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Merton trucks case presentation. .pptx
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- 2. TREY research 2 Contents • Contribution marginalanalysis • Problem 1 • Problem 2 • Problem 3 • Problem 4
- 3. TREY research Add a footer3 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
- 4. TREY research Add a footer4 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
- 5. TREY research Add a footer5 Answer report Sensitivity report
- 6. TREY research Add a footer6 (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. :-
- 7. TREY research Add a footer7 (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). :-
- 8. TREY research Add a footer8 (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 :-
- 9. TREY research Add a footer9 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:-
- 10. TREY research Add a footer10 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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- 12. TREY research Add a footer12 (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:
- 13. TREY research Add a footer13 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.
- 14. TREY research Add a footer14 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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