MERTON TRUCK COMPANY

CASE SOLUTION

HARSHID DESAI AMRUT MODY SCHOOL OF MANAGEMEMNT ROLL NO. 03

Merton Truck Company

Calculating contribution for each truck, Contribution for model 101 = selling price (direct mat. + direct labour + variable o/h) = 39000 (24000 + 4000 + 8000) = Rs. 3000/Contribution for model 102 = selling price (direct mat. + direct labour + variable o/h) = 38000 (20000 + 4500 + 8500) = Rs. 5000/-

Decisions variables: x1 = number of model 101 trucks produced, x2 = number of model 102 trucks produced, The algebraic formulation is: Max. 3000.x1 + 5000.x2, Constrains, 1.x1 + 2.x2 2.x1 + 2.x2 2.x1 + ..+ 3.x2 x1, x2 0.

4000, 6000, 5000, 4500,

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Q.1 (A) Find best products mix for

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As it is been clearly shown in the above result that there is no change in the contribution and also optimum solution is derived by not producing the any truck of model 103. (b) Contribution on model 103 would be as high as Rs.2350/- before it becomes worthwhile to produce. We can see that in above table in maximum objective coeff. Column . If we further increase the contribution by even Rs. 1/- it becomes worth while to produce the new model. That is clearly been seen in below table.

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Q.4 Sol. As we seen in question no. 2 there is no use of increasing the capacity of engine assembly department over 500 machine hours. As another constrains i.e. stamping and assembly of model 102 does not allow you to produce the higher no. of units. In addition to that direct labor cost increase by 50% which further reduces company s contribution. And optimum solution would remain same as of having 4500 machine hours. So x1 and x2 would be 1500 nos trucks. It is shown in graph below.

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In above graph contribution decreased for model 101 and model 102 by 600 and 1200 respectively due the increased cost of direct labour by 50%.

Q.5 Sol. As asked in the question model 101 truck has to atleast 3 times the model 102 trucks. So introducing 5th constrain as follow, >> x1 3x2 >> x1 3x2 0

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Adding constrain to the problem and solving using TORA we