or sol7 inventory control

7/30/2019 Or Sol7 Inventory Control

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TTN-DoMS, IIT-MADRAS

T.T. NarendranDepartment of Management Studies

Indian Institute of Technology Madras


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Problem 1Let Q* be the optimal order quantity in the sample EOQ model

(constant rate of demand, instantaneous supply, no shortages)

Let the actual order quantity Q be given by

Q=K.Q*, K>0

a) Derive an expression for the ratio of the actual total

cost/unit time to the optimal cost.

b) If the holding cost/unit/unit time is overestimated by 20

percent, what is the percentage increase in the total cost

over the optimal cost?


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a) Q*=(2DC0/CC)

Q = k.Q*

TC = DC0/kQ*+kQ*CC/2

TC = DC0CC/ k(2DCo) + k(2DCo) CC/2 CC

= (DC0CC)/(2).k + k(DC0CC)/2

TC = (DC0CC/2) (k+1/k)

TC* = (DC0CC/2)

TC/TC* = (k+1/k)

b) If CC = 1.2 CC*

TC = 2DC(1.2) CC

TC/TC* = 1.2


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M/s Gujjubhai Shah and company is an old firm where scientific

inventory management is totally unknown. For a certain raw

material that they buy, it is estimated that each purchase order

would cost Rs.200/-. The cost of the raw material is Rs.30 per

kg; the annual inventory carrying cost is 10 per cent of the cost

per kg; the monthly requirement is 100 kg.

The ordering quantity has been arbitrarily fixed as 440kg. If

this ordering quantity is to be optimum by fluke, what should be

the value of the shortage cost, assuming that shortages can be

allowed?

Problem 2


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Co = Rs.200 ; C = Rs.30

I = 14% per year

D = 100 kg/Month = 1200 Kg/year

Q = 440 ; CS = ?

Q = (2DCo/Cc) (1 + Cc + Cs)

440 = (2*1200*200/4.2)(1 + 4.2/Cs)

Cs = 6.052


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Marchand and Wadichand Fabrics run a huge show room forReadymade Garments. The Show room is famous for garments madeof a certain variety of cotton. The firm has a quarterly requirementsof 2000 metres length of this variety of cloth, which costs Rs.38/-per metre.

The proprietors have made a precise estimate of their ordering cost -

thirty two rupees, thirty paise per order. The carrying cost per year isestimated to be 8.5 per cent of the cost per metrelength. Marchands son Lalit Chand, who has recently joined thebusiness after completing his MBA, insists that they would procurethe cloth in economic order quantities. However, their supplier, NaiduGaru Textile Mills, Coimbatore, is not willing to sell less than 2000metres of the cloth per order.

What is the discount on the unit cost that Lalit Chand should demand

in order to give up his insistence on EOQ?

Problem 3


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Demand = 2000 m/quarter

D = 8000 per year

C = Rs.38

Co = Rs.32.30

i = 8.5%

Cc= 38*8.5/100 = 3.23

Q* = (2DCo/Cc)

= 2*8000*32.30/3.23 = 400

TC =(2DCoCc) + 8000*38 = 305292

Let discount be d%

Cc = 38*8.5(1-d)/100

= Rs.3.23(1-d)


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TC (With discount) = 8000*32.3/2000 + 8000*38*(1-d)

+ 2000*38*8.5*(1-d)/2*100

=129.2+3230(1-d)+304000(1-d)

307230(1-d) = 305292 - 129.2

1-d = 0.9937

d = 0.0063

i.e. 0.63% discount


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The Central Supplied Unit, IIT, Madras is consideringquotations from three vendors to whom it has to place ordersfor rice.

Eskay Provision Stores quotes Rs.120/- per bag irrespective ofthe quantity ordered. Beavy and Company will accept ordersonly for 800 or more bags but quotes a price of Rs.108/- perbag. Bellam Trading Co. will accept orders only for 1000 ormore bags at Rs.100/- per bag.

The total requirement of the twelve hostels, for whom the CSUbuys, is 3000 bags per semester. Inventory carrying costs are

20 per cent of unit cost and ordering cost for the CSU Rs.400/-per order. If you were the Hostel Affairs Secretary, whichvendor should you recommend to the Warden, CSU, for placingthe orders and for what quantity per order?

Problem 4


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D = 3000 per semester

= 6000 per year

Co = Rs.400

i = 20% = 0.2

Vendor 1 C = Rs.120

EOQ = (2*6000*400/0.2*120)

= 447.21

TC = (2DCoCc) = (2*6000*400*0.2*120) = 10733.126


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Vendor 2 C = Rs.108

Q = 800 or more

EOQ = (2*6000*400/0.2*108) = 471.40

TC = (6000/800)*400 + (800/2)*0.2*108 = 3000 + 8640

=11640

Vendor 3 C = Rs.100

Q = 1000

EOQ = (2*6000*400/0.2*100) = 489.89

At Q = 1000 , TC = (6000/1000)*400 + 1000*0.2*100/2

=2400 + 10000 = 12400

Hence Vendor1 is chosen with an order quantity of 447.21 bags


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A company carries an inventory composed of two products A and Bfor which there is a uniform yearly demand of 10,000 and 20,000

units respectively. Product A costs the company Rs.2/unit topurchase, Rs.1/unit/year to store, and Rs.100 in administrativecosts every time it is reordered. Product B costs the companyRs.4/unit to purchase, Rs.2/unit/year to store, and Rs.200 inadministrative costs for each reorder.

a) Set up the total cost equations for each productseparately and solve for their optimum order quantities.

b) Owing to a liquidity problem, the company requires thatthe average yearly investment in inventory for the two productstaken together be less than or equal to Rs.6000/-. Is the answer

found in part (a) consistent with this added constraint?c) Find the optimum solution when the average yearly

investment in inventory must be less than or equal to Rs.5000.Average yearly investment is the rupee value of the averageamount of inventory present.

Problem 5


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D = 1000 D2 = 20000C1 = Rs.2 C2 = Rs.4CC1 = Rs.1 CC2 = Rs.2C

O1= Rs.100 C

O2= Rs.200

a) Q1* = ( 2*10000*100/1) Q2

* = ( 2*20000*200/2)= 1414.2 = 2000

TC1

= D1

CO1

/Q1

+ Q1*

CC1

/2 TC2

= D2

CO2

/Q2

+ Q2*

CC2

/2

Q1* = (2D1CO1/ CC1) Q2

* = (2D2CO2/ CC2)

b) Amount invested in inventory (average)

= Q1C1/2 + Q2C2/2 = (1414.2*2)/2 + (2000*4)/2

= 1414.2 + 4000 = 5414.2 6000

Answer found in part (a) is consistent with the added constraint


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c) If inventory 5000, it is binding. We setup the lagrangean function

L = D1CO1/Q1 + Q1CC1/2 + D2CO2/Q2 + Q2 CC2/2

+ (Q1C1/2 + Q2C2/2 6000)

= 1000*100/Q1 + 1*Q1/2 + 2000*200/Q2 + 2*Q2/2+ (2*Q1/2 + 4*Q2/2 - 6000)

L/Q1=0 gives

-1000*100/Q12 + + = 0

1000*100/Q12 = ( + )

L/Q1 = 0

Q1 =

100000/( +

)

L/Q2 = 0 Q2 = 400000/(1 + )

L/ = 0 Q1 + 2Q2 = 6000


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Let ,

= 0 Q1 = 1414.2 Q2 = 2000 Q1 + 2Q2 = 5414.2

= 1.0 Q1 = 816.5 Q2 = 1414.2 Q1 + 2Q2 = 3644.9

= 0.2 Q1 = 1195.2 Q2 = 1825.7 Q1 + 2Q2 = 4846.7

= 0.15 Q1 = 1240.3 Q2 = 1865.01 Q1 + 2Q2 = 4970.37

= 0.13 Q1 = 1259.88 Q2 = 1881.4 Q1 + 2Q2 = 5022.7

= 0.14 Q1 = 1250 Q2 = 1873.17 Q1 + 2Q2 = 4996.34

The order quantities are 1250 and 1873 respectively.


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A shop sells two products that it orders periodically from one supplier.The demand rates for both products are predictable and constant.

Demand rates and inventory costs are given by the following table:

a) Find the optimum values of the inventory cost per month and thelength of both inventory cycles under the assumption that bothinventory processes are managed separately.b) The shop finds that they could save on the order cost of product B

if both products were ordered together. This course of action can beevaluated by finding the minimum total inventory cost per month,subject to the constraint that the two inventory periods be equal.Write the objective function and the constraint. Show how to solve this byLagrange Multiplier method.

Problem 6

Product Demand rate Order cost Holding cost

Units/month Rs./order Rs./Unit/Month

A 100 50 25

B 300 50 3


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Product A D1 = 100 per month

CO1 = Rs.50

CC1 = 2.50

Q1* = (2D1CO1/ CC1) = ( 2*100*50/2.5) = 63.24

N = D/Q = 100/63.24 = 1.581

TC1 = (2DCo/Cc) = (2*100*50*2.5) = 158.1139

Product B D2 = 300

CO2 = Rs.50

CC2 = 3.0

Q2* = ( 2*300*50/3) =100

N = 3

TC2 = 300 Time periods are different


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b) We assume that we can save order cost of B if they are ordered together.The periods have to be equal D1/ Q1 = D2/ Q2

L = (D1/Q1)*50 + Q1* CC1/2 + Q2* CC2/2 + (D2/Q2 - D1/Q1)

L/Q1 = 0 gives -50D1/Q12 + CC1/2 + (D1/Q1

2) = 0

Q1 = (2D1 (-+50)/ CC1) = 80(50-)

L/Q2 = 0 gives CC2/2 -

(D2/Q2

2

) = 0

Q2 = 2D2/CC2 = 200

L/ = 0 gives 100/Q1 = 300/Q2

Q1/Q2 = 3

200 = -720 + 720*50 = 720*50/920 = 39.13

Q1 = 29.49 Q2 = 88.46

P bl 7


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Messrs. Ghatilkar and Pillkarni Chemicals Private Ltd., own a plant in Chembur

which processes three chemicals in liquid form for consumption within the

factory itself at a uniform rate. The following are the particulars relating to the

three items:

Item Annual consumption in litres Cost per Processing Time for

Litre (Rs.) 100 Litres

1. 36,000 20 2 hours 13 1/3 min.2. 30,000 40 3 hours 20 min.

3. 50,000 20 1 hour 36 min.

Setting up the plant to start production of any of these items takes 8 hours for

every set up. However, during set-up the plant is not productively engaged; the

estimated cost of the idleness of the plant due to set up is Rs.100/- per hour.

The company works on a two-shifts-5 day week basis for 50 weeks, the

remaining 2 weeks being the annual shut down period for maintenance work.

Each shift is of 8 hours duration.

Problem 7


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The inventory carrying charges per year are 10 per cent of the

cost per litre for each chemical.

The chemicals processed by the plant are collected and stored

in 10-litres cans. On an average how many such cans should be

made available, if the company desires to turn out these

chemicals in economic manufacturing quantities?

Suppose there are only 500 cans available, and assuming that

the average requirement of cans should never exceed this

number, how should the economic manufacturing quantities be

modified to satisfy this restriction?


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Three items A,B,C

Item A Item B

D = 36000 D = 30000

CO = Rs.100 per hour *8hrs = Rs800 CO = Rs.800

C = 20 C = 40

i = 10% i = 10%

CC = Rs.2 CC = Rs.4

P = 350*16*60*3*100/400 P = 350*16*60*100/200

= 25200 litres = 168000 litres

Q = {2DCO

/ CC

(1-D/P)} Q = {2*30000*800/2(1-30/168)}

= {2*3600*800/2(1-36/252)} = 3822.132 litres

= 5796.55 litres

No. of cans = 5796.55/20 = 289.83 cans No. of cans = 191.11 cans


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Item C

D = 50000

CO= Rs.100 per hour *8hrs = Rs.800C = 20

CC = Rs.2

P = 350*16*60*100/96 = 350000 litres

No. of cans = 341.56 cans

Total cans required = 289.83 + 191.11 + 341.56 = 822.5

Only 500 cans are available


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Lagrangean L = 3600*800/Q1 + (Q1)2*0.857/2 + 30000*800/Q2+ Q2*4*0.821/2 + 50000*800/Q3 + (Q3)2*0.857/2+ (Q1+ Q2 + Q3/20 500)

L/Q1 = 0 -36000*800/Q12 + 0.857 + /20 = 0

36000*800/Q12 = (17.14 + )/20

Q1 = {36000*800*20/(17.14 + )}Q2 = {30000*800*20/(32.84 + )}Q3 = {50000*800*20/(17.14 + )}Q1 + Q2 + Q3 = 10000

Solving we get = 34Q1 = 3356Q2 = 2680Q3 = 3955


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M/s Kanjivellam Enterprises Pvt. Limited own a medium-sized

factory divided into 3 manufacturing shops. The factory makesthree products A, B and C.

Manufacturing one unit of A requires 2 hours at Shop I, 1 hour

at Shop II and 2 hours at Shop III.

Manufacturing one unit of B requires 3 hours at Shop II and 2

hours at Shop III.Manufacturing one unit of C requires 1 hour at Shop I and 2

hours at Shop III.

The available capacities for shops, I, II, and III are 90 machine-

hrs. 120 machine-hours and 220 machine-hours respectively.The profits per unit of A, B and C are Rs.5/-, Rs.5/- and Rs.3/-

respectively. What is the optimal weekly product-mix that

maximises the total profit of M/s Kanjivellam Enterprises Pvt.

Ltd.,?

Problem 8


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Let x1, x2, x3 be quantities of A, B, and C respectively

Maximize 5x1 + 5x2 + 3x32x1 +x3 90

x1 +3x2 1202x1 +2x2 +2x3 220x1, x2 , x3 0P reqd = 2x1 +2x3 = 150/weekQ reqd = 3x2 = 105/week

R reqd = 3x3 = 180/week

Item P Item Q

D =150*50 = 7500 D = 105*50 = 5250

CO = 50 CO= 50i = 20% of Rs.5 = Rs.1 CC = 20% of Rs4.20 = Rs.0.84Q = {2DCO/CC Q =790.57

= {2*7500*50/1} = 866.02QC/2 = 2165.06 QC/2 = 1660.20

Item R


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Item R

D = 180*50 = 9000

CO = 50

CC = 20% of Rs.3 = Rs.0.60Q = {2DCO/ CC = {2*9000*50/0.6} = 1224.75

QC/2 = 1837.117

Budget = 2165.06 + 1660.20 + 1837.12

= 5662.373

Violates budget restriction

New values of order quantities areItem P = 866.02*5000/5662.373 = 764.71

Item Q = 790.57*5000/5662.373 = 698.09

Item R = 1224.75*5000/5662.373 = 1081.48


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Messrs. Spoilda Soil Fertilizer Company has undertaken a

contract to supply 50 tons of fertilizer at the end of the firstmonth, 70 tons at the end of the second month and 90 tons at

the end of the third month. The cost of producing X tons of

fertilizer in any month is given by

(4500X + 20X2

) RupeesThe company can produce more amount of fertilizer in any

month and supply it in the next month. However, there is an

inventory carrying cost of Rs.400/- per ton per month.

Formulate the problem of finding the optimal level of productionin each of the three periods and the total cost incurred as a

dynamic programming problem and solve.

Assume initial and final inventory to be equal to Zero.

Problem 9


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This problem can be solved by forward recursion.

Stage : Each monthStates : Inventory carried from the previous month

Decision Variable: Quantity to be produced each month

Criterion : Minimize cost

X1 X2 X3

Month 1 Month 2 Month 3

Initial Inventory = 0 S1 S2


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n = 1; 2 more months remaining:f1(0,X1) = 20X1

2 + 4500X1f1*(0) = min {20X1

2 + 4500X1}

= 20(50 + S1)2 + 4500(50 + S1)= 20(2500 + S1

2 + 100S1) + 225000 + 4500S1= 275000 + 20S1

2 + 6500S1 at X1* = 50 + S1

n = 2; 1 more month remaining:f2(S1,X2) = 20X2

2 + 4500X2 + 400S1 + f1*(0)

= 20X22 + 4500X2 + 400S1 + 275000 + 20S1

2 + 6500S1f2*(S1) = min {20X2

2 + 4500X2 + 400S1 + 275000 + 20S12

+ 6500S1}

= 20(70 + S2 S1)2 + 4500(70 + S2 S1) + 400S1

+ 275000 + 20S12 + 6500S1

= 688000 + 40S12 + 20S2

2 - 400S1 + 7300S2 - 40S1S2at X2* = 70 + S2 - S1


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n = 3; last month:f3(S2,X3) = 20X3

2 + 4500X3 + 400S2 + f2*(S1)

= 20X32 + 4500X3 + 400S2 + 688000 + 40S1

2 + 20S22

- 400S1

+ 7300S2

- 40S1S

2f3*(S2) = min {20X3

2 + 4500X3 + 400S2 + 688000 + 40S12 + 20S2

2

- 400S1 + 7300S2 - 40S1S2}

= 20(90 - S2)2 + 4500(90 S2) + 400S2 + 688000 + 40S1

2

+ 20S22 - 400S1 + 7300S2 - 40S1S2

= 1255000 + 40S12 + 40S22 - 40S1S2 - 400S1 - 400S2at X3* = 90 - S2

Differentiating with S1 and S2 and solving the equations we get,

S1 = 10 and S2 = 10

Hence, X3* = 80 Total cost = 12,51,000

X2* = 70

X1* = 60


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Consider the problem of determining economic lot sizes for four

different items. Assume that the demand occurs at a constantrate over time. The constant order quantity for the jth item is

denoted by Qj. The total storage space available is A, where

each unit of jth item occupies an area aj.

The objective is to find the Qjs which minimize the total cost

subject to the total area constraint. The total cost is of the form.Uj

TC = j=14

-------- + VjQj , Qj > 0

Qj

Where Uj and Vj are constants.

Formulate the problem as a dynamic programming model.

Assume, if necessary, that Qj is discrete.

Problem 10


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Stage : Each item

State :Area remaining for the last n items

Decision Variable: How much to orderCriterion : Minimize cost

n = 1; last item:

f1(S1,Q1) = + V1Q1

f1*(S1) = Min + V1Q1

= Min ,

U1Q1 U1

Q10 Q1 S1a1

U1Q1

S1a1


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n = 2; two more items:

f2(S2,Q2) = + V2Q2 + f1*(S2

Q2a2)

In general, the recursive relation is

fn(Sn,Qn) = + VnQn + fn-1*(Sn Qnan)fn*(Sn) = Min + VnQn + fn-1*(Sn Qnan)

For the final stage (n = 4),

Sn = A [The total storage space available ]

UnQn Un

Qn0 Qn Snan

U2Q2


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Thank You !!!

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