantity.

31

61

Step 1: Determine the largest (cheapest) feasible EOQ and list all the candidates

8823 .$=c cost unit cheapest the Consider

62

97622 .$=c cost unit Consider

32

63

2031 .$=c cost unit Consider

64

Step 2: Find cost given by each candidate and choose the best candidate

=

++=

=

=

++=

=

++=

22

222

2

33

333

3

2

2

2

cQKQh

Q

cQKQh

Q

cQKQh

Q ii

iiii

λλ

λλ

λλ

TC

candidate For

TC

candidate For

TC cost total compute candidate each For

33

65

=

==

=

++=

=

*Q

cQKQh

Q

quantity, order optimal an Hence,

quantity order the for cost annual total minimum The

TC

candidate For

11

111

1

2λ

λ

66


Lesson 13

Reading: Section 4.7 , pp. 214-217 (4th Ed.), pp. 205-208 (5th

Ed.)

Exercise: 22, 24, pp. 220-221(4th Ed.), p. 211 (5th Ed.)

34

67

Outline

• Resource Constrained Multi-Product Inventory Systems– Fund constraint– Space constraint

LESSON 14INVENTORY MODELS (DETERMINISTIC)RESOURCE CONSTRAINED SYSTEMS

68

Resource Constrained Multiple Product Inventory Systems

• So far, we have discussed inventory control models assuming that there is only one product of interest.

• Often, there can be multiple products that may compete for the same resource such as fund, space, etc.

• In such a case, the EOQ solutions may be satisfactory if the fund/space required by the EOQ solutions is less than that available.

• However, The EOQ solutions cannot be implemented if the fund/space required by the EOQ solutions is more than that available.

35

69

• An important observation on the resource constrained models is the following:If the following ratio

is the same for all products, an optimal order quantity of each product can be obtained by reducing its EOQ value by a constant multiplication factor.

• We shall discuss two cases:– Fund constraint (satisfies the above condition)– Space constraint (may not satisfy the condition)

yearper unit per cost Holdingunit per required Resource

Resource Constrained Multiple Product Inventory Systems

70

Fund ConstraintSame Interest Rate For All Products

• If the same interest rate is applied on all products, the ratio

is the same for all products. • Consequently, an optimal order quantity of each

product can be obtained by reducing its EOQ value by a constant multiplication factor.

Rate Interest Rate) restunit)(Inte per (Costunit per Cost

yearper unit per cost Holdingunit per Cost

1==

36

71

Fund ConstraintSame Interest Rate For All Products

Steps1. Compute the EOQ values and the total investment

required by the EOQ lot sizes. If the investment required does not exceed the budget constraint, stop.

2. Reduce the lot sizes proportionately. To do this, multiply each EOQ value by the constant multiplier,

Sizes Lot EOQby Required Investment TotalBudget=m

72

Example: Fund ConstraintSame Interest Rate For All Products

Example 4: A vegetable stand wants to limit the investment in inventory to a maximum of $300. The appropriate data are as follows:

Tomatoes Lettuce ZucchiniAnnual demand 1000 1500 750(in pounds)Cost/pound $0.29 $0.45 $0.25The ordering cost is $5 in each case and the annual interest rate is 25%. What are the optimal quantities that should be purchased?

37

73

==

===

===

==

===

===

22

2

2222

22

11

1

1111

11

2

2

2

1

Qc

hK

EOQQ

Ichi

Qc

hK

EOQQ

Ichi

required Fund

Lettuce, for nComputatio

required Fund

Tomatoes, for nComputatio

λ

λ

74==

==

==

===

===

required fund Totalavailable Fund

factor, tionmultiplica constant theby value EOQ each reduce

required, than less is available fund Since available Fund

required fund Total

required Fund

Zucchini, for nComputatio

m

Qc

hK

EOQQ

Ichi

33

3

3333

33

2

3

λ

38

75

==

==

==

33

22

11

mQQ

mQQ

mQQ

*

*

*

:Zucchini forquantity order Optimal

.

:Lettuce forquantity order Optimal

:Tomatoes forquantity order Optimal

76

Example: Fund ConstraintSame Interest Rate For All Products

Tomatoes Lettuce ZucchiniIndex, i 1 2 3Annual Demand, λi 1000 1500 750Ordering/Set-up Cost, Ki 5 5 5Holding cost/unit/year, hi 0.0725 0.1125 0.0625Cost/unit, wi 0.29 0.45 0.25Lot sizes, QiFund requiredTotal fund requiredFund available 300Constant multiplier, mQi reduced proportionately, Q'i

39

77

Space Constraint

• For each product, compute the following ratio

• If the above ratio is the same for all products, a procedure similar to the one for the budget constraint may be applied.

• Assume that the above ratios are different for different products (a reason may be that space requirement is not necessarily proportional to costs).

yearper unit per cost Holdingunit per required Space

78

Steps1. Compute the EOQ values and the total space

required by the EOQ lot sizes. If the space required does not exceed the space constraint, stop.

2. By trial and error, find a value of such that the space required by the following lot sizes equals the space available:

Space Constraint

θ

ii

iii wh

KQ

θλ

22*

+=

40

79

Space Constraint

iwih

i

iKiQ

i

i

i

i

i

product for unit per required Space product for yearper unit per cost Holding

product for demand Annual

product for cost up-Set product forquantity Order

Where,

===λ

==

80

Example: Space Constraint

Example 5: A vegetable stand has exactly 500 square feet of space. The appropriate data are as follows:

Tomatoes Lettuce ZucchiniAnnual demand 1000 1500 750(in pounds)Space required 0.5 0.4 1(square feet/pound)Cost/pound $0.29 $0.45 $0.25The ordering cost is $5 in each case and the annual interest rate is 25%. What are the optimal quantities that should be purchased?

41

81

2 Step to goavailable, than more is required space Since

required space Total

Zucchini,

Lettuce,

Tomatoes,space more require values EOQ if Check 1: Step

=++=

===

===

===

===

===

===

332211

3

3333

33

2

2222

22

1

1111

11

2

3

2

2

2

1

QwQwQw

hK

EOQQ

Ichi

hK

EOQQ

Ichi

hK

EOQQ

Ichi

λ

λ

λ

,

,

,

82

( )

( )

( )

( )0262.00625.0

41.3467373.075052

00.121

1181.01125.015.3657373.0

15005240.02

1

0609.00725.039.3717373.0

10005250.02

1

221

7373.0

23

22

21

2

=

−

×××

×=θ

=

−

×××

×=θ

=

−

×××

×=θ

−

×λ

=θ

===

θ

ii

ii

ii h

EOQmK

w

m678.16

500sizes lot EOQ theby reuired Space

available Spacevalues possible of range the Find

:2 and 1 Steps between step optional An

42

83

( )( )

( )( )

0.1181 and 0.0262 between search to sufficient is it Hence,

on bound upper An

on bound lowerA values possible of range the Find

:)(continued 2 and 1 Steps between step optional An

θ==

θθθ=θ

==

θθθ=θ

θ

1181.00262.0,1181.0,0609.0max

,,max

0262.00262.0,1181.0,0609.0min

,,min

321

321

84value? trial the decrease or increase you Will:Question


required Space

required Space

required Space

of value Trial :2 Step

=

==

=+

=

==

=+

=

==

=+

=

=

33

33

333

22

22

222

11

11

111

22

22

22

Qw

whK

Q

Qw

whK

Q

Qw

whK

Q

θλ

θλ

θλ

θ

43

85value? trial the decrease or increase you Will:Question


required Space

required Space

required Space


=

==

=+

=

==

=+

=

==

=+

=

=

33

33

333

22

22

222

11

11

111

22

22

22

Qw

whK

Q

Qw

whK

Q

Qw

whK

Q

θλ

θλ

θλ

θ

86process? error and trial the stop youdo When:Question


required Space

required Space

required Space


=

==

=+

=

==

=+

=

==

=+

=

=

33

33

333

22

22

222

11

11

111

22

22

22

Qw

whK

Q

Qw

whK

Q

Qw

whK

Q

θλ

θλ

θλ

θ

44

87


Trial value of θTomatoes Lettuce Zucchini

Index, i 1 2 3Annual Demand, λi 1000 1500 750Ordering/Set-up Cost, Ki 5 5 5Holding cost/unit/year, hi 0.0725 0.1125 0.0625Space requirement/unit, wi 0.5 0.4 1Lot sizes, QiSpace requiredTotal space requiredSpace available 500Conclusion

88


Trial value of θ 0.1Tomatoes Lettuce Zucchini

Index, i 1 2 3Annual Demand, λi 1000 1500 750Ordering/Set-up Cost, Ki 5 5 5Holding cost/unit/year, hi 0.0725 0.1125 0.0625Space requirement/unit, wi 0.5 0.4 1Lot sizes, Qi 240.77 279.15 169.03Space required 120.39 111.66 169.03Total space required 401.0748Space available 500Conclusion Decrease trial value (why?)

45

89



Index, i 1 2 3Annual Demand, λi 1000 1500 750Ordering/Set-up Cost, Ki 5 5 5Holding cost/unit/year, hi 0.0725 0.1125 0.0625Space requirement/unit, wi 0.5 0.4 1Lot sizes, Qi 328.80 341.66 270.50Space required 164.40 136.66 270.50Total space required 571.5639Space available 500Conclusion Increase trial value (why?)

90



Index, i 1 2 3Annual Demand, λi 1000 1500 750Ordering/Set-up Cost, Ki 5 5 5Holding cost/unit/year, hi 0.0725 0.1125 0.0625Space requirement/unit, wi 0.5 0.4 1Lot sizes, Qi 294.41 319.66 224.93Space required 147.21 127.86 224.93Total space required 499.9997Space available 500Conclusion ?

46

91

==

==

==

33

22

11

QQ

QQ

QQ

*

*

*

:Zucchini forquantity order Optimal

:Lettuce forquantity order Optimal

:Tomatoes forquantity order Optimal


92


Lesson 14

Reading: Section 4.8 , pp. 221-225 (4th Ed.), pp. 212-215

Exercise: 26, 28, p. 225 (4th Ed.), p. 215

47

93

Outline

• EOQ Model for Production Planning– The multi-product inventory control model with a

finite production rate – An example showing the problem with separate

EPQ computation– The procedure– An example


EOQ MODEL FOR PRODUCTION PLANNING

94

EOQ Model for Production Planning

• This model is an extension of the EPQ model• Consider the problem of producing many products in

a single facility. The facility may produce only one product at a time.

• In each production cycle there is only one setup for each product, and the products are produced in the same sequence in each production cycle. This assumption is called the rotation cycle policy.

• For example, if there are three products A, B and C, then a production sequence under the rotation cycle policy is A, B, C, A, B, C, ….

48

95


• The goal is to determine the optimal production quantities of various products produced in each cycle and the optimal length of the cycle.

• Finding optimal production quantity of each product separately using the EPQ formula

may not give a good solution because a production quantity may not be large enough to meet the demand between two production runs of the product.

'

2*

j

jjj h

KQ

λ=

96


• For example, suppose that there are three products A, B and C, then a production sequence under the rotation cycle policy is A, B, C, A, B, C, ….

• The production quantity of product A obtained from the EPQ formula may not be large enough to meet the demand during the production run of products B and C.

• The next example elaborates on the problem of using EPQ formula separately for each product.

49

97

Example 6: Tomlinson Furniture has a single lathe for turning the wood for various furniture pieces including bedposts, rounded table legs, and other items. Two products and some relevant information appear below:

Annual Setup Time Unit Annual Piece Demand (hours) Cost ProductionJ-55R 18,000 1.2 $20 33,600H-223 24,000 0.8 35 52,800Worker time for setup is valued at $85 per hour, and holdingcosts are based on a 20 percent annual interest charge. Assume 8 hours per day and 240 days per year.

ExampleProblem with Separate EPQ Computation

Optional

98

ExampleProblem with Separate EPQ Computation

Find the optimal production quantities separately for each product and show that production quantity of H-223 is not large enough to meet the demand between two production runs of H-223.

( ) /unit/year

/unit/yeartime workertime up-set

:55R-J Product

8571.1$5357.0141'

5357.0600,33000,18

4$2020.0102$852.1

=−=

λ

−=

==λ

=×===×=×=

Phh

P

IchK

Optional

50

99

( )

met) demand (downtime days days

(days) for lastsinventory Maximum

days day per hr hr/ slide) next the (see days

223-H of time setup223-H for Uptime required (days), downtime Minimum

days Uptime,

units Inventory,Max

units EPQ

:slide) previous the from (continued 55R-J Product

3027.47047.8

2400001,188509.652

3027488020274

0439.10240600,331404.406,1

85.6525357.0114.406,11

1404.406,18571.1

000,181022'

2

*

1

*

*

>=

=λ

=

=+=

+=

===

=−=

λ

−=

=××

=λ

==

H.

..

PQ

T

PQH

hK

Q

Optional

100

( )

( )

days Uptime,

units Inventory,Max

units EPQ

/unit/year

/unit/year

time workertime up-set:223-H Product

2027.4240800,525848.924

32.5044545.0158.9241

5848.9248182.3

000,24682'

2

8182.3$4545.0171'

4545.0800,52000,24

7$3520.0

68$858.0

*

1

*

*

===

=−=

λ

−=

=××

=λ

==

=−=

λ

−=

==λ

=×==

=×=×=

PQ

T

PQH

hK

Q

Phh

P

Ich

K

Optional

51

101

produced. is 55R-J product whenmet be cannot 223-H Product

for demand produced, are quantities EPQ If :Conclusion

met) be cannot demand (downtime days days

(days) for lastsinventory Maximumdays day per hr hr/ days

55R-J of time setup55R-J for Uptime required (days), downtime Minimum

:slide) previous the from (continued 223-H Product

1939.100432.5

240000,243190.504

1939.1082.10439.10

<=

=λ

=

=+=+=

H

Optional

102

J-55R H-223Annual production rate (units/year) 33600 52800Annual Demand rate (units/year) 18000 24000Setup time (hour) 1.2 0.8K, Setup cost, $Unit cost, $ 20 35h, Holding cost/unit/yeardemand/productionh'

Separate EPQ solutionQ*Maximum inventoryUptime (days)Minimum downtime (days)Maximum inventory lasts for (days)

Optional

52

103

J-55R H-223Annual production rate (units/year) 33600 52800Annual Demand rate (units/year) 18000 24000Setup time (hour) 1.2 0.8K, Setup cost, $ 102 68Unit cost, $ 20 35h, Holding cost/unit/year 4 7demand/production 0.5357 0.4545h' 1.8571 3.8182

Separate EPQ solutionQ* 1406.1404 924.5848Maximum inventory 652.8509 504.3190Uptime (days) 10.0439 4.2027Minimum downtime (days) 4.3027 10.1939Maximum inventory lasts for (days) 8.7047 5.0432

Optional

104

• The previous example shows that if production quantities of different products are computed separately, then demand of every product may not be met.

• Therefore, all the products must be considered at the same time.

• To solve the integrated problem, first, the cycle time is computed. For each product , the production quantity is the demand of the product during the cycle time. If is the annual demand of product


T

TQ jj λ=

jjQ

jλ j

53

105

• Let T be the cycle time and Tj be the production time of product j

• Let sj be the setup time of product j and n be the number of products


j

jjjjjj PT

TTPTQ

λλ =⇒==

time Idle+++++++= nn TsTsTsT L2211

106

∑

∑

∑∑∑∑

=

=

====

−≥⇒

≤+⇒≤+⇒

≤++++++⇒

+++++++=

n

j j

j

n

jj

n

j j

jn

jj

n

j

jn

j

j

nn

nn

P

sT

Ps

TTT

Ts

TT

Ts

TT

Ts

TT

Ts

TsTsTsT

1

1

1111

2211

2211

1

11

1

1

λ

λ

L

L time Idle


54

107

• Two rules for T*

• T* is the maximum of the two. T* = max (Cycle1, Cycle2)


∑

∑

=

=

λ−

=≥n

j j

j

n

jj

P

sT

1

1*

1Cycle1

∑

∑

=

=

λ=≥ n

jjj

n

jj

h

KT

1

1*

'

2Cycle2

108

ExampleEOQ Model for Production Planning

Example 7: Tomlinson Furniture has a single lathe for turning the wood for various furniture pieces including bedposts, rounded table legs, and other items. Two products and some relevant information appear below:

Annual Setup Time Unit Annual Piece Demand (hours) Cost ProductionJ-55R 18,000 1.2 $20 33,600H-223 24,000 0.8 35 52,800Worker time for setup is valued at $85 per hour, and holdingcosts are based on a 20 percent annual interest charge. Assume 8 hours per day and 240 days per year. Find the optimal production quantities.

55

109

=+=

=

=+==

=+==

=+==

∑∑

∑∑

∑∑

∑∑

==

==

==

==

2211

2

11

2

2

1

12

11

21

2

11

21

2

11

λλ

λλ

λλλλ

''

''

hh

hh

PPPP

KKKK

ssss

jjj

n

jjj

j j

jn

j j

j

jj

n

jj

jj

n

jj

110

( )==

==

=−

=

∑

∑

∑

∑

=

=

=

=

Cycle2Cycle1,

Cycle2

Cycle1

max

'

*T

h

K

P

s

n

jjj

n

jj

n

j j

j

n

jj

1

1

1

1

2

1

λ

λ

56

111

(check) 223-H of time) setup uptime55R-J of downtime


Uptime Downtime

Uptime,

Inventory,Max

:55R-J Product

+>=

==

=−=

==

=

−=

==

(

*

*

*

**

λ

λ

λ

HTP

QT

PQH

TQ

1

1

112

(check) 55R-J of time) setup uptime223-H of downtime


Uptime Downtime

Uptime,

Inventory,Max

:223-H Product

+>=

==

=−=

==

=

−=

==

(

*

*

*

**

λ

λ

λ

HTP

QT

PQH

TQ

1

1

57

113

J-55R H-223 TotalK, Setup cost, $ 102 68 170h, Holding cost/unit/year 4 7demand/production 0.5357 0.4545 0.9903h' 1.8571 3.8182h'(demand)

EOQ Model for Production PlanningCycle1 in days Cycle2 in daysCycle time=max(cycle1, cycle2) in daysQ* = cycle time demandMaximum inventoryUptime (days)Downtime (days)Maximum inventory lasts for (days)

114


Lesson 15

Reading: Section 4.9 , pp. 226-229 (4th Ed.), pp. 215-220 (5th

Ed.)

Exercise: 29, 30 pp. 230-231(4th Ed.), pp. 219-220 (5th Ed.)

58

115

Outline

• Single-Period Models– Discrete Demand– Continuous Demand, Uniform Distribution– Continuous Demand, Normal Distribution

• Multi-Period Models – Given a Q, R Policy, Find Cost– Optimal Q, R Policy without Service Constraint– Optimal Q, R Policy with Type 1 Service Constraint– Optimal Q, R Policy with Type 2 Service Constraint

LESSON 16: INVENTORY MODELS (STOCHASTIC)

116

Stochastic Inventory Control Models Inventory Control with Uncertain Demand

• In Lessons 16-20 we shall discuss the stochastic inventory control models assuming that the exact demand is not known. However, some demand characteristics such as mean, standard deviation and the distribution of demand are assumed to be known.

• Penalty cost, : Shortages occur when the demand exceeds the amount of inventory on hand. For each unit of unfulfilled demand, a penalty cost of is charged. One source of penalty cost is the loss of profit. For example, if an item is purchased at $1.50 and sold at $3.00, the loss of profit is $3.00-1.50 = $1.50 for each unit of demand not fulfilled.

p

p

59

117

• More on penalty cost, – Penalty cost is estimated differently in different situation.

There are two cases:1. Backorder - if the excess demand is backlogged and

fulfilled in a future period, a backorder cost is charged. Backorder cost is estimated from bookkeeping, delay costs, goodwill etc.

2. Lost sales - if the excess demand is lost because the customer goes elsewhere, the lost sales is charged. The lost sales include goodwill and loss of profit margin. So, penalty cost = selling price - unit variable cost + goodwill, if there exists any goodwill.

p

Stochastic Inventory Control Models Inventory Control with Uncertain Demand

118

Single- and Multi- Period Models

• Stochastic models are classified into single- and multi-period models.

• In a single-period model, items are received in the beginning of a period and sold during the same period. The unsold items are not carried over to the next period.

• The unsold items may be a total waste, or sold at a reduced price, or returned to the producer at some price less than the original purchase price.

• The revenue generated by the unsold items is called the salvage value.

60

119


• Following are some products for which a single-period model may be appropriate: – Computer that will be obsolete before the next order– Perishable products– Seasonal products such as bathing suits, winter

coats, etc.– Newspaper and magazine

• In the single-period model, there remains only one question to answer: how much to order.

120


• In a multi-period model, all the items unsold at the end of one period are available in the next period.

• If in a multi-period model orders are placed at regular intervals e.g., once a week, once a month, etc, then there is only one question to answer: how much to order.

• However, we discuss Q, R models in which it is assumed that an order may be placed anytime. So, as usual, there are two questions: how much to order and when to order.

61

121

Single-Period Models

• In the single-period model, there remains only one question to answer: how much to order.

• An intuitive idea behind the solution procedure will now be given: Consider two items A and B. – Item A

• Selling price $900• Purchase price $500• Salvage value $400

– Item B• Selling price $600• Purchase price $500• Salvage value $100

122


• Item A– Loss resulting from unsold items = 500-400=$100/unit – Profit resulting from items sold = $900-500=$400/unit

• Item B– Loss resulting from unsold items = 500-100=$400/unit – Profit resulting from items sold = $600-500=$100/unit

• If the demand forecast is the same for both the items, one would like to order more A and less B.

• In the next few slides, a solution procedure is discussed that is consistent with this intuitive reasoning.

62

123


• First define two terms:

• Loss resulting from the items unsold (overage cost)co= Purchase price - Salvage value

• Profit resulting from the items sold (underage cost)cu = Selling price - Purchase price

• The QuestionGiven costs of overestimating/underestimating demand and the probabilities of various demand sizes how many units will be ordered?

124

• Demand may be discrete or continuous. The demand of computer, newspaper, etc. is usually an integer. Such a demand is discrete. On the other hand, the demand of gasoline is not restricted to integers. Such a demand is continuous. Often, the demand of perishable food items such as fish or meat may also be continuous.

• Consider an order quantity Q• Let p = probability (demand<Q)

= probability of not selling the Qth item.• So, (1-p) = probability of selling the Qth item.

Single-Period Models (Discrete Demand)Decision Rule

63

125

• Expected loss from the Qth item = • Expected profit from the Qth item = • So, the Qth item should be ordered if

• Decision Rule (Discrete Demand):– Order maximum quantity Q such that

where p = probability (demand<Q)

uo

u

uo

ccc

por

cppc

+≤

−≤

,

)1(

opc( ) ucp−1

uo

u

ccc

p+

≤

Single-Period Models (Discrete Demand)Decision Rule

126

Example 1: Demand for cookies:Demand Probability of Demand1,800 dozen 0.052,000 0.102,200 0.202,400 0.302,600 0.202,800 0.103,000 0,05

Selling price=$0.69, cost=$0.49, salvage value=$0.29a. Construct a table showing the profits or losses for each

possible quantity (Self study)b. What is the optimal number of cookies to make?c. Solve the problem by marginal analysis.

Single-Period Models (Discrete Demand)

Note: The demand is discrete. So, the demand cannot be other numbers e.g., 1900 dozens, etc.

64

127

Demand Prob Prob Expected Revenue Revenue Total Cost Profit(dozen) (Demand) (Selling Number From From Revenue

all the units) Sold Sold Unsold Items Items

1800 0.052000 0.12200 0.22400 0.32600 0.22800 0.13000 0.05

a. Sample computation for order quantity = 2200:Expected number sold = 1800×Prob(demand=1800) +2000×Prob(demand=2000) + 2200×Prob(demand≥2200) = 1800(0.05)+2000(0.1)+2200(0.2+0.3+0.2+0.1+0.05) = 1800(0.05)+2000(0.1)+2200(0.85) = 2160 Revenue from sold items=2160(0.69)=$1490.4Revenue from unsold items=(2200-2160)(0.29)=$11.6

Single-Period Models (Discrete Demand)Self Study

128

Demand Prob Prob Expected Revenue Revenue Total Cost Profit(dozen) (Demand) (Selling Number From From Revenue

all the units) Sold Sold Unsold Items Items

1800 0.05 1 1800 1242.0 0.0 1242 882 3602000 0.1 0.95 1990 1373.1 2.9 1376 980 3962200 0.2 0.85 2160 1490.4 11.6 1502 1078 4242400 0.3 0.65 2290 1580.1 31.9 1612 1176 4362600 0.2 0.35 2360 1628.4 69.6 1698 1274 4242800 0.1 0.15 2390 1649.1 118.9 1768 1372 3963000 0.05 0.05 2400 1656.0 174.0 1830 1470 360


a. Sample computation for order quantity = 2200:Total revenue=1490.4+11.6=$1502 Cost=2200(0.49)=$1078Profit=1502-1078=$424

b. From the above table the expected profit is maximized for an order size of 2,400 units. So, order 2,400 units.

Self Study

65

129

c. Solution by marginal analysis:

Order maximum quantity, Q such that

Demand, Q Probability(demand) Probability(demand<Q), p1,800 dozen 0.052,000 0.102,200 0.202,400 0.302,600 0.202,800 0.103,000 0,05

( ) =+

≤<=0cc

cQp

u

udemandyProbabilit

==

o

u

cc


130

• In the previous slide, one may draw a horizontal line that separates the demand values in two groups. Above the line, the values in the last column are not more than

The optimal solution is the last demand value above the line . So, optimal solution is 2,400 units.


=+ 0ccc

u

u

66

131

Continuous Distribution

• Often the demand is continuous. Even when the demand is not continuous, continuous distribution may be used because the discrete distribution may be inconvenient.

• For example, suppose that the demand of calendar can vary between 150 to 850 units. If demand varies so widely, a continuous approximation is more convenient because discrete distribution will involve a large number of computation without any significant increase in accuracy.

• We shall discuss two distributions:– Uniform distribution– Continuous distribution

132


• First, an example on continuous approximation.• Suppose that the historical sales data shows:Quantity No. Days sold Quantity No. Days sold

14 1 21 1115 2 22 916 3 23 617 6 24 318 9 25 219 11 26 120 12

67

133


• A histogram is constructed with the above data and shown in the next slide. The data shows a good fit with the normal distribution with mean = 20 and standard deviation = 2.49.

• There are some statistical tests, e.g., Chi-Square test, that can determine whether a given frequency distribution has a good fit with a theoretical distribution such as normal distribution, uniform distribution, etc. There are some software, e.g., Bestfit, that can search through a large number of theoretical distributions and choose a good one, if there exists any. This topic is not included in this course.

134


Mean = 20Standard deviation = 2.49

68

135


The figure below shows an example of uniform distribution.

136

• The decision rule for continuous demand is similar to the decision rule for the discrete demand

• Decision Rule (Continuous Demand):– Order quantity Q such that

where p = probability (demand≤Q)– Note the following difference

• The word “maximum”• The “= “ in the formula for p• The “≤ “ in the definition of p

Single-Period Models (Continuous Demand)Decision Rule

uo

u

ccc

p+

=

69

137

• There is a nice pictorial interpretation of the decision rule. Although we shall discuss the interpretation in terms of normal and uniform distributions, the interpretation is similar for all the other continuous distributions.

• The area under the curve is 1.00. Identify the vertical line that splits the area into two parts with areas

The order quantity corresponding to the vertical line is optimal.


( ) right the on

and left the on

uo

o

uo

u

ccc

p

cccp

+=−

+=

1

138

• Pictorial interpretation of the decision rule for uniform distribution (see Examples 2, 3 and 4 for application of the rule):


850150

Demand

Prob

abili

ty

Area

uo

u

ccc

p

+=

= ( )

uo

o

ccc

p

+=

−= 1 Area

*Q

70

139

• Pictorial interpretation of the decision rule for normal distribution (see Examples 3 and 4 for application of the rule):


Prob

abili

ty

Demand

Area

uo

u

ccc

p

+=

=

*Q

( )

uo

o

ccc

p

+=

−= 1

Area

140

Example 2: The J&B Card Shop sells calendars. The once-a-year order for each year’s calendar arrives in September. The calendars cost $1.50 and J&B sells them for $3 each. At the end of July, J&B reduces the calendar price to $1 and can sell all the surplus calendars at this price. How many calendars should J&B order if the September-to-July demand can be approximated bya. uniform distribution between 150 and 850

Single-Period Models (Continuous Demand)

71

141

Solution to Example 2:

Overage costco = Purchase price - Salvage value =

Underage costcu = Selling price - Purchase price =


142

p ≤ =

Now, find the Q so that p = probability(demand<Q) =

Q* = a+p(b-a) =

uo

u

ccc+

Area =

850150 Demand

Prob

abili

ty

Area=

Q*


72

143

Example 3: The J&B Card Shop sells calendars. The once-a-year order for each year’s calendar arrives in September. The calendars cost $1.50 and J&B sells them for $3 each. At the end of July, J&B reduces the calendar price to $1 and can sell all the surplus calendars at this price. How many calendars should J&B order if the September-to-July demand can be approximated byb. normal distribution with µ = 500 and σ=120.


144

Solution to Example 3: co =$0.50, cu =$1.50 (see Example 2)

p≤ = = 0.750.501.501.50

+uo

u

ccc+


73

145


Now, find the Q so that p = 0.75

146

Example 4: A retail outlet sells a seasonal product for $10 per unit. The cost of the product is $8 per unit. All units not sold during the regular season are sold for half the retail price in an end-of-season clearance sale. Assume that the demand for the product is normally distributed with µ = 500 and σ = 100.a. What is the recommended order quantity?b. What is the probability of a stockout?c. To keep customers happy and returning to the store later, the owner feels that stockouts should be avoided if at all possible. What is your recommended quantity if the owner is willing to tolerate a 0.15 probability of stockout?d. Using your answer to part (c), what is the goodwill cost you are assigning to a stockout?

Single-Period Models (Continuous Demand)Self Study

74

147

Solution to Example 4:a. Selling price=$10,

Purchase price=$8Salvage value=10/2=$5cu =10 - 8 = $2, co = 8-10/2 = $3

p≤ = = 0.4

Now, find the Q so that p = 0.4or, area (1) = 0.4Look up Table A-1 for Area (2) = 0.5-0.4=0.10

322+uo

u

ccc+

Prob

abili

ty

Demandµ=500

Area=0.40

z = 0.255

σ = 100

Area=0.10

(1)

(2)

(3)


148

z = 0.25 for area = 0.0987z = 0.26 for area = 0.1025So, z = 0.255 (take -ve, as p = 0.4 <0.5) for area = 0.10So, Q*=µ+ z σ=500+(-0.255)(100)=474.5 units.

b. P(stockout)=P(demand≥Q)=1-P(demand<Q)=1-p=1-0.4=0.6


75

149

c. P(stockout)=Area(3)=0.15Look up Table A-1 for Area (2) = 0.5-0.15=0.35z = 1.03 for area = 0.3485z = 1.04 for area = 0.3508

So, z = 1.035 for area = 0.35So, Q*=µ+ z σ=500+(1.035)(100)=603.5 units.

Prob

abili

ty

Demandµ=500

Area=0.35

z = 1.035

σ = 100Area=0.15

(1)

(2)

(3)


150

d. p=P(demand<Q)=1-P(demand≥Q)=1-P(stockout)=1-0.15=0.85

For a goodwill cost of gcu =10 - 8+g = 2+g, co = 8-10/2 = $3

Now, solve g in p = = =0.85

Hence, g=$15.

uo

u

ccc+ 32

2++

+)( gg


76

151


Lesson 16


Exercise: 8a, 12a, and 12b, pp. 256-258 (4th Ed.), pp. 248-250 (5th Ed.)

152

Outline

• Multi-Period Models – Lot size-Reorder Point (Q, R) Systems– Notation, Definition and Some Formula– Example: Given a Q, R Policy, Find Cost

LESSON 17: INVENTORY MODELS (STOCHASTIC)INTRODUCTION TO THE Q,R SYSTEMS

77

153

Lot Size - Reorder Point (Q,R) Systems

• In the simple EOQ model, demand is known and fixed. However, often demand is random. The lot size-reorder point (Q, R) systems allow random demand.

• There are two decision variables in a (Q, R) system:– Order quantity, Q and– Reorder point, R

• The Q, R policy is as follows:– When the level of on-hand inventory hits reorder

point, R place an order with lot size Q.

154


• In the simple EOQ model, R is the demand during the lead time.

• However, in presence of random demand, R usually includes a safety stock, in addition to the expected demand during the lead time. So,

Reorder point, R = lead-time demand + safety stock

78

155


• In the simple EOQ model, only holding cost and ordering costs are considered.

• In presence of random demand, the demand may sometimes be too high and exceed the inventory on hand. The result is stock-out.

• For each unit of shortage, a penalty cost p is charged. See Lesson 16 for more information on penalty cost.

Penalty cost = p per unit.

156


• The goal of a lot size-reorder point system is to find Qand R so that the total annual holding cost, ordering cost and stock-out cost is minimized.

• The current lesson only covers how to compute cost from a given policy.

• The next three lessons address the question how to find optimal Q and R so that the total annual cost is minimized.

79

157


Whenever the inventory onhand hits R, a quantity Q is ordered.

158


Too high lead-time demand may cause stock-outs. Safety stock reduces the chance ofstock-outs.

80

159


Safety Stock

Lead-Time Demand

The reorder point is computedfrom the lead-time demand and the safety stock.

160


Safety Stock

Lead-Time Demand

Goal: Find Q and R such that total annual holding cost, orde-ring cost and stock-out cost is minimized.

81

161

t

ttp

y

y

σ==σ

=σλ==µ

==

demand time-lead of deviation standard

demand annual of deviation standard demand time-lead mean

yearin time lead unit per cost out-stock

(Q,R) PolicyNotation, Definition and Some Formula

162


( )

( )( )

( )zLn

zLzF

zzF

Rz

σ==

==

=σ

µ−=

cycle per units out-stock 786781- pp. 4,- ATable from available

function, loss edstandardiz the time-lead during out stocking not ofy Probabilit

786781- pp. 4,- ATable from available of left the on area the

curve, normal the under area cumulative the ,

82

163


Prob

abili

ty

Lead-time Demand

( )zF Area

=

Rµ

σµ−

=R

z

( )zF1- Area

=

Prob(stockout)

Prob(no stockout)

164

Qn

Q

RQQ

R

Q

R

λ=

λ=

µ−+=+=

µ−=

=

µ−=

shortages of number annual Expected

cycles or orders of number annual Expected

stocksafety inventory Average

stocksafety inventory, Average

regular inventory, Average

stockSafety

22

2


83

165

Qnp

QKRhhQ

Qnp

QK

Rh

hQ

λ+λ+µ−+=

λ=

λ=

µ−=

=

)(2

)(2

cost annual Total

cost out-stock Annual

cost ordering Annual

stocksafety cost, holding Annual

regular cost, holding Annual


166

• Type 1 service– Type 1 service level, α is the probability of not

stocking out during the lead time.

– F(z) is available from Table A-4, pp. 781-786• Type 2 service

– Type 2 service level is measured by fill rate, β which is the proportion of demands that are met from stock

Qn

−= 1β

( )σ

µα

−==

RzzF ,


84

167

Example - Given A Q,R Policy, Find Cost

Annual demand for number 2 pencils at the campus store is normally distributed with mean 2,000 and standard deviation 300. The store purchases the pencils for 10 cents and sells them for 35 cents each. There is a two-month lead time from the initiation to the receipt of an order. The store accountant estimates that the cost in employee time for performing the necessary paper work to initiate and receive an order is $20, and recommends a 25 percent annual interest rate for determining holding cost. The cost of a stock-out is the cost of lost profit plus an additional 20 cents per pencil, which represents the cost of loss of goodwill. Currently, a (Q,R) system with Q= 1500, R = 500 is used.

168

Find

a. The safety stock


stockSafety

demand,time-Lead

time,Lead

=µ−=

=λτ=µ

=τ

R

85

169

b. The average inventory level

c. The expected annual number of orders


stocksafety inventory Average22

=

µ−+=+= RQQ

=

λ=

QcyclesorordersofnumberannualExpected

170

d. The probability of not stocking out during the lead-time

e. The expected number of units stock-out per cycle


( ) 786)781-pp.4,-ATable(Seetime-leadduringoutstockingnotofy Probabilit

==

=σ

µ−=

zF

Rz

( )786)781-pp.4,-ATable(See

zLn

==

σ=

86

171

f. The fill rate

g. The expected annual number of shortages


1 =−=βQnrate,fillThe

shortagesofnumberannualExpected

=

λ=Qn

172

h.The holding cost per unit per year and penalty cost per unit.


cost,Penalty cost,Holding

=

==

pIch

87

173

i. The average annual holding cost associated with this policy.


)(

2

==µ−=

==

costholdingannualTotal

stocksafety cost,holdingAnnual

regularcost,holdingAnnual

Rh

hQ

174

j. The total annual cost associated with this policy.


)(2

=

λ+λ+µ−+=

=λ=

=λ

=

Qnp

QKRhhQ

Qnp

QK

costannualTotal

costout-stockAnnual

costorderingAnnual

88

175


Lesson 17

Reading: Section 5.4, pp. 259-262 (4th Ed.), pp. 250-254 (5th

Ed.)

Exercise: 13b (use the result of 13a), p. 271 (4th Ed.), p. 261 (5th Ed.)

176

Outline

• Multi-Period Models – Lot size-Reorder Point (Q, R) Systems

• Optimization without service– Procedure– Example

LESSON 18: INVENTORY MODELS (STOCHASTIC)Q,R SYSTEMS

OPTIMIZATION WITHOUT SERVICE

89

177

Goal: Given find (Q,R) to minimize total cost

Step 1: Take a trial value of Q = EOQStep 2: Find a trial value of R = where and

are respectively mean and standard deviation of the lead-time demand and is the normal distribution variate corresponding to the area on the right, 1-F(z) = see Table A-4, pp. 835-841

Step 3: Find the expected number of stock-outs per cycle, where is the standardized loss function available from Table A-4, pp. 835-841

σµ z+ µ σ

z

λpQh /

)(zLn σ= )(zL

pKh ,,,,τλ

Procedure to find the Optimal (Q,R) Policy Without Any Service Constraint

178

Procedure to find the Optimal (Q,R) Policy Without Any Service Constraint

Step 4: Find the modified

Step 5: Find the modified value of R = where is the recomputed value of the normal distribution variate corresponding to the area on the right, 1-F(z) = see Table A-4, pp. 835-841

Step 6: If any of modified Q and R is different from the previous value, go to Step 3. Else if none of Q and Ris modified significantly, stop.

[ ]Knph

Q +=λ2

σµ z+ z

λpQh /

90

179

Example - Optimal (Q,R) Policy

Annual demand for number 2 pencils at the campus store is normally distributed with mean 2,000 and standard deviation 300. The store purchases the pencils for 10 cents and sells them for 35 cents each. There is a two-month lead time from the initiation to the receipt of an order. The store accountant estimates that the cost in employee time for performing the necessary paper work to initiate and receive an order is $20, and recommends a 25 percent annual interest rate for determining holding cost. The cost of a stock-out is the cost of lost profit plus an additional 20 cents per pencil, which represents the cost of loss of goodwill. Find an optimal (Q,R) policy

180


=

=

===

==

===

=

τσσ

λτµτ

σλ

y

y

p

Ich

K

demand, time-lead of deviation Standard

demand, time-lead Mean

time, Lead

demand, annual of deviation Standard demand, annual Mean

cost,Penalty

cost, Holding

cost, ordering Fixed

91

181


Iteration 1 Step 1:

===hK

Qλ2

EOQ

Step 2:

==−λp

QhzF )(1

=z (Table A-4)

=+= σµ zR

182

Example - Optimal (Q,R) Policy Step 3: =)(zL (Table A-4)

== )(zLn σ Step 4:

[ ] =+= Knph

Qλ2

Question: What are the stopping criteria?

Step 5: ==−λp

QhzF )(1

=z (Table A-4)

=+= σµ zR

92

183

Example - Optimal (Q,R) Policy Iteration 2Step 3: =)(zL (Table A-4)

== )(zLn σ Step 4:

[ ] =+= Knph

Qλ2

Question: Do the answers converge?

Step 5: ==−λp

QhzF )(1

=z (Table A-4)

=+= σµ zR

184

Fixed cost (K ) Note: K , h , and p Holding cost (h ) are input dataPenalty cost (p )Mean annual demand (λ) inputLead time (τ) in years input dataLead time demand parameters:

Mean, µ <--- computedStandard deviation, σ input data

Iteration 1 Iteration 2Step 1 Q=Step 2 Area on the right=1-F (z )

z =R =

Step 3 L(z )=n =

Step 4 Modified Q=Step 5 Area on the right=1-F (z )

z =Modified R =

σµ z+

σµ z+

EOQ

λpQh /

λpQh /

Table A1/A4

Table A1/A4

Table A4

)( zLσ[ ] hKnp /+λ2

93

185


Lesson 18


Ed.)

Exercise: 13a, p. 271 (4th Ed.), p. 261

186

LESSON 19: INVENTORY MODELS (STOCHASTIC)Q,R SYSTEMS

OPTIMIZATION WITH SERVICE

Outline

• Multi-Period Models – Lot size-Reorder Point (Q, R) Systems

• Optimization with service– Procedure for Type 1 Service– Procedure for Type 2 Service– Example

94

187

Optimization With Service

• In Lesson 18, we discuss the procedure of finding an optimal Q, R policy without any service constraint and using a stock-out penalty cost of p per unit.

• Managers often have difficulties to estimate p.• A substitute for stock-out penalty cost, p. is service

level.• In this lesson we shall not use stock-out penalty cost,

p. Instead , we shall assume that a service level must be met. Next slide defines two major types of service levels.

188

Optimization With Service

• Type 1 service– The probability of not stocking out during the lead

time is denoted by α. In problems with Type 1 service, α is specified e.g., α = 0.95

• Type 2 service– Fill rate, β: The proportion of demands that are

met from stock is called filled rate and is denoted by β. In problems with Type 2 service, β is specified e.g., β = 0.999

95

189

Procedure to Find the Optimal (Q,R) Policy with Type 1 Service

Goal: Given find (Q,R) to minimize total costFirst, find mean of the lead-time demand, andstandard deviation of the lead-time demand,

Step 1: Set Q = EOQStep 2: Find z for which area on the left, F(z) =αStep 3: Find R =

ατλ ,,,, Kh

σµ z+

λτµ =τλσ =

190

Goal: Given find (Q,R) to minimize total costFirst, find mean of the lead-time demand, andstandard deviation of the lead-time demand,

Step 1: Take a trial value of Q = EOQStep 2: Find expected number of shortages per cycle,

standardized loss function, and the standard normal variate z from Table A-4, pp. 835-841. Find a trial value of R= σµ z+

,/)( σnzL =

βτλ ,,,, Kh


),( β−= 1Qn

λτµ =τλσ =

96

191

Step 3: Find area on the right, 1-F(z) from Table A-1 or A-4, pp. 835-841

Step 4: Find the modified

Step 5: Find expected number of shortages per cycle,standardized loss function,

and the standard normal variate z from Table A-4, pp. 835-841. Find the modified R=

Step 6: If any of modified Q and R is different from the previous value, go to Step 3. Else if none of Q and Ris modified significantly, stop.


2121 )))(/((/))(/( zFnhKzFnQ −++−= λ

),( β−= 1Qn ,/)( σnzL =

σµ z+

192

Example - Optimal (Q,R) Policy with Service

Annual demand for number 2 pencils at the campus store is normally distributed with mean 2,000 and standard deviation 300. The store purchases the pencils for 10 cents and sells them for 35 cents each. There is a two-month lead time from the initiation to the receipt of an order. The store accountant estimates that the cost in employee time for performing the necessary paper work to initiate and receive an order is $20, and recommends a 25 percent annual interest rate for determining holding cost.

97

193

a. Find an optimal (Q,R) policy with Type 1 service, α=0.95 and Q=EOQ

b. Find an optimal (Q,R) policy with Type 2 service, β=0.999 and using the iterative procedure


194


=

=

===

===

=

τσσ

λτµτ

σλ

y

y

Ich

K

demand, time-lead of deviation Standard

demand, time-lead Mean

time, Lead

demand, annual of deviation Standard demand, annual Mean

cost, Holding

cost, ordering Fixed

98

195

a. Type 1 service, α = 0.95 Step 1. Q = EOQ =Step 2. Find z for which area on the left, F(z) = α = 0.95

Step 3. R =

b. Type 2 service, β=0.999 This part is solved with the iterative procedure as shown next.

=+ σµ z


196


Iteration 1 Step 1:

===hk

Qλ2

EOQ

Step 2:

=−= )( β1Qn

==σn

zL )(

=z (Table A-4)

=+= σµ zR

99

197


Step 3: =− )(zF1 Step 4:

2

12

1

−

++−

=)()( zF

nhK

zFnQ λ

=

Question: What are the stopping criteria?

198


Step 5:

=−= )( β1Qn

==σn

zL )(

=z (Table A-4)

=+= σµ zR

100

199


Step 3: =− )(zF1 Step 4:

2

12

1

−

++−

=)()( zF

nhK

zFnQ λ

=

Question: Do the answers converge?

Iteration 2

200


Step 5:

=−= )( β1Qn

==σn

zL )(

=z (Table A-4)

=+= σµ zR

101

201

Fixed cost (K ) Note: K , and hHolding cost (h ) are input dataMean annual demand (λ) input dataLead time (τ) in years input dataLead time demand parameters:

Mean, µ <--- computedStandard deviation, σ input data

Type 2 service, fill rate, β input dataIteration 1 Iteration 2

Step 1 Q =Step 2 n =

L(z)=z =R =

Step 3 Area on the right=1-F(z )Step 4 Modified Q =Step 5 n =

L(z)=z =R =

2121 )))(/((/))(/( zFnhKzFn −++− λ

EOQ)( β−1Q

σ/n

σµ z+41-835 pp. A1/A4,Table

41-835 pp. A1/A4,Table

σ/n

σµ z+41-835 pp. A1/A4,Table

)( β−1Q

202

(Q,R) Systems

Remark

• We solve three versions of the problem of finding an optimal (Q,R) policy– No service constraint – Type 1 service– Type 2 service

• All these versions may alternatively and more efficiently solved by Excel Solver. This is discussed during the tutorial.

102

203

Multiproduct Systems

• Pareto effect– A concept of economics applies to inventory systems– Rank the items in decreasing order of revenue

generated– Item group A: top 20% items generate 80% revenue– Item group B: next 30% items generate 15% revenue– Item group A: last 50% items generate 5% revenue

204

Multiproduct Systems

• Exchange curves– Parameters like K and I are not easy to measure– Instead of assigning values to such parameters show

the trade off between holding cost and ordering cost for a large number of values of K/I

– The effect of changing the ratio K/I is shown by plotting holding cost vs ordering cost

– Similarly, instead of assigning a value to type 2 service level β, one may show the trade off between cost of safety stock and expected number of stock-outs.

103

205


Lesson 19


Ed.)Section 5.7 (skim) pp. 275-280 (4th Ed.), pp. 265-270 (5th Ed.)

Exercise: 16 and 17, p. 271 (4th Ed.), p. 262 (5th Ed.)

1

Lesson 20: Stochastic Inventory Control Finding An Optimal (Q, R) Policy Using Excel Solver

This lesson illustrates the procedure of finding an optimal (Q,R) policy using Excel Solver. Download the file Lesson_20_stochastic_inventory_w03_331.xls from the course website and follow the instruction given below. (A) Exercise on Normal Distribution

1. Find area from z • Area on the left of z = NORMSDIST(z)

2. Find z from area • z = NORMSINV(Area on the left of z)

3. Find area from x • Area on the left of x = NORMDIST(x,µ,σ,TRUE)

4. Find x from area • x = NORMINV(Area on the left of x,µ,σ)

5. Find L(z) from z • L(z) =NORMDIST(z,0,1,FALSE)-z*(1-NORMDIST(z,0 ,1,TRUE))

Question A Find

1. area on the left of z = 2.5 2. z corresponding to the area on the left = 0.4 3. the area on the left if x=600, µ=300 and σ=120 4. x if the area on the left =0.4, µ=300 and σ=120 5. L(z) if z=2.5

A B1 1. Find area from z2 z= 2.53 Area on the left of z = NORMSDIST(z) ?45 2. Find z from area 6 Area on the left of z = 0.47 z = NORMSINV(Area on the left of z) ?89 3. Find area from x

10 x= 60011 mu= 30012 sigma= 12013 Area on the left of x = NORMDIST(x,mu,sigma,TRUE) ?1415 4. Find x from area 16 Area on the left of x 0.417 mu= 30018 sigma= 12019 x = NORMINV(Area on the left of x,mu,sigma) ?2021 5. Find L(z) from z22 z= 2.523 L(z) =NORMDIST(z,0,1,FALSE)-z*(1-NORMDIST(z,0,1,TRUE)) ?

2

(B) Exercise on (Q,R) policy for a given pair of Q and R

• Inputs: µ = mean of the lead-time demand and σ = standard deviation of the lead-time demand

1. Find probability of stock-out • Probability of no stock-out = Area on the left of R

= NORMDIST(R,µ,σ,TRUE) 2. Find proportion of shortages and fill rate

• z = (R-µ)/σ • L(z) =NORMDIST(z,0,1,FALSE)-z*(1-NORMDIST(z,0,1,TRUE)) • Number of units of demand not met in each cycle, n = σ*L(z) • Proportion of shortages = n/Q and fill rate = 1- n/Q

3. Find costs • Further inputs required: λ, h, K, p

• Holding cost, regular inventory = h*Q/2 • Holding cost, safety stock = h*(R-µ) • Ordering cost = K*λ/Q • Stock-out cost = (proportion of shortages)*λ*p • Total cost = Holding cost (regular & safety) + ordering cost + stock-

out cost For Parts B and C: Annual demand for number 2 pencils at the campus store is normally distributed with mean 2,000 and standard deviation 300. The store purchases the pencils for 10 cents and sells them for 35 cents each. There is a two-month lead time from the initiation to the receipt of an order. The store accountant estimates that the cost in employee time for performing the necessary paper work to initiate and receive an order is $20, and recommends a 25 percent annual interest rate for determining holding cost. The cost of a stock-out is the cost of lost profit plus an additional 20 cents per pencil, which represents the cost of loss of goodwill.

Question B For Q=1500 and R=500 find

1. probability of stock-out 2. fill rate 3. total inventory costs

Fixed cost (K) 20Holding cost (h) 0.025Penalty cost (p) 0.45Mean annual demand (lambda) 2000Lead time (tau) in years 0.166666667Lead time demand parameters:

mu 333.3333333sigma 122.4744871

3

(C) Exercise on finding an optimal ( Q,R) policy using Excel Solver

• Minimize total cost • Decision variables: Q and R • Constraints

• Case 1: (Section 5.4) no constraint • Case 2: (Section 5.5, Type 1 service) probability of no stock-out

= α • Case 3: (Section 5.5, Type 2 service) fill rate = β i.e., proportion

of shortages = 1 -β • Both Q and R are non-negative

Question C Find an optimal (Q,R) policy (using Excel Solver) with 1. no constraint on α,β 2. α=0.95 and β unrestricted 3. β=0.96 and α unrestricted

A B C D E F12 Ordering cost (K) 203 Holding cost (h) 0.0254 Penalty cost (p) 0.455 Mean annual demand (lambda) 20006 Lead time (tau) in years 0.166677 Lead time demand parameters:8 mu 333.3339 sigma 122.47410 Probability(stockout)11 =1-NORMDIST(R,mu,sigma,TRUE)12 Order quantity, Q 1500 ?13 Reorder point, R 500 Fill Rate =1-n/Q14 ?1516 z=(R-mu)/sigma ? Holding cost, regular = h*Q/2 ?17 L(z) ? Holding cost, safety = h* (R-mu) ?18 n=sigma*L(z) ? Ordering cost = K*lambda/Q ?19 proportion of shortages=n/Q ? Stock-out cost = n/Q*lambda*p ?20 Fill Rate =1-n/Q ? Total cost ?

A B C D E F1 Inputs2 Ordering cost (K) 203 Holding cost (h) 0.0254 Penalty cost (p) 0.455 Mean annual demand (lambda) 20006 Lead time (tau) in years 0.166677 Lead time demand parameters:8 mu 333.3339 sigma 122.474 Constraints10 Probability(no stockout) Alpha11 Decision Variables =NORMDIST(R,mu,sigma,TRUE)12 Order quantity, Q 1500 >= 0.9513 Reorder point, R 500 Fill Rate =1-n/Q Beta14 >= 0.9615 Computation for fill rate16 z=(R-mu)/sigma Holding cost, regular = h*Q/217 L(z) Holding cost, safety = h* (R-mu)18 n=sigma*L(z) Ordering cost = K*lambda/Q19 proportion of shortages=n/Q Stock-out cost = n/Q*lambda*p20 Fill Rate =1-n/Q Total cost

4

Excel Solver

1. Add-in Solver, if needed: Click on the “Tools” menu. If the “Solver” is not one of the items of the “Tools” menu, choose the “Add-Ins.” Click on the box adjacent to “Solver Add-In” (a check mark will appear) and click “OK.”

2. Define objective: From the “Tools” menu, choose “Solver.” You get the

“Solver Parameters” window. Click on the little arrow adjacent to the box “Set Target Cell.” Click on the Cell containing total cost and click again on the arrow. Click on the circle adjacent to “Min.”

3. Decision variables: The cell addresses for the decision variables are entered

in the box below “By Changing Cells:.” The decision variables are the cells containing Q and R.

4. Constraints: The constraints are ente red in the box below “Subject to the

Constraints.” Click on “Add” and you get the “Add Constraint” window. Provide the information on constraint (LHS, inequality, RHS) and click on “OK.”

5. Non-negativity: From the “Solver Parameters” window click on “Options.” You

get the “Solver Options” window. Click on a box adjacent to “Assume Non-Negativity.” Click on “OK.”

6. Solve: Finally, from the “Solver Parameters” window click on “Solve.” A

“Solver Results” window pops up. Click on “OK.”

1

Outline

• Hierarchy of Production Decisions• MRP and its importance• Input and Output of an MRP system• MRP Calculation• Lot Sizing• Lot Sizing with Capacity Constraint

LESSON 21: MATERIAL REQUIREMENTS PLANNING

2

• The next slide presents a schematic view of the aggregate production planning function and its place in the hierarchy of the production planning decisions.

• Forecasting: First, a firm must forecast demand for aggregate sales over the planning horizon.

• Aggregate planning: The forecasts provide inputs for determining aggregate production and workforce levels over the planning horizon.

• Master production schedule (MPS): Recall, that the aggregate production plan does not consider any “real” product but a “fictitious” aggregate product. The MPS translates the aggregate plan output in terms of specific production goals by product and time period. For example,


3


Aggregate Planning


Inventory Control


Vehicle Routing

Forecast of Demand

4

suppose that a firm produces three types of chairs: ladder-back chair, kitchen chair and desk chair. The aggregate production considers a fictitious aggregate unit of chair and find that the firm should produce 550 units of chairs in April. The MPS then translates this output in terms of three product types and four work-weeks in April. The MPS suggests that the firm produce 200 units of desk chairs in Week 1, 150 units of ladder-back chair in Week 2, and 200 units of kitchen chairs in Week 3.

• Material Requirements Planning (MRP): A product is manufactured from some components or subassemblies. For example a chair may require two back legs, two front legs, 4 leg supports, etc. While forecasting, aggregate plan


5

Hierarchy of Production DecisionsMaster Production Schedule

Ladder-back chair

Kitchen chair

Desk chair

1 2

April May

790790

3 4 5 6 7 8

200200

150150

120120

200200

150150

200200

120120

Aggregate production plan for chair family

550550

200200

6

and MPS consider the volume of finished products, MRP plans for the components, and subassemblies. A firm may obtain the components by in-house production or purchasing. MRP prepares a plan of in-house production or purchasing requirements of components and subassemblies.

• Scheduling: Scheduling allocates resource over times in order to produce the products. The resources include workers, machines and tools.

• Vehicle Routing: After the products are produced, the firm may deliver the products to some other manufacturers, or warehouses. The vehicle routing allocates vehicles and prepares a route for each vehicle.


7

Hierarchy of Production DecisionsMaterials Requirement Planning

Back slats

Leg supports

Seat cushion

Seat-frameboards

Frontlegs

Backlegs

8

Material Requirements Planning

• The demands for the finished goods are obtained from forecasting. These demands are called independent demand.

• The demands for the components or subassemblies depend on those for the finished goods. These demands are called dependent demand.

• Material Requirements Planning (MRP) is used for dependent demand and for both assembly and manufacturing

• If the finished product is composed of many components, MRP can be used to optimize the inventory costs.

9

• Next two slides explain the importance of an MRP system. The first one shows inventory levels when an MRP system is not used. The next one shows the same when an MRP system is used.

• The chart at the top shows inventory levels of the finished goods and the chart on the bottom shows the same of the components.

• If the production is stopped (like it is at the beginning of the chart), the finished goods inventory level decreases because of sales. However, the component inventory level remains unchanged. When the production resumes, the finished goods inventory level increases, but the component inventory level decreases.

Importance of an MRP System

10

Inventory without an MRP System


11

Inventory with an

MRP System


12

• Without an MRP system:– Component is ordered at time A, when the inventory

level of the component hits reorder point, R– So, the component is received at time B. – However, the component is actually needed at time

C, not B. So, the inventory holding cost incurred between time B and C is a wastage.

• With an MRP system:– We shall see in this lesson that given the production

schedule of the finished goods and some other information (see the next slide), it is possible to predict the exact time, C when the component will be required. Order is placed carefully so that it is received at time C.


13

• MRP Inputs:– Master Production Schedule (MPS): The MPS of the

finished product provides information on the net requirement of the finished product over time.

– Bill of Materials: For each component, the bill of materials provides information on the number of units required, source of the component (purchase/ manufacture), etc. There are two forms of the bill of materials:

• Product Structure Tree: The finished product is shown at the top, at level 0. The components assembled to produce the finished product is shown at level 1 or below. The sub-components used to produce the components at level 1 is

MRP Input and Output

14

MRPcomputerprogram

MRPcomputerprogram

Bill ofMaterials

file

Bill ofMaterials

file

Inventoryfile

Inventoryfile



ReportsReportsTo Production To Purchasing

ForecastsOrders


15

shown at level 2 or below, and so on. The number in the parentheses shows the requirement of the item. For example, “G(4)” implies that 4 units of G is required to produce 1 unit of B.The levels are important. The net requirements of the components are computed from the low levels to high. First, the net requirements of the components at level 1 is computed, then level 2, and so on.


16

• Bill of Materials: For each item, the name, number, source, and lead time of every component required is shown on the bill of materials in a tabular form.

– Inventory file: For each item, the number of units on hand is obtained from the inventory file.

• MRP Output:– Every required item is either produced or purchased.

So, the report is sent to production or purchasing.


17

Bill of Materials: Product Structure Tree

Level 1

Level 0

Level 2

Level 3

18

BILL OF MATERIALS Product Description: Ladder-back chair Item: A

ComponentItem Description

QuantityRequired

Source

B Ladder-back 1 ManufacturingC Front legs 2 PurchaseD Leg supports 4 PurchaseE Seat 1 Manufacturing

Bill of Materials

19

BILL OF MATERIALS Product Description: Seat Item: E

ComponentItem Description

QuantityRequired

Source

H Seat frame 1 ManufacturingI Seat cushion 1 Purchase

Bill of Materials

20

Component Units in LeadInventory time

(weeks)

Seat Subassembly 25 2

Seat frame 50 3

Seat frame boards 75 1

On Hand Inventory and Lead time

21

• Now, the MRP calculation will be demonstrated with an example.

• Suppose that 150 units of ladder-back chair is required.• The previous slide shows a product structure tree with

seat subassembly, seat frames, and seat frame boards. For each of the above components, the previous slide also shows the number of units on hand.

• The net requirement is computed from top to bottom. Since 150 units of ladder-back chair is required, and since 1 unit of seat subassembly is required for each unit of ladder-back chair, the gross requirement of seat-subassembly is 150×1 =150 units. Since there are 25 units of seat-subassembly in the inventory, the net requirement of the seat-subassembly is 150-25 = 125

MRP Calculation

22

units. Since 1 unit of seat frames is required for each unitof seat subassembly, the gross requirement of the seat frames is 125×1 = 125 units. (Note that although it follows from the product structure tree that 1 unit of seat frames is required for each unit of ladder-back chair, the gross requirement of seat frames is not 150 units because each of the 25 units of seat-subassembly also contains 1 unit of seat frames.) Since there are 50 units of seat frames in the inventory, the net requirement of the seat frames is 125-50 = 75 units. The detail computation is shown in the next two slides.

• A similar logic is used to compute the time of order placement.

MRP Calculation

23

MRP Calculation

UnitsQuantity of ladder-back chairs to be produced 150Gross requirement, seat subassemblyLess seat subassembly in inventory 25 Net requirement, seat subassemblyGross requirement, seat framesLess seat frames in inventory 50 Net requirement, seat framesGross requirement, seat frame boardsLess seat frame boards in inventory 75 Net requirement, seat frame boards

Assume that 150 units of ladder-back chairs are to be produced at the end of week 15

24

WeekComplete order for seat subassembly 14Minus lead time for seat subassembly 2 Place an order for seat subassembly Complete order for seat framesMinus lead time for seat frames 3 Place an order for seat frames Complete order for seat frame boardsMinus lead time for seat frame boards 1 Place an order for seat frame boards

Assume that 150 units of ladder-back chairs are to be produced at the end of week 15 and that there is a one-week lead time for ladder-back chair assembly

MRP Calculation: Time of Order Placement

25

• Scheduled Receipts: – Items ordered prior to the current planning period

and/or– Items returned from the customer

• Lot-for-lot (L4L)– Order quantity equals the net requirement – Sometimes, lot-for-lot policy cannot be used. There

may be restrictions on minimum order quantity or order quantity may be required to multiples of 50, 100 etc.

MRP Calculation: Some Definitions

26

Example 1: Each unit of A is composed of one unit of B, two units of C, and one unit of D. C is composed of two units of D and three units of E. Items A, C, D, and E have on-hand inventories of 20, 10, 20, and 10 units, respectively. Item B has a scheduled receipt of 10 units in period 1, and C has a scheduled receipt of 50 units in Period 1. Lot-for-lot (L4L) is used for Items A and B. Item C requires a minimum lot size of 50 units. D and E are required to be purchased in multiples of 100 and 50, respectively. Lead times are one period for Items A, B, and C, and two periods for Items D and E. The gross requirements for A are 30 in Period 2, 30 in Period 5, and 40 in Period 8. Find the planned order releases for all items.

MRP Calculation

27

Level 0

Level 1

Level 2

MRP Calculation

28

MRP Calculation

Period 1 2 3 4 5 6 7 8 9 10GrossRequirementsScheduledreceiptsOn hand fromprior periodNetrequirementsTime-phased NetRequirementsPlanned orderreleases

Item

A

LT=

Q=

Planned orderdelivery

29

MRP Calculation

1WK

L4L


Item

A

LT=

Q=


30 30 40

20

All the information above are given.

30

MRP Calculation

1WK

L4L


Item

A

LT=

Q=


30 30 40

--

20 20

20 units are just transferred from Period 1 to 2.

31

MRP Calculation

1WK

L4L


Item

A

LT=

Q=


30 30 40

--

20 20

10

10

10

10

The net requirement of 30-20=10 units must be ordered in week 1.

32

MRP Calculation

1WK

L4L


Item

A

LT=

Q=


30 30 40

--

20 20 0 0 0

10

10

10

10

On hand in week 3 is (20+10)-30=0 unit.

33

MRP Calculation

1WK

L4L


Item

A

LT=

Q=


30 30 40

--

20 20 0 0 0

10 30

3010

10 30

10 30


34


Item

A

LT=

Q=


MRP Calculation

30 30 40

1WK

L4L

--

20 20 0 0 0 0 0 0

10 30 40

30 4010

10 30 40

10 30 40


35


Item

A

LT=

Q=


MRP Calculation

30 30 40

1WK

L4L

--

20 20 0 0 0 0 0 0 0 0

10 30 40

30 4010

10 30 40

10 30 40


36

MRP Calculation


Item

B

LT=

Q=


Exercise

37

MRP Calculation


Item

C

LT=

Q=


Exercise

38

MRP Calculation


Item

D

LT=

Q=


Exercise

39

MRP Calculation


Item

E

LT=

Q=


Exercise

40


Lesson 21Reading:

Section 7.1 pp. 355-364 (4th Ed.), pp. 346-358 (5th

Ed.)Exercise:

4 and 9 pp. 364-366 (4th Ed.), pp. 356-358 (5th Ed.)

41

Outline

• Lot Sizing• Lot Sizing Methods

– Lot-for-Lot (L4L)– EOQ– Silver-Meal Heuristic– Least Unit Cost (LUC)– Part Period Balancing

LESSON 22: MATERIAL REQUIREMENTS PLANNING: LOT SIZING

42

Lot-Sizing• In Lesson 21

– We employ lot for lot ordering policy and order production as much as it is needed.

– Exception are only the cases in which there are constraints on the order quantity.

– For example, in one case we assume that at least 50 units must be ordered. In another case we assume that the order quantity must be a multiple of 50.

• The motivation behind using lot for lot policy is minimizing inventory. If we order as much as it is needed, there will be no ending inventory at all!

43

Lot-Sizing• However, lot for lot policy requires that an order be placed

each period. So, the number of orders and ordering cost are maximum.

• So, if the ordering cost is significant, one may naturally try to combine some lots into one in order to reduce the ordering cost. But then, inventory holding cost increases.

• Therefore, a question is what is the optimal size of the lot? How many periods will be covered by the first order, the second order, and so on until all the periods in the planning horizon are covered. This is the question of lot sizing. The next slide contains the statement of the lot sizing problem.

44

Lot-Sizing• The lot sizing problem is as follows: Given net

requirements of an item over the next T periods, T >0, find order quantities that minimize the total holding and ordering costs over T periods.

• Note that this is a case of deterministic demand. However, the methods learnt in Lessons 11-15 are not appropriate because – the demand is not necessarily the same over all

periods and– the inventory holding cost is only charged on ending

inventory of each period

45

Lot-Sizing• Although we consider a deterministic model, keep in mind

that in reality the demand is uncertain and subject to change.

• It has been observed that an optimal solution to the deterministic model may actually yield higher cost because of the changes in the demand. Some heuristic methods give lower cost in the long run.

• If the demand and/or costs change, the optimal solution may change significantly causing some managerial problems. The heuristic methods may not require such changes in the production plan.

• The heuristic methods require fewer computation steps and are easier to understand.

• In this lesson we shall discuss some heuristic methods. The optimization method is discussed in the text, Appendix 7-A, pp 406-410 (not included in the course).

46

Lot-Sizing• Some heuristic methods:

– Lot-for-Lot (L4L): • Order as much as it is needed. • L4Lminimizes inventory holding cost, but maximizes

ordering cost.– EOQ:

• Every time it is required to place an order, lot size equals EOQ.

• EOQ method may choose an order size that covers partial demand of a period. For example, suppose that EOQ is 15 units. If the demand is 12 units in period 1 and 10 units in period 2, then a lot size of 15 units covers all of period 1 and only (15-12)=3 units of period 2. So, one does not save the ordering cost of period 2, but carries some 3 units in

47


the inventory when that 3 units are required in period 2. This is not a good idea because if an order size of 12 units is chosen, one saves on the holding cost without increasing the ordering cost!

• So, what’s the mistake? Generally, if the order quantity covers a period partially, one can save on the holding cost without increasing the ordering cost. The next three methods, Silver-Meal heuristic, least unit cost and part period balancing avoid order quantities that cover a period partially. These methods always choose an order quantity that covers some K periods, K >0.

• Be careful when you compute EOQ. Express both holding cost and demand over the same period. If the holding cost is annual, use annual demand. If the holding cost is weekly, use weekly demand.

48


– Silver-Meal Heuristic• As it is discussed in the previous slide, Silver-Meal heuristic

chooses a lot size that equals the demand of some K periods in future, where K>0.

• If K =1, the lot size equals the demand of the next period. • If K =2, the lot size equals the demand of the next 2 periods.• If K =3, the lot size equals the demand of the next 3 periods,

and so on.• The average holding and ordering cost per period is computed

for each K=1, 2, 3, etc. starting from K=1 and increasing K by 1 until the average cost per period starts increasing. The best K is the last one up to which the average cost per period decreases.

49


– Least Unit Cost (LUC) • As it is discussed before, least unit cost heuristic

chooses a lot size that equals the demand of some K periods in future, where K>0.

• The average holding and ordering cost per unit is computed for each K=1, 2, 3, etc. starting from K=1 and increasing K by 1 until the average cost per unit starts increasing. The best K is the last one up to which the average cost per unit decreases.

• Observe how similar is Silver-Meal heuristic and least unit cost heuristic. The only difference is that Silver-Meal heuristic chooses K on the basis of average cost per period and least unit cost on average cost per unit.

50


– Part Period Balancing • As it is discussed before, part period balancing

heuristic chooses a lot size that equals the demand of some K periods in future, where K>0.

• Holding and ordering costs are computed for each K=1, 2, 3, etc. starting from K=1 and increasing Kby 1 until the holding cost exceeds the ordering cost. The best K is the one that minimizes the (absolute) difference between the holding and ordering costs.

• Note the similarity of this method with the Silver-Meal heuristic and least unit cost heuristic. Part period balancing heuristic chooses K on the basis of the (absolute) difference between the holding and ordering costs.

51

• Some important notes– Inventory costs are computed on the ending inventory.– L4L minimizes carrying cost– Silver-Meal Heuristic, LUC and Part Period Balancing

are similar– Silver-Meal Heuristic and LUC perform best if the costs

change over time– Part Period Balancing perform best if the costs do not

change over time– The problem extended to all items is difficult to solve

Lot-Sizing

52

Example 2: The MRP gross requirements for Item A are shown here for the next 10 weeks. Lead time for A is three weeks and setup cost is $10. There is a carrying cost of $0.01 per unit per week. Beginning inventory is 90 units.

Week Gross requirements Week Gross requirements1 30 6 802 50 7 203 10 8 604 20 9 2005 70 10 50

Determine the lot sizes.

Lot-Sizing

53

Period 1 2 3 4 5 6 7 8 9 10GrossRequirements

30 50 10 20 70 80 20 60 200 50

BeginningInventory

90 60 10 0

NetRequirements

0 0 0 20

Time-phased NetRequirementsPlanned orderReleasePlannedDeliveriesEndingInventory

60 10 0

Lot-Sizing: Lot-for-Lot

Use the above table to compute ending inventory of various periods.

54


30 50 10 20 70 80 20 60 200 50

BeginningInventory

90 60 10 0

NetRequirements

0 0 0 20


60 10 0


20

Week 4 net requirement = 20 > 0. So, an order is required.

55


30 50 10 20 70 80 20 60 200 50

BeginningInventory

90 60 10 0

NetRequirements

0 0 0 20


60 10 0


20

20

20

A delivery of 20 units is planned for the 4th period..

56


30 50 10 20 70 80 20 60 200 50

BeginningInventory

90 60 10 0

NetRequirements

0 0 0 20


60 10 0


0

70

20

20

0

20

The net requirement of the 5th period is 70 periods.

Exercise

57

Lot-Sizing: EOQ• First, compute EOQ

– Annual demand is not given. Annual demand is estimated from the known demand of 10 weeks.

– Compute annual holding cost per unit

units/year

502006020807020105030

r weeks/yea weeks10 over demand Total

demand, annual Estimated

068,3

5210590

5210

5210

=

×=

×+++++++++

=

×=

λ

/year$0.52/unit/unit/week == 01.0$h

58

Lot-Sizing: EOQ• First, compute EOQ

• Therefore, whenever it will be necessary to place an order, the order size will be 344 units. This will now be shown in more detail.

units EOQ

/unit/year$/order

units/year

34451.34352.0

068,31022

52.010$068,3

≈=××

=λ

=

===λ

hK

hK

59


30 50 10 20 70 80 20 60 200 50

BeginningInventory

90 60 10 0

NetRequirements

0 0 0 20


60 10 0

Lot-Sizing: EOQ


60


30 50 10 20 70 80 20 60 200 50

BeginningInventory

90 60 10 0

NetRequirements

0 0 0 20


60 10 0

Lot-Sizing: EOQ

20


61


30 50 10 20 70 80 20 60 200 50

BeginningInventory

90 60 10 0

NetRequirements

0 0 0 20


60 10 0

Lot-Sizing: EOQ

20

344

344

Order size = EOQ = 344, whenever it is required to place an order.

62


30 50 10 20 70 80 20 60 200 50

BeginningInventory

90 60 10 0

NetRequirements

0 0 0 20


60 10 0

Lot-Sizing: EOQ

20

344

324

344

324

Week 5 b. inv=344-20=324>70= gross req. So, no order is required.

Exercise

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j 1 2 3 4 5 6 7r j 20 70 80 20 60 200 50 Per

H. Ord. PeriodOrder for weeks Q 4 5 6 7 8 9 10 Cost Cost Cost1 week, week 4

2 weeks, weeks 4 to 53 weeks, weeks 4 to 64 weeks, weeks 4 to 75 weeks, weeks 4 to 86 weeks, weeks 4 to 9

7 weeks, weeks 4 to 10

Units in the inventory at the end of Week

Lot-Sizing: Silver-Meal-Heuristic

The order is placed for K periods, for some K>0. Use the above table to find K.

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j 1 2 3 4 5 6 7r j 20 70 80 20 60 200 50 Per






20 0.00 10 10.0

If K=1, order is placed for 1 week and the order size = 20. Then, the ending inventory = inventory holding cost =0. The order cost = $10. Average cost per period = (0+10)/1=$10.

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j 1 2 3 4 5 6 7r j 20 70 80 20 60 200 50 Per






20 0.00 10 10.090 70 0.70 10 5.35

If K=2, order is placed for 2 weeks and the order size = 20+70=90.Then, inventory at the end of week 4 = 90-20=70 and holding cost =70 ×0.01. = 0.70. Average cost per period = (0.70+10)/2=$5.35.

Exercise

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30 50 10 20 70 80 20 60 200 50

BeginningInventory

90 60 10 0

NetRequirements

0 0 0 20


60 10 0



67


30 50 10 20 70 80 20 60 200 50

BeginningInventory

90 60 10 0

NetRequirements

0 0 0 20


60 10 0


20


Exercise

68

j 1 2 3 4 5 6 7r j 20 70 80 20 60 200 50

H. Ord. UnitOrder for weeks Q 4 5 6 7 8 9 10 Cost Cost Cost1 week, week 4




Lot-Sizing: Least Unit Cost


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j 1 2 3 4 5 6 7r j 20 70 80 20 60 200 50






20 0.00 10 .500

If K=1, order is placed for 1 week and the order size = 20. Then, the ending inventory = inventory holding cost =0. The order cost = $10. Average cost per unit = (0+10)/20=$0.50

70

j 1 2 3 4 5 6 7r j 20 70 80 20 60 200 50






20 0.00 10 .50090 70 0.70 10 .119

If K=2, order is placed for 2 weeks and the order size = 20+70=90.Then, inventory at the end of week 4 = 90-20=70 and holding cost =70 ×0.01. = 0.70. Average cost per unit = (0.70+10)/90=$0.119.

Exercise

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30 50 10 20 70 80 20 60 200 50

BeginningInventory

90 60 10 0

NetRequirements

0 0 0 20


60 10 0


72



30 50 10 20 70 80 20 60 200 50

BeginningInventory

90 60 10 0

NetRequirements

0 0 0 20


60 10 0

20


Exercise

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Lot-Sizing: Part Period Balancing

j 1 2 3 4 5 6 7r j 20 70 80 20 60 200 50

H. Ord. DiffOrder for weeks Q 4 5 6 7 8 9 10 Cost Cost1 week, week 4





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j 1 2 3 4 5 6 7r j 20 70 80 20 60 200 50

H. Ord. DiffOrder for weeks Q 4 5 6 7 8 9 10 Cost Cost1 week, week 4




20 0.00 10 10.090 70 0.70 10 9.30

170 150 80 2.30 10 7.70190 170 100 20 2.90 10 7.10250 230 160 80 60 5.30 10 4.70450 430 360 280 260 200 15.30 10 5.30

NOT COMPUTED1 week, week 92 weeks, weeks 9 to 10

200 0.00 10 10.0250 50 0.50 10 9.50

The above computation is similar to that of the Silver-Meal heuristic. The primary difference is that the (absolute) difference betweenholding and ordering cost is shown in the last column.

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30 50 10 20 70 80 20 60 200 50

BeginningInventory

90 60 10 0

NetRequirements

0 0 0 20


60 10 0


76



30 50 10 20 70 80 20 60 200 50

BeginningInventory

90 60 10 0

NetRequirements

0 0 0 20


60 10 0

200

20 200

250 250

230 160 80 60 0 50 0

250 250

230 160 80 60 0 50

The computation is similar to that of the Silver-Meal heuristic.

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• Lot-for-Lot– See the last slide entitled “lot-sizing: lot-for-lot”– Number of orders: 7– Ordering cost = 7 × $10/order = $70– Holding cost = (60+10) × $0.01/unit/week = $0.70– Total cost = 70+0.70 =$70.70

• EOQ– See the last slide entitled “lot-sizing: EOQ”– Number of orders: 2– Ordering cost = 2 × $10/order = $20– Holding cost = (60 +10 +324 +254 +174 +154 +94

+237 +187) × $0.01/unit/week = $14.94– Total cost = 20+14.94 =$34.94

Cost Comparison

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• Silver-Meal Heuristic– See the last slide entitled “lot-sizing: Silver-Meal

heuristic”– Number of orders: 2– Ordering cost = 2 × $10/order = $20– Holding cost = (60 +10 +230 +160 +80 +60 +50) ×

$0.01/unit/week = $6.50– Total cost = 20+6.50 =$26.50

• Least Unit Cost– See the last slide entitled “lot-sizing: least unit cost”– Number of orders: 2– Ordering cost = 2 × $10/order = $20– Holding cost = (60 +10 +430 +360 +280 +260 +200) ×


Cost Comparison

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• Part-Period Balancing– See the last slide entitled “lot-sizing: part-period

balancing”– Number of orders: 2– Ordering cost = 2 × $10/order = $20– Holding cost = (60 +10 +230 +160 +80 +60 +50) ×


• Conclusion: In this particular case, Silver-Meal heuristic and part period balancing yield the least total holding and ordering cost of $26.50 over the planning period of 10 weeks.

Cost Comparison

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Lesson 22Reading:

Section 7.2-7.3 pp. 366-375 (4th Ed.), pp. 358-366 (5th Ed.)

Exercise: 17 and 25 pp. 371-373, 375 (4th Ed.), pp. 363, 366

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Outline

• Lot Sizing with Capacity Constraints– Order Partial Requirements– Checking Feasibility– Lot Shifting Technique– An Improvement Procedure

LESSON 23: MATERIAL REQUIREMENTS PLANNING: LOT SIZING WITH CAPACITY

CONSTRAINTS

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Lot Sizing with Capacity Constraints

• In Lessons 21 and 22 we have assumed that there is no capacity constraint on production. However, often, the production capacity is limited.

• In this lesson we assume that it is required to develop a production plan (i.e., production quantities of various periods) that minimizes total inventory holding and ordering costs.

• Capacity constraints make the problem more realistic.

• At the same time, capacity constraints make the problem difficult.

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Lot Sizing with Capacity Constraints

• We shall discuss – the lot shifting technique, a heuristic procedure

that constructs a production plan, and – another procedure that improves a given

production plan.• At times, capacity may be so low that it may not be

possible to meet the demand of all periods. We shall discuss a procedure to check feasibility.

• First, a property that is new for the problems with capacity constraints.

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Order Partial Requirements

• Recall from Lesson 22, that if a lot-sizing solution includes an order size that covers a period only partially, then we can reduce the holding cost without increasing the ordering cost. So, Silver-Meal heuristic, least unit cost and part-period balancing consider order sizes that equals demand of K periods in future, for some K>0.

• As it is shown by the next example, the above does not hold if there are some capacity constraints. It may be essential to order partial requirement of a period.

85


Example 3Production Requirement Production Capacity

June 10 12July 10 8• Without capacity constraint, June production quantity

must include either all or none of the July production requirement– If production is ordered only in June, produce all in

June– If production is ordered in both June and July,

produce June requirement in June and July requirement in July

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• With capacity constraint, June production quantity may include a part of the July production requirement– If production is ordered only in June, produce all in

June– If production is ordered in both June and July,

June production quantity must include all of the June requirement and 2 units of the July requirement

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Checking Feasibility

• As it was discussed before, sometimes capacity may be so low that it may not be possible to meet the demand of all periods.

• So, given demand over the planning horizon and the corresponding capacity constraints, we ask if there exists a feasible solution that meets all the demand. This is the feasibility problem.

• The procedure is stated in the next slide and then the procedure is illustrated with an example.

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• Procedure to check feasibility: – For every period, compute the cumulative

requirement and the cumulative capacity. • If for every period, the cumulative capacity is

larger than (or equal to) the cumulative requirement, then there exists a feasible solution.

• Else, if there is a period in which cumulative capacity is smaller than the cumulative requirement, then there will be a shortage in that period, and, therefore, there is no feasible solution.

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Example 4Production Production

Requirement Capacity June 10 15 July 14 11 August 15 12 September 16 17

Question: Is it possible to meet the production requirements of all the months?

90

Production ProductionRequirement Cumulative Capacity Cumulative

June 10 10 15 15July 14 24 11 26August 15 39 12 38September 16 55 17 55

Answer: The August requirement cannot be met even after full production in June, July and August. Hence, it is not possible to meet the production requirements of all the months.


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Lot Shifting Technique

• Lot shifting technique constructs a feasible production plan, if there exists one or provides a proof that there is no feasible solution.

• Lot shifting method is a heuristic. The production plan obtained from the lot shifting technique is not necessarily optimal. It is possible to improve the production plan.

• An improvement procedure will be discussed after the discussion on the lot shifting technique.

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• The lot shifting method repeatedly does the following:– Find the first period with less capacity.

• If possible, back-shift the excess capacity to some prior periods. Continue.

• If it is not possible to back-shift the excess capacity to some prior periods, stop. There is no feasible solution.

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Requirement Capacity June 10 30July 14 13August 15 13September 16 17

Question: Find a feasible production plan

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Production ProductionRequirement Capacity

June 10 30July 14 13 (less capacity) August 15 13September 16 17

Rule: Find the first period with less capacity. The first period with shortage is July when the capacity = 13 < 14 = production requirement.

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June 10 30 July 14 13 13 (less capacity) August 15 13September 16 17

July production requirement is 14 units which is 1 unit more than the capacity of 13 unit. So, this 1 unit must be produced in some earlier month. There is only one month before July.

96



June 10 11 30 (back-shift)July 14 13 13 (less capacity) August 15 13September 16 17

Rule: Back-shift the excess requirement to prior periods. One unit excess demand of July is back-shifted to June. So, the June production is 10+1=11 units.

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June 10 11 30July 14 13 13August 15 13 13 (less capacity)September 16 17

Rule: Find the first period with less capacity

98



June 10 11 13 30 (back-shift)July 14 13 13August 15 13 13 (less capacity)September 16 17

Rule: Back-shift the excess requirement to prior periods

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An Improvement Procedure

• Now, a procedure will be discussed that can sometimes find an alternative production plan that may reduce the total holding and ordering cost over the planning horizon.

• Keep in mind that it is guaranteed that whenever, there is an improved plan, the procedure will be able to identify that plan. Sometimes, the procedure may fail to identify an improved plan, although there may actually exist one.

100



Requirement Capacity June 13 30July 13 13August 13 13September 16 17

Question: Is it possible to improve the plan if K= $50, h=$2/unit/month?

101


June 13 30July 13 13August 13 13September 16 17

Improvement procedure:Start from the last period and work backwards. In each iteration, back-shift all units of the period under consideration to one or more previous periods if additional holding cost is less than the ordering cost saved.


102

Production ProductionRequirement Capacity Excess

June 13 30 17July 13 13 0August 13 13 0September 16 17 1

Improvement procedure:Start from the last period and work backwards. In each iteration, back-shift all units of the period under consideration to one or more previous periods if additional holding cost is less than the ordering cost saved.


103



Back-shift September production to June? Additional holding cost = (16)(2)(3) =$96 > 50 = KSo, do not back-shift September production to June.


104



Back-shift August production to June? Additional holding cost = (13)(2)(2) =$52 > 50 = KSo, do not back-shift August production to June.


105



Back-shift July production to June? Additional holding cost = (13)(2)(1) =$26 < 50 = KSo, yes, back-shift July production to June.


106



Final Plan The above is the result of the improvement procedure.


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Lesson 23Reading: Section 7.4 pp. 375-379 (4th Ed.), pp. 366-369

(5th Ed.)Exercise: 28 p. 380 (4th Ed.), p. 369 (5th Ed.)

Lesson 24Reading: Section 7.6 pp. 387-395 (4th Ed.) pp. 377-384

(5th Ed.)Exercise: 37 p. 395 (4th Ed.), p. 384 (5th Ed.)

instructor mohammed fazle baki

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