Hard Further Maths Transportation problems Questions

Challenging, exam-style Further Maths Transportation problems questions with worked solutions. Stretch yourself on the hardest transportation, improvement-index, optimality, stepping-stone problems.

transportationimprovement-indexoptimalitystepping-stonecostshadow-costs
Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
The table below shows the cost (in pounds) of transporting one unit from each supply point SiS_i to each demand point DjD_j, together with the supplies and demands. D1D2D3SupplyS13487S283510S33375Demand4711\begin{array}{c|ccc|c} & D_1 & D_2 & D_3 & \text{Supply} \\ \hline S_1 & 3 & 4 & 8 & 7 \\ S_2 & 8 & 3 & 5 & 10 \\ S_3 & 3 & 3 & 7 & 5 \\ \hline \text{Demand} & 4 & 7 & 11 & \end{array} Starting from the north-west corner solution, perform one stepping-stone iteration and find the resulting total cost.
Show worked solution

Worked solution

  1. Apply the north-west corner rule

    allocate min⁡(si,dj) starting at the top-left cell\text{allocate } \min(s_i, d_j) \text{ starting at the top-left cell}

    Each step fills the current top-left cell and crosses out an exhausted supply or demand.

  2. Record the allocations in the order they are made

    (S1,D1)→4, (S1,D2)→3, (S2,D2)→4, (S2,D3)→6, (S3,D3)→5(S_1, D_1) \to 4,\ (S_1, D_2) \to 3,\ (S_2, D_2) \to 4,\ (S_2, D_3) \to 6,\ (S_3, D_3) \to 5

    Moving right when a demand remains and down when a supply remains gives the initial solution.

  3. Write out the initial allocation

    x11=4x_{11} = 4, x12=3x_{12} = 3, x22=4x_{22} = 4, x23=6x_{23} = 6, x33=5x_{33} = 5

    These are the occupied cells of the north-west corner basic feasible solution.

  4. Set up the shadow-cost equations on the occupied cells

    u1+v1=3,u1+v2=4,u2+v2=3,u2+v3=5,u3+v3=7u_1 + v_1 = 3,\quad u_1 + v_2 = 4,\quad u_2 + v_2 = 3,\quad u_2 + v_3 = 5,\quad u_3 + v_3 = 7

    Each occupied cell gives one equation ui+vj=ciju_i + v_j = c_{ij}.

  5. Solve the equations taking u1=0u_1 = 0

    u1=0,u2=−1,u3=1,v1=3,v2=4,v3=6u_1 = 0,\quad u_2 = -1,\quad u_3 = 1,\quad v_1 = 3,\quad v_2 = 4,\quad v_3 = 6

    With u1=0u_1 = 0 the remaining potentials follow one occupied cell at a time.

  6. Compute the improvement index of every unoccupied cell

    I13=8−(0)−(6)=2,I21=8−(−1)−(3)=6,I31=3−(1)−(3)=−1,I32=3−(1)−(4)=−2I_{13} = 8 - (0) - (6) = 2,\quad I_{21} = 8 - (-1) - (3) = 6,\quad I_{31} = 3 - (1) - (3) = -1,\quad I_{32} = 3 - (1) - (4) = -2

    The improvement index is Iij=cij−ui−vjI_{ij} = c_{ij} - u_i - v_j.

  7. Shift θ\theta units around the stepping-stone loop

    θ=4 at entering cell (S3,D2)\theta = 4 \text{ at entering cell } (S_3, D_2)

    The entering cell gains θ\theta units and the minus cells lose them.

  8. Cost the new allocation

    new cost=101+(4)(−2)=93\text{new cost} = 101 + (4)(-2) = 93

    The cost changes by θ\theta times the entering improvement index.

  9. Recall what a transportation problem asks

    min⁡∑i∑jcijxij\min \sum_i \sum_j c_{ij} x_{ij}

    The aim is to meet every demand from the available supplies at least total cost.

  10. Recall the balance condition

    ∑isi=∑jdj\sum_i s_i = \sum_j d_j

    A transportation problem can only be solved once total supply equals total demand.

  11. Recall how a dummy restores balance

    dummy supply or demand=∣∑isi−∑jdj∣\text{dummy supply or demand} = \left|\sum_i s_i - \sum_j d_j\right|

    A dummy row or column with zero costs absorbs the surplus, leaving a balanced problem.

  12. Recall the number of occupied cells needed

    occupied cells=m+n−1\text{occupied cells} = m + n - 1

    A non-degenerate basic feasible solution of an m×nm \times n problem fills exactly m+n−1m+n-1 cells.

  13. Recall the north-west corner rule

    start at the top-left cell and allocate as much as possible\text{start at the top-left cell and allocate as much as possible}

    Each allocation exhausts a supply or a demand, and we move right or down accordingly.

  14. Recall the shadow-cost equations

    ui+vj=cij on every occupied cellu_i + v_j = c_{ij} \text{ on every occupied cell}

    Fixing u1=0u_1 = 0 makes the potentials uiu_i and vjv_j uniquely determined.

  15. Recall the improvement index

    Iij=cij−ui−vjI_{ij} = c_{ij} - u_i - v_j

    The improvement index measures the change in cost from sending one unit through an empty cell.

  16. State the new total cost

    new cost=93\text{new cost} = 93

    This is the cost after one stepping-stone iteration.

Answer
9393
Question 2
9 markschallenging
The table below shows the cost (in pounds) of transporting one unit from each supply point SiS_i to each demand point DjD_j, together with the supplies and demands. D1D2D3SupplyS18336S26694S33857Demand3131\begin{array}{c|ccc|c} & D_1 & D_2 & D_3 & \text{Supply} \\ \hline S_1 & 8 & 3 & 3 & 6 \\ S_2 & 6 & 6 & 9 & 4 \\ S_3 & 3 & 8 & 5 & 7 \\ \hline \text{Demand} & 3 & 13 & 1 & \end{array} Use the north-west corner method and the stepping-stone method to find the minimum total transportation cost.
Show worked solution

Worked solution

  1. Apply the north-west corner rule

    allocate min⁡(si,dj) starting at the top-left cell\text{allocate } \min(s_i, d_j) \text{ starting at the top-left cell}

    Each step fills the current top-left cell and crosses out an exhausted supply or demand.

  2. Record the allocations in the order they are made

    (S1,D1)→3, (S1,D2)→3, (S2,D2)→4, (S3,D2)→6, (S3,D3)→1(S_1, D_1) \to 3,\ (S_1, D_2) \to 3,\ (S_2, D_2) \to 4,\ (S_3, D_2) \to 6,\ (S_3, D_3) \to 1

    Moving right when a demand remains and down when a supply remains gives the initial solution.

  3. Write out the initial allocation

    x11=3x_{11} = 3, x12=3x_{12} = 3, x22=4x_{22} = 4, x32=6x_{32} = 6, x33=1x_{33} = 1

    These are the occupied cells of the north-west corner basic feasible solution.

  4. Set up the shadow-cost equations on the occupied cells

    u1+v1=8,u1+v2=3,u2+v2=6,u3+v2=8,u3+v3=5u_1 + v_1 = 8,\quad u_1 + v_2 = 3,\quad u_2 + v_2 = 6,\quad u_3 + v_2 = 8,\quad u_3 + v_3 = 5

    Each occupied cell gives one equation ui+vj=ciju_i + v_j = c_{ij}.

  5. Solve the equations taking u1=0u_1 = 0

    u1=0,u2=3,u3=5,v1=8,v2=3,v3=0u_1 = 0,\quad u_2 = 3,\quad u_3 = 5,\quad v_1 = 8,\quad v_2 = 3,\quad v_3 = 0

    With u1=0u_1 = 0 the remaining potentials follow one occupied cell at a time.

  6. Compute the improvement index of every unoccupied cell

    I13=3−(0)−(0)=3,I21=6−(3)−(8)=−5,I23=9−(3)−(0)=6,I31=3−(5)−(8)=−10I_{13} = 3 - (0) - (0) = 3,\quad I_{21} = 6 - (3) - (8) = -5,\quad I_{23} = 9 - (3) - (0) = 6,\quad I_{31} = 3 - (5) - (8) = -10

    The improvement index is Iij=cij−ui−vjI_{ij} = c_{ij} - u_i - v_j.

  7. Move to a better solution through the entering cell

    θ=3 at (S3,D1),I31=−10\theta = 3 \text{ at } (S_3, D_1), \quad I_{31} = -10

    The stepping-stone cycle reduces the cost by θ∣Iij∣\theta |I_{ij}|.

  8. Cost the improved allocation

    110+(3)(−10)=80110 + (3)(-10) = 80

    One iteration is enough here to reach the optimum.

  9. Confirm optimality of the new solution

    every improvement index of the new solution≥0\text{every improvement index of the new solution} \ge 0

    No empty cell can reduce the cost further, so this allocation is optimal.

  10. Recall what a transportation problem asks

    min⁡∑i∑jcijxij\min \sum_i \sum_j c_{ij} x_{ij}

    The aim is to meet every demand from the available supplies at least total cost.

  11. Recall the balance condition

    ∑isi=∑jdj\sum_i s_i = \sum_j d_j

    A transportation problem can only be solved once total supply equals total demand.

  12. Recall how a dummy restores balance

    dummy supply or demand=∣∑isi−∑jdj∣\text{dummy supply or demand} = \left|\sum_i s_i - \sum_j d_j\right|

    A dummy row or column with zero costs absorbs the surplus, leaving a balanced problem.

  13. Recall the number of occupied cells needed

    occupied cells=m+n−1\text{occupied cells} = m + n - 1

    A non-degenerate basic feasible solution of an m×nm \times n problem fills exactly m+n−1m+n-1 cells.

  14. Recall the north-west corner rule

    start at the top-left cell and allocate as much as possible\text{start at the top-left cell and allocate as much as possible}

    Each allocation exhausts a supply or a demand, and we move right or down accordingly.

  15. State the minimum total cost

    Cmin⁡=80C_{\min} = 80

    This is the least possible total transportation cost.

Answer
8080
Question 3
9 markschallenging
The table below shows the cost (in pounds) of transporting one unit from each supply point SiS_i to each demand point DjD_j, together with the supplies and demands. D1D2D3SupplyS14978S22475S33984Demand1115\begin{array}{c|ccc|c} & D_1 & D_2 & D_3 & \text{Supply} \\ \hline S_1 & 4 & 9 & 7 & 8 \\ S_2 & 2 & 4 & 7 & 5 \\ S_3 & 3 & 9 & 8 & 4 \\ \hline \text{Demand} & 1 & 11 & 5 & \end{array} Find the north-west corner solution, compute its improvement indices with u1=0u_1 = 0, and state whether it is optimal.
Show worked solution

Worked solution

  1. Apply the north-west corner rule

    allocate min⁡(si,dj) starting at the top-left cell\text{allocate } \min(s_i, d_j) \text{ starting at the top-left cell}

    Each step fills the current top-left cell and crosses out an exhausted supply or demand.

  2. Record the allocations in the order they are made

    (S1,D1)→1, (S1,D2)→7, (S2,D2)→4, (S2,D3)→1, (S3,D3)→4(S_1, D_1) \to 1,\ (S_1, D_2) \to 7,\ (S_2, D_2) \to 4,\ (S_2, D_3) \to 1,\ (S_3, D_3) \to 4

    Moving right when a demand remains and down when a supply remains gives the initial solution.

  3. Write out the initial allocation

    x11=1x_{11} = 1, x12=7x_{12} = 7, x22=4x_{22} = 4, x23=1x_{23} = 1, x33=4x_{33} = 4

    These are the occupied cells of the north-west corner basic feasible solution.

  4. Set up the shadow-cost equations on the occupied cells

    u1+v1=4,u1+v2=9,u2+v2=4,u2+v3=7,u3+v3=8u_1 + v_1 = 4,\quad u_1 + v_2 = 9,\quad u_2 + v_2 = 4,\quad u_2 + v_3 = 7,\quad u_3 + v_3 = 8

    Each occupied cell gives one equation ui+vj=ciju_i + v_j = c_{ij}.

  5. Solve the equations taking u1=0u_1 = 0

    u1=0,u2=−5,u3=−4,v1=4,v2=9,v3=12u_1 = 0,\quad u_2 = -5,\quad u_3 = -4,\quad v_1 = 4,\quad v_2 = 9,\quad v_3 = 12

    With u1=0u_1 = 0 the remaining potentials follow one occupied cell at a time.

  6. Compute the improvement index of every unoccupied cell

    I13=7−(0)−(12)=−5,I21=2−(−5)−(4)=3,I31=3−(−4)−(4)=3,I32=9−(−4)−(9)=4I_{13} = 7 - (0) - (12) = -5,\quad I_{21} = 2 - (-5) - (4) = 3,\quad I_{31} = 3 - (-4) - (4) = 3,\quad I_{32} = 9 - (-4) - (9) = 4

    The improvement index is Iij=cij−ui−vjI_{ij} = c_{ij} - u_i - v_j.

  7. Recall what a transportation problem asks

    min⁡∑i∑jcijxij\min \sum_i \sum_j c_{ij} x_{ij}

    The aim is to meet every demand from the available supplies at least total cost.

  8. Recall the balance condition

    ∑isi=∑jdj\sum_i s_i = \sum_j d_j

    A transportation problem can only be solved once total supply equals total demand.

  9. Recall how a dummy restores balance

    dummy supply or demand=∣∑isi−∑jdj∣\text{dummy supply or demand} = \left|\sum_i s_i - \sum_j d_j\right|

    A dummy row or column with zero costs absorbs the surplus, leaving a balanced problem.

  10. Recall the number of occupied cells needed

    occupied cells=m+n−1\text{occupied cells} = m + n - 1

    A non-degenerate basic feasible solution of an m×nm \times n problem fills exactly m+n−1m+n-1 cells.

  11. Recall the north-west corner rule

    start at the top-left cell and allocate as much as possible\text{start at the top-left cell and allocate as much as possible}

    Each allocation exhausts a supply or a demand, and we move right or down accordingly.

  12. Recall the shadow-cost equations

    ui+vj=cij on every occupied cellu_i + v_j = c_{ij} \text{ on every occupied cell}

    Fixing u1=0u_1 = 0 makes the potentials uiu_i and vjv_j uniquely determined.

  13. Recall the improvement index

    Iij=cij−ui−vjI_{ij} = c_{ij} - u_i - v_j

    The improvement index measures the change in cost from sending one unit through an empty cell.

  14. Recall the optimality test

    optimal  ⟺  Iij≥0 for every unoccupied cell\text{optimal} \iff I_{ij} \ge 0 \text{ for every unoccupied cell}

    If any improvement index is negative, cost can still be reduced.

  15. Recall the stepping-stone cycle

    +  −  +  −  around a closed loop+ \; - \; + \; - \; \text{around a closed loop}

    The loop starts at the entering cell and turns only at occupied cells.

  16. Recall how much is shifted round the loop

    θ=min⁡{allocations at the minus cells}\theta = \min\{\text{allocations at the minus cells}\}

    Shifting θ\theta keeps every allocation non-negative and empties one cell.

  17. State whether the solution is optimal

    not optimal: I13=−5<0\text{not optimal: } I_{13} = -5 < 0

    The improvement indices alone decide whether a solution is optimal.

Answer
not optimal: I13=−5<0\text{not optimal: } I_{13} = -5 < 0
Question 4
9 markschallenging
The table below shows the cost (in pounds) of transporting one unit from each supply point SiS_i to each demand point DjD_j, together with the supplies and demands. D1D2D3SupplyS18339S22337S35954Demand1712\begin{array}{c|ccc|c} & D_1 & D_2 & D_3 & \text{Supply} \\ \hline S_1 & 8 & 3 & 3 & 9 \\ S_2 & 2 & 3 & 3 & 7 \\ S_3 & 5 & 9 & 5 & 4 \\ \hline \text{Demand} & 1 & 7 & 12 & \end{array} Find the north-west corner solution and its improvement indices (with u1=0u_1 = 0). Which empty cell should enter the solution next?
Show worked solution

Worked solution

  1. Apply the north-west corner rule

    allocate min⁡(si,dj) starting at the top-left cell\text{allocate } \min(s_i, d_j) \text{ starting at the top-left cell}

    Each step fills the current top-left cell and crosses out an exhausted supply or demand.

  2. Record the allocations in the order they are made

    (S1,D1)→1, (S1,D2)→7, (S1,D3)→1, (S2,D3)→7, (S3,D3)→4(S_1, D_1) \to 1,\ (S_1, D_2) \to 7,\ (S_1, D_3) \to 1,\ (S_2, D_3) \to 7,\ (S_3, D_3) \to 4

    Moving right when a demand remains and down when a supply remains gives the initial solution.

  3. Write out the initial allocation

    x11=1x_{11} = 1, x12=7x_{12} = 7, x13=1x_{13} = 1, x23=7x_{23} = 7, x33=4x_{33} = 4

    These are the occupied cells of the north-west corner basic feasible solution.

  4. Set up the shadow-cost equations on the occupied cells

    u1+v1=8,u1+v2=3,u1+v3=3,u2+v3=3,u3+v3=5u_1 + v_1 = 8,\quad u_1 + v_2 = 3,\quad u_1 + v_3 = 3,\quad u_2 + v_3 = 3,\quad u_3 + v_3 = 5

    Each occupied cell gives one equation ui+vj=ciju_i + v_j = c_{ij}.

  5. Solve the equations taking u1=0u_1 = 0

    u1=0,u2=0,u3=2,v1=8,v2=3,v3=3u_1 = 0,\quad u_2 = 0,\quad u_3 = 2,\quad v_1 = 8,\quad v_2 = 3,\quad v_3 = 3

    With u1=0u_1 = 0 the remaining potentials follow one occupied cell at a time.

  6. Compute the improvement index of every unoccupied cell

    I21=2−(0)−(8)=−6,I22=3−(0)−(3)=0,I31=5−(2)−(8)=−5,I32=9−(2)−(3)=4I_{21} = 2 - (0) - (8) = -6,\quad I_{22} = 3 - (0) - (3) = 0,\quad I_{31} = 5 - (2) - (8) = -5,\quad I_{32} = 9 - (2) - (3) = 4

    The improvement index is Iij=cij−ui−vjI_{ij} = c_{ij} - u_i - v_j.

  7. Pick the most negative improvement index

    min⁡ijIij=−6 at (S2,D1)\min_{ij} I_{ij} = -6 \text{ at } (S_2, D_1)

    The empty cell with the most negative index enters the solution.

  8. Recall what a transportation problem asks

    min⁡∑i∑jcijxij\min \sum_i \sum_j c_{ij} x_{ij}

    The aim is to meet every demand from the available supplies at least total cost.

  9. Recall the balance condition

    ∑isi=∑jdj\sum_i s_i = \sum_j d_j

    A transportation problem can only be solved once total supply equals total demand.

  10. Recall how a dummy restores balance

    dummy supply or demand=∣∑isi−∑jdj∣\text{dummy supply or demand} = \left|\sum_i s_i - \sum_j d_j\right|

    A dummy row or column with zero costs absorbs the surplus, leaving a balanced problem.

  11. Recall the number of occupied cells needed

    occupied cells=m+n−1\text{occupied cells} = m + n - 1

    A non-degenerate basic feasible solution of an m×nm \times n problem fills exactly m+n−1m+n-1 cells.

  12. Recall the north-west corner rule

    start at the top-left cell and allocate as much as possible\text{start at the top-left cell and allocate as much as possible}

    Each allocation exhausts a supply or a demand, and we move right or down accordingly.

  13. Recall the shadow-cost equations

    ui+vj=cij on every occupied cellu_i + v_j = c_{ij} \text{ on every occupied cell}

    Fixing u1=0u_1 = 0 makes the potentials uiu_i and vjv_j uniquely determined.

  14. Recall the improvement index

    Iij=cij−ui−vjI_{ij} = c_{ij} - u_i - v_j

    The improvement index measures the change in cost from sending one unit through an empty cell.

  15. Recall the optimality test

    optimal  ⟺  Iij≥0 for every unoccupied cell\text{optimal} \iff I_{ij} \ge 0 \text{ for every unoccupied cell}

    If any improvement index is negative, cost can still be reduced.

  16. Select the entering cell

    entering cell=(S2,D1)\text{entering cell} = (S_2, D_1)

    This cell reduces the total cost fastest per unit re-routed.

Answer
entering cell=(S2,D1)\text{entering cell} = (S_2, D_1)
Question 5
9 markschallenging
The table below shows the cost (in pounds) of transporting one unit from each supply point SiS_i to each demand point DjD_j, together with the supplies and demands. D1D2D3SupplyS12487S276710S37484Demand1911\begin{array}{c|ccc|c} & D_1 & D_2 & D_3 & \text{Supply} \\ \hline S_1 & 2 & 4 & 8 & 7 \\ S_2 & 7 & 6 & 7 & 10 \\ S_3 & 7 & 4 & 8 & 4 \\ \hline \text{Demand} & 1 & 9 & 11 & \end{array} Use the north-west corner method and the stepping-stone method to find the minimum total transportation cost.
Show worked solution

Worked solution

  1. Apply the north-west corner rule

    allocate min⁡(si,dj) starting at the top-left cell\text{allocate } \min(s_i, d_j) \text{ starting at the top-left cell}

    Each step fills the current top-left cell and crosses out an exhausted supply or demand.

  2. Record the allocations in the order they are made

    (S1,D1)→1, (S1,D2)→6, (S2,D2)→3, (S2,D3)→7, (S3,D3)→4(S_1, D_1) \to 1,\ (S_1, D_2) \to 6,\ (S_2, D_2) \to 3,\ (S_2, D_3) \to 7,\ (S_3, D_3) \to 4

    Moving right when a demand remains and down when a supply remains gives the initial solution.

  3. Write out the initial allocation

    x11=1x_{11} = 1, x12=6x_{12} = 6, x22=3x_{22} = 3, x23=7x_{23} = 7, x33=4x_{33} = 4

    These are the occupied cells of the north-west corner basic feasible solution.

  4. Set up the shadow-cost equations on the occupied cells

    u1+v1=2,u1+v2=4,u2+v2=6,u2+v3=7,u3+v3=8u_1 + v_1 = 2,\quad u_1 + v_2 = 4,\quad u_2 + v_2 = 6,\quad u_2 + v_3 = 7,\quad u_3 + v_3 = 8

    Each occupied cell gives one equation ui+vj=ciju_i + v_j = c_{ij}.

  5. Solve the equations taking u1=0u_1 = 0

    u1=0,u2=2,u3=3,v1=2,v2=4,v3=5u_1 = 0,\quad u_2 = 2,\quad u_3 = 3,\quad v_1 = 2,\quad v_2 = 4,\quad v_3 = 5

    With u1=0u_1 = 0 the remaining potentials follow one occupied cell at a time.

  6. Compute the improvement index of every unoccupied cell

    I13=8−(0)−(5)=3,I21=7−(2)−(2)=3,I31=7−(3)−(2)=2,I32=4−(3)−(4)=−3I_{13} = 8 - (0) - (5) = 3,\quad I_{21} = 7 - (2) - (2) = 3,\quad I_{31} = 7 - (3) - (2) = 2,\quad I_{32} = 4 - (3) - (4) = -3

    The improvement index is Iij=cij−ui−vjI_{ij} = c_{ij} - u_i - v_j.

  7. Move to a better solution through the entering cell

    θ=3 at (S3,D2),I32=−3\theta = 3 \text{ at } (S_3, D_2), \quad I_{32} = -3

    The stepping-stone cycle reduces the cost by θ∣Iij∣\theta |I_{ij}|.

  8. Cost the improved allocation

    125+(3)(−3)=116125 + (3)(-3) = 116

    One iteration is enough here to reach the optimum.

  9. Confirm optimality of the new solution

    every improvement index of the new solution≥0\text{every improvement index of the new solution} \ge 0

    No empty cell can reduce the cost further, so this allocation is optimal.

  10. Recall what a transportation problem asks

    min⁡∑i∑jcijxij\min \sum_i \sum_j c_{ij} x_{ij}

    The aim is to meet every demand from the available supplies at least total cost.

  11. Recall the balance condition

    ∑isi=∑jdj\sum_i s_i = \sum_j d_j

    A transportation problem can only be solved once total supply equals total demand.

  12. Recall how a dummy restores balance

    dummy supply or demand=∣∑isi−∑jdj∣\text{dummy supply or demand} = \left|\sum_i s_i - \sum_j d_j\right|

    A dummy row or column with zero costs absorbs the surplus, leaving a balanced problem.

  13. Recall the number of occupied cells needed

    occupied cells=m+n−1\text{occupied cells} = m + n - 1

    A non-degenerate basic feasible solution of an m×nm \times n problem fills exactly m+n−1m+n-1 cells.

  14. Recall the north-west corner rule

    start at the top-left cell and allocate as much as possible\text{start at the top-left cell and allocate as much as possible}

    Each allocation exhausts a supply or a demand, and we move right or down accordingly.

  15. State the minimum total cost

    Cmin⁡=116C_{\min} = 116

    This is the least possible total transportation cost.

Answer
116116

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