Minimization Case Of Simplex Method
CAPITAL ‘M’, THIS METHOD IS KNOWN AS BIG ‘M’ METHOD.
By using the above concept, let us write the equations of the inequalities of the problem.
Minimise Z = 3 x 2.5 y 0p 0q M A1 M A2 S.T. Objective Function.
2 x 4 y – 1 p 1A1 = 40
3 x 2 y – 1 q 1 A2 = 50 Structural Constraints. And x, y, p, q, A1, A2 all ≥ 0 Non negativity Constraint.
Simplex format of the above is:
Minimise Z = 3x 2.5y 0p 0q MA1 MA2 S.T.
2x 4y – 1p 0q 1A1 0A2 = 40
3x 2y 0p – 1q 0A1 1A2 = 50
Note: As the variables A1 and A2 are basis variables, their Net evaluation is zero.
Now take 6M and 5M, 6 M is greater and if we subtract 2.5 from that it is negligible. Hence –6m will be the lowest element. The physical interpretation is if patient purchases Y now, his cost will be reduced by an amount 6M. In other words, if the patient does not purchase the Y at this point, his penalty is 6M, i.e., the opportunity cost is 6M. As the non-basis variable Y has highest opportunity cost (highest element with negative sign), Y is the incoming variable. Hence, the column under Y is key column. To find the outgoing variable, divide requirement column element by key column element and find the replacement ratio. Select the lowest ratio, i.e., here it is 10, falls in first row, hence A1 is the outgoing variable.
To transfer key row, divide all the elements of key row by key number (= 4).
40/4 = 10, 2/4 = 0.5, –1/4 = – 0.25, 0/25 = 0, 1/25 = 0.25, 0/4 = 0To transfer non-key row elements: New row element = old row element – corresponding Key row element × (Key column number/key number).
50 – 40 × 2/4 = 30
3 – 2 × 0.5 = 2
2 – 4 × 0.5 = 0
0 – (–1) × 0.5 = 0.5 .
–1 – 0 × 0.5 = – 1
0 – 1 × 0.5 = – 0.5
1 – 0 × 0.5 = 1
Note:
- The elements under A1 and A2 i.e., artificial variable column are negative versions of elements under artificial variable column.
- The net evaluation row elements of basis variables are always zero. While writing the second tables do not change the positions of the rows).