Lesson 4D.1.1-4D.1.2

4D.1.1-4D.1.2 Initial solutions and the stepping-stone method Quiz: Pearson Edexcel Further Maths, Unit 40

20 questions

Written by Revision Ninja · Last reviewed 11 October 2026

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Lesson 4D.1.1-4D.1.2, Initial solutions and the stepping-stone method: 20 multiple choice questions for the Pearson Edexcel Further Maths (9FM0), Unit 40: Transportation problems, written with Revision Ninja.

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The 20 questions

  1. What is the north-west corner method used for in a transportation problem?

    • Finding an initial basic feasible solution.
    • Finding the maximum flow through a network.
    • Allocating each job to exactly one worker at minimum cost.
    • Finding the optimal solution directly without any iteration.

  2. In the north-west corner method, which cell is considered first?

    • The lower right-hand cell.
    • The cell with the lowest cost.
    • The upper left-hand cell.
    • The cell in the last row of the largest column.

  3. In the north-west corner method, what limits the amount sent along a route?

    • The cost of the cell in pounds.
    • The number of occupied cells already in the table.
    • The smaller of the remaining supply and remaining demand for that cell.
    • The larger of the remaining supply and remaining demand.

  4. A transportation problem has m rows and n columns. When is its solution degenerate?

    • When the total cost is zero.
    • When the number of occupied cells is greater than m + n - 1.
    • When m equals n.
    • When the number of occupied cells is less than m + n - 1.

  5. How many occupied cells does a non-degenerate basic solution of a 3 by 4 transportation problem need?

    • 5
    • 7
    • 12
    • 6

  6. In the stepping-stone method, what does the improvement index I_ij of an unoccupied cell represent?

    • The number of units that can be sent into the cell.
    • The shadow cost of the row only.
    • C_ij - R_i - K_j, the change in total cost per unit sent into that cell.
    • C_ij + R_i + K_j, the total cost of the route.

  7. In the shadow cost calculation, R1 is set to zero. Why is this allowed?

    • Row one always carries zero supply in an optimal solution.
    • It ensures every improvement index is positive.
    • It is an arbitrary choice; any starting value gives the same improvement indices.
    • It makes all column shadow costs equal to the costs.

  8. The north-west corner solution for supplies A = 20, B = 30, demands P = 10, Q = 25, R = 15, with costs A: 4, 6, 8 and B: 5, 3, 7, allocates AP = 10, AQ = 10, BQ = 15 and BR = 15. What is its total cost?

    • 260
    • 230
    • 240
    • 250

  9. How many cells are occupied in that north-west corner solution, and is it degenerate?

    • Three, so it is degenerate.
    • Five, so it is not degenerate.
    • Four, so it is degenerate.
    • Four, so it is not degenerate.

  10. For the same north-west corner solution with R_A = 0, what is the shadow cost R_B for row B?

    • 5
    • 3
    • -5
    • -3

  11. In the same solution, what is the improvement index of unoccupied cell A-R (cost 8)?

    • -2
    • 0
    • -10
    • 2

  12. Which cell should enter the solution next in that solution, if the aim is to reduce cost?

    • Row B, column Q
    • Row A, column R
    • Row A, column P
    • Row B, column P

  13. Using the stepping-stone loop for cell A-R, what is the largest amount θ that can be moved around the loop?

    • 15
    • 20
    • 10
    • 5

  14. After one improvement from A-R with θ = 10, which cell leaves the solution?

    • Row A, column P
    • Row A, column Q
    • Row B, column Q
    • Row B, column R

  15. What is the total cost after the improvement from A-R with θ = 10?

    • 230
    • 260
    • 240
    • 220

  16. A 3 by 3 problem has 4 occupied cells in a basic solution. Is it degenerate, and what must be done?

    • It is not degenerate because 4 is even.
    • It is not degenerate, so no change is needed.
    • It is degenerate; a zero allocation is placed in an unoccupied cell so the solution has 5 occupied cells.
    • It is degenerate; one occupied cell must be removed to leave 3.

  17. Why must the stepping-stone loop alternate between plus and minus cells?

    • So that all costs in the loop are reduced to zero.
    • So that row and column totals stay unchanged after moving θ units.
    • So that the improvement index equals the value of θ.
    • So that the loop closes on a single occupied cell only.

  18. In a minimising transportation problem, which unoccupied cell should enter the solution?

    • The cell with the most negative improvement index.
    • The cell with the largest positive improvement index.
    • The cell whose index is closest to zero but not zero.
    • The cell with the smallest positive improvement index.

  19. After the single improvement (AP = 10, AR = 10, BQ = 25, BR = 5), what is the improvement index of cell A-Q?

    • 0
    • -2
    • 4
    • 2

  20. Is the improved solution (AP = 10, AR = 10, BQ = 25, BR = 5) optimal for minimisation?

    • No, because cell A-Q has a negative index.
    • Yes, because all its improvement indices are non-negative.
    • No, because the solution is degenerate and needs a dummy.
    • No, because cell B-P has an index of -2.

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