Optimal solutions:

Consider the following linear programing problem:

  Max Z=5x1+3x2+x3

Subject to the constraints:

  x1+x2+3x3≤6

  5x1+3x2+6x3≤15

  x1,x2,x3 ≥0

Calculate the optimum solution for the above linear programming problem by using the simplex algorithm as follows:

• The constraints are ≤ constraints, now it is necessary to convert it to an equality constraint by adding slack variables s₁, s₂ to the two constraints. The standard form of linear programming problem is as shown below:

  Max Z=5x1+3x2+x3+0s1+0s2

  x1+x2+3x3+s1=6

  5x1+3x2+6x3+s2=15

  x1,x2,x3,s1,s2≥0

The initial simplex table is as follows:

• Choose base variables by observing that which values form an identity matrix in the table. Here X₄, X₅ variable values form an identity matrix. Take the corresponding C_j values of X₄, X₅ as C_b values.

   Zj= CbXb  Zb= (0×6) + (0×15)  Zb= 0Z1= CbX1 Z1= (0×1)+(0×5) Z1= 0

• Calculate the rest of variables and shown in the initial table:

  C_j   5 3 1 0 0

Base C_b X_b X₁ X₂ X₃ X₄ X₅

X₄ 0 6 1 1 3 1 0

X₅ 0 15 5 3 6 0 1

Z_j 0 0 0 0 0 0 0

Z_j-C_j   0 -5 -3 -1 0 0

• From the above simplex table, observe Zj−Cj values. -5 is the most negative number and therefore the negative entry is -5.
• Calculate the ratio value by using the following formula.

Ratio=Right hand side value fo the constraintCoefficient of entering variable in the constraint

  C_j   5 3 1 0 0  

Base C_b X_b X₁ X₂ X₃ X₄ X₅ Ratio= X_b/ X₁

X₄ 0 6 1 1 3 1 0 (6/1)=6

X₅ 0 15 5 3 6 0 1 (15/5)=3

Z_j 0 0 0 0 0 0 0  

Z_j-C_j   0 -5 -3 -1 0 0  

R2→15R2

R1→R1−R2

The resultant iteration table is as shown below:

  C_j   5 3 1 0 0

Base C_b X_b X₁ X₂ X₃ X₄ X₅

X₄ 0 3 0 0.4 1.8 1 -0.2

X₁ 5 3 1 0.6 1.2 0 0.2

Z_j   15 5 3 6 0 1

Z_j-C_j   15 0 0 5 0 1

Since, the last row Z_j-C_j contains all positive entries, the solution is optimal. Therefore, the value decision variables and Max Z is,

x1= 5 x2= 0        x3= 0        Max Z = 5x1+3x2+x3 Max Z = (5×0)+(3×5)+(1×0)Max Z = 15

The optimal maximized Z value is 15.