Transportation Problems Questions and Answers

Answer: D

It has mn decision variables and (m + n) constraints.

Let \(x_{ij}\) be the quantity shipped from source i to destination j. There is one such variable for every source-destination pair, so \(m \times n\) variables.

Each source has a supply constraint (what it ships cannot exceed, or must equal, its supply), giving m constraints, and each destination has a demand constraint, giving n more: \(m + n\) in all, plus non-negativity.

The other options confuse the two counts: mn constraints would mean one per route, and \(m + n\) variables would mean one per location rather than per route.

Answer: D

Vogel's approximation method first finds the penalty of each row and column (the difference of the two smallest costs). The largest penalty, 3, is in column D4 (costs 4, 1, 9), so the cheapest cell there, S2-D4, takes 10 units. Next the largest penalty is 2, and column D1 is chosen, which gives x11 = 20. Then row S3 has the largest penalty and takes 20 units in D2, x32 = 20, followed by x13 = 10, x22 = 20 and x23 = 20. The cost is

The solution has x32 = 20, while options (A) to (C) show x32 = 10, which leaves 10 units of S3 unassigned, so none of them lists the solution correctly. The keyed answer is none of the above.

Answer: A

A basic feasible solution to a transportation problem with m origins and n destinations is said to be degenerate if the number of positive allocations is less than m + n - 1. The answer is (A).

When there are exactly m + n - 1 positive allocations the solution is non-degenerate (B), so that is the opposite case. 'Unbounded' (C) and 'unbalanced' (D) do not describe the number of allocations: a transportation problem is unbalanced when total supply differs from total demand.

Answer: B

The total cost by Vogel's approximation method is 80. Supply is 34 and demand is 30, so a dummy destination with demand 4 and zero cost is added. The allocations made, in order, are F3-dummy 4, F1-W2 4, F2-W5 8, F2-W4 4, F1-W4 4, F3-W1 4 and F3-W3 6.

\[ 4(2) + 8(1) + 4(2) + 4(2) + 4(6) + 6(4) = 80 \]

The values 76, 90 and 96 are not the cost of this allocation. The answer is (B).