MERTON TRUCK COMPANY
1) No. of Model 101= x, no. of model 102=y.Unit contribution/model= S.P – (Variable costs + Fixed cost).For Model 101, variable cost=36000 /unit,For Model 102, variable cost=33000/unit.Fixed cost for both the model= 8600000 Therefore, total contribution z=39000x+38000y-36000x-33000y-8600000Z=3000x+5000y-8600000 The Constraints are:Engine assembly: x + 2y <= 4000Metal stamping: 2x + 2y <=6000Model 101 assembly: 2x <=5000Model 102 assembly: 3y <= 4500x,y>=0.Solving, x=2000, y=1000, z=2400000(b)If engine assembly capacity is increased by 1 unit to 4001, optimal productmix is x=1999, y=1001. By sensitivity analysis, shadow price for engineassembly is 2000 for allowable increase/decrease of available constraint by500. Hence, extra unit of capacity is worth 2000. Therefore, increase incontribution z is 2000
(c) Since, allowable increase of available constraint of engine assembly is500,hence capacity increase by 100 will lead to increase in contribution= shadowprice (engine assembly capacity) * increase in avl. Constraint (<500) = 2000* 100 = 100 times as contribution increase in (b) Also if we use solver, forengine assembly constraint= 4100, x = 1900, y=1100, z= 2600000. Therefore , increase in contribution = 200000.
(d) Maximum allowable increase
2) Since shadow price is 2000, maximum rent Merton can give to supplier is2000 for an hour of engine assembly capacity utilization. Also since allowableincrease is 500, maximum number of engine hours it should rent is 500.3) For Model 103, we introduce a 3 rd
constraint w.Contribution by w is given as 2000.Hence now, z=3000x + 5000y + 2000w-8600000.As per given data, machine-hour/truck for model 103 are 0.8, 1.5 and 1 onengine assembly line, metal stamping line and Model 101 line respectively.Hence, new constraints:Engine assembly: x + 2y + .8w <= 4000Metal stamping: 2x + 2y + 1.5w <= 6000
Model 101 assembly : 2x + w < = 5000Model 102 assembly : 3y <=...
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