Question

Linear Programming Problem A manufacturer of three models of tote bag must determine the production plan...

Linear Programming Problem A manufacturer of three models of tote bag must determine the production plan for the next quarter. The specifics for each model are shown in the following table. Model Revenue ($ per item) Cutting (hours per item) Sewing (hours per item) Packing (hours per item)

A $8.75 .10 .05 .20

B $10.50 .15 .12 .20

C $11.50 .20 .18 .20

Time available in the three production departments are: Cutting 450 hours, Sewing 550 hours, Packing 450 hours. Based on market research, the company wants to make at least 300 model A's, 400 model B's and 250 model C's but no more than no more than 1200 of any one model Costs in each department are: Cutting $10.00 per hour, Sewing $12.50 per hour, Packing $9.50 per hour Write the Linear Programming formulation for this problem to help determine the best production mix of the three models that will maximize profit.

Linear Programming

Linear Programming is the process of determining the optimum combination of 2 or more variables that are linearly related by the construction of a mathematical model. The variables normally work under a set of restrictions (constraints) such as limited material, labor, time, or money.

Answer and Explanation:

Decision Variables

The first step in any linear optimization problem is to determine the decision variables. These are what we want to find out. For this example they would be:

a: # of totes of model A to make

b: # of totes of model B to make

c: # of totes of model C to make

Optimization Function

Then we can build the optimization function. This is what we want to either maximize or minimize. For this example we want to maximize profit (P) so we must first determine profit for each item. To determine profit we subtract the cost from the revenue

ItemAProfit=$8.75−(0.10∗$10+0.05∗$12.50+0.20∗$9.50)=$8.75−($1+$0.625+$1.90)=$8.75−$3.53=$5.22ItemAProfit=$8.75−(0.10∗$10+0.05∗$12.50+0.20∗$9.50)=$8.75−($1+$0.625+$1.90)=$8.75−$3.53=$5.22

ItemBProfit=$10.50−(0.15∗$10+0.12∗$12.50+0.20∗$9.50)=$10.50−($1.50+$1.50+$1.90)=$10.50−$4.90=$5.60ItemBProfit=$10.50−(0.15∗$10+0.12∗$12.50+0.20∗$9.50)=$10.50−($1.50+$1.50+$1.90)=$10.50−$4.90=$5.60

ItemCProfit=$11.50−(0.20∗$10+0.18∗$12.50+0.20∗$9.50)=$11.50−($2+$2.25+$1.90)=$11.50−$6.15=$5.35ItemCProfit=$11.50−(0.20∗$10+0.18∗$12.50+0.20∗$9.50)=$11.50−($2+$2.25+$1.90)=$11.50−$6.15=$5.35

We can now build our optimization formula as:

P max = 5.22a + 5.60b + 5.35c

Constraints

Finally we detetermine the constraints.

There is a limit on the hours of each department so we can express this as:

Cutting Hours:

0.10a + 0.15b + 0.20c <= 450

Sewing Hours:

0.05a + 0.12b + 0.18c <= 550

Packing Hours:

0.20a + 0.18b + 0.20c <= 450

There are also restrictions on the amount to ake for each model:

For Model A

300 <= a <=1200

For Model B

400 <= b <=1200

For Model C

250 <= c <=1200

Finally there is an implied non-negativity constraint since it is not possible to produce negative amounts of tote bags. This can be expressed as:

a, b, c >= 0

Write Linear Program in Excel

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