Resource Allocation Model – Maximisation Case
Problem 3.15: Solve the following L.P.P
Maximize Z = 1a + 2b + 3c – 1d S.t.
1a + 2b + 3c = 15 2a + 1b + 5c = 20
1a + 2b + 1c + 1f = 10 and a, b, c, f all are ≥ 0.
In this problem given constraints are equations rather than inequalities. Also by careful examination, we can see that in the third equation variable ‘f ’ exists and it also exists in objective function. Hence we consider it as a surplus variable and we add two more slack variables ‘d’ and ‘e’ is added to first and second equations. But the cost coefficient in objective function for variables d and e will be –M. When we use big M in maximization problem, we have to use –M in objective function. While solving the problem, all the rules related to solving maximization problem will apply. Hence now the simplex format of the problem is as follows:
Maximize Z = 1a + 2b + 3c – 1d – Me – Mf s.t.
1a + 2b + 3c + 1d + 0e + 0f = 15 2a + 1b + 5c + 0d + 1e + 0f = 20
1a + 2b + 1c + 0d + 0e + 1f = 10 and a, b, c, d, e, f all = 0.
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Table I. a = 0, b = 0, c = 0, d = 15, e = 20, f = 10 and Z = Rs. 0.
Problem Profit Cj Capacity, 1 2 3 – 1 – M – M Replacement
variable Rs. units a b c d e f ratio
d – 1 15 1 2 3 1 0 0 5
e – M 20 2 1 5 0 1 0 4
f – M 10 1 2 1 0 0 1 10
Net 2 + 3M 4 + 3M 4 + 8M 0 0 0
evaluation.
Table II. A = 0, b = 0, c = 4, d = 0, e = 0, f = 0., Z = Rs. 3 × 4 = Rs. 12.
Problem Profit Cj Capacity, 1 2 3 – 1 – M – M Replacement
Variable Rs. units a b c d e f Ratio
d – 1 6 3/5 9/5 0 1 0 0 30/9
c 3 4 2/5 1/5 1 0 0 0 20
f – M 3 – 1/5 7/5 0 0 1 1 15/7
Net 2/5 + M /5 16/5 + 7/5M 0 M – M 0
evaluation.
TableL: III. a = 0, b = 15/7, c = 25/7, d = 15/7, e = 0, f = 0, Z = Rs. 107 – 15/7 = Rs.92.
Problem Profit Profit 1 2 3 – 1 – M – M
variable Rs. capacity a b c d e f
d – 1 15/7 6/7 0 0 1 0 0
c 3 25/7 3/7 0 1 0 – 5/7 – 5/7
b 2 15/7 – 1/7 1 0 0 5/7 5/7
Net 6/7 0 0 0 – M + 5/7 – M + 5/7
evaluation
Linear Programming Models : Solution by Simplex Method 8181818181 Table: IV. a =5/2, b = 5/2, c = 5/2, d = 0, e = 0, f = 0, Z = Rs. 15/–
Problem Profit Profit 1 2 3 – 1 – M – M
Variable Rs. capacity a b c d e f
a 1 5/2 1 0 0 7/6 0 0
c 3 5/2 0 0 1 – 1/2 – 5/7 – 5/7
b 2 5/2 0 1 0 1/6 5/7 5/7
0 0 0 – 4/12 – M + 5/7 – M = 5/7
As all the elements of net evaluation row are either zeros or negative elements, the solution is optimal.
Z = Rs. 15/–. And a = b = c = 5/2.
Problem 3.16: Solve the given l.p.p: Simplex format is:
Maximize 4x + 3y s.t. Maximize Z = 4x + 3y + 0S + 0p + 0q – MA1 – MA2 s.t.
1x + 1y ≤ 50 1x +1y+ 1S + 0p + 0q + 0A1 + 0A2 = 50 1x + 2y ≥ 80 1x + 2y +0S – 1p + 0q + 1A1 + 0A2 = 80
3x + 2y ≥ 140 3x + 2y + 0S + 0p – 1q + 0A1 + 1A2 = 140 And both x and y ≥ 0 and x, y, S, p, q, A1 and A2 all ≥ 0
Table: I. X = 0, y, = 0, S = 50, p = 0 q = 0, A1 = 80, A2 = 140 and Z = Rs. 220 M.
Problem Profit Profit: 4 3 0 0 0 –M –M Replace–
variable Rs. capacity units x y S p q A1 A2 ment ratio
S 0 50 1 1 1 0 0 0 0 50
A1 – M 80 1 2 0 – 1 0 1 0 40
A2 – M 140 3 2 0 0 – 1 0 1 70
Net 4 + 4M 3 + 4M 0 M M 0 0
evaluation.
Now in the net evaluation row the element under variable ‘x’ is 4 + 4M is greater than the element 3 + 3M. But if we take ‘x’ as the incoming variable we cannot send artificial variable out first. Hence we take ‘y’ as the incoming variable, so that A1 go out first.
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Table: II. X = 0, y = 40, S = 10, A1 = 0, A2 = 60, p = 0, q = 0 and Z = Rs. 120 – 60 M.
Problem Profit Profit: 4 3 0 0 0 – M – M Replace–
variable Rs. capacity units x y S p q A1 A2 ment ratio
S 0 10 0.5 0 1 0.5 0 – 0.5 0 20
y 3 40 0.5 1 0 – 0.5 0 0.5 0 80
A2 – M 60 2 0 0 2 – 1 – 2 1 30
Net 2.5 + 8M 0 0 1.5 – 2M – M – 1.5 – 3M 0 evaluation.
Table: III. x = 20, y = 30, p = 0, q = 0, A1 = 0, A2 = 20, Z = Rs. 170/–
Problem Profit Profit: 4 3 0 0 0 – M – M Replace–
variable Rs. capacity units x y S p q A1 A2 ment ratio
x 4 20 1 0 2 1 0 – 1 0
y 3 30 0 1 – 1 – 1 0 1 0
A2 – M 20 0 0 – 4 0 – 1 0 1
Net 0 0 – 5 –4M – 1 – M – M = 1 0
evaluation.
In the last table though basis variables have the opportunity cost as 0, still artificial variable A2 exists in the problem, hence the original problem has no feasible solution.
X = 20, Y = 30.
Problem 3.17: Solve the given l.p.p. Simplex version of the problem is:
Maximize Z = 1a + 1.5b + 5c + 2d s.t. Maximize Z = 1a + 1.5b + 5c + 2d + 0S1 + 0S2 – MA1 – MA2 s.t.
3a + 2b + 1c + 4d ≤ 6 3a + 2b +1c + 4d + 1S1 + 0S2 + 0A1 + 0A2 = 6 2a + 1b + 5c + 1d ≤ 4 2a +1b + 5c + 1d + 0S1 +1S2 + 0A1 + 0A2 = 4 2a + 6b – 4c + 8d = 0 2a + 6b – 4c + 8d + 0S1 + 0S2 +1A1 + 0A2 = 0 1a + 3b – 2c + 4d = 0 1a +3b – 2c + 4d + 0S1 + 0S2 + 0A1 + 1A2 = 0 And a, b, c, d all ≥ 0 and a, b, c, d, S1, S2, A1, A2 all ≥ 0
Linear Programming Models : Solution by Simplex Method 8383838383 Table I. A = 0, b = 0, c = 0, d = 0, S1 = 6, S2 = 0, A1 = 0, A2 = 0 and Z = Rs. 0.00
Problem Profit Cj Capacity 1 1.5 5 2 0 0 –M –M Replace–
variable Rs. units a b c d S1 S2 A1 A2 ment ratio
S1 0 6 3 2 1 4 1 0 0 0 6/4 = 1.5
S2 0 4 2 1 5 1 0 1 0 0 4/1 = 4
A1 – M 0 2 6 – 4 8 0 0 1 0 0
A2 – M 0 1 3 – 2 4 0 0 0 1 0
Net 1 + 3M 1.5 + 9M 5 – 6M 2 + 12M 0 0 0 0 evaluation
Note: Both A1 and A2 have the same replacement ratio. That is to say there is a tie in outgoing variables. This type of situation in Linear Programming Problem is known as DEGENERACY. To Solve degeneracy refer to the rules stated earlier.
Now let us remove as and when a surplus variable goes out, which will ease our calculation and also save time.
Table: II. a = 0, b = 0, c = 0, d = 0, S1 = 6, S2 = 4, A1 = 0, A2 = 0, Z = Rs. 0/–
Problem Profit Cj Capacity 1 1.5 5 2 0 0 – M – M Replace–
variable Rs. units a b c d S1 S2 A1 A2 ment ratio
S1 0 6 2 – 1 3 0 1 0 0
S2 0 4 7/4 1/4 11/2 0 0 1 0
A1 – M 0 0 0 0 0 0 0 1
d 2 0 1/4 3/4 – 1/2 1 0 0 0
Net 1/2 0 6 0 0 0 0
evaluation
In the given problem the inequalities number 3 and 4 i.e., 2a + 6b – 4c + 8d = 0 and 1a + 3b – 2c + 4d = 0 appears to be similar. If you carefully examine, we see that
2a + 6b – 4c + 8d = 0 is 2 × (1a + 3b – 2c + 4d = 0). Hence one of them may be considered as redundant and cancelled. i.e., third constraint is double the fourth constraint hence we can say it is not independent. Hence we can eliminate the third row and the column under A1 from the tableau. For identifying the redundancy, one need not wait for final optimal table.
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Table: III. (Second reduced second table) S1 = 6, S2 = 4, d = 0, a = 0, b = 0, c = 0, Z = Rs. 0
Problem Profit Cj Capacity 1 1.5 5 2 0 0 – M – M Replace–
variable Rs. units a b c d S1 S2 A1 A2 ment ratio
S1 0 6 2 – 1 3 0 1 0 2
S2 0 4 7/4 1/4 11/2 0 0 1 8/11
d 2 0 1/4 3/4 – 1/2 1 0 0 —
Net 1/2 0 6 0 0 0
evaluation.
Table: IV. S1 = 42/11, C = 5, d = 4/11 and Z = Rs. 4.36.
Problem Profit Cj Capacity 1 1.5 5 2 0 0 – M – M Replace–
variable Rs. units a b c d S1 S2 A1 A2 ment ratio
S1 0 42/11 23/22 – 25/22 0 0 1 – 6/11
c 5 8/11 7/22 1/22 1 0 0 2/11
d 2 4/11 9/22 17/22 0 1 0 1/11
Net – 31/22 – 3/11 0 0 0 – 12/11 evaluation
As all the elements of net evaluation row are either zeros or negative elements the solution at this stage is optimal. Hence c = 8/11, d = 4/11 and Z = Rs. 48/11 = Rs. 4.36.
3.10. ARTIFICIAL VARIABLE METHOD OR TWO PHASE METHOD
In linear programming problems sometimes we see that the constraints may have ≥, ≤ or = signs. In such problems, basis matrix is not obtained as an identity matrix in the first simplex table; therefore, we introduce a new type of variable called, the artificial variable. These variables are fictitious and cannot have any physical meaning. The introduction of artificial variable is merely to get starting basic feasible solution, so that simplex procedure may be used as usual until the optimal solution is obtained. Artificial variable can be eliminated from the simplex table as and when they become zero i.e, non–basic. This process of eliminating artificial variable is performed in PHASE I of the solution. PHASE II is then used for getting optimal solution. Here the solution of the linear programming problem is completed in two phases, this method is known as TWO PHASE SIMPLEX METHOD. Hence, the two–phase method deals with removal of artificial variable in the fist phase and work for optimal solution in the second phase. If at the end of the first stage, there still remains artificial variable in the basic at a positive value, it means there is no feasible solution for the problem given. In that case, it is not
Linear Programming Models : Solution by Simplex Method 8585858585 necessary to work on phase II. If a feasible solution exists for the given problem, the value of objective function at the end of phase I will be zero and artificial variable will be non-basic. In phase II original objective coefficients are introduced in the final tableau of phase I and the objective function is optimized.
Problem 3.18: By using two phase method find whether the following problem has a feasible