Discussion
Consider the following Linear Programming Problem :
Maximize : \(Z = 5x_{1} + 6x_{2} + 4x_{3}\)
Subject to \(3x_{1} + x_{2} + 4x_{3} \le 60\)
\(2x_{1} + 3x_{2} + 3x_{3} \le 75\)
\(x_{1}, x_{2}, x_{3} \ge 0\)
The value of the decision variables after the first iteration will be :
Answer: B
After the first iteration, \(x_1 = 0, x_2 = 25, x_3 = 0\).
Add slack variables and start from \(x_1 = x_2 = x_3 = 0\). For a maximisation, the entering variable is the one with the largest coefficient in Z, which is \(x_2\) (6). The ratio test decides how far it can increase:
\[ \begin{aligned} \text{row 1:} &\ 60 / 1 = 60 \\ \text{row 2:} &\ 75 / 3 = 25 \end{aligned} \]
The smaller ratio, 25, is in the second constraint, so its slack leaves and \(x_2\) enters at 25, with the other decision variables still 0. Z becomes 150.
\(x_1 = 20\) is what you get if \(x_1\) enters (ratio 60/3), and 37.5 is \(x_1\)'s ratio in the second row. \(x_3 = 25\) is wrong because \(x_3\)'s binding ratio would be 60/4 = 15.
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