Resource Allocation Model – Maximisation Case
Problem 3.20: This can be written as:
2. The existence of big M avoids the use of digital computer for calculations
3.11. DEGENERACY IN LINEAR PROGRAMMING PROBLEMS
The degeneracy in linear programming problems and the methods of solving degeneracy, if it exists, are discussed earlier in the chapter. To recollect the same a brief discussion is given below:
While improving the basic feasible solution to achieve optimal solution, we have to find the key column and key row. While doing so, we may come across two situations. One is Tie and the other is Degeneracy.
The tie occurs when two or more net evaluation row elements of variables are equal. In maximization problem, we select the highest positive element to indicate incoming variable and in minimization we select lowest element to indicate incoming variable (or highest numerical value with negative sign).
When two or more net evaluation row elements are same, to break the tie, we select any one of them to indicate incoming variable and in the next iteration the problem of tie will be solved.
To select the out going variable, we have to select the lowest ratio or limiting ratio in the replacement ratio column. Here also, some times during the phases of solution, the ratios may be equal. This situation in linear programming problem is known as degeneracy. To solve degeneracy, the following methods are used:
1. Select any one row as you please. If you are lucky, you may get optimal solution, otherwise the problem cycles.
OR
2. Identify the rows, which are having same ratios. Say for example, S1 and S3 rows having equal ratio. In such case select the row, which contains the variable with smaller subscript.
That is select row containing S1 as the key row. Suppose the rows of variable x and z are having same ratio, then select the row-containing x as the key row.
3. (a) Divide the elements of unit matrix by corresponding elements of key column. Verify the ratios column-wise in unit matrix starting from left to right. Once the ratios are unequal, the degeneracy is solved. Select the minimum ratio and the row containing that element is the key row. (This should be done to the rows that are in tie).
(b) If the degeneracy is not solved by 3 (a), then divide the elements of the main matrix by the corresponding element in the key column, and verify the ratios. Once the ratios are unequal, select the lowest ratios. (This should be done only to rows that are in tie).
Problem 3.21: A company manufactures two product A and B. These are machined on machines X and Y. A takes one hour on machine X and one hour on Machine Y. Similarly product B takes 4 hours on Machine X and 2 hours on Machine Y. Machine X and Y have 8 hours and 4 hours as idle capacity.
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The planning manager wants to avail the idle time to manufacture A and B. The profit contribution of A is Rs. 3/– per unit and that of B is Rs.9/– per unit. Find the optimal product mix.
Solution: Simplex format is:
Maximize Z = 3a + 9b s.t. Maximize Z = 3a + 9b + 0S1 + 0S2 s.t.
1a + 4b ≤ 8 1a + 4b + 1S1 + 0S2 = 8
1a + 2b ≤ 4 both a and b are ≥ 0 1a + 2b + 0S1 + 1S2 = 4 and a, b, S1, S2 all ≥ 0.
Table: I. A = 0, b = 0, S1 = 8, S2 = 4 and Z = Rs.0/–
Problem Profit Cj 3 9 0 0 Replacement
variable Rs. Capacity a b S1 S2 ratio
S1 0 8 1 4 1 0 8/4 = 2
S2 0 4 1 2 0 1 4/2 = 2
Net evaluation 3 9 0 0
Now to select the out going variable, we have to take limiting ratio in the replacement ratio column. But both the ratios are same i.e. = 2. Hence there exists a tie as an indication of degeneracy in the problem. To solve degeneracy follow the steps mentioned below:
(i) Divide the elements of identity column by column from left to right by the corresponding key column element.
Once the ratios are unequal select the lowest ratio and the row containing that ratio is the key row.
In this problem, for the first column of the identity (i.e. the S1 column) the ratios are: 1/4, and 0/2. The lowest ratio comes in row of S2. Hence S2 is the outgoing variable. In case ratios are equal go to the second column and try.
Table: II. a = 0, b = 2, S1 = 0, S2 = 0, Z = Rs. 18/–
Problem Profit Cj 3 9 0 0 Replacement
variable Rs. Capacity a b S1 S2 ratio
S1 0 0 1 – 2 – 1 0
b 9 2 0 1/2 1/2 1
Net evaluation 0 – 9/2 – 3/2 0
Optimal solution is b = 2 and Profit is 2 × 9 = Rs. 18/–
Linear Programming Models : Solution by Simplex Method 9393939393
Problem 3.22: Simplex version is:
Maximize Z = 2x + 1y s.t Maximize Z = 2x + 1y + 0S1 + 0S2 + 0S3 s.t.
4x + 3y ≤ 12 4x + 3y + 1S1 + 0S2 + 0S3 = 12
4x + 1y ≤ 8 4x + 1y + 0S1 + 1S2 + 0S3 = 8
4x – 1y ≤ 8 4x – 1y + 0S1 + 0S2 + 1S3 = 8
Both x and y are ≥ 0 x, y, S1, S2, S3 all ≥ 0 Table: I. x = 0, y = 0, S1 = 12, S2 = 8, S3 = 8 and Z = Rs. 0/–
Problem Profit Cj 2 1 0 0 0 Replacement
variable Rs. Capacity units x y S1 S2 S3 ratio
S1 0 12 4 3 1 0 0 12/4 = 3
S2 0 8 4 1 0 1 0 8/4 = 2
S3 0 8 4 – 1 0 0 1 8/4 = 2
Net evaluation 2 1 0 0 0
As the lowest ratio (= 2) is not unique, degeneracy occurs. Hence divide the elements of the identity column by column from left to right and verify the ratios.
In this problem, the elements of the first column of identity under S1 (for 2nd and 3rd row) are 0,0. If we divide them by respective key column element, the ratios are 0/4 and 0/4 which are equal, hence we cannot break the tie.
Now try with the second column i.e., column under S2. The elements of 2nd and 3rd row are 1 and 0. If we divide them by respective elements of key column, we get 1/4 and 0/4. The ratios are unequal and the lowest being zero for the third row. Hence S3 is the outgoing variable.
Table: II. x = 2, y = 0, S1 = 4, S2 = 0, S3 = 0, and Z = Rs. 4/–
Problem Profit Cj 2 1 0 0 0 Replacement
variable Rs. Capacity units x y S1 S2 S3 ratio
S1 0 4 0 4 1 0 – 1 1/4
S2 0 0 0 2 0 1 – 1 0/2
X 2 2 1 – 1/4 0 0 1/4 —
Net evaluation 0 3/2 0 0 – 1/2
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Table: III. x = 2, y = 0, S1 = 4, S2 = 0, S3 = 0, Z = Rs. 4/–
Problem Profit Cj 2 1 0 0 0 Replacement
variable Rs. Capacity units x y S1 S2 S3 ratio
S1 0 4 0 0 1 – 2 1 4/1
y 1 0 0 1 0 1/2 – 1/2 —
x 2 2 1 0 0 1/8 1/8 2 × 8/1 = 16
Net evaluation 0 0 0 – 3/4 1/4
Table: IV. x = 3/2, y = 2, S1 = 0, S2 = 0, S3 = 4 and Z = Rs. 3 + 2 = Rs. 5/–
Problem Profit Cj 2 1 0 0 0 Replacement
variable Rs. Capacity units x y S1 S2 S3 ratio
S3 0 4 0 0 1 – 2 1
y 1 2 0 1 1/2 – 1/2 0
x 2 3/2 1 0 – 1/8 3/8 0
Net evaluation 0 0 – 1/4 – 1/4 0
As the elements of net evaluation row are either zeros or negative elements, the solution is optimal.
x = 3/2 and y = 2, and Z = Rs. 2 × 3/2 + 2 × 1 = Rs. 5/–
Shadow price = ¼ × 12 + ¼ × 8 = Rs. 5/–.
3.12.1. Special Cases
Some times we come across difficulties while solving a linear programming problem, such as alternate solutions and unbound solutions. Let us solve some problems of special nature.
Problem. 3.23:
Solve the given l.p.p. by big – M method. Simplex version is:
Maximize Z = 6a + 4b s.t. Maximize Z = 6a + 4b + 0S1 + 0S2 + 0S3 – MA 2a + 3b ≤ 30 2a + 3b + 1S1 + 0S2 + 0S3 + 0A = 30 3a + 2b ≤ 24 3a + 2b + 0S1 + 1S2 +0S3 + 0A = 24 1a + 1b ≥ 3 and x, y both ≥ 0. 1a + 1b + 0S1 + 0S2 – 1S3 + 1A = 3
a, b, S1, S2, S3, A all ≥ 0.
Linear Programming Models : Solution by Simplex Method 9595959595
Table: I. a = 0, b = 0, S1 = 30, S2 = 24, S3 = 0, A = 3 and Z = Rs. 3M
Problem Profit Cj 6 4 0 0 0 – M Replacement
variable Rs. Capacity units a b S1 S2 S3 A ratio.
S1 0 30 2 3 1 0 0 0 30/2 = 15
S2 0 24 3 2 0 1 0 0 24/3 = 8
A – M 3 1 1 0 0 – 1 1 3/1 = 3
Net 6 + M 4 + M 0 0 – M 0
evaluation
Table II. a = 3, b = 0, S1 = 24, S2 = 15, S3 = 0, A = 0, Z = Rs. 18/– (out going column eliminated)
Problem Profit Cj 6 4 0 0 0 – M Replacement
variable Rs. Capacity units a b S1 S2 S3 A ratio
S1 0 24 0 1 1 0 2 24/2 = 12
S2 0 15 0 – 1 0 1 3 15/3 = 5
A 6 3 1 1 0 0 – 1 —
Net 0 – 2 0 0 6
evaluation
Table: III. a = 8, b = 0, S1 = 14, S2 = 5, S3 = 0, A = 0, Z = Rs. 48
Problem Profit Cj 6 4 0 0 0 – M Replacement
variable Rs. Capacity units a b S1 S2 S3 A ratio
S3 0 14 0 5/3 1 – 2/3 0 42/5
S2 0 5 0 – 1/3 0 1/3 0 —
a 6 8 1 2/3 0 1/3 0 24/2 = 12
Net 0 0 0 – 3 0
evaluation
In the above table, as all the net evaluation elements are either zeros or negative elements the optimal solution is obtained. Hence the answer is: a = 8 and the profit Z = Rs. 48/–. But the net evaluation element in the column under 'b' is zero. This indicates that the problem has alternate solution.
If we modify the solution we can get the values of basis variables, but the profit Z will be the same.
This type of situation is very helpful to the production manager and marketing manager to arrange the
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production schedules and satisfy the market demands of different segments of the market. As the profits of all alternate solutions are equal, the manager can select the solution, which is more needed by him. Now let us workout the alternate solution.
Table: IV. a = 12 /5, b = 42 / 5 , S1 = 0, S2 = 0 , S3 = 39/5, A = 0, Z = Rs. 48/–
All the elements of net evaluation rows are either zeros or negative elements so the solution is Optimal. The answer is : a = 12/5, b = 42/5 and Z = 12/5 × 6 + 42/5 × 4 = Rs. 48/– . Shadow price = element in column under S2 multiplied by element on the R.H.S of second inequality i.e., 24 × 2 = Rs.
48/–. Once we get one alternate solution then any number of solutions can be written by using the following rule.
(a) One alternate value of the basis variable is: First value × d + second value × (1–d). In the given example, the first value of ‘a’ is 8 and second value of ‘a’ is 12/5. Hence next value is 8 d + (1 – d) × 12/5. Like this we can get any number of values. Here ‘d’ is any positive fraction number: for example 2/5, 3/5 etc. It is better to take d = 1/2, so that next value is 1/2 × first value + 1/2 × second value. Similarly the values of other variables can be obtained.
3.12.2. Unbound Solutions
In linear programming problem, we come across certain problem, where feasible region is unbounded i.e., the value of objective function can be increased indefinitely. Then we say that the solution is UNBOUND. But it is not necessary, however, that an unbounded feasible region should yield an unbounded value of the objective function. Let us see some examples.
Problem 3.24: Maximize Z = 107a + 1b + 2c s.t 14a +1b – 6c + 3d = 7
16a + ½ b + 6c ≤ 5
3a – 1b – 6c ≤ 0 and a, b, c and d all ≥ 0.
We find that there is variable ‘d’ in the first constraint, with coefficient as 3. Second and third constraints do not have variable ‘d’. Hence we can divide the first constraint by 3 we can write as: 14/
3 a + 1/3 b – 6/3 c + 3/3 d = 7/3. If we write like this, we can consider the variable 'd' as slack variable.
Linear Programming Models : Solution by Simplex Method 9797979797
Simplex version is
Maximize Z = 107a + 1b + 2c + 0d + 0S1 + 0S2 s.t.
14/3 a + 1/3b – 2c + 1d + 0S1 + 0S2 = 7/3 16a + ½ b – 6c + 0d + 1S1 + 0S2 = 5 3a – 1b – 1c + 0d + 0S1 + 1S2 = 0 and a, b, c, d, S1, S2 all ≥ 0.
TableL: I. a = 0, b = 0, c = 0, d = 7/3, S1 = 5, S2 = 0 and Z = Rs. 0.
Problem Cost Cj 107 1 2 0 0 0 Replacement
variable Rs. Capacity units a b c d S1 S2 ratio
d 0 7/3 14/3 1/3 – 2 1 0 0 7/14
S1 0 5 16 1/2 6 0 1 0 5/16
S2 0 0 3 – 1 – 1 0 0 1 0/3
Net evaluation 107 1 2 0 0 0
Table: II. a = 0, b = 0, c = 0, d = 7/3, S1 = 5, S2 = 0 and Z = Rs.0/–
Problem Cost Cj 107 1 2 0 0 0 Replacement
variable Rs. Capacity units a b c d S1 S2 ratio
d 0 7/3 0 17/9 – 4/9 1 0 – 14/9 —
S1 0 5 0 35/6 – 2/3 0 1 – 16/3 —
a 107 0 1 – 1/3 – 1/3 0 0 1/3 —
Net evaluation 0 110/3 113/3 0 0 107/3
As the elements of incoming variable column are negative we cannot workout replacement ratios and hence the problem cannot be solved. This is an indication of existence of unbound solution to the given problem.
Problem 3.25: Simplex version is:
Maximize Z = 6x – 2y s.t. Maximize Z = 6x – 2y + 0S1 + 0S2 s.t.
2x – 1y ≤ 2 2x – 1y + 1S1 + 0S2 = 2
1x + 0y ≤ 4 and both x and y are ≥ 0. 1x + 0y + 0S1 + 1S2 = 4