QUANTITATIVE TECHNIQUES IN BUSINESS - Daily Math Guide

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## Posted by : Allan_Dell on Tuesday, January 15, 2013 | 7:51 PM

Problems intended for graphical and Simplex method:

Problems in Graphical Method:

Problem 1:

A tailor shop makes two products, dress and pants which must be processed through assembly and Finishing department. Assembly department is available for 24 hours in every production period, while the finishing department is available for 38 hours of work. Manufacturing one dress requires 2 hours in assembly and 3 hours in the finishing department. Each pants requires 3 hours in the assembly and 5 hours in the finishing department. One dress contributes \$ 150 as profit and pants \$ 100. The problem is to determine the number of dress and pants to make production period in order to maximize the profit.

Problem 2:

A steel producer makes two types of steel, regular and special.A ton of regular steel requires two hours  in the open-hearth furnace and three hours in the soaking pit. A ton of special steel requires two hours in the open-hearth furnace and five hours in the soaking pit. The open-hearth furnace is available for eight hours per day and the soaking pit is fifteen hours a day. The profit on a ton of regular steel is \$4000 and it is \$ 6000 on a ton of special steel. Determine how many tons of each type of steel should be made to maximize the profit considering that demand on regular steel is at least one ton.

Simplex Method:

Maximization:

## Steps for constructing the Simplex solution:

1.      From the problem condition, set up the constraints.

2.      Convert the inequality explicit constraints to equations by adding slack variables.

3. Enter the numerical coefficients and variables in the Simplex Tables.

4.Calculate the C j and Z j values.

5. Determine the optimum column or entering variable by choosing the most positive value   in the C j – Z j row.
6. Divide the quantity column values by the non-zero and non-negative entries in the optimum   column. The smallest quotient belongs to the pivotal row.

7. Compute the values for the replacing row by dividing all entries by the pivot. Enter the result in the next table.

8. Compute the new entries for the remaining rows by reducing the optimum column entries to zero.

9. Calculate and Z j values.

10. If there is a positive entry in the j  minus  Z j  row, return to step 5. The final solution has been obtained if there is no positive value in the – Z j row.

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