Discussion
Consider the following Linear Programming Problem :
Maximize : \(Z = 8x_{1} + 5x_{2}\)
Subject to \(2x_{1} + x_{2} \le 500\)
\(x_{1} \le 150\)
\(x_{2} \le 250,\ x_{1}, x_{2} \ge 0\)
The value of decision variable after the first iteration will be :
Answer: B
After the first iteration, \(x_{1} = 150,\ x_{2} = 0\).
Add slack variables \(s_{1}, s_{2}, s_{3}\) to the three constraints. The starting basis is all slacks, with \(x_{1} = x_{2} = 0\).
In a maximisation, the entering variable is the one with the largest positive coefficient in Z, which is \(x_{1}\) (8 against 5). The minimum ratio test picks the leaving row:
\[ \begin{aligned} \text{row 1: } & 500 / 2 = 250 \\ \text{row 2: } & 150 / 1 = 150 \\ \text{row 3: } & x_{1} \text{ absent, no ratio} \end{aligned} \]
The smallest ratio is 150, so \(s_{2}\) leaves and \(x_{1}\) enters at 150; \(x_{2}\) is still non-basic, so it stays 0. (0, 250) would result from bringing \(x_{2}\) in first, and (250, 0) ignores the constraint \(x_{1} \le 150\). (150, 200) is where the second iteration lands, when \(x_{2}\) enters.
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