Metron truck case presentation. .. .

1. Merton trucks case

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Contents
• Contribution marginal analysis
• Problem 1
• Problem 2
• Problem 3
• Problem 4

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Contribution Marginal Analysis
Model 101 Model 102
Direct Material 24000 20000
Direct labour
Engine assembly 1200 2400
Metal stamping 800 600
Final assembly 2000 1500
Total Direct labour 2000 4500
Overheads
Engine assembly 2100 4000
Metal stamping 2400 2000
Final assembly 3500 2500
Total overheads 8000 8500
Variable cost/unit 36000 33000
Selling price/unit 39000 38000
Contribution margin
(SP-VC)
3000 5000

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Problem 1
(A).Find the best product mix for Merton ?
Solution:
Let the no. of units of model 101 =x
and the no. of units of model 102 =y
Objective
Max. Z= 3000x +5000y
Constraints
x+2y <=4000
2x+2y<=6000
2x <=5000
3y <=4500
x >=0
y >=0

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Answer report Sensitivity report

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(B). What would be the best product mix if Engine assembly capacity were raised by 1 unit,
from 4000-4001 machine hours? What is the unit of capacity worth?
The best product mix if engine assembly were raised by 1 unit from 4,000 to 4,001
machine-hours is 1,999 Model 101 trucks and 1,001 Model 102 trucks. The extra unit
of capacity is worth $2,000.
:-

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(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 contribution
would be 100 times that in part (b).
If the capacity were to be increased from 4,000 to 4,100, the contribution increases by
$2,00,000 which is 100 times that in part (b).
:-

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(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 ?
As seen from Sensitivity analysis table, allowable increase for engine assembly is
500 calculated below:
Finding the value of x and y using equation
2x + 2y = 6000
3y = 4500
We get x= 1500, y=1500
Substituting the value of x and y in the below equation,
= 1x + 2y
= 1*(1500) + 2*(1500) = 4500
Therefore, additional units can be added is (4500 – 4000 = 500).
New product mix:
Variables: Model 101 Model102
1500 1500
Max. profit $3400000
Solution :-

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Problem 2
Increase in one unit of capacity increases the contribution by $ 2000. Company can
use this alternative by renting 500 machine hours till which contribution increases
after that there is no change in contribution of increased unit in capacity. Also
company should be willing to pay $ 2000 for a rented machine hour.
New product mix:
Variables:
Model 101 Model 102 Objective: Max Profit
1500 1500 3400000
Solution:-

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Problem 3
Let number of Model 101 Truck manufactured = X
Let number of Model 102 Truck manufactured = Y
Let number of Model 103 Truck manufactured = Z
Objective: Max. Z= 3000x+5000y+2000z
Constraints:
EA x+2y+0.8z <=4000
MS 2x+2y+1.5z<=6000
MA 101 2x+z <=5000
MA 102 3y <=3000
Non.n x >=0
Non.n y >=0
Non.n z >=0

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(a) As represented in above table, in order to maximize their profit Merton should not produce any 103 Truck.
(b) Reduced cost for Model 103 is 350. This shows that the contribution of model 103 must be >= (2000+350)
= 2350 to be worth to be produced.
Problem 4.
Solution:-
From Sensitivity analysis it is clear that maximum 500 units of
engine capacity can be added. Therefore, new constraint will
be
x + 2y <= 4500 (Engine Assembly Constraint)
Upon solving LP the product mix we get is:

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Variables: Model 101 Model102
1500 1500
Max. profit: 3400000
New contribution 3400000
Inc. in fixed assets 750000
Inc. in labour 250*3600 900000
Labour cost saving 500*1200 600000
Net contribution 2350000
S0 additional revenue is -50000$,hence Metron should not assemble engines on
overtime.

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Problem 5
Let no. of model 101 is x
Let no. of model 102 is y
Objective: Max. Z=3000x +5000y
Constraints:
EA x+2y <=4000
MS 2x+2y <=6000
MA 101 2x <=5000
MA 102 3y <= 3000
Product ratio –x+3y =0

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ThankYou
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