Further Mathematics · Pearson Edexcel 9FM0 · Option D2 · Topic 1

Transportation problems

Transportation problems: moving goods from sources to destinations at minimum cost.

  • 3 sub-topics
  • Taught in Year 13
  • 47 practice questions
  • 3 animated lessons
Practice questions (47) Lessons Worksheets All Edexcel Further Mathematics topics

What the specification says

Every sub-topic Pearson Edexcel can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 1.1Transportation: initial solutionY13
  2. 1.2Transportation: improving the solutionY13
  3. 1.3Transportation as a linear programY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 47 questions in this topic, free, no account needed.

Transportation problems

Not needed for Year 12Easy

Decision variables in a transportation problem · Year 13 · 1 mark

  1. A transportation problem has 3 supply points and 3 demand points. How many decision variables does its linear programming formulation have?
    (1)

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  1. There is one variable for each route: the number of units sent from each supply point to each demand point.BB: 9

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Transportation problems

Not needed for Year 12Medium

Transportation: an initial solution · Year 13 · 6 marks

Three warehouses AA, BB and CC supply three shops XX, YY and ZZ. The table shows the cost in £ of moving one unit, with the supplies and demands. XYZsupplyA158629B612724C741335demand182248\begin{array}{c|ccc|c} & \text{X} & \text{Y} & \text{Z} & \text{supply} \\ \hline \text{A} & 15 & 8 & 6 & 29 \\ \text{B} & 6 & 12 & 7 & 24 \\ \text{C} & 7 & 4 & 13 & 35 \\ \hline \text{demand} & 18 & 22 & 48 & \end{array}
  1. (a)
    Explain why the problem is balanced.
    (1)
  2. (b)
    Use the north-west corner method to find an initial solution, and state its cost.
    (4)
  3. (c)
    Explain whether the solution is degenerate.
    (1)
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Part (a)1 mark

  1. Total supply = total demand = 88.

Part (b)4 marks

  1. Start top left, use up each supply or demand in turn.AXAX=18, AY{}= 18, \ AY=11, BY{}= 11, \ BY=11, BZ{}= 11, \ BZ=13, CZ{}= 13, \ CZ=35{}= 35
  2. 18×1518 \times 15+11×8{}+ 11 \times 8+11×12{}+ 11 \times 12+13×7{}+ 13 \times 7+35×13{}+ 35 \times 13=1036{}= 1036

Part (c)1 mark

  1. 5 cells are used; a non-degenerate solution needs 3+3−1=53 + 3 - 1 = 5. It is not degenerate.

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Transportation problems

Not needed for Year 12Hard

Transportation: improving the solution · Year 13 · 6 marks

The table shows a transportation problem. XYZsupplyA931344B371536C1071039demand233858\begin{array}{c|ccc|c} & \text{X} & \text{Y} & \text{Z} & \text{supply} \\ \hline \text{A} & 9 & 3 & 13 & 44 \\ \text{B} & 3 & 7 & 15 & 36 \\ \text{C} & 10 & 7 & 10 & 39 \\ \hline \text{demand} & 23 & 38 & 58 & \end{array} The north-west corner method gives the initial solution AXAX=23, AY{}= 23, \ AY=21, BY{}= 21, \ BY=17, BZ{}= 17, \ BZ=19, CZ{}= 19, \ CZ=39{}= 39.
  1. (a)
    Find the shadow costs, taking the shadow cost of AA as 0.
    (3)
  2. (b)
    Find the improvement indices of the unused cells, and state whether the solution is optimal.
    (3)
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Part (a)3 marks

  1. Ri+KjR_i + K_j=Cij for each used cell{}= C_{ij} \text{ for each used cell}
  2. RA=0, RB=4, RC=−1R_A = 0, \ R_B = 4, \ R_C = -1
  3. KX=9, KY=3, KZ=11K_X = 9, \ K_Y = 3, \ K_Z = 11

Part (b)3 marks

  1. I=C−R−KI = C - R - K.IAZI_{AZ}=2, IBX{}= 2, \ I_{BX}=−10, ICX{}= -10, \ I_{CX}=2, ICY{}= 2, \ I_{CY}=5{}= 5
  2. Not optimal: the most negative index is in cell BX, which should enter.

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