LESSON 2 Simplex Method

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) x y

Assembly Dept. (hrs/pc.) 4 2

Finishing Dept. (hrs/pc.) 2 4

Profit (Php/pc) 180 100

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

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: Contribution to profit column Variable column Constant column

Cj

180

100

0

0

Objective coef. row

Prod

Qty

x

y

S1

S2

0

S1

60

4

2

1

0

0

S2

48

2

4

0

1

Variable row Constraint coefficients

Zj Cj-Zj To complete the table, follow the following steps: 1. Solve for the Zj row : 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

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 Cj-Zj

0

0 180

0 100

0 0

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 Cj-Zj

0

0 180

0 100

0 0

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

Pivotal row

Cj

0 0

180

100

0

0

Prod

Qty

x

y

S1

S2

S1 S2

60 48

60/4=15 48/2=24

60/2=30 4

1 0

0 1

Zj Cj-Zj

0

0 180

0 100

0 0

0 0

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. 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

TABLE 2

Cj

180

100

0

0

Prod

Qty

x

y

S1

S2

60/4 = 15

4/4 = 1

2/4 =½

1/4

0/4 = 0

Replacing Row

180

X

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

Sub tract

values vertically)

48 2 4 0 1

-

The x or S1 row of Table 2 15 1 ½ ¼ 0

S2 in Table 2 (2) (2) (2) (2) (2)

= = = = =

18 0 3 -½ 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

0

10

-45

0

Cj-Zj

Highest non-negative integer Optimum column The next is Table 3 TABLE 3

Cj

Row to be replaced

180

Replacing Row

100

180

100

0

0

Prod

Qty

x

y

S1

S2

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 15 1 ½ ¼ 0 TABLE 3

-

The y-row of Table 3 6 0 1 -3/2 1/3

Cj

(½) (½) (½) (½) (½)

= = = = =

x-row in Table 3 3 1 0 1 1/6

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

Cj-Zj

2760

0

0

-330

-33.33

No more positive integer

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?