As has been said earlier, the solution values of the primal can be read directly from the optimal solution table of the dual. The reverse of this also is true, The following two properties of primaldual should be understood.
PrimalDual Property 1. If feasible solution exists for both primal and dual problems, then both problems n optimal solution for which the objective function values are equal. A peripheral relationship is that, problem has an unbounded solution, its. dual has no feasible solution.
PrimalDual Property 2. The optimal values for decision variables in one problem are read from row (0) of the optimal table for the other problem. The following steps are involved in reading the solution values be primal from the optimal solution table of the dual:
Step I. The slacksurplus variables in the dual problem are associated with the basic variables of the primal in the optimal solution. Hence, these slacksurplus variables have to be identified in the dual problem.
Step II. Optimal value of basic primal variables can be directly read from the elements in the index row corresponding to the columns of the slacksurplus variables with changed signs.
Step III. Values of the slack variables of the primal can be read from the index row under the nonbasic variables of the dual solution with changed signs.
Step IV. Value of the objective function is same for primal and dual problems.
Step I. Convert the minimisation into a maximisation problem.
Maximise Z*= 2y_{1} – 3y_{2} + 5 y_{3}
Step II. Make RHS of constraints positive,
– 2 y_{1} – 4 y_{2} + y_{3} ≥ – 2 is rewritten as
2 y_{1} + 4 y_{2} – y_{3} ≤ 2
Step III. Make the problem as N + S coordinates problem
Maximise Z* = 2 y_{1} – 3 y_{2} + 5 y_{3} + 0S_{1} + 0S_{2} + 0S_{3} – MA_{1} – MA_{3}
Subject to – 2y_{1} + 3 y_{2} – S_{1} + A_{1} = 5
2 y_{1} + 4 y_{2} – y_{2} + S_{2} = 2
y_{1} + 3 y_{3} – S_{3} + A_{3} = 3
y_{1}, y_{2}, y_{3}, S_{1}, S_{2}, S_{3} , A_{1}, A_{3} ≥ 0
Step IV. Make N coordinates assume 0 values,
y_{1} = y_{2} = y_{3}=S_{1} = S_{3} =0,
A_{1} = 5, S_{2} = 2, A_{3} = 3 is the basic feasible solution, This can be represented in the table as follows.
Initial Solution
C_{j} 
2 
3 
5 
0 
0 
0 
M 
M 
Min Ratio 

C_{B} 
Basic Variables 
Solution Variables 
y_{1} 
y_{2} 
y_{3} 
S_{1} 
S_{2} 
S_{3} 
A_{1} 
A_{2} 

M 
A_{1} 
5 
2 
3 
0 
1 
0 
0 
1 
0 

0 
S_{2} 
2 
2 
4 
1 
0 
1 
0 
0 
0 

M 
A_{2} 
3 
1 
0 
3 
0 
0 
1 
0 
1 

Z_{j} 


M 
3M 
3M 
M 
0 
M 
M 
M 

(C_{j}Z_{j}) 

2M 
3+3M 
5+3M 
M 
0 
M 
0 
0 

Step V. C_{j} – Z_{j} is + ve under some columns, it is not the optimal solution, Perform the optimality test.
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