material requirements planning

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 8: 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,

Hierarchy of Production Decisions

3


Aggregate Planning

Master Production Schedule

Inventory Control

Operations Scheduling

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 Decisions Materials 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

Master Production

Schedule

Master Production

Schedule

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 Lead Inventory 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 1501 =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 unit of seat subassembly, the gross requirement of the seat frames is 1251 = 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 subassembl y 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

READING AND EXERCISES

Lesson 8

Reading:

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 9: 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

Lot-Sizing

• Some heuristic methods:

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

Lot-Sizing

• Some heuristic methods:– 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

Lot-Sizing

• Some heuristic methods:– 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

Lot-Sizing

• Some heuristic methods:– 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 K by 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 requirements

1 30 6 80

2 50 7 20

3 10 8 60

4 20 9 200

5 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

5210

590

5210

5210

/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

52.0

10$

068,3

h

K

h

K

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

63

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.

64

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.

65

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

66


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


69

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

71



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

73

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





74


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 between holding and ordering cost is shown in the last column.

75



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.

77

• 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

78

• 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

79

• 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

80


Lesson 9

Reading:

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

81

Outline

• Lot Sizing with Capacity Constraints– Order Partial Requirements– Checking Feasibility– Lot Shifting Technique– An Improvement Procedure

LESSON 10: MATERIAL REQUIREMENTS PLANNING: LOT SIZING WITH CAPACITY

CONSTRAINTS

82

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.

83

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.

84

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 3

Production Requirement Production Capacity

June 10 12

July 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

86


• 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

87

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.

88


• 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 4

Production 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


Requirement Cumulative Capacity Cumulative

June 10 10 15 15

July 14 24 11 26

August 15 39 12 38

September 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.


91

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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Example 5



June 10 30

July 14 13

August 15 13

September 16 17

Question: Find a feasible production plan

94




June 10 30

July 14 13 (less capacity)

August 15 13

September 16 17

Rule: Find the first period with less capacity. The first period with shortage is July when the capacity = 13 < 14 = production requirement.

95




June 10 30

July 14 13 13 (less capacity)

August 15 13

September 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.

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June 10 11 30 (back-shift)

July 14 13 13 (less capacity)

August 15 13

September 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 30

July 14 13 13

August 15 13 13 (less capacity)

September 16 17

Rule: Find the first period with less capacity

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June 10 11 13 30 (back-shift)

July 14 13 13

August 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


Example 6



June 13 30

July 13 13

August 13 13

September 16 17

Question: Is it possible to improve the plan if K= $50, h=$2/unit/month?

101



June 13 30

July 13 13

August 13 13

September 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


Requirement Capacity Excess

June 13 30 17

July 13 13 0

August 13 13 0

September 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.


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June 13 30 17

July 13 13 0

August 13 13 0

September 16 17 1

Back-shift September production to June?

Additional holding cost = (16)(2)(3) =$96 > 50 = K

So, do not back-shift September production to June.


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June 13 30 17

July 13 13 0

August 13 13 0

September 16 17 1

Back-shift August production to June?

Additional holding cost = (13)(2)(2) =$52 > 50 = K

So, do not back-shift August production to June.


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June 13 30 17

July 13 13 0

August 13 13 0

September 16 17 1

Back-shift July production to June?

Additional holding cost = (13)(2)(1) =$26 < 50 = K

So, yes, back-shift July production to June.


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June 26 30 4

July 0 13 13

August 13 13 0

September 16 17 1

Final Plan

The above is the result of the improvement procedure.


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Lesson 10

Reading: Section 7.4 pp. 375-379 (4th Ed.), pp. 366-369 (5th Ed.)

Exercise: 28 p. 380 (4th Ed.), p. 369 (5th Ed.)

Lesson 11

Reading: Section 7.6 pp. 387-395 (4th Ed.) pp. 377-384 (5th Ed.)


108

Outline

• Just-in-Time (JIT)• Examples of Waste• Some Elements of JIT

LESSON 11: JUST-IN-TIME

109

• Producing only what is needed and when it is needed

• A philosophy

• An integrated management system

Just-in-time

110

• Theme: eliminate all waste including the ones caused by:– inventory management – supplier selection– defective parts– scheduling of production and delivery– information system

Just-in-time

111

Examples Of Waste

• Watching a machine run• Waiting for parts• Counting parts• Overproduction• Moving parts over long

distances• Storing inventory• Looking for tools• Machine breakdown• Rework

112

Some Elements Of JIT

1. Focused factory networks 2. Grouped Technology: Cellular layouts 3. Quality at the source

4. Flexible resources5. Pull production system6. Kanban production control7. Small-lot production and purchase8. Quick setups9. Supplier networks

113

Cellular Layouts

• Group dissimilar machines in a cell to produce a family of parts

• Reduce setup time and transit time• Send work in one direction through the cell (resembling

a small assembly line)• Adjust cycle time by changing worker paths

114

Cellular Layouts

Enter

Worker 1

Worker 2Worker

3

Exit

Key: Product routeWorker route

Machines

115

Original Process Layout

12

1

2

3

4

5

6 7

8

9

10

11

A B C Raw materials

Assembly

116

Part Routing Matrix

Parts 1 2 3 4 5 6 7 8 9 10 11 12A x x x x xB x x x xC x x xD x x x x xE x x xF x x xG x x x xH x x x

Machines

117

Part Routing Matrix - Reordered

Parts 1 2 4 3 5 6 7 8 9 10 11 12A x x x x xB x x x xC x x xD x x x x xE x x xF x x xG x x x xH x x x

Machines

118

Parts 1 2 4 3 5 6 7 8 9 10 11 12A x x x x xD x x x x xB x x x xC x x xE x x xF x x xG x x x xH x x x

Machines


119

Parts 1 2 4 8 3 5 6 7 9 10 11 12A x x x x xD x x x x xB x x x xC x x xE x x xF x x xG x x x xH x x x

Machines


120

Parts 1 2 4 8 3 5 6 7 9 10 11 12A x x x x xD x x x x xF x x xB x x x xC x x xE x x xG x x x xH x x x

Machines


121

Parts 1 2 4 8 10 3 5 6 7 9 11 12A x x x x xD x x x x xF x x xB x x x xC x x xE x x xG x x x xH x x x

Machines


122

Parts 1 2 4 8 10 3 6 9 5 7 11 12A x x x x xD x x x x xF x x xC x x xG x x x xB x x x xE x x xH x x x

Machines


123

Cellular Layout Solution

12

12 3

4

5

6

7

8 910

11

A BCRaw materials

Cell1 Cell 2 Cell 3

Assembly

124

Advantages of Cellular Layouts

• Reduced material handling and transit time

• Reduced setup time

• Reduced work-in-process inventory

• Better use of human resources

• Easier to control

• Easier to automate

125

Disadvantages of Cellular Layouts

• Inadequate part families

• Poorly balanced cells

• Expanded training and scheduling of workers

• Increased capital investment

126

Quality At The Source

• Jidoka is the authority to stop a production line• Andon lights signal quality problems• Undercapacity scheduling allows for planning,

problem solving & maintenance• Visual control makes problems visible• Poka-yoke prevents defects

127

Kaizen

• Continuous improvement• Requires total employment involvement• The essence of JIT is the willingness of workers to

• spot quality problems,• halt production when necessary,• generate ideas for improvement,• analyze problems, and • perform different functions

128

Flexible Resources

• Multifunctional workers

• General purpose machines

• Study operators & improve operations

129

Flexible Resources

130

Flexible Resources

Worker 1 Worker 2

Worker 3

Cell 1

Cell 5Cell 3

Cell 2

Cell 4

131

Pull Production System

• In a push system, a schedule is prepared in advance and as soon as one process completes its work, its products are sent to the next process.

• In a pull system, workers take the parts or materials from the preceding stations as needed.Workers at the preceding stations may produce the next unit only after their outputs are taken by the workers in the subsequent processes.

• Although the concept of pull production seems simple, it can be difficult to implement. Kanbans are introduced to implement the pull system.

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Kanban Production Control

• A kanban is a card that indicates quantity of production

• Kanbans maintain the discipline of pull production– - A production kanban authorizes production– - A withdrawal kanban authorizes the movement of goods

MachiningM-2

AssemblyA-4

Part no.: 7412Description: Slip rings

From : To:

Box capacity 25

Box Type A

Issue No. 3/5

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The Origin Of Kanban

Q = order quantityR = reorder point = demand during lead time

Bin 1 Bin 2

Q - RR

Reorder Card

Kanban

a. Two-bin inventory system b. Kanban Inventory System

134

A Single-Kanban System

135

A Single-Card Kanban System

Consider the fabrication cell that feeds two assembly lines.

1. As an assembly line needs more parts, the kanban card for those parts is taken to the receiving post and a full container of parts is removed from the storage area.

2. The receiving post accumulates cards for both assembly lines and sequences the production of replenishment parts.

136

A Dual Kanban System

137

1. When the number of tickets on the withdrawal kanban reaches a predetermined level, a worker takes these tickets to the store location.

2. The workers compares the part number on the production ordering kanban at the store with the part number on the withdrawal kanban.

3. The worker removes the production ordering kanban from the containers, places them on the production ordering kanban post, and places the withdrawal kanbans in the containers.

4. When a specified number of production ordering kanbans have accumulated, work center 1 proceeds with production.

A Two-Card System

138

5. The worker transports parts picked up at the store to work center 2 and places them in a holding area until they are required for production.

6. When the parts enter production at work center 2, the worker removes the withdrawal kanbans and places them on the withdrawal kanban post.

Kanban Squares

X X X

XX

X

Flow of workFlow of information

Kanban Racks

407 409 410 412

408

411

Signal Kanban

407

408

409

407 408 409

Kanban Post Office

65 66 67 68 69 70 71

72 73 74 75 76 77 78

79 80 81 82 83 84 85

86 87 88 89 90 91 92

93 94 95 96 97 98 99

100 101 102 103 104 105 106

Types Of Kanbans

• Kanban Square– marked area designed to hold items

• Signal Kanban– triangular kanban used to signal production at the

previous workstation• Material Kanban

– used to order material in advance of a process• Supplier Kanban

– rotates between the factory and supplier

Determining Number Of Kanbans

where– y = number of kanbans or containers– = average demand over some time period– L = lead time to produce parts– w = safety stock, usually 10% of the demand during

lead time– a = container size

avg.demand during lead time + safety stockcontainer size

No. of kanbans =

a

wLDy

D

Kanban Calculation Example

Problem statement:

= 150 bottles per hour L = 30 minutes = 0.5 hours

= (150)(0.5) = 75 w = 10% of

a = 25 bottles

Solution:

Round up to 4 (allow some slack)

or down to 3 (force improvement)

D

LDLD

containers Kanban .))(.(

3325

751075

a

wLDy

146

Small-Lot Production

• Requires less space & capital investment

• Moves processes closer together

• Makes quality problems easier to detect

• Makes processes more dependent on each other

147

Small-lot Production

and Purchase

148

Reducing Setup Time

• Preset desired settings

• Use quick fasteners• Use locator pins• Prevent

misalignments• Eliminate tools• Make movements

easier

149

Trends In Supplier Policies

1. Locate near to the customer

2. Use small, side loaded trucks and ship mixed loads

3. Consider establishing small warehouses near to the customer or consolidating warehouses with other suppliers

4. Use standardized containers and make deliveries according to a precise delivery schedule

5. Become a certified supplier and accept payment at regular intervals rather than upon delivery

150

Benefits Of JIT

1. Reduced inventory

2. Improved quality

3. Lower costs

4. Reduced space requirements

5. Shorter lead time

6. Increased productivity

7. Greater flexibility

8. Better relations with suppliers

9. Simplified scheduling and control activities

10. Increased capacity

11. Better use of human resources

12. More product variety

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Lesson 11

Reading: Section 7.6 pp. 387-395 (4th Ed.) pp. 377-384 (5th Ed.)


material requirements planning

Documents

production planning

aggregate planning

plan of inhouse production

master production schedule

fictitious aggregate

aggregate plan output

aggregate sales

requirements of components