Merton Truck

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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 would now be as follows: 2y <= 4000

Running the linear programming module again would produce the result as given below:

Objective Function Value = 12000000.000 (Total Contribution)

Product Mix:
The following optimum product mix is obtained for the given constraints:

Variable | Value| Reduced Costs|
x| 1500| 0.000|
y| 1500| 0.000|

Slack/Shadow Prices:
The constraints and their shadow prices are as obtained below: Constraint| Slack/Surplus (Machine-hours)| Dual/Shadow Prices ($) | Engine Assembly| 1000| 0|
Metal Stamping| 0| 1500.000|
Model 101 Assembly| 2000| 0.000|
Model 102 Assembly| 0| 666.667|

Lower/Upper Limits:

Coefficient/Constraint| Lower Limit| Current Value| Upper Limit| x| 0| 3000| 5000|
y| 3000| 5000| No Upper Limit|
Engine Assembly Capacity| 3000| 4000| No Upper Limit|
Metal Stamping Capacity| 3000| 6000| 8000|
Model 101 Assembly| 3000| 5000| No Upper Limit|
Model 102 Assembly| 1500| 4500| 6000|

Model 102 is outsourced:
The product mix does not change even if we change the engine assembly capacity to X<=4000

It would remain as shown in the table shown in section 6.6.1.1 The company could rent the capacity at a maximum of shadow price of engine assembly capacity i.e., $ 2000 per machine-hour Since, the upper limit of engine assembly capacity is 4500 machine-hours; the company could envisage renting out 500 units or either Model 101 or Model 102 trucks

Solution 3 :

Solution 3 :

Problem says that there is a consideration of introducing Model 103 truck which will require following Machine Hrs at different stage of production:

Engine Assembly : 0.8 Machine Hrs / truck
Metal Stamping : 1.5 Machine Hrs / truck
It can be manufactured with Model 101 Assembly at a rate of 4 Machine Hrs/truck. Contribution of Model 103 will be : 2000

| Model 101(X)| Model 102(Y)| Model03(Z’)| Constraint Sign| Machine Hrs|...
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