Free Essay

Merton Truck Company

In: Business and Management

Submitted By saske9800
Words 1574
Pages 7
Operations Research
Assignment 1: Merton Truck Company

Submitted to
Professor Sumit Kumar

By
Sneh Chandel
PGPX 2015-16
Roll No – 1503010

On
27/01/2015

Problem 1 1. (a)
The best product mix can be found out by plotting out the corner points of graph obtained to maximize function p = 3000x + 5000y – 8600000
The above function was derived by subtracting cost price of ‘Truck model 101’ & ‘Truck Model 102’ from sales price of ‘Truck model 101’ & ‘Truck Model 102’.
Sales price is quoted in text as $39,000 for ‘Model 101’ and $38,000 for ‘Model 102’
Cost Price of truck was obtained from following table: | | | Model 101 | | | Model 102 | Direct Materials | | | $24000 | | | $20000 | Direct Labor | | | | | | | | Engine assembly | 1200 | | | 2400 | | | Metal Stamping | 800 | | | 600 | | | Final Assembly | 2000 | | | 1500 | | Total Direct Labor | | | $4000 | | | $4500 | Overhead | | | | | | | | Engine Assembly | 2525 | | | 4850 | | | Metal stamping | 3480 | | | 3080 | | | Final Assembly | 6200 | | | 3500 | | | | | 12205 | | | 11430 | Total | | | 40205 | | | 35930 |

& the below table to calculate the variable overhead for individual trucks:

Department | Variable overhead/unit 101 | Variable overhead/unit 102 | Engine assembly | 2100 | 4000 | Metal stamping | 2400 | 2000 | Model 101 Assembly | 3500 | | Model 102 Assembly | | 2500 | Total | 8000 | 8500 |

Therefore Total costs of the trucks come out to be Direct Material + Direct Labor + Variable overhead
i.e Total cost for ‘model 101’ is $24,000 + $4,000 + $,8000 = $36,000
i.e Total cost for ‘model 102’ is $20,000 + $4,500 + $,8,500 = $33,000
Profit Contribution: | Model 101 | Model 102 | Sales Price | $39,000 | $38,000 | Cost Price | $36,000 | $33,000 | Profit Contribution | $3,000 | $5,000 |

Therefore Profit Equation is P = 3000x + 5000y
However, we also need to account for Fixed overhead per month which is given by: Department | Fixed Overhead/month $million | Engine assembly | 1.7 | Metal stamping | 2.7 | Model 101 Assembly | 2.7 | Model 102 Assembly | 1.5 | Total | 8.6 |

Therefore Profit Equation becomes P = 3000x + 5000y – 8,600,000
The variables for the profit equation are constrained by the following manufacturing equations:

Department | Machine -hours101 | Machine hours 102 | Machine hours available/month | | | | | Engine Assembly | 1 | 2 | 4000 | Metal Stamping | 2 | 2 | 6000 | Model 101 Assembly | 2 | - | 5000 | model 102 Assembly | - | 3 | 4500 |

or: x + 2y <= 4000
2x + 2y <= 6000
2x <= 5000
3y <= 4500
To form the boundary for figure obtained, we will also be considering x = 0 & y = 0.
Graph obtained as per linear programing via online grapher: http://www.quickmath.com/webMathematica3/quickmath/graphs/inequalities/advanced.jsp#c=plot_advancedgraphinequalities&v1=x%2B2y%3C%3D4000&v2=2x%2B2y%3C%3D6000&v3=2x%3C%3D5000&v4=3y%3C%3D45000&v7=x&v8=y&v9=0&v10=3000&v11=0&v12=3000&v13=purple&v15=forest-green&v16=dark-turquoise&v19=1

The corner points of the graph have been obtained by using online calculator: http://www.zweigmedia.com/RealWorld/LPGrapher/lpg.html

Vertex Lines Through Vertex Value of Objective
(2000,1000) x+2y = 4000; 2x+2y = 6000 2400000 Maximum
(1000,1500) x+2y = 4000; 3y = 4500 1900000
(2500,500) 2x+2y = 6000; 2x = 5000 1400000
(2500,0) 2x = 5000; y = 0 -1100000
(0,1500) 3y = 4500; x = 0 -1100000
(0,0) x = 0; y = 0 0

Therefore, as per corner points obtained via graph, maximum profitability can be obtained at product mix x = 2000 & y = 1000.

Problem 1 1. (b)
Best product mix can be obtained by manipulating constraint function for maximum profitability function: p = 3000x + 5000y – 8600000 which are x + 2y <= 4001
2x + 2y <= 6000
2x <= 5000
3y <= 4500
Graph Obtained:

Corner Points are given as :

Vertex Lines Through Vertex Value of Objective
(1999,1001) x+2y = 4001; 2x+2y = 6000 2402000 Maximum
(1001,1500) x+2y = 4001; 3y = 4500 1903000
(2500,500) 2x+2y = 6000; 2x = 5000 1400000
(2500,0) 2x = 5000; y = 0 -1100000
(0,1500) 3y = 4500; x = 0 -1100000
(0,0) x = 0; y = 0 0

Shadow price calculation
Shadow price of the assembly is 2000 $
We arrive at this value by subtracting the profitability obtained at points (2000,1000) & (1999, 1001), which is $2,402,000 - $2,400,000 = $2000

Problem 1 1. ©
If assembly unit capacity is increased from 4000 to 4100 hours, then maximum profitability with these constraints can be given by: p = 3000x + 5000y – 8,600,000 subject to x + 2y <= 4100
2x + 2y <= 6000
2x <= 5000
3y <= 4500

Corner points are as per given below:
Vertex Lines Through Vertex Value of Objective
(1900,1100) x+2y = 4100; 2x+2y = 6000 2600000 Maximum
(1100,1500) x+2y = 4100; 3y = 4500 2200000
(2500,500) 2x+2y = 6000; 2x = 5000 1400000
(2500,0) 2x = 5000; y = 0 -1100000
(0,1500) 3y = 4500; x = 0 -1100000
(0,0) x = 0; y = 0 0

Contribution increase to profitability by increasing 100 engine assembly hours can therefore be given by subtracting maximum profitability at constraints x + 2y = 4000 & x + 2y = 4100, which comes out to be:
2,600,000 – 2,400,000 = $200,000
Shadow price of the assembly is 200,000$
Therefore, if assembly unit capacity is increased from 4000 to 4100, then extra unit Engine assembly capacity is worth 200,000$

Problem 1 1. (d)
Number of swaps possible without leading to drop in profits can be given by solving the below two equations for Q: x + 2y = 4000 + Q x + y = 3000
Solution of which gives Q as 500
The maximum number of swaps that can be obtained can be found while keeping production capacity at maximum. Reducing 101 assembly does not lead to a constraint because each swap leads to two unused hours. However, the number of swaps is constrained by the number of 101s being produced currently, i.e., 2000, which means that we cannot give up more than we are producing. Therefore the number of swaps we can do is 500.

Problem 2
The maximum rent we can pay to outsource engine assembly operation should not exceed the cost incurred to assemble engine within assembly line.
Therefore, maximum rent should not exceed the cost as found in 1.(b) which is $2000.
Maximum number of machine hours to be rented out can be considered to be same as maximum number of swaps possible for engine assembly.

Problem 3

Product mix after having introduced model 103 can be found out by plotting on graph the profitability equation: P = 3000x + 5000y + 2000z – 8,600,000
Using online linear programing via website: http://www.zweigmedia.com/RealWorld/simplex.html
Maximize p = 3000x + 5000y + 2000z – 8,600,000 subject to constraints as per given below x + 2y + 0.8z <= 4000
2x + 2y + 1.5z <= 6000
2x + 1z <= 5000
3y <= 4500

Optimal Solution for above equation can be calculated as: p = 2,400,000; x = 2000, y = 1000, z = 0
Therefore, the corner points does not contain z as a variable.
3 (a) According to data presented above, merton should not produce model 103 trucks.
3 (b) Applying marginal analysis to equation p = 3000x + 5000y + 2000z – 8,600,000, we get a rough estimate that profit starts to get affected above the value of 2,400,000 only when coefficient of z is increased to value greater than 2350 keeping x and y at their optimum production value.

Problem 4
New profitability equation is established in accordance to new constraints proposed.
Costing for 101’s, 102’s and 101’s and 102’s on over time can be given by below table: | | | Model 101 | | | Model 102 | | Overtime 101’s | | Overtime 102’s | | | Direct Materials | | | $24000 | | | $20,000 | | $24,000 | | $20,000 | | | Direct Labor | | | | | | | | | | | | | | Engine assembly | 1200 | | | 2400 | | | 1800 | | 3600 | | | | Metal Stamping | 800 | | | 600 | | | 800 | | 600 | | | | Final Assembly | 2000 | | | 1500 | | | 2000 | | 1500 | | | Total Direct Labor | | | $4000 | | | $4500 | | $4600 | | 5700 | | | Variable Overhead | | | $8,000 | | | $8,500 | | $8,000 | | $8,500 | | | Total | | | $36,000 | | | $33,000 | | $36,600 | | $34,200 | | |

Therefore profit contribution from overtime 101’s comes out to be (39000 – 36600) $2,400 & contribution from 102’s comes out to be (38000 – 34200) $3,800 after having subtracted cost price calculated from given general sales price.
Fixed overhead cost can be given as original costs + $0.75 m which is $9,350,000
Profitability equation can be given as Maximize p = 3000x + 5000y + 2400u + 3800v subject to x + 2y <= 4000 u + 2v <= 2000
2x + 2y + 2u + 2v <= 6000
2x + 2u <= 5000
3y + 3v <= 4500
Optimal solution for corner points can be given as Optimal Solution: p = 11,700,000; x = 1500, y = 1250, u = 0, v = 250
Here, we find that doing overtime is not viable as it results to an increase in fixed overhead of $.75 million, which is not enough to compensate the increase in profit out overtime which is $.7 million.
Problem 5
Model 101’s can be kept at least equal to three times the no. of model 102’s by introducing the proper constraints to the profitability equation, i.e
We need to Maximize p = 3000x + 5000y subject to x + 2y <= 4000
2x + 2y <= 6000
2x <= 5000
3y <= 4500 x - 3y >= 0 (new condition)
The solution to the above problem gives us a graph where the corner point will give us the optimal product mix.

Optimal Solution: p = 10500000; x = 2250, y = 750
Ans: Optimal Product Mix: no. of 101 trucks produced: 2250 no. of 102 trucks produced: 750

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