Optimization Mathods Sporks

Operations Management: Optimization Method

Homework 1

To solve the problem we will use a linear programming model. The decision variables of the model are:

x1 = Total number of sporks produced in thousands

x2 = Total number of packets produced in thousands

x3 = Total number of school packs produced in thousands

y1 = Total number of sporks sold as sporks in thousands

y2 = Total number of packets sold as packets in thousands

y3 = Total number of school packs sold as school packs in thousands.

We can now find relations between the variables as we are told that each packet needs one spork, and each school pack needs ten sporks.

the total number of sporks sold is: y1 = x1 − x2 − 10x3. (E1)

the total number of packets sold is: y2 = x2 − 100x3 (E2)

the total number of school packs sold : y3 = x3. (E3)

we can now obtain the following system:

x1 = y1 + y2 + 110y3 (E4)

x2 = y2 + 100y3 (E5)

x3 = y3. (E6)

We can easily determine the objective function with the information in the problem (with the selling prices and the direct costs of each of the units produced). The total profit is:

Total profit = 5y1 + 15y2 + 300y3 − 2.5x1 − 4x2 − 8x3. (E7)

With the equation (E1) (E2) and (E3), we express the total profit with only the variable “x” with a simple transformation:

Total profit = 5x1−5x2−50 x3+15 x2−1500 x3+300 x3− 2.5x1 − 4x2 − 8x3.

= 2.5x1 + 6x2 − 1258x3. (E8)

We must introduce the imposition of 200 hours of production time. Only the sporks need production time in the molding area. Sporks needs 0.8 hours per 1000 produced:

0.8x1 ≤ 200 (E9)

For packaging-area capacity we have:

1.5x2 + 2.5x3 ≤ 200, (E10)

For supervisory time we have:

0.2x1 + 0.5x2 + 0.5x3 ≤ 200. (E11)

Then we have to add the non-negativity imposition, so with the information given in the problem we have:

x1 − x2 − 10x3 ≥ 0 (E12)

x2 − 100x3 ≥ 0 (E13)

x3 ≥ 0 (E14)...