Merton Truck

MERTON TRUCK COMPANY

Sol 1 :

Given :
Selling Price od Model 101 truck : 39000
Selling Price of Model 102 truck : 38000

We know,
Contribution C = SP – VC

VC for Model 101 :
Direct Material + Direct Labor + Variable Overhead : 24000 + 4000 + 8000 = $36000

VC for Model 102:
Direct Material + Direct Labor + Variable Overhead : 20000+ 4500+8500 = $33000

Let no of Model 101 produced be X
Let no of Model 102 produced be Y

Z= (39000-36000)X + (38000=33000)Y
Z=3000X + 5000Y

So objective is to Maximize Z

Constraints :

1 | Engine Assembly | X + 2Y <= 4000 | 2 | Metal Smapling | 2X + 2Y <= 6000 | 3 | Model 101 Assembly | 2X <=5000 | 4 | Model 102 Assembly | 3Y <=4500 | 5 | Min no | X >= 0 | 6 | Min no | Y >=0 |

Solving for C with above constraints, we get :

X = 2000 and Y = 1000
Corresponding C will be : 2000(3000) + 5000(1000) = $ 1100000

So best product mix is manufacturing of 2000 Model 101 truck and 1000 Model 102 truck.

Sol 1.B :

Changing Engine Assembly capacity from 4000 to 4001 :

X + 2Y <= 4001 Solving for C with new constraints :

We get value of X and Y as :
X = 1999 and Y = 1001
Corresponding C will be $1100200
Extra Unit of capacity of Engine Assemble is : 1100000-1002000 = $ 2000 (i.e Shadow price of Engine Assembly ) .

Sol 1.c :

If the engine capacity is increased to 4100, constraint eq will become : X+2Y<= 4100

Solving fr C will new constraints we get value as :
X= 1900 and Y = 1100

C will be 11200000

Thus it can clearly seen that value of C has been increased from 1002000 to 1120000 which is 100 times. Graph in this will look like :

Solution 2 :
The company could rent additional capacity up to a maximum of the $ 2000 per machine-hour (this is the opportunity cost of 1 machine-hour of engine assembly capacity).
Case1: Model 101 is outsourced:
The engine capacity constraint equation