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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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