Maximization Case Of Simplex Method
Table 2. To do this the following procedure is used.
Step 1: To Write the incoming variable ‘b’ in place of outgoing variable S_{2}. Enter the profit of ‘b’ in profit column. Do not alter S_{1} and S_{3}. While doing so DO NOT ALTER THE POSITION OF THE ROWS.
Step 2: DIVIDING THE ELEMENTS OF OLD COLUMN BY KEY COLUMN ELEMENTS obtains capacity column elements.
Step 3: Transfer of key row:
DIVIDE ALL ELEMENTS OF KEY ROW BY RESPECTIVE KEY COLUMN NUMBER.
Step 4: Transfer of NonKey rows: NEW ROW NUMBER = (old row number – corresponding key row number) × fixed ratio. Fixed ratio = Key column number of the row/key number.
Step 5: Elements of Net evaluation row are obtained by: Objective row element at the top of the row – Σ key column element × profit column element.
Step 6: Select the highest positive element in net evaluation row or highest opportunity cost and mark the column by an arrow to indicate key column (incoming variable).
Step 7: Find the replacement ratios by dividing the capacity column element in the row by key column element of the same row and write the ratios in replacement ratio column. Select the limiting (lowest) ratio and mark with a tick mark to indicate key row (outgoing variable). The element at the intersection of key column and key row is known as key number. Continue these steps until we get:
 For maximisation problem all elements of net evaluation row must be either zeros or negative elements.

For Minimisation problem, the elements of net evaluation row must be either zeros or positive elements.
\1. Transfer of Key row: 2000/10, 5 /10, 10 /10, 0 /10, 1 / 10, 0 / 10
2. Transfer of Non key rows:
Rule:
(Old row Number – corresponding key row number) – Key column number / key number = new row no.
1st row. 2500 – 2000 × 6/10 = 1300 2nd row: 500 – 2000 × 2/10 = 100
10 – 10 × 6/10 = 0
6 – 10 × 6/10 = 0 1 – 5 × 2/10 = 0
1 – 0 × 6/10 = 1 2 – 10 × 2/10 = 0
0 – 1 × 6/10 = – 0.6 0 – 0 × 2/10 = 0
0 – 0 × 6/10 = 0 0 – 1 × 2/10 = – 0.2
1 – 0 × 2 / 10 = 1
Replacement ratios: 1300/7 = 185.7, 200/0.5 = 400, 100/0 = Infinity.
Net evaluation row elements = Column under 'a' = 23 – (7 × 0 0.5 × 32 0 × 0) = 23 – 16 = 7
'b' = 32 – (0 × 0 1 × 32 0 × 0) = 32 – 32 = 0
S_{1}= 0 – (1 × 0 0 × 32 0 × 0) = 0
S_{2}= 0 – (– 0.6 × 0 0.1 × 32 –0.2 × 0) = – 3.2
S_{3}= 0 – (0 × 0 0 × 32 1 × 0) = 0
This concept is used to check whether the problem is done correctly or not. To dothis MULTIPLY THE ELEMENTS IN NET EVALUATION ROW UNDER SLACK VARIABLES WITH THE ORIGINAL CAPACITY CONSTRAINTS GIVEN IN THE PROBLEM AND FIND THE SUM OF THE SAME.