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Transportation problems questions
Several warehouses, several shops, and a cost for every route between them. The first job is any feasible plan at all, and the north-west corner method supplies one without looking at the costs.
6 original questions · 24 marks · the transportation problems notes · Decision Mathematics 2
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A transportation problem has supplies of 40, 30 and 50, and demands of 35, 45 and 25. State what must be done before any method is applied, giving the figure involved.
Worked answer
Total supply is 120 and total demand is 105, so the problem is unbalanced. Add a dummy destination of demand 15, reachable from every source at a cost of zero. B1 for the two totals, B1 for the dummy destination, B1 for a demand of 15. Whichever source ends up supplying the dummy is the one whose goods are not sent anywhere.State how many cells a solution to a 4 by 5 transportation problem should use, and what it means if it uses fewer.
Worked answer
4 + 5 − 1 = 8 cells. Using fewer means the solution is degenerate, which happens when a supply and a demand run out together. M1 for 4 + 5 − 1, A1 for the cell count, B1 for degeneracy. A zero allocation is recorded in a suitable cell to restore the count, without moving any goods.Supplies are A 40, B 30, C 50 and demands are X 35, Y 45, Z 40. Costs per unit are AX 8, AY 5, AZ 6, BX 4, BY 9, BZ 7, CX 3, CY 6, CZ 5. Find the north-west corner solution and its cost.
Worked answer
AX takes min(40, 35) = 35, using up X. AY then takes A's remaining 5, using up A. BY takes 30, using up B. CY takes the last 10 that Y needs, and CZ takes 40.
Allocations: AX 35, AY 5, BY 30, CY 10, CZ 40.
Cost = 35(8) + 5(5) + 30(9) + 10(6) + 40(5) = 280 + 25 + 270 + 60 + 200 = 835. M1 for starting at the north-west corner, A1 A1 for the allocations, M1 for the cost sum, A1 for 835.For that solution, confirm the cell count is right and comment on the quality of the answer.
Worked answer
It uses five cells and 3 + 3 − 1 = 5, so the count is right and no zero allocation is needed. The cost is poor. The method never looks at a cost at all, and it has put 30 units on BY at 9 per unit, the dearest route in the table. B1 for the cell count, B1 for the comment on the cost. Its only purpose is to hand the improvement machinery a feasible plan to start from.A source in a different problem cannot supply one particular destination at all. Explain how this is handled, and why a cost of zero would be wrong.
Worked answer
Give that cell a cost large enough that no improvement method would ever choose it, and say in your working that you have done so. A cost of zero would make the route look like the cheapest in the table, so the method would fill it first: the forbidden route needs to be made unattractive, not free. B1 for a prohibitively large cost, B1 for saying so in the working, B1 for why zero is wrong. A dummy destination is the opposite case, where zero is right because nothing is actually moved.Supplies are A 30, B 40 and C 20; demands are X 30, Y 25 and Z 20. Costs per unit are AX 5, AY 8, AZ 6, BX 7, BY 4, BZ 9, CX 6, CY 5 and CZ 3. Prepare the problem for solution, find the north-west corner solution and its cost, and comment on the number of cells used.
Worked answer
Supply totals 90 and demand totals 75, so the problem is unbalanced. Add a dummy destination of demand 15, with a cost of zero from every source.
North-west corner on the 3 by 4 table: AX takes min(30, 30) = 30, which exhausts row A and column X together. Move on to BY, which takes min(40, 25) = 25, then BZ takes the remaining 15 of B's supply, then CZ takes the 5 that Z still needs, and CD takes 15.
Cost = 30(5) + 25(4) + 15(9) + 5(3) + 15(0) = 150 + 100 + 135 + 15 + 0 = 400.
Only five cells are in use against 3 + 4 − 1 = 6, so the solution is degenerate. It happened because row A and column X ran out at the same moment. Record a zero allocation in AY, next along the route, which restores the count without moving any goods and lets every shadow cost be found later.
B1 for the imbalance, B1 for the dummy demand of 15, M1 for the north-west corner method, A1 A1 for the allocations, M1 for the cost, A1 for 400, B1 for the degeneracy and the zero allocation.
Balance the problem before starting and count the cells after finishing. Both steps carry marks, and a degenerate table left uncorrected makes the shadow costs impossible.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise transportation problems one question at a time
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