In any simplex table, if corresponding to any negative ∆j, all elements of the…

2012

In any simplex table, if corresponding to any negative ∆j, all elements of the column are negative or zero, the solution under the test is

Answer: B. unbounded solutionConcept — what the entering column and the minimum-ratio test decide. In the simplex method the net-evaluation row Δj = Zj − Cj says whether a non-basic…

  1. A.

    degenerate solution

  2. B.

    unbounded solution

  3. C.

    alternative solution

  4. D.

    non-existing solution

Show answer & explanation

Correct answer: B

Concept — what the entering column and the minimum-ratio test decide. In the simplex method the net-evaluation row Δj = ZjCj says whether a non-basic variable can still improve the objective: a negative Δj marks column j as an improving direction. How far that variable may be increased is decided separately, by the minimum-ratio test. The ratio bi / aij is a genuine bound only when aij > 0, because only a positive coefficient forces the corresponding basic variable to fall as the entering variable rises. An improving direction along which nothing can be forced down is a direction along which the objective can be pushed forever.

Application to this table.

  1. The negative Δj identifies column j as improving, so xj is eligible to enter the basis.

  2. The leaving variable is found from the ratios bi / aij, formed only over rows whose aij > 0.

  3. Every element of this column is negative or zero, so no row qualifies: the set of admissible ratios is empty and no leaving variable exists.

  4. Increase xj by any amount t ≥ 0. Each basic variable moves to bi − t·aij. With aij ≤ 0 that quantity never decreases, so every basic variable stays non-negative however large t becomes — feasibility is never lost.

  5. The objective changes by −t·Δj, which grows without limit as t → ∞. The problem therefore has an unbounded solution: no finite optimum exists.

Cross-check and contrast.

  • Numerical check: maximise Z = 2x1 + x2 subject to x1x2 ≤ 10, x1, x2 ≥ 0. The x2 column carries the single constraint coefficient −1 while x2 still improves Z, so x2 can be raised indefinitely (−x2 ≤ 10 holds for every x2 ≥ 0) and Z → ∞.

  • Degeneracy is a different condition altogether: it means a basic feasible solution in which at least one basic variable equals zero. In a table it arises when the smallest admissible ratio is itself zero, and a tie among the smallest ratios leaves the next basis degenerate; either way the ratio test must have at least one positive aij to be carried out at all — the very thing this column lacks. It describes a basic variable stuck at zero, not an objective function without a limit.

  • Multiple optima are read from an already-optimal table in which a non-basic variable has Δj = 0, not from a column that is still improving.

  • Infeasibility — no solution existing at all — is detected at termination of the Big-M or two-phase method, when an artificial variable is still positive in the basis; it cannot be read from the sign pattern of one column of an otherwise feasible table.

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