Merton Truck Free Essays

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| | 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 f or 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. We will write a custom essay sample on Merton Truck or any similar topic only for you Order Now 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. . 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 requir e 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| Engine Assembly| 1| 2| 0. 8| =| 4000| Metal Stampling| 2| 2| 1. 5| =| 6000| Model 101 Assembly| 2| | 4| =| 5000| Model 102 Assembly| | 3| | =| 4500| Min no| 1| | | =| 0| Min no| | 1| | =| 0| Min no| | | 1| +| 0| Z will be 3000X+5000Y+2000Z Solving for Z with above constraints , we get X = 1900 Y = 1100 Z’=0 Z = 11200000 As it can be clearly seen at optimal solution Model 103 production should be 0. Thus they should not produce Model 103 Trucks. (B)Contribution from Model 103 should be $349(Reduced cost as seen above) more so as to make me worthwhile to produce. Solution 5: New constraint is given as Model 101 production should be at least 3 times the no of Model 103. It can be written in equation form as : X – 3Y = 0 Solving for Z adding above constraint in the existing constraints , we get X= 2250 Y=750 Z=10500000 Graph for the new constraints is : It can be clearly seen with the introduction of new constraint there is a loss of opportunity. How to cite Merton Truck, Papers