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Linear Programming Model

Question 1: Formulate a linear programming model for the make-or-buy decision for Cleveland Stapler that will meet the 5,000-unit demand at a minimum total cost.

Answer:

First we need to compare relevant costing of each component. Under this we should consider only variable cost since fixed cost has to be incurred either we go or purchase option or Buy option

PARTICULAR

BASE

STAPLE CATRIDGE

HANDLE

DIRECT LABOUR

20000

30000

40000

DIRECT MATERIAL

10000

20000

30000

VARIABLE COST

10000

10000

10000

LABOUR COST

2500(5000*0.25)

1250

1000

MACHINE HRS COST

500 (5000*0.10)

750

600

TOTAL VARIABLE COST

43000

62000

81600

No of Units

5000

5000

5000

Cost per unit

8.6 (43000/5000)

12.4

16.32

 

PURCHASE PRICE

11

8

20

DECISION

MAKE

BUY

Make

Note: Rent and Depreciation costs have been ignored or decision making as these have to be incurred in any case.

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SAY IF 2000 LABOUR HRS(limiting factor) ARE AVAILABLE THEN FIRM SHOULD FOLLOW BELOW MENTIONED STRATEGY

Cost per unit

8.6 (43000/5000)

12.4

16.32

 

PURCHASE PRICE

11

8

20

INCREMENTAL SAVING FROM MANUFACTURING

2.4'

-4.4

3.68

LabourHrs /Unit

0.50

0.20

0.20

Incremental saving Per unit of limiting factor

4.8

 

18.4

Production rank

2

NA

1

In case oflabour being limiting factor, there is benefit in manufacturing whole 5000 units of handle first in-house

Handle

5000 units x 0.20

1000 hrs required

5000 units produced

Base

5000 units x 0.50

2500 hrs required

Subject to maximum 1000 hrs available(2000-1000). Thus maximum base can be produced is 1000 hrs / 0.50 total 2000 units can be produced

Final equation if labourhrs are limiting factor

PARTICUALRS

BASE

STAPLE CATRIDGE

HANDLE

UNITS REQUIRED

5000

5000

5000

PRODUCTION/BUY OUT

PRODUCE 2000 UNITS AND BUY 3000 UNITS FROM OUTSIDE

BUY ALL 5000 UNITS

PRODUCE 5000 UNITS

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