LESSON 2
The SIMPLEX METHOD of Linear Programming Maximization Method
The simplex method of linear programming was developed by George B. Dantzig of Stanford University.
It is a repetitive optimizing technique and it repeats the process of mathematically moving from an extreme point to another extreme point (in the graphical method) until an optimal solution is reached.
The simple method can handle an infinite number of variables.
i. Simplex Maximization Problems
The method of solving a maximization problem is different from minimization in the simplex method.
Steps in Solving a Maximization Problem
1. Set up the constraints from the conditions of the problem.
2. Convert the inequality explicit constraints to equations by adding slack variables when the constraints contains “”. But if the constraint has a “” symbol, convert first the symbol to “” by multiplying the inequality by negative 1 and then add the slack variable. If the problems contain the “=” symbol in the constraints, just add a slack variable.
3. Enter then numerical coefficients and variables in the simplex table. 4. Calculate Cj and Zj values.
5. Determine the optimum column or entering by choosing the most positive value in the Cj – Zj 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 value for the replacing row by dividing all entries by the pivot. Enter the result in the next column.
8. Compute the new entries for the remaining rows by reducing the optimum column entries to zero (entries in the constrain rows.)
9. Calculate Cj and Zj values. Compute also for Cj – Zj row.
10. If there is a positive entry in the Cj-Zj row, return to step 5. The final solution has been obtained if there is no positive value in the Cj-Zj row.
EXAMPLE PROBLEM: Consider the Margan Furniture Problem:
The Margan Furniture makes two products: tables and chairs, which must be processed through assembly and finishing departments. Assembly department is available for 60 hours in every production period, while the finishing department is available for 48 hours of work. Manufacturing one table requires 4 hours in the assembly and 2 hours in the finishing. Each chair requires 2 hours in the assembly and 4 hours in the finishing. One table contributes P180 to profit, while a chair contributes P100. The problem is to determine the number of tables and chairs to make per production period in order to maximize the profit.
I Given:
Let x = the number of pieces of tables y = the number of pieces of chairs
II Tabulation:
Product (Qty) Assembly Dept.
(hrs/pc.) Finishing Dept.(hrs/pc.) Profit (Php/pc)
x 4 2 180
y 2 4 100
Step 1:
Consider the Margan Furniture Problem: Maximize: Z = 180x +100y Subject to: 4x + 2y 60 2x + 4y 48 x 0, y 0 Step 3:
Simplex Table Organization The initial Simplex Table will be:
Cj 180 100 0 0 Prod Qty x y S1 S2 0 S1 60 4 2 1 0 0 S2 48 2 4 0 1 Zj Cj-Zj
To complete the table, follow the following steps:
1. Solve for the Zjrow : each Cj value is multiplied to every value in the constant column. (See the table below)
2. Solve for the Cj - Zj row: Under each column subtract the value of Zj from the value of Cj. (See the table below)
Cj 180 100 0 0 Prod Qty x y S1 S2 0 S1 60 4 2 1 0 0 S2 48 2 4 0 1 (a) Zj 0(60)=0 0(48)=0 0 0(4)=0 0(4)=0 0(2)=0 0(4)=0 0(1)=0 0(0)=0 0(0)=0 0(1)=0 (b) Cj-Zj 180-0 =180 100-0 =100 0-0=0 0-0=0 Step 2:
New Program with Slack Variables: Maximize: Z= 180x +100y + 0S1 + 0S2
Subject to: 4x + 2y + S1 = 60
2x + 4y +S2 = 48
x 0, y 0 S1 0, S2 0
Objective coef. row Variable row
Constraint coefficients Contribution to profit column
Variable column
Constant column
The initial simplex table should have this final content: Cj 180 100 0 0 Prod Qty x y S1 S2 0 S1 60 4 2 1 0 0 S2 48 2 4 0 1 Zj 0 0 0 0 0 Cj-Zj 180 100 0 0
From the Cj-Zj row choose the highest non negative integer. The column where the number is located becomes the optimum column.
TABLE 1 Cj 180 100 0 0 Prod Qty x y S1 S2 0 S1 60 4 2 1 0 0 S2 48 2 4 0 1 Zj 0 0 0 0 0 Cj-Zj 180 100 0 0
Highest non negative integer
Optimum column
From the Qty column, divide each quantity with the value found in optimum column under the same row. Choose the smaller qoutient. The row where this quotient belongs becomes the pivotal row. TABLE 1 Cj 180 100 0 0 Prod Qty x y S1 S2 Pivotal row 0 S1 60 60/4=15 60/2=30 1 0 0 S2 48 48/2=24 4 0 1 Zj 0 0 0 0 0 Cj-Zj 180 100 0 0
Pivot the table in a counter clockwise direction at the intersection of the pivotal row and the optimum column or at the pivot.
The next is Table 2
The smaller quotient is between 15 and 24 is 15. Therefore the row where 15 belongs is the pivotal row and 4 is the pivot.
TABLE 2 Cj 180 100 0 0 Prod Qty x y S1 S2 Replacing Row 180 X 60/4 = 15 4/4 = 1 2/4 =½ 1/4 0/4 = 0 Row to be replaced 0 S2 Zj Cj-Zj
The pivotal row will now have the following contents by dividing every element in the constant column with the pivot. The objective is to change the pivot to 1 in the next table.
Solve for S2 row (the Row to be replaced)
From S2 of
Table 1 (copy the values vertically)
Sub
tract The x or Srow of 1 Table 2 S2 in Table 2 48 - 15 (2) = 18 2 - 1 (2) = 0 4 - ½ (2) = 3 0 - ¼ (2) = -½ 1 - 0 (2) = 1
Table 2 will have this values
TABLE 2 Cj 180 100 0 0 Prod Qty x y S1 S2 15 / (1/2) = 30 180 X 15 1 ½ 1/4 0/4 = 0 18/3 = 6 Pivotal row 0 S2 18 0 3 -1/2 1 Zj 2700 180 90 45 0 Cj-Zj 0 10 -45 0
Highest non-negative integer
Optimum column The next is Table 3
TABLE 3 Cj 180 100 0 0 Prod Qty x y S1 S2 Row to be replaced 180 Replacing Row 100 Y 6 0 1 -3/2 1/3 Zj Cj-Zj
Solve for x-row, the Row to be replaced From y-row of
Table 2 The y-rowof Table 3 x-row inTable 3
15 - 6 (½) = 3 1 - 0 (½) = 1 ½ - 1 (½) = 0 ¼ - -3/2 (½) = 1 0 - 1/3 (½) = 1/6 TABLE 3 Cj 180 100 0 0 Prod Qty x y S1 S2 Row to be replaced 180 x 12 1 0 1 1/6 Replacing Row 100 Y 6 0 1 -3/2 1/3 Zj 2160 180 100 330 33.33 No more positive integer Cj-Zj 2760 0 0 -330 -33.33
12, the number of x product to be manufactured 6, the number of y product to be manufactured 2760, the maximum profit
ASSIGNMENT: Answer the following using the simplex method.
1.
A computer system manufacturer has just introduced two time-sharing programs for the generations of a wide range statistical output. Preliminary market research indicates that each hour of usage of SPPS will result in P400 profit, and each hour of STAA will result in P360 profit for the company.The company is capable of producing a combined total of 1,000 hours per month for both programs. In addition, production requires processing in two divisions, programming and storage. There is a maximum monthly budget of P400,000 for programming and P600,000 for storage. Each hour of SPPS uses P200 of programming budget and P500 of storage budget, while each hour of STAA uses P800 of programming budget and P1,000 of storage budget. What combination of SPPS and STAA should the manufacturer produce in order to maximize the company’s profit?
