Merton Truck Company

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  ≤ 4000,

2.x1 + 2.x2  ≤ 6000,

2.x1 +……… ≤ 5000,

……..+ 3.x2  ≤ 4500,

x1, x2 ≥ 0.

Q.1 (A) Find best products mix for Merton.

Sol. Solving by graphical method as follow.

[pic 1]

Best product mix for Merton would be

Model 101 – 2000 trucks

Model 102 – 1000 trucks

Q.1 (B) What would be the best product mix if engine assembly capacity were raised by one unit, from 4000 to 4001 machine hour? What is the extra unit of capacity worth?

Sol.  Solving by graphical method as follow.

[pic 2]

As shown in the graph best product mix after raising the capacity by one unit of machine-hour would be model-101 1999 trucks and model-102 1001 trucks.

Extra unit of capacity worth is (11002000 – 11000000) Rs. 2000.

Q.1 (C) Assume that a second additional unit of engine assembly capacity is worth the same as the first. Verify that if the capacity were increased to 4100 machine hours, then the increase in the contribution would be 100 times that in part (B).

Sol.

[pic 3]

If we increase the one unit in machine hour capacity increases the contribution by Rs. 2000/- as we seen in Q.1 (b). whereas if we increase the capacity by 100 units of machine hour i.e from 4000 to 4100 contribution increases by Rs. 200000/- . which is (2000 * 100 times) 100 times of Rs. 2000/-. Hence it is been verified.

Q.1 (D) How many units of engine assembly capacity can be added before there is a change in the value of an additional unit of capacity?

Sol.

[pic 4]

As shown in the graph above (line 1) i.e. constrain 1 which is engine assembly machine hour, we can push it in the opposite direction of origin till it passes through intersection point of (line 2 and line 4) 2nd and 4th constrain i.e. stamping and model 102 assembly.

        To derive that point of intersection we need to calculate as below.

Finding the value of x1 and x2 using equation

>> 2.(x1) + 2.(x2) = 6000

>>               3.(x2) = 4500

So we get,

x2 = 1500, and x1 = 1500

[pic 5]

With the further increase in unit of capacity of engine assembly machine hour (from 4500 to 4501) there is no change in the contribution.

So no. of units can be added calculated below,

Substituting the value of x1 and x2 in the below equation,

= 1.(x1) + 2.(x2)

= 1.(1500) + 2.(1500)

= 4500

Therefore, unit can be added is (4500 – 4000 = 500).

Q.2

Sol.

Company should adopt this alternative as we seen in Q1 (b) and Q (d). increase in one unit of capacity inceases contribution by Rs. 2000/- and company should rent 500 machine hours till which contribution increases after that there is no change in contribution of increased unit in capacity. So it is obvious that company should be willing to pay Rs. 2000/- for a rented machine hour.

Q.3

Sol.

Decisions variables:

x1 = number of model 101 trucks produced,