LINEAR PROGRAMMING

Simplex Method

Adiel Aislin S. Tagumpay Dominic G. Jason Justine Ara Guzman Jon Edison C. Montejo

Simplex MaximizationSteps in Solving Maximization Problems1.Set up the objective function and the constraints of the problem.2.Convert the objective function and the explicit constraints to equation form (standard form). Follow the following rules in converting the LP model to standard form:

Symbol in the constraint To put into standard form

<Add a slack variable (S1) that represents the amount by which the right-hand side of the constraint exceed its left-hand side.

=Add an artificial variable (A1) or extraneous variable, which serves a starting basic variable. This variable has no physical meaning in the problem hence it is called artificial .

>Subtract a surplus variable (S1), which represents the excess amount of the left-hand side of the constraint over the right-hand side, and add an artificial variable (A1).

The artificial variables in the optimum or final solution should be zero provided a feasible solution exists. The reasonable way to do this is to penalize these variables in the objective function by using a very small number coefficient, -M, (M for million), or -100 if the numerical coefficient found in the constraints and objective function are all less than or equal to 10, let it equal to -1000 if the numerical coefficients are all less than or equal to 100, let it equal to -10000, if the numerical coefficients are all less than or equal to 1000,and so on.

3. Set up the initial tableau. The number of rows = no. of constraints +4 The number of columns = no. decision variable + no. of slacks/surplus variables + no. of

artificial variables + 4

3.1. Fill up the first and second rows of the initial tableau.

a) The variable row (second row) contains the variable of the objective function. List the variables in order, beginning with the decision, followed be the slack variables then the artificial variables (the subscripts follow the constraint number)

b) The Cj row (first row) takes the numerical coefficients of the variables in the variable row.

3.2. Put the entries in the first three columns of the initial tableau.

a) The second column is the solution column (also called basis or variable column).

b) the first column is the contribution to profit column (Ci), which takes the

coefficients of the variables in the solution column.

c) The third column is the quantity column or the constant column, which takes the

constant terms or the right-hand values of the constraints.

4. Select the optimum column, the column with the most positive value in the Ci – Zi row of the initial table

5. Select the pivot in the pivotal row.

6. Use the pivot to clear the optimum column in the normal manner. This gives the next tableau.

7. Repeat steps 4-6 until there are no more negative numbers in the bottom row (with the possible expectation of the Answer column.)

Writing LP Model to Equation Form

Example:

Maximize Z= 5x+12ySubject to 2x+3y≤80

3x+4y≤100 x,y≥0

Solution:

2x+3y+S1=803x+4y+S2=100

Objective functionMax Z= 5x+12y+0S1+0S2

Since the inequalities involve less than, simply add slack variable.

For constraint 1, 2x+3y≤80, add S1. The subscript 1 indicates that the slack variable is added to the left side of the 1st constraint.

Thus, constraint 1 becomes2x+3y+S1=80

Doing the same process to constraint 2, then it becomes

3x+4y+S2=100

The objective function should also contain the slack variable added in the constraints. But since the slack variable has no contribution to the objective function, its numerical coefficient must be zero. Thus the objective function becomes

Max Z= 5x+12y+0S1+0S2

Obtaining the Final Solution of LPP

Example1. Maximize Z= 90x + 110y Subject to 6x + 4y ≤ 120 ►Explicit constraint

3x + 10y ≤ 180 ►Explicit constraint

x ≥ 0, y ≥ 0 ►implicit constraint

New program with slack variables (simplified):Maximize Z= 90x + 110y +0S1 + 0S2

Subject to 3x + 2y + S1 = 60 3x + 10y + S2 = 180

x,y,S1,S2 ≥ 0

Cj

90 110 0 0 Qi

CjSolution Quantity x y S1 S2

0 S1 60 3 2 1 0 0

S2 180 3 10 0 1 0

Zj 0 0 0 0 0

Cj - Zj 90 110 0 0

Objective coef. Row

Variable row

ConstraintCoefficient rows

Table 1

Iteration 1

i. Determine the optimum column and the pivotal row, and then highlight them.ii. The intersection of these column and row is the pivot element.iii. Determine the Qi column.

60 ÷ 2 = 30180 ÷ 10 = 18

Table 2

Iteration 2i. Pivotal row: second constraint coefficient rowChange the outgoing variable by the entering variable (reflect corresponding coefficients in the Ci Col.) Pivot element in the replacing pivot row should be equal to one (1), thus divide each entry in this row by the pivot element (10)

Table 3

ii. Remaining rows1. First coefficient row

The intersection with the optimum column should be ZERO.There are several algebraic manipulations that can be used to make it

zero, one is to apply formula 7.1.

The solution is not yet final at this point since there is still positive entry in the Cj – Zj row

Table 5

Table 4

Since the entries in the Cj – Zj are zeros and negative numbers, then the last iteration gives the optimum solution.

Thus optimum values of x = 10, y = 15 Z = 2550

Table 6

Simplex Minimization

Steps in solving Minimization Problems

The steps in solving minimization problems by simplex method are the same as in maximization except for steps 2, 4, and 7.

2. Convert the objective function and the explicit constraints to equation form (standard form).

Follow the following rules in converting the LP model to standard form:

Symbols in the constraint To put into standard form

≤ Add a slack variable (S1)

= Subtract a surplus variable (S!) and add an artificial variable (A1)

≥ Add an artificial variable (A1)

4. Select the optimum column, column with the most negative value in the Cj-Zj row of the initial table.7. Repeat steps 4-6 until there are no more positive numbers in the bottom row (with the possible exception of the Answer column.

Cj 6 4 M 0 0 M Qi

Ci Solution Quantity x y A1 S2 S3 A3

M A1 10 1 2 1 0 0 0 5

0 S2 4 0 1 0 1 0 0 4

M A3 18 3 2 0 0 -1 1 19

Zj 28M 4M 4M M 0 -M M

Cj - Zj 6 – 4M 4 – 4M 0 0 M 0

Iteration 1

Cj 6 4 M 0 0 M Qi

Ci Solution Quantity x y A1 S2 S3 A3

M A1 2 1 0 1 -2 0 0 2

4 y 4 0 1 0 1 0 0 ---

M A3 10 3 0 0 -2 -1 1 3.33

Zj 16 + 12M 4M 4 M 4 – 4M -M M

Cj - Zj 6 – 4M 0 0 -4 + 4M M 0

Iteration 2

Cj 6 4 M 0 0 M

Ci Solution Quantity x y A1 S2 S3 A3

6 x 4 1 0 -1/2 0 -1/2 1/2

4 y 3 0 1 3/4 0 1/4 -1/4

0 S2 1 0 0 -3/4 1 -1/4 1/4

Zj 36 6 4 0 0 -2 2

Cj - Zj 0 0 M 0 2 M - 2

Final tableau

From the last table, The Cj – Zj has zeros and positive entries, thus it gives the optimumSolution. The solution are x = 4,y = 3and the minimum value of Z = 36.