1. This condition occurs when the problem has incompatible constraints.
An Infeasible Problem
This condition occurs when the problem has incompatible constraints.
Saturday, December 08, 2018

2. Final simplex table shows optimal solution as all Cj-Zj
+ve or zero in case of minimization and
–ve or zero in case of maximization.
Observing the solution base, we find that an artificial variable is present as a basic variable.
Both of these values are totally meaningless since the artificial variable has no meaning.
Hence, in such a situation, it is said that LPP has got an infeasible solution. 8
Saturday, December 08, 2018

3. Example Solve the following LPP Max. Z= 4X1+ 3X2 subject to X1+X2 ≤ 50
where as X1, X2 ≥ 0
Saturday, December 08, 2018

4. Introduce slack variable in ≤ constraint and
surplus and artificial variables and assign 0 co-efficient to surplus and slack variable & ‘-M’ to artificial variable.
Max. Z = 4X1 + 3X2 + 0S1 + 0S2 + 0S3 – MA1 – MA2
subjected to:
X1 + X2 + S1 + 0S2 + 0S3 + 0A1 + 0A2 = 50
X1 + 2X2 + 0S1 - S2 + 0S3 + A1 + 0A2 = 80
3X1+ 2X2 + 0S1 + 0S2 - S3 + 0A1 + A2 = 140
Where X1, X2, S1, S2, S3, A1, A2 ≥ 0

5. Simplex Table I Cj → Basic var. Sol. value X1 X2 S1 S2 S3 A1 A2
Contribution per unit 4 3 -M ↓
Basic var.
Sol. value X1 X2 S1 S2 S3 A1 A2
MIN. RATIO 50 1 80 2 -1
140 3*
140/3 → Zj
-220M
-4M M
Cj-Zj
4+4M ↑
3+4M
-2M

6. Simplex Table II Cj → Contribution per unit Basic var. Sol. value X14 3 -M ↓
Basic var.
Sol. value X1 X2 S1 S2 S3 A1
MIN. RATIO
10/3
1/3* 1
1/3
10 →
100/3
4/3 -1 25
140/3
2/3
-1/3 70 Zj M
8-4M M
-M-4
Cj-Zj
4M+1 ↑
M+4 3 3 3 3 3

7. Simplex Table III Cj → Basic var. Sol. Value X1 X2 S1 S2 S3 A1 10 1 20
Contribution per unit 4 3 -M ↓
Basic var.
Sol. Value X1 X2 S1 S2 S3 A1 10 1 20 -4 -1 40 -2 Zj
190-20M
1+4M M
M-1
Cj-Zj
-1-4M
1-M

8. Since Cj-Zj row contains all elements –ve or zero we are having optimum solution.
Artificial variable is present as a basic variable the given problem has infeasible solution.
Saturday, December 08, 2018

9. Example 2 Cj → 15 25 -M BASIC VAR. SOL. VALUE X1 X2 S1 S2 A1 A2 20 7 6-M ↓
BASIC VAR.
SOL. VALUE X1 X2 S1 S2 A1 A2 20 7 6 -1 1 30 8 5 18 3 -2 ZJ
-38M
-10M
-4M M
Cj-Zj
15+10M
25+4M

10. Solution : Write the original problem represented by the above table:
In the given problem, there are two decision variables X1 and X2 with objective function co-efficients equal to 15 and 25 respectively.
There are three constraints with RHS values as 20, 30 and 18 and involving ≥,≤ and = signs respectively. This is indicated by the slack, surplus and artificial variables.
The objective function is of maximisation type because the artificial variables bear negative co-efficient (-M) in the objective function.

11. The LPP is formulated as follows :
Maximize Z = 15x1+ 25x2
subjected to 7X1+ 6X2 ≥ 20
8X1+ 5X2 ≤ 30
3X1-2X2 = 18
Where x1, x2, s1, s2, a1, a2 ≥0
Now, introduce slack, surplus and artificial variables and the resultant LPP is as below :
Max. Z= 15X1+25X2+0S1+0S2-MA1-MA2
Subjected to : 7X1+6X2-S1+A1 = 20
8X1+5X2+0S1+1S2 = 30
3X1-2X2+0S1+0S2+A2 = 18
Where as X1,X2,S1,S2,A1,A2 ≥ 0

12. Simplex Table 1 Cj → 15 25 -M ↓ Basic Var. Sol. Value X1 X2 S1 S2 A1-M ↓
Basic Var.
Sol. Value X1 X2 S1 S2 A1 A2
MIN. RATIO 20 7* 6 -1 1
20/7 → 30 8 5
30/8 18 3 -2 Zj
-38M
-10M
-4M
Cj-Zj
15+10M ↑
25+4M +M

13. Simplex Table 2 Cj → 15 25 -M ↓ Basic Var. Sol. Value X1 X2 S1 S2 A1-M ↓
Basic Var.
Sol. Value X1 X2 S1 S2 A1 A2
MIN. RATIO
20/7 1
6/7
-1/7 -
50/7
-13/7
8/7*
50/8 →
66/7
-32/7
3/7 22 Zj
300-66M
90+32M
-15-3M
Cj-Zj
85-32M
15+3M 7 7 7 ↑ 7 7

14. Simplex Table 3 Cj → 15 25 ↓ Basic Var. Sol. Value X1 X2 S1 S2 15/4 1↓
Basic Var.
Sol. Value X1 X2 S1 S2
15/4 1
5/8
1/8
25/4
-13/8
7/8 -M A2
27/4
-31/8
-3/8 Zj
225-27M
75+31M
15+ 3M
Cj-Zj
125-3M
-15 -3M 4 8 8 8 8

15. Since all the values of Cj-Zj are 0 or –ve so this is the final table but the artificial variable still exist so the problem has an infeasible solution.