Consider the following transportation problem :DestinationsSourcesD1D2D3D4S1641514S2892716S343625610154The initial transportation cost by Vogel's method is :

Vogel's method gives an initial cost of 114.Supply 14 + 16 + 5 = 35 equals demand 6 + 10 + 15 + 4 = 35, so the problem is balanced. At each step, penalty = difference of the two lowest costs in a row or column; allocate to the cheapest cell of the line with the largest penalty.Largest penalty 5 (row S2): cell S2-D3, cost 2, allocate 15; D3 done, S2 has 1 left.Largest penalty 3 (column D4): cell S3-D4, cost 2, allocate 4; D4 done, S3 has 1 left.Largest penalty 2 (row S1, tie with column D1): cell S1-D2, cost 4, allocate 10; D2 done, S1 has 4 left.Only D1 remains (demand 6): S1 gives 4 at cost 6, S2 gives 1 at cost 8, S3 gives 1 at cost 4.\[ \begin{aligned} \text{Cost} &= 15(2) + 4(2) + 10(4) \\ &\quad + 4(6) + 1(8) + 1(4) \\ &= 30 + 8 + 40 + 24 + 8 + 4 = 114 \end{aligned} \]Breaking the tie in step 3 by column D1 instead leads to the same total. 124, 128 and 152 are what cruder starting rules such as the north-west corner method give.

Transportation Problems Questions and Answers

Q: For a transportation problem, let \(c_{ij}\) denote the unit cost of the cell \((i, j)\). Assume that \(c_{11} = 10\), \(c_{12} = 12\), \(c_{21} = 11\) and \(c_{22} = 13\). Let \(\alpha_i\) and \(\beta_j\), \(i, j = 1, 2, 3\), represent the simplex multipliers associated with any basis corresponding to the unit cost \(c_{ij}\). Assume that \(\alpha_1 = x\), \(\alpha_2 = x + 1\), \(\beta_1 = y\) and \(\beta_2 = y + 2\). The relative cost coefficient \(d_{ij}\) is the difference between the current solution and the new improved solution.If \(x = 4\), \(c_{13} = 19\) and \(\beta_3 = y + 5\), then which one of the following is TRUE?

The answer is \(y = 6\) and \(d_{13} = 4\).In the u-v (MODI) method the simplex multipliers satisfy \(\alpha_i + \beta_j = c_{ij}\) on the basic cells, and the relative cost of a non-basic cell is \(d_{ij} = c_{ij} - \alpha_i - \beta_j\). With \(x = 4\), cell (1,1) gives\[ \begin{aligned} \alpha_1 + \beta_1 &= 4 + y = 10 \\ y &= 6 \end{aligned} \]and the other three given cells agree: \(4 + 8 = 12\), \(5 + 6 = 11\), \(5 + 8 = 13\). Then \(\beta_3 = y + 5 = 11\), and\[ d_{13} = 19 - 4 - 11 = 4. \]\(d_{13} = 7\) comes from using \(\beta_2 = 8\) in place of \(\beta_3\), and \(y = 5\) fails the condition on cell (1,1). A positive \(d_{13}\) means bringing cell (1,3) into the basis would not lower the cost.

Q: Consider the following tansportation problem :F1F2F3F4W1193030107W2703040609W34087030185871434(Columns: Factory F1 to F4; rows: Warehouse W1 to W3.)Let Xij be the allocation in the (i, j)th cell. The solution (x11 = 5, x14 = 2, x22 = 2, x23 = 7, x32 = 6, x34 = 12) is :

The given allocation is the Optimal solution.It uses 6 cells, equal to m + n - 1 = 3 + 4 - 1, so it is a basic solution. Test it with the u-v (MODI) method, setting \(u_{1} = 0\) and using \(u_{i} + v_{j} = c_{ij}\) in the occupied cells:\[ \begin{aligned} v_{1} &= 19,\ v_{4} = 10,\ u_{3} = 20 \\ v_{2} &= -12,\ u_{2} = 42,\ v_{3} = -2 \end{aligned} \]The reduced costs \(c_{ij} - u_{i} - v_{j}\) of the empty cells are 42, 32, 9, 8, 1 and 52, all positive, so no reallocation can lower the cost and the solution is optimal (total cost 863). It is not the Vogel or least cost starting solution: both of those first fill cell (3, 2), the lowest-cost cell at 8, with all 8 units, not 6.

Q: The number of non-negative variables in a basic feasible solution to a \(m \times n\) transportation problem is :

A basic feasible solution has \(m + n - 1\) basic variables.A transportation problem with m sources and n destinations has \(m + n\) constraints (one per row total and one per column total). Because total supply equals total demand, one of these equations follows from the others, so only \(m + n - 1\) are independent, and a basic solution has that many basic variables, the allocated cells. A non-degenerate basic feasible solution has exactly \(m + n - 1\) positive allocations.mn is the total number of cells, not the number in a basis. \(m + n\) ignores the redundant constraint, and \(m + n + 1\) has no basis at all.

Q: Consider the transportation problemMin. \(z = \sum_{i=1}^{m}\sum_{j=1}^{n} c_{ij}x_{ij}\)Subject to :\(\sum_{j=1}^{n} x_{ij} = a_{i}, i = 1, 2, ....., m,\)\(\sum_{i=1}^{m} x_{ij} = b_{j}, j = 1, 2, ....., n\) and \(x_{ij} \ge 0\)The existence of a feasible solution of the transportation problem is possible :(I) if \(\sum_{i=1}^{m} a_{i} = \sum_{j=1}^{n} b_{j}\)(II) if and only if \(\sum_{i=1}^{m} a_{i} = \sum_{j=1}^{n} b_{j}\)Then which of the following is true ?

Both statements are true.If total supply equals total demand, a feasible solution always exists: for example \(x_{ij} = a_{i}b_{j}/T\) with \(T = \sum a_{i} = \sum b_{j}\) satisfies every row and column constraint and is non-negative. So (I) holds.Conversely, with equality constraints, summing all row constraints gives \(\sum\sum x_{ij} = \sum a_{i}\) and summing all column constraints gives \(\sum\sum x_{ij} = \sum b_{j}\), so a feasible solution forces \(\sum a_{i} = \sum b_{j}\). Hence the condition is necessary as well as sufficient, and (II) holds too.

Q: A dairy firm has three plants \(P_{1}, P_{2}, P_{3}\) located in a state. They daily milk production at each plant (in thousand liters) is \(P_{1} \to 6\), \(P_{2} \to 1\), \(P_{3} \to 10\). Each day, the firm must fulfil the needs of its four distribution centers \(D_{1}, D_{2}, D_{3}, D_{4}\). The minimum requirement at each center (in thousand liters) is \(D_{1} \to 7\), \(D_{2} \to 5\), \(D_{3} \to 3\), \(D_{4} \to 2\). Cost in hundred of rupees of shipping one thousand liters from each plant to each distribution center is given below :PlantD1D2D3D4P123117P21061P358159Then the initial basic feasible by North-West Corner method is :

The North-West Corner solution is degenerate, with cost ₹ 11,600.Supply is 6 + 1 + 10 = 17 and demand is 7 + 5 + 3 + 2 = 17, so the problem is balanced. Starting at the top-left cell:P1 to D1: 6 (P1 exhausted, D1 needs 1 more)P2 to D1: 1 (P2 exhausted and D1 satisfied at the same time)P3 to D2: 5, P3 to D3: 3, P3 to D4: 2\[ \begin{aligned} &6(2) + 1(1) + 5(8) + 3(15) + 2(9) \\ &= 12 + 1 + 40 + 45 + 18 = 116 \end{aligned} \]The cost is 116 hundred rupees, ₹ 11,600. A basic solution needs \(m + n - 1 = 3 + 4 - 1 = 6\) occupied cells, but only 5 are used, because a row and a column were satisfied together at the second step. So the solution is degenerate, and the options calling it non-degenerate or costing ₹ 12,000 are wrong.

Q: Consider the following transportation problem :WarehouseFactoryW1W2W3W4F112346F243208F30221104686A solution to the problem is given below :62646Which of the following is true about the solution ?

The solution is degenerate and optimal.Supply (6 + 8 + 10 = 24) equals demand (4 + 6 + 8 + 6 = 24). A basic solution of a 3 × 4 problem needs m + n - 1 = 6 occupied cells, but this one has only 5: (F1, W2) = 6, (F2, W3) = 2, (F2, W4) = 6, (F3, W1) = 4 and (F3, W3) = 6. So it is degenerate.To test optimality, place a zero allocation (epsilon) in cell (F1, W1) and solve \(u_i + v_j = c_{ij}\) on the six basic cells. Taking \(u_3 = 0\) gives \(v_1 = 0, v_3 = 2, u_2 = 0, v_4 = 0, u_1 = 1, v_2 = 1\).The reduced costs \(c_{ij} - u_i - v_j\) of the empty cells are then 0 (F1, W3), 3 (F1, W4), 4 (F2, W1), 2 (F2, W2), 1 (F3, W2) and 1 (F3, W4). None is negative, so the solution is optimal, with total cost 12 + 4 + 0 + 0 + 12 = 28. Optimality can be tested once the degeneracy is resolved, so the last option is wrong too.

Q: The number of basic variables in an \((m \times n)\) transportation table are :

There are \(m + n - 1\) basic variables.A transportation problem with \(m\) sources and \(n\) destinations has \(m + n\) supply and demand constraints, but because total supply equals total demand one of them is redundant. So only \(m + n - 1\) constraints are independent, and a basic feasible solution has \(m + n - 1\) basic variables (allocated cells).\(m + n\) ignores the redundancy, \(m + n + 1\) has no basis, and \(mn\) is the total number of cells, most of which are non-basic. A solution with fewer than \(m + n - 1\) positive allocations is called degenerate.

In the transportation problem, the stepping stone method is used to:

A.Optimize the initial feasible solution.
B.Resolve unbalanced transportation problems.
C.Identify degeneracy in the solution.
D.Determine the initial feasible solution.

Exam Year(s)

DSSSB - 2023,

Answer: A

The stepping stone method is used to optimise the initial feasible solution of a transportation problem. It checks the unused cells for an opportunity cost, by tracing a closed path of stones, and moves allocation to those cells if cost can be reduced. The initial solution itself is found by methods such as the north-west corner rule or the Vogel approximation method, so the fourth option is the wrong one, and unbalanced problems are handled by adding a dummy row or column.

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The following table gives the unit transportation costs, the supply at each origin and the demand of each destination for a transportation problem.

Origin / Destination	D₁	D₂	D₃	D₄	Supply
O₁	3	4	8	7	60
O₂	7	3	7	6	80
O₃	3	9	3	4	100
Demand	40	70	50	80

Let \(x_{ij}\) denote the number of units to be transported from origin \(i\) to destination \(j\). If the u-v method is applied to improve the basic feasible solution given by \(x_{12} = 60,\ x_{22} = 10,\ x_{23} = 50,\ x_{24} = 20,\ x_{31} = 40\) and \(x_{34} = 60\), then the variables entering and leaving the basis, respectively, are

A.\(x_{11}\) and \(x_{24}\)
B.\(x_{13}\) and \(x_{23}\)
C.\(x_{14}\) and \(x_{24}\)
D.\(x_{33}\) and \(x_{24}\)

Exam Year(s)

GATE Mathematics - 2012,

Answer: A

The entering variable is \(x_{11}\) and the leaving variable is \(x_{24}\): \(x_{11}\) and \(x_{24}\).

Find \(u_{i}, v_{j}\) from \(u_{i} + v_{j} = c_{ij}\) on the basic cells, with \(u_{1} = 0\): \(v_{2} = 4\), \(u_{2} = -1\), \(v_{3} = 8\), \(v_{4} = 7\), \(u_{3} = -3\), \(v_{1} = 6\).

Reduced costs \(c_{ij} - u_{i} - v_{j}\) on the non-basic cells:

\[ \begin{aligned} x_{11} &: 3 - 0 - 6 = -3 \\ x_{33} &: 3 + 3 - 8 = -2 \\ x_{13}, x_{14} &: 0 \\ x_{21} &: 2,\quad x_{32}: 8 \end{aligned} \]

The most negative is \(x_{11}\), so it enters. Its loop is \(x_{11}(+), x_{12}(-), x_{22}(+), x_{24}(-), x_{34}(+), x_{31}(-)\). The minus cells hold 60, 20 and 40, so \(\theta = 20\) and \(x_{24}\) leaves.

The options entering \(x_{13}\) or \(x_{14}\) are wrong because those cells have zero reduced cost and would not lower the cost, and \(x_{33}\), at -2, is less negative than \(x_{11}\), at -3.

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For a transportation problem, let \(c_{ij}\) denote the unit cost of the cell \((i, j)\). Assume that \(c_{11} = 10\), \(c_{12} = 12\), \(c_{21} = 11\) and \(c_{22} = 13\). Let \(\alpha_i\) and \(\beta_j\), \(i, j = 1, 2, 3\), represent the simplex multipliers associated with any basis corresponding to the unit cost \(c_{ij}\). Assume that \(\alpha_1 = x\), \(\alpha_2 = x + 1\), \(\beta_1 = y\) and \(\beta_2 = y + 2\). The relative cost coefficient \(d_{ij}\) is the difference between the current solution and the new improved solution.

If \(x = 4\), \(c_{13} = 19\) and \(\beta_3 = y + 5\), then which one of the following is TRUE?

A.\(y = 6\) and \(d_{13} = 7\)
B.\(y = 6\) and \(d_{13} = 4\)
C.\(y = 5\) and \(d_{13} = 7\)
D.\(y = 5\) and \(d_{13} = 4\)

Exam Year(s)

GATE Mathematics - 2026,

Answer: B

The answer is \(y = 6\) and \(d_{13} = 4\).

In the u-v (MODI) method the simplex multipliers satisfy \(\alpha_i + \beta_j = c_{ij}\) on the basic cells, and the relative cost of a non-basic cell is \(d_{ij} = c_{ij} - \alpha_i - \beta_j\). With \(x = 4\), cell (1,1) gives

\[ \begin{aligned} \alpha_1 + \beta_1 &= 4 + y = 10 \\ y &= 6 \end{aligned} \]

and the other three given cells agree: \(4 + 8 = 12\), \(5 + 6 = 11\), \(5 + 8 = 13\). Then \(\beta_3 = y + 5 = 11\), and

\[ d_{13} = 19 - 4 - 11 = 4. \]

\(d_{13} = 7\) comes from using \(\beta_2 = 8\) in place of \(\beta_3\), and \(y = 5\) fails the condition on cell (1,1). A positive \(d_{13}\) means bringing cell (1,3) into the basis would not lower the cost.

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Consider the following tansportation problem :

	F₁	F₂	F₃	F₄
W₁	19	30	30	10	7
W₂	70	30	40	60	9
W₃	40	8	70	30	18
	5	8	7	14	34

(Columns: Factory F₁ to F₄; rows: Warehouse W₁ to W₃.)

Let X_ij be the allocation in the (i, j)^th cell. The solution (x₁₁ = 5, x₁₄ = 2, x₂₂ = 2, x₂₃ = 7, x₃₂ = 6, x₃₄ = 12) is :

A.Initial solution by Vogel's method
B.Optimal solution
C.Initial solution by least cost method
D.Suboptimal solution which can be improved

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The given allocation is the Optimal solution.

It uses 6 cells, equal to m + n - 1 = 3 + 4 - 1, so it is a basic solution. Test it with the u-v (MODI) method, setting \(u_{1} = 0\) and using \(u_{i} + v_{j} = c_{ij}\) in the occupied cells:

\[ \begin{aligned} v_{1} &= 19,\ v_{4} = 10,\ u_{3} = 20 \\ v_{2} &= -12,\ u_{2} = 42,\ v_{3} = -2 \end{aligned} \]

The reduced costs \(c_{ij} - u_{i} - v_{j}\) of the empty cells are 42, 32, 9, 8, 1 and 52, all positive, so no reallocation can lower the cost and the solution is optimal (total cost 863). It is not the Vogel or least cost starting solution: both of those first fill cell (3, 2), the lowest-cost cell at 8, with all 8 units, not 6.

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The number of non-negative variables in a basic feasible solution to a \(m \times n\) transportation problem is :

A.\(m + n - 1\)
B.\(mn\)
C.\(m + n\)
D.\(m + n + 1\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

A basic feasible solution has \(m + n - 1\) basic variables.

A transportation problem with m sources and n destinations has \(m + n\) constraints (one per row total and one per column total). Because total supply equals total demand, one of these equations follows from the others, so only \(m + n - 1\) are independent, and a basic solution has that many basic variables, the allocated cells. A non-degenerate basic feasible solution has exactly \(m + n - 1\) positive allocations.

mn is the total number of cells, not the number in a basis. \(m + n\) ignores the redundant constraint, and \(m + n + 1\) has no basis at all.

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Consider the transportation problem

Min. \(z = \sum_{i=1}^{m}\sum_{j=1}^{n} c_{ij}x_{ij}\)

Subject to :
\(\sum_{j=1}^{n} x_{ij} = a_{i}, i = 1, 2, ....., m,\)
\(\sum_{i=1}^{m} x_{ij} = b_{j}, j = 1, 2, ....., n\) and \(x_{ij} \ge 0\)

The existence of a feasible solution of the transportation problem is possible :

(I) if \(\sum_{i=1}^{m} a_{i} = \sum_{j=1}^{n} b_{j}\)
(II) if and only if \(\sum_{i=1}^{m} a_{i} = \sum_{j=1}^{n} b_{j}\)

Then which of the following is true ?

A.Only (I) is true
B.Only (II) is true
C.Both are true
D.None of them is true

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

Both statements are true.

If total supply equals total demand, a feasible solution always exists: for example \(x_{ij} = a_{i}b_{j}/T\) with \(T = \sum a_{i} = \sum b_{j}\) satisfies every row and column constraint and is non-negative. So (I) holds.

Conversely, with equality constraints, summing all row constraints gives \(\sum\sum x_{ij} = \sum a_{i}\) and summing all column constraints gives \(\sum\sum x_{ij} = \sum b_{j}\), so a feasible solution forces \(\sum a_{i} = \sum b_{j}\). Hence the condition is necessary as well as sufficient, and (II) holds too.

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Consider the following transportation problem :

	Destinations
Sources	D₁	D₂	D₃	D₄
S₁	6	4	1	5	14
S₂	8	9	2	7	16
S₃	4	3	6	2	5
	6	10	15	4

The initial transportation cost by Vogel's method is :

A.128
B.152
C.124
D.114

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

Vogel's method gives an initial cost of 114.

Supply 14 + 16 + 5 = 35 equals demand 6 + 10 + 15 + 4 = 35, so the problem is balanced. At each step, penalty = difference of the two lowest costs in a row or column; allocate to the cheapest cell of the line with the largest penalty.

Largest penalty 5 (row S2): cell S2-D3, cost 2, allocate 15; D3 done, S2 has 1 left.
Largest penalty 3 (column D4): cell S3-D4, cost 2, allocate 4; D4 done, S3 has 1 left.
Largest penalty 2 (row S1, tie with column D1): cell S1-D2, cost 4, allocate 10; D2 done, S1 has 4 left.
Only D1 remains (demand 6): S1 gives 4 at cost 6, S2 gives 1 at cost 8, S3 gives 1 at cost 4.

\[ \begin{aligned} \text{Cost} &= 15(2) + 4(2) + 10(4) \\ &\quad + 4(6) + 1(8) + 1(4) \\ &= 30 + 8 + 40 + 24 + 8 + 4 = 114 \end{aligned} \]

Breaking the tie in step 3 by column D1 instead leads to the same total. 124, 128 and 152 are what cruder starting rules such as the north-west corner method give.

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A dairy firm has three plants \(P_{1}, P_{2}, P_{3}\) located in a state. They daily milk production at each plant (in thousand liters) is \(P_{1} \to 6\), \(P_{2} \to 1\), \(P_{3} \to 10\). Each day, the firm must fulfil the needs of its four distribution centers \(D_{1}, D_{2}, D_{3}, D_{4}\). The minimum requirement at each center (in thousand liters) is \(D_{1} \to 7\), \(D_{2} \to 5\), \(D_{3} \to 3\), \(D_{4} \to 2\). Cost in hundred of rupees of shipping one thousand liters from each plant to each distribution center is given below :

Plant	D1	D2	D3	D4
P1	2	3	11	7
P2	1	0	6	1
P3	5	8	15	9

Then the initial basic feasible by North-West Corner method is :

A.Non-degenerate and cost associated with this solution is ₹ 11,600
B.degenerate and cost associated with this solution is ₹ 12,000
C.non-degenerate and cost associated with this solution is ₹ 12,000
D.degenerate and cost associated with this solution is ₹ 11,600

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

The North-West Corner solution is degenerate, with cost ₹ 11,600.

Supply is 6 + 1 + 10 = 17 and demand is 7 + 5 + 3 + 2 = 17, so the problem is balanced. Starting at the top-left cell:

P1 to D1: 6 (P1 exhausted, D1 needs 1 more)
P2 to D1: 1 (P2 exhausted and D1 satisfied at the same time)
P3 to D2: 5, P3 to D3: 3, P3 to D4: 2

\[ \begin{aligned} &6(2) + 1(1) + 5(8) + 3(15) + 2(9) \\ &= 12 + 1 + 40 + 45 + 18 = 116 \end{aligned} \]

The cost is 116 hundred rupees, ₹ 11,600. A basic solution needs \(m + n - 1 = 3 + 4 - 1 = 6\) occupied cells, but only 5 are used, because a row and a column were satisfied together at the second step. So the solution is degenerate, and the options calling it non-degenerate or costing ₹ 12,000 are wrong.

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Consider the following transportation problem :

	Warehouse
Factory	W₁	W₂	W₃	W₄
F₁	1	2	3	4	6
F₂	4	3	2	0	8
F₃	0	2	2	1	10
	4	6	8	6

A solution to the problem is given below :

	6
		2	6
4		6

Which of the following is true about the solution ?

A.Degenerate but not optimal
B.Degenerate and optimal
C.Non-degenerate and optimal
D.Optimality cannot be tested

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The solution is degenerate and optimal.

Supply (6 + 8 + 10 = 24) equals demand (4 + 6 + 8 + 6 = 24). A basic solution of a 3 × 4 problem needs m + n - 1 = 6 occupied cells, but this one has only 5: (F1, W2) = 6, (F2, W3) = 2, (F2, W4) = 6, (F3, W1) = 4 and (F3, W3) = 6. So it is degenerate.

To test optimality, place a zero allocation (epsilon) in cell (F1, W1) and solve \(u_i + v_j = c_{ij}\) on the six basic cells. Taking \(u_3 = 0\) gives \(v_1 = 0, v_3 = 2, u_2 = 0, v_4 = 0, u_1 = 1, v_2 = 1\).

The reduced costs \(c_{ij} - u_i - v_j\) of the empty cells are then 0 (F1, W3), 3 (F1, W4), 4 (F2, W1), 2 (F2, W2), 1 (F3, W2) and 1 (F3, W4). None is negative, so the solution is optimal, with total cost 12 + 4 + 0 + 0 + 12 = 28. Optimality can be tested once the degeneracy is resolved, so the last option is wrong too.

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10.

The number of basic variables in an \((m \times n)\) transportation table are :

A.\(m + n - 1\)
B.\(m + n + 1\)
C.\(mn\)
D.\(m + n\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

There are \(m + n - 1\) basic variables.

A transportation problem with \(m\) sources and \(n\) destinations has \(m + n\) supply and demand constraints, but because total supply equals total demand one of them is redundant. So only \(m + n - 1\) constraints are independent, and a basic feasible solution has \(m + n - 1\) basic variables (allocated cells).

\(m + n\) ignores the redundancy, \(m + n + 1\) has no basis, and \(mn\) is the total number of cells, most of which are non-basic. A solution with fewer than \(m + n - 1\) positive allocations is called degenerate.

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