# Linear programming model

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The Eastside Manufacturing Company produces four different aircraft components from fabricated sheet metal for several major aircraft companies. The manufacturing process consists of four operations - stamping, assembly, finishing, and packaging. The processing time per unit for each of the operations and total available hours per year to produce these components are given below.

Operation Component (hr/unit)

1 2 3 4 Total Hours per Year

Stamping 0.06 0.17 0.10 0.14 700

Assembly 0.18 0.20 -- 0.14 700

Finishing 0.07 0.20 0.08 0.12 800

Packaging 0.09 0.12 0.07 0.15 600

The sheet metal required for each component, the estimated annual demand, and the profit contribution provided by each component, are as follows:

Component Sheet Metal

(ft2) Estimated Annual Demand Profit Contribution

$

1 2.6 2,600 90

2 1.4 1,800 100

3 2.5 4,100 80

4 3.2 1,200 120

The company has 15,000 square feet of fabricated metal delivered each month. The company has the following prioritized production goals for these components:

? Avoid the use of overtime given the low profit margins.

? Meet all component demand.

? Achieve an annual profit contribution of at least $700,000

? Avoid additional material orders - pricing is dependent upon annual contracts with fixed amounts and delivery schedules

a. Formulate a linear programming model to determine the quantity of each component to produce to achieve the company's goals.

Answer:The decision variables are: x1 the quantity of the component 1 produced and sold

x2 the quantity of the component 2 produced and sold

x3 the quantity of the component 3 produced and sold

x4 the quantity of the component 4 produced and sold

First we write the entire program as a linear program

The objective function without any goal achievement requirement is maximization of profit

The profit as an objective function is 90x1 + 100x2 + 80x3 + 120x4 Eq(1)

The operation time availability constraints for each operation is as given below

0.06x1 + 0.17x2 + 0.10x3 + 0.14x4 <= 700 Eq(2) Stamping operation

0.18x1 + 0.20x2 + 0.14x4 <= 700 Eq(3) Assembly operation

0.07x1 + 0.20x2 + 0.08x3 + 0.12x4 <= 800 Eq(4) Finishing operation

0.09x1 + 0.12x2 + 0.07x3 + 0.15x4 <= 600 Eq(5) Packaging operation

The sheet metal availability constraint is

2.6x1 + 1.4x2 + 2.5x3 + 3.2x4 <= 15000 Eq(6) Sheet metal availability

Annual demand requirement is

x1 >= 2600 Eq(7) Component 1

x2 >= 1800 Eq(8) Component 2

x3 >= 4100 Eq(9) Component 3

x4 >= 1200 Eq(10) Component 4

Now we write goal program considering each of the goals in the order of the priority. Each goal is given priority order based on importance.

P1: Avoid the use of overtime given the low profit margins

Let us define the deviation variables for this goal. There are four constraints for operations hours, one each for each department. Let us say d1+ and d1- be the deviation variables for the amount by which the total operations hours exceeds and is less than the target value of 700 for stamping operation. Similarly the deviation variables for other constraints for operations time are written

The goal equation for stamping operation is

0.06x1 + 0.17x2 + 0.10x3 + 0.14x4 - d1+ + d1- = 700

The goal equation for assembly operation is

0.18x1 + 0.20x2 + 0.14x4 - d2+ + d2- = 700

The goal equation for finishing operation is

0.07x1 + 0.20x2 + 0.08x3 + 0.12x4 - d3+ + d3- = 800

The goal equation for packaging operation is

0.09x1 + 0.12x2 + 0.07x3 + 0.15x4 - d4+ + d4- = 600

Next flexible constraint is raw material availability, as we have to avoid additional orders. Let us say d5+ and d5- be the deviation variables when raw material required exceeds or less than the target value of 15000. The goal equation for this constraint is

2.6x1 + 1.4x2 + 2.5x3 + 3.2x4 - d5+ + d5- = 15000

Note that we have to avoid overtime and hence all the exceed deviation variables needs to be minimized. The linear program P1 at priority level 1 is

Minimize: d1+ + d2+ + d3+ + d4+

Subject to

x1 >= 2600

x2 >= 1800

x3 >= 4100

x4 >= 1200

90x1 + 100x2 + 80x3 + 120x4 >= 700000

2.6x1 + 1.4x2 + 2.5x3 + 3.2x4 - d5+ + d5- = 15000

All the variables including the deviation variables and decision variables are non-negative

We solve the above linear program and obtain optimal solution. The values of the deviation variables in the objective function are noted down. In the next level program the values of the deviation variables in the above are included as hard constraints.

P2: Avoid additional material orders

The objective is to minimize the over usage of the materials and hence we have

Minimize: d5+

Subject to

x1 >= 2600

x2 >= 1800

x3 >= 4100

x4 >= 1200

90x1 + 100x2 + 80x3 + 120x4 >= 700000

d1+ = optimal value in P1

d2+ = optimal value in P1

d3+ = optimal value in P1

d4+ = optimal value in P1

The non-negativity condition.

b. Determine the solution that will best achieve the company's goals in component production, including the quantity of each component produced, the annual profit contribution attained, and the levels of goal achievement.

Answer: Copy-paste the P1 in any solver package and obtain value of deviation variables as mentioned above. Include the optimal values of the deviation variables in P1 as hard constraints in O2 and again solve. The solution obtained in P2 is optimal considering the goals.

c. Determine how the solution would be affected if the first two goal priorities were reversed.

Answer: In this case P2 becomes P1 without any constraints on d1+, d2+, d3+, and d4+. Solve it and obtain optimal value at d5+. Use this value as constraint in next level problem where d1+, d2+, d3+, and d4+ becomes decision variables.

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