# Simplex Method Examples

Get ready for a few solved examples of simplex method in operations research. In this section, we will take linear programming (LP) maximization problems only.

Do you know how to divide, multiply, add, and subtract? Yes. Then there is a good news for you. About 50% of this technique you already know.

Simplex Method Example-1, Example-2

Now, open the door and windows of your mind and concentrate on the following example.

"Minds are like parachutes; they work best when open." -Lord Thomas Dewar

## Simplex Method: Example 1

Maximize z = 3x1 + 2x2

subject to

-x1 + 2x2 ≤ 4
3x1 + 2x2 ≤ 14
x1 – x2 ≤ 3

x1, x2 ≥ 0

Solution.

First, convert every inequality constraints in the LPP into an equality constraint, so that the problem can be written in a standard from. This can be accomplished by adding a slack variable to each constraint. Slack variables are always added to the less than type constraints.

Converting inequalities to equalities

-x1 + 2x2 + x3 = 4
3x1 + 2x2 + x4 = 14
x1 – x2 + x5 = 3
x1, x2, x3, x4, x5 ≥ 0

Where x3, x4 and x5 are slack variables.

Since slack variables represent unused resources, their contribution in the objective function is zero. Including these slack variables in the objective function, we get

Maximize z = 3x1 + 2x2 + 0x3 + 0x4 + 0x5

Initial basic feasible solution

Now we assume that nothing can be produced. Therefore, the values of the decision variables are zero.
x1 = 0, x2 = 0, z = 0

When we are not producing anything, obviously we are left with unused capacity
x3 = 4, x4 = 14, x5 = 3

We note that the current solution has three variables (slack variables x3, x4 and x5) with non-zero solution values and two variables (decision variables x1 and x2) with zero values. Variables with non-zero values are called basic variables. Variables with zero values are called non-basic variables.

Simplex Method: Table 1

On small screens, scroll horizontally to view full calculation

a11 = -1, a12 = 2, a13 = 1, a14 = 0, a15 = 0, b1 = 4
a21 = 3, a22 = 2, a23 = 0, a24 = 1, a25 = 0, b2 = 14
a31= 1, a32 = -1, a33 = 0, a34 = 0, a35 = 1, b3 = 3

Calculating values for the index row (zj – cj)

z1 – c1 = (0 X (-1) + 0 X 3 + 0 X 1) - 3 = -3
z2 – c2 = (0 X 2 + 0 X 2 + 0 X (-1)) - 2 = -2
z3 – c3 = (0 X 1 + 0 X 0 + 0 X 0) - 0 = 0
z4 – c4 = (0 X 0 + 0 X 1 + 0 X 0) - 0 = 0
z5 – c5 = (0 X 0 + 0 X 0 + 0 X 1) – 0 = 0

Choose the smallest negative value from zj – cj (i.e., – 3). So column under x1 is the key column.
Now find out the minimum positive value
Minimum (14/3, 3/1) = 3
So row x5 is the key row.
Here, the pivot (key) element = 1 (the value at the point of intersection).
Therefore, x5 departs and x1 enters.

We obtain the elements of the next table using the following rules:

1. If the values of zj – cj are positive, the inclusion of any basic variable will not increase the value of the objective function. Hence, the present solution maximizes the objective function. If there are more than one negative values, we choose the variable as a basic variable corresponding to which the value of zj – cj is least (most negative) as this will maximize the profit.

2. The numbers in the replacing row may be obtained by dividing the key row elements by the pivot element and the numbers in the other two rows may be calculated by using the formula:

 New number= old number- (corresponding no. of key row) X (corresponding no. of key column) pivot element

Calculating values for table 2

x3 row

a11 = -1 – 1 X ((-1)/1) = 0
a12 = 2 – (-1) X ((-1)/1) = 1
a13 = 1 – 0 X ((-1)/1) = 1
a14 = 0 – 0 X ((-1)/1) = 0
a15 = 0 – 1 X ((-1)/1) = 1
b1 = 4 – 3 X ((-1)/1) = 7

x4 row

a21 = 3 – 1 X (3/1) = 0
a22 = 2 – (-1) X (3/1) = 5
a23 = 0 – 0 X (3/1) = 0
a24 = 1 – 0 X (3/1) = 1
a25 = 0 – 1 X (3/1) = -3
b2 = 14 – 3 X (3/1) = 5

x1 row

a31 = 1/1 = 1
a32 = -1/1 = -1
a33 = 0/1 = 0
a34 = 0/1 = 0
a35 = 1/1 = 1
b3 = 3/1 = 3

Table 2

Use horizontal scrollbar to view full calculation

cj 3 2 0 0 0
cB Basic variables
B
x1 x2 x3 x4 x5 Solution values
b (= XB)
0 x3 0 1 1 0 1 7
0 x4 0 5 0 1 -3 5
3 x1 1 -1 0 0 1 3
zj-cj   0 -5 0 0 3

Calculating values for the index row (zj – cj)

z1 – c1 = (0 X 0 + 0 X 0 + 3 X 1) - 3 = 0
z2 – c2 = (0 X 1 + 0 X 5 + 3 X (-1)) – 2 = -5
z3 – c3 = (0 X 1 + 0 X 0 + 3 X 0) - 0 = 0
z4 – c4 = (0 X 0 + 0 X 1 + 3 X 0) - 0 = 0
z5 – c5 = (0 X 1 + 0 X (-3) + 3 X 1) – 0 = 3

Key column = x2 column
Minimum (7/1, 5/5) = 1
Key row = x4 row
Pivot element = 5
x4 departs and x2 enters.

Calculating values for table 3

x3 row

a11 = 0 – 0 X (1/5) = 0
a12 = 1 – 5 X (1/5) = 0
a13 = 1 – 0 X (1/5) = 1
a14 = 0 – 1 X (1/5) = -1/5
a15 = 1 – (-3) X (1/5) = 8/5
b1 = 7 – 5 X (1/5) = 6

x2 row

a21 = 0/5 = 0
a22 = 5/5 = 1
a23 = 0/5 = 0
a24 = 1/5
a25 = -3/5
b2 = 5/5 = 1

x1 row

a31 = 1 – 0 X (-1/5) = 1
a32 = -1 – 5 X (-1/5) = 0
a33 = 0 – 0 X (-1/5) = 0
a34 = 0 – 1 X (-1/5) = 1/5
a35 = 1 – (-3) X (-1/5) = 2/5
b3 = 3 – 5 X (-1/5) = 4

Don't convert the fractions into decimals, because many fractions cancel out during the process while the conversion into decimals will cause unnecessary complications.

Simplex Method: Final Optimal Table

cj 3 2 0 0 0
cB Basic variables
B
x1 x2 x3 x4 x5 Solution values
b (= XB)
0 x3 0 0 1 -1/5 8/5 6
2 x2 0 1 0 1/5 -3/5 1
3 x1 1 0 0 1/5 2/5 4
zj-cj   0 0 0 1 0

Since all the values of zj – cj are positive, this is the optimal solution.
x1 = 4, x2 = 1
z = 3 X 4 + 2 X 1 = 14.

The largest profit of Rs.14 is obtained, when 1 unit of x2 and 4 units of x1 are produced. The above solution also indicates that 6 units are still unutilized, as shown by the slack variable x3 in the XB column.

Real life complex applications usually involve hundreds of constraints and thousands of variables. So virtually these problems can not be solved manually. For solving such problems, you will have to rely on employing an electronic computer.