Skip to content

CMA Intermediate · Operations Management and Strategic Management · Optimum Allocation of Resources - LPP

In a maximisation simplex table, the Cj − Zj row reads: x1 = 5, x2 = 8, s1 = 0, s2 = −2. Which variable should enter the basis and is the solution optimal?

x2 should enter the basis and the solution is not yet optimal. In a maximisation problem, optimality requires every Cj − Zj to be zero or negative. Positive values exist for x1 and x2, and the largest positive value, 8, belongs to x2, so it is chosen.

  1. Ax2 enters; the solution is not optimalCorrect
  2. Bx1 enters; the solution is not optimal
  3. Cs2 enters; the solution is not optimal
  4. DNo variable enters; the solution is optimal

Explanation

In maximisation, the solution is optimal only when all Cj − Zj values are ≤ 0. Here x1 and x2 are positive, so it is not optimal. The largest positive value is 8, belonging to x2, so x2 enters. Choosing x1 is not the standard rule, and s2 with −2 would reduce profit.

Did you get it right without looking?

One question tells you little. A timed set on Optimum Allocation of Resources - LPP shows your real accuracy, how long you take and where you lose marks.

Practise Optimum Allocation of Resources - LPPFree diagnostic mock

More Optimum Allocation of Resources - LPP questions