System Engineering 3(3+0)

Lesson 2 Simplex method problems

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.

Iteration 1:

	cj	3	2	0	0	0
cB	Basic variables B	x1	x2	x3	x4	x5	Solution values b (=XB)
0	x3	-1	2	1	0	0	4
0	x4	3	2	0	1	0	14
0	x5	1	-1	0	0	1	3
zj-cj		-3	-2	0	0	0

 

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 (z_j – c_j)

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 = (0x 0 + 0 x 0 + 0 x 1) – 0 = 0

Choose the smallest negative value from z_j – c_j (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 z_j – c_j 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 z_j – c_j 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:

 

Calculating values for Iteration 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

Iteration 2:

	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 (z_j – c_j)

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

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

Final iteration:

	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

 

Result:

Since all the values of zj – c_j are positive, this is the optimal solution.
x₁ = 4, x₂ = 1
                              Max 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 X_B column.