Illustration 2.7 (Fixed charge transportation problem – Single item)
Consider the data in Illustration 2.2 with additional fixed cost of transporting from i to j. The
fixed cost is given in Table 2.21. Find the optimum solution to the fixed cost transportation
problem with minimum sum of fixed costs and the variable transportation costs.
Table 2.21 – Fixed cost of transporting from i to j
200
300
600
320
500
200
In the fixed cost transportation problem there is a fixed cost of transporting from a given
source to a destination. This cost could include the fixed cost of hiring a vehicle, toll charges
etc. This cost is in addition to the unit transportation cost incurred. The fixed cost is incurred
only when there is transportation between the source and destination (arc). The
formulation is as follows:
Let Xij be the number of TV sets transported from warehouse i to retailer j.
Let Yij = 1 if there is transportation between i and j; = 0 otherwise.
The objective is to minimize the total cost of transportation. This is given by
2
Minimize
3
2
3
 f Y   C X
i 1 j 1
ij ij
i 1 j 1
ij
ij
The quantity of sets supplied by a warehouse should not exceed the availability. This is given
by
n
X
j 1
ij
 ai
The quantity of sets received by a retailer should meet the demand. This is given by
m
X
i 1
ij
 bj
There is transportation in an arc (warehouse-distributor) only when it is chosen. This is given
m
by
X
i 1
ij
 MYij where M is large and positive. Yij takes binary values.
The formulation has 12 variables out of which 6 are binary. It has 11 constraints. This is
solved optimally using the solver. The output is given in Table 2.22
Table 2.22 – Output from solver
x11
x12
x13
y11
y12
y13
4
6
5
x21
80
0
x22
x23
3.6E-12 2.1E-11
y21
y22
y23
5
7
8
1
0
0
200
300
600
320
500
200
60
90
0
1
1
obj fn
2360
cons
80
150
<=
<=
100
150
80
3.6E-12
<=
<=
2000
0
80
>=
80
2.1E-11
<=
0
60
>=
60
0
<=
0
90
>=
90
60
<=
2000
90
<=
2000
The optimum solution is given by Y11 = Y22 = Y33= 1; X11 = 80, X22 = 60 and X33 = 90 with cost =
2360.
Bottleneck transportation problem
The bottleneck transportation problem (Garfinkel and Rao, 1971) finds a feasible solution
that minimizes the maximum value of the individual costs. This is relevant in situations
where the Cij values represent transportation times and we transport all the items at the
same time. The formulation is as follows:
Let Xij be the number of items sets transported from warehouse i to retailer j.
Let Yij = 1 if there is transportation between i and j; = 0 otherwise.
The objective is to minimize the maximum cost of transportation. This is given by
Minimize u
u  cijYij
The above constraint minimizes the maximum of the costs
The quantity of sets supplied by a warehouse should not exceed the availability. This is given
by
n
X
j 1
ij
 ai
The quantity of sets received by a retailer should meet the demand. This is given by
m
X
i 1
ij
 bj
There is transportation in an arc (warehouse-distributor) only when it is chosen. This is given
m
by
X
i 1
ij
 MYij where M is large and positive. Yij takes binary values.
(The objective function call also be written as Minimize (Maximum Cij where Xij > 0. One of
the ways of formulating this problem is by introducing the binary Y ij variables)
hIllustration 2.8
Consider the data shown in Table 2.23. Find the optimum solution to the bottleneck
transportation problem where we minimize the maximum cost?
Table 2.23 – Transportation costs
4
3
5
100
5
7
8
150
80
60
90
The above formulation was solved using the excel solver and the optimum solution is given
in Table 2.24
Table 2.24 – Optimum solution from the solver
u
x11
x21
x12
x22
x13
x23
y11
y21
y12
y22
y13
y23
4
5
3
7
5
8
7
8.3E-27
10
90
0
1
1
200
300
600
0
80
50
3.5E-26
1
1
0
320
500
200
obj fn
7
cons
100
130
<=
<=
100
150
8.3E-27
10
<=
<=
0
2000
80
>=
80
90
<=
2000
60
>=
60
80
<=
2000
90
>=
90
50
<=
2000
3.5E-26
<=
0
7
<=
0
7
<=
3
7
<=
5
7
<=
5
7
7
<=
<=
7
0
The solution is X12 = 10, X13 = 90, X21 = 80 and X22 = 50. The corresponding unit
transportation costs are 3, 5, 5, 7 and the maximum cost is 7. The total cost is 1230.
If we had solved it as a normal transportation problem to minimize the total cost, the
solution would have been X12 = 60, X13 = 40, X21 =50, X23 = 80 with total cost = 1180. The unit
costs corresponding to the allotted cells are 3, 5, 5, 8 and the maximum cost is 8.
We can solve the bottleneck transportation problem from the optimum solution to the
minimum cost transportation problem. Since the maximum cost is 8 and we wish to
minimize this, we can set all the unit costs ≥ 8 to a large value (say 100) and solve the
transportation problem. We set C23 to M (say 200) and solve the problem to get the solution
X12 = 10, X13 = 90, X21 = 80 and X22 = 50 with cost = 1230.
Now the maximum cost is 7. We now set all costs ≥ 7 to a large value (say 100) and solve the
transportation problem. We set C22 to M (say 200) and solve the problem to get the solution
X12 = 60, X13 = 40, X21 = 80 and X23 = 50 with cost = 10780. We observe that there are
allocation to the cells with cost = M (200). We used the value 200 since we cannot input M
into the solver. The fact that there is allocation to cells with cost = M indicates that the
solution is actually infeasible if we had not allotted to the cells with cost M.
The solution to the bottleneck transportation problem is 7 since that is the lowest value that
we had before we encountered infeasibility (or allocation to cells with cost M).
Minimax transportation problem
The minimax transportation problem (Ahuja, 1986) tries to minimize the maximum
individual cost of transportation. It adjusts the transportation quantities such that the
maximum among the CijXij values is minimized. It is applicable when each of the
transportation is made by a different person and we wish to ensure that the revenues are
balanced or that one person does not gain too much revenue. The bottleneck transportation
problem is formulated as follows:
Let Xij be the quantity transported from warehouse (supply) i to destination (demand) j.
There are m supply points and n destinations.
Minimize u
Subject to
n
X
j 1
ij
 ai i
ij
 b j j
m
X
i 1
u  Cij X ij
Xij integer
Illustration 2.9
Consider the profit matrix given in Table 2.25 and the corresponding profit maximization
transportation problem. Solve a minimax transportation problem to minimize the maximum
revenue through the individual transportation problem. If the three retail stores belong to
three different people, solve a relevant transportation problem that balances the profits
earned by the three retailers?
Table 2.25 – Profit matrix
R1
R2
R3
F1
1.7
2.6
2.3
F2
1.35
2.26
1.78
F3
-1000 3.04
2.51
F4
1.73
2.71
2.2
210
230
200
200
160
180
150
The minimax transportation problem is formulated as
Minimize u
Subject to
X11 + X12 + X13 ≤ 200
X21 + X22 + X23 ≤ 160
X31 + X32 + X33 ≤ 180
X41 + X42 + X43 ≤ 150
X11 + X21 + X31 + X41 ≥ 210
X12 + X22 + X32 + X42 ≥ 230
X13 + X23 + X33 + X43 ≥ 200
u  Cij X ij
Xij ≥ 0 and integer
The optimum solution to the minimax transportation problem is X11 = 55, X12 = 68, X13 = 77,
X21 = 71, X22 = 37, X23 = 62, X32 = 59, X33 = 71, X41 = 84, X42 = 66, u = 179.36. The total profit =
1401.18
The profits received by the three retailers are 1.7X11 + 1.35X21 + 1.73X41, 2.6X12 + 2.26X22 +
3.04X32 + 2.71X42 and 2.3X13 + 1.78X23 + 2.51X33 + 2.2X43. We wish to balance these
quantities. Let u and v be the maximum and minimum of these. The problem is to
Minimize u - v
Subject to
X11 + X12 + X13 ≤ 200
X21 + X22 + X23 ≤ 160
X31 + X32 + X33 ≤ 180
X41 + X42 + X43 ≤ 150
X11 + X21 + X31 + X41 ≥ 210
X12 + X22 + X32 + X42 ≥ 230
X13 + X23 + X33 + X43 ≥ 200
u ≥ 1.7X11 + 1.35X21 + 1.73X41
u ≥ 2.6X12 + 2.26X22 + 3.04X32 + 2.71X42
u ≥ 2.3X13 + 1.78X23 + 2.51X33 + 2.2X43
v ≤ 1.7X11 + 1.35X21 + 1.73X41
v ≤ 2.6X12 + 2.26X22 + 3.04X32 + 2.71X42
v ≤ 2.3X13 + 1.78X23 + 2.51X33 + 2.2X43
Xij ≥ 0 and integer.
The optimum solution to the problem is X11 = 110, X12 = 70, X13 = 20, X22 = 160, X33 = 180, X41
= 150 with u = 543.6, V = 446.5 and u – v = 97.1. The total cost is 1487.9
Transportation problem with side constraints
This version of the transportation problem has additional constraints over and above the
usual transportation constraints. If we are concerned with both time and cost, we would
minimize the total cost subject to constraints on time and vice versa. A typical
transportation problem with side constraint would be as follows:
Let Xij be the quantity transported from warehouse (supply) i to destination (demand) j.
There are m supply points and n destinations.
m
The objective is to Minimize
n
 C X
i 1 j 1
ij
ij
Subject to
n
X
j 1
ij
 ai i
ij
 b j j
m
X
i 1
m
n
 t
i 1 j 1
ij
X ij  T
Xij integer
Illustration 2.10
Consider the transportation problem between three supply points and three destinations.
The unit cost of transportation is given in Table 2.26and the time per unit transported is
given in Table 2.27. The supplies in the three points are 280, 270 and 250 units. The
demands are 300, 240 and 260 units. The total time should be within 5000 units and the
cost within 17000.
Table 2.26 – Cost of transportation
R1
R2
R3
W1
15
24
23
W2
24
22
19
W3
28
26
25
Table 2.27- Unit Time for transportation
R1
R2
R3
W1
6
3
5
W2
4
5
7
W3
8
9
6
Solve the problem to minimize the total cost of transportation. If there is a restriction on
total time to xx what is the increase in cost? Solve a bottleneck transportation problem to
minimize the maximum time? Add the side constraint and solve the problem to find the
increase in cost? If the time matrix represented the total time to transport any quantity
between i and j, how does the problem get modified?
The transportation problem to minimize total cost is given by
Minimize 15X11 + 24X12 + 23X13 + 24X21 + 22X22 + 19X23 + 28X31 + 26X32 + 25X33
Subject to
X11 + X12 + X13 = 280
X21 + X22 + X23 = 270
X31 + X32 + X33 = 250
X11 + X21 + X31 = 300
X12 + X22 + X32 = 240
X13 + X23 + X33 = 260
Xij ≥ 0
The optimum solution is given by X11 = 280, X22 = 10, X23 = 260, X31 = 20, X32 = 230 and the
total cost is 15900. The total time associated with this solution is 5780. The present solution
violates the time constraint. We add the side constraint
6X11 + 3X12 + 5X13 + 4X21 + 5X22 + 7X23 + 8X31 + 9X32 + 6X33 ≤ 5000
The optimum solution including the side constraint is X11 = 280, X22 = 166, X23 = 104, X31 = 20,
X32 = 74, X33 = 156 and the total cost is 16212. The increase in cost is 312.
We now solve a minimax transportation problem for the given situation (with side
constraints).
Minimize u
Subject to
X11 + X12 + X13 = 280
X21 + X22 + X23 = 270
X31 + X32 + X33 = 250
X11 + X21 + X31 = 300
X12 + X22 + X32 = 240
X13 + X23 + X33 = 260
u ≥ 6X11, u ≥ 3X12, u ≥ 5X13, u ≥ 4X21, u ≥ 5X22, u ≥ 7X23, u ≥ 8X31, u ≥ 9X32 and u ≥ 6X33
Xij ≥ 0
The optimum solution to the balanced transportation problem is X11 = 104, X12 = 109, X13 =
67, X21 = 119, X22 = 62, X23 = 89, X31 = 77, X32 = 69, X33 = 104 and the total cost is 18178 with
the minimum value of the maximum time is 624. This solution violates the cost constraint.
Adding the side constraint
15X11 + 24X12 + 23X13 + 24X21 + 22X22 + 19X23 + 28X31 + 26X32 + 25X33 ≤ 17000 we get the
optimum solution X11 = 197, X12 = 83, X21 = 76, X22 = 26, X23 = 168, X31 = 27, X32 = 131, X33 = 92
and the total cost is 16997 with the minimum value of the maximum time is 1182.
We now solve a bottleneck transportation problem for the given situation (with side
constraints).
If the time matrix represents the total time to transport any quantity, we wish to minimize
the maximum time. We define Yij = 1 if a positive quantity is transported between i and j
Minimize u
Subject to
X11 + X12 + X13 = 280
X21 + X22 + X23 = 270
X31 + X32 + X33 = 250
X11 + X21 + X31 = 300
X12 + X22 + X32 = 240
X13 + X23 + X33 = 260
X11 ≤ 1000Y11, X12 ≤ 1000Y12, X13 ≤ 1000Y13, X21 ≤ 1000Y21, X22 ≤ 1000Y22, X23 ≤ 1000Y23, X31 ≤
1000Y31, X32 ≤ 1000Y32, X33 ≤ 1000Y33,
u ≥ 6Y11, u ≥ 3Y12, u ≥ 5Y13, u ≥ 4Y21, u ≥ 5Y22, u ≥ 7Y23, u ≥ 8Y31, u ≥ 9Y32 and u ≥ 6Y33
The optimum solution is given by X11 = 270, X13 = 10, X21 = 30, X22 = 240, X33 = 250 and the
total cost is 16530 with the minimum value of the maximum time is 6.
This solution meets the side constraint and hence is optimal when the side constraint is
considered.
Exclusionary side constraints
The exclusionary side constraint problem is where transportation to a retailer is prohibited
from a given pair of suppliers simultaneously.
If simultaneous transportation from suppliers 1 and 3 is prohibited for retailer (demand) 2,
the constraint will be X12*X32 = 0 where Xij are the quantities supplied from i to j. If we use Yij
= 1 if some quantity is supplied from i to j, the non linear constraint in X can be made linear
through Xij ≤ M Yij and adding the specific exclusivity constraint Y12 + Y32 ≤ 1.
Illustration 2.11
The supplies are 100 and 150 while the demands are 80, 60 and 90. The unit cost of
transportation is given in Table 2.28. There is a restriction that we cannot simultaneously
transport from both the warehouses to retailer 3. Solve the problem?
Table 2.28 – Unit cost of transportation
R1
R2
R3
W1
4
3
5
W2
5
7
8
The problem is first solved as a transportation problem.
Minimize 4X11 + 3X12 + 5X13 + 5X21 + 7X22 + 8X23
Subject to
X11 + X12 + X13 ≤ 100
X21 + X22 + X23 ≤ 150
X11 + X21 ≥ 80
X12 + X22 ≥ 60
X13 + X23 ≥ 90
Xij ≥ 0
The optimum solution is X12 = 60, X13 = 40, X21 = 80, X23 = 50 and the total cost is 1180. It
violates the condition that we cannot simultaneously transport from both the warehouses
to retailer 3. We nnow add the Yij variables and the constraints Xij ≤ M Yij and the specific
exclusivity constraint Y13 + Y23 ≤ 1.
The optimum solution to the exclusivity constrained transportation problem is X 12 = 10, X13 =
90, X21 = 80, X22 = 50 and the total cost is 1230. The cost increases by 50.
Multi period transportation problems
In this application we consider transportation of items in multiple periods. There are m
supplies and n demand points. The unit cost of transportation is Cij. There are k periods. The
supply at time period t is ait. The demand at retailer j for period t is bjt.
Let Xijt be the quantity transported from supply i to demand j in time period t. The objective
function is to
Minimize ∑𝑖 ∑𝑗 ∑𝑡 𝐶𝑖𝑗 𝑋𝑖𝑗𝑡
Subject to
∑ 𝑋𝑖𝑗𝑡 ≤ 𝑎𝑖𝑡
𝑗
∑𝑖 ∑𝑡𝑙 𝑋𝑖𝑗𝑙 ≥ ∑𝑡𝑙 𝑏𝑙 for t = 1 to k
Xijt ≥ 0.
It is assumed that the supply and demand quantities are such that feasible solutions exist.
Illustration 2.12
The unit cost of transportation between two supply points and three demand points is given
in Table 2.29. The supply quantities in two periods are 100 and 150 in period 1 and 80 and
120 in period 2. The demand quantities in two periods are 80, 60 and 70 in period 1 and 70
60 and 80 in period 2. Solve a multi period transportation problem.
Table 2.29 – Unit cost of transportation
R1
R2
R3
W1
4
3
5
W2
5
7
8
The formulation is as follows:
Let Xij1 and Xij2 be the quantities supplied in the two periods.
The objective is to Minimize 4X111 + 3X121 + 5X131 + 5X211 + 7X221 + 8X231 + 4X112 + 3X122 + 5X132
+ 5X212 + 7X222 + 8X232
Subject to
X111 + X121 + X131 ≤ 100
X211 + X221 + X231 ≤ 150
X112 + X122 + X132 ≤ 80
X212 + X222 + X232 ≤ 120
X111 + X211 ≥ 80
X121 + X221 ≥ 60
X131 + X231 ≥ 70
X111 + X211 + X112 + X212 ≥ 150
X121 + X221 + X122 + X222 ≥ 120
X131 + X231 + X132 + X232 ≥ 150
Xij ≥ 0
The optimum solution to the above LP is given by X121 = 60, X131 = 40, X211 = 90, X231 = 30,
X122 = 60, X132 = 20, X212 = 60, X232 = 60. The minimum cost is 2130. It is observed that while
total supply exceeds total demand in period 1, it is lesser than total demand in period 2. The
sum of all the supplies exceeds the total demand for all periods. Due to this, we transport
excess in period 1 to demand point 1.
(It is also customary to add an inventory cost if excess is transported in a period. This has
not been included in the above formulation).
Fixed Charge Transportation problem (multiple items)
The basic transportation problem assumes that only one item is transported from a set of
supply points to a given set of destination points. When we have multiple items, we can
solve a transportation problem for each item and add the costs. This approach does not
provide the optimum if we have a fixed charge for each arc. The multiple item
transportation problem with fixed charge fij and unit transportation cost Cij between nodes i
and j is given as follows:
Let Xijk be the quantity of item k transported between supply point i and demand point j. The
objective is to
m
Minimize
n
m
n
p
 f Y   C X
i 1 j 1
ij ij
i 1 j 1 k 1
ij
ijk
Subject to
n
X
j 1
ijk
 aik i, k
ijk
 b jk j , k
m
X
i 1
Xijk ≤ 10000Yij
Xij integer and Yij binary.
Illustration 2.13
Consider the transportation of a product from three source points to four destinations. The
unit cost of transportation is given in Table 2.30 and the fixed cost of transportation is given
in Table 2.31
Table 2.30 – Cost of unit transportation
R1
R2
R3
R4
W1
W2
W3
18
21
24
24
22
26
22
21
25
17
15
16
Table 2.31 – fixed cost of transportation
R1
R2
R3
R4
W1
240
180
265
188
W2
147
158
170
180
W3
187
198
165
175
The quantities available in three supply points are 80, 70 and 50 and the demands are 40, 60
and 30 and 70. Consider a second item with supply 120, 170, 160 and demands 100, 80, 70
and 200. Solve the fixed charge transportation for each item and by combining the items.
Determine the decrease in total cost?
The optimum solution considering the first item only is given by is X12 = 60, X14 =20, X21 = 40,
X23 = 30, X34 = 50, Y12 = Y14 = Y21 = Y23 = Y34 = 1 and the total cost is 4910.
The optimum solution considering the second item only is given by is X11 = 100, X12 =20, X22
= 60, X23 = 70, X24 = 40, X34 = 160, Y11 = Y12 = Y22 = Y23 = Y24 = Y34 = 1 and the total cost is 9333.
We solve a combined problem where we define Yij = 1 if we choose to transport items from i
to j. Since we have two items, instead of defining the variables as Xijk we define them as Xij
for item 1 and Zij for item 2. The formulation is
Minimize 240Y11 + 180Y12 + 265Y13 + 188Y14 + 147Y21 + 158Y22 + 170Y23 + 180Y24 + 187Y31 +
198Y32 + 165Y33 + 175Y34 + 18X11 + 24X12 + 22X13 + 17X14 + 21X21 + 22X22 + 21X23 + 15X24 +
24X31 + 26X32 + 25X33 + 16X34 + 18Z11 + 24Z12 + 22Z13 + 17Z14 + 21Z21 + 22Z22 + 21Z23 + 15Z24 +
24Z31 + 26Z32 + 25Z33 + 16Z34
Subject to
X11 + X12 + X13 + X14 = 80
X21 + X22 + X23 + X24 = 70
X31 + X32 + X33 + X34 = 50
X11 + X21 + X31 = 40
X12 + X22 + X32 = 60
X13 + X23 + X33 = 30
X14 + X24 + X34 = 70
Z11 + Z12 + Z13 + Z14 = 120
Z21 + Z22 + Z23 + Z24 = 170
Z31 + Z32 + Z33 + Z34 = 160
Z11 + Z21 + Z31 = 100
Z12 + Z22 + Z32 = 80
Z13 + Z23 + Z33 = 70
Z14 + Z24 + Z34 = 200
X11 ≤ 1000Y11; X12 ≤ 1000Y12; X13 ≤ 1000Y13; X14 ≤ 1000Y14; X21 ≤ 1000Y21; X22 ≤ 1000Y22; X23 ≤
1000Y23; X24 ≤ 1000Y24; X31 ≤ 1000Y31; X32 ≤ 1000Y32; X33 ≤ 1000Y33; X34 ≤ 1000Y34;
Z11 ≤ 1000Y11; Z12 ≤ 1000Y12; Z13 ≤ 1000Y13; Z14 ≤ 1000Y14; Z21 ≤ 1000Y21; Z22 ≤ 1000Y22; Z23 ≤
1000Y23; Z24 ≤ 1000Y24; Z31 ≤ 1000Y31; Z32 ≤ 1000Y32; Z33 ≤ 1000Y33; Z34 ≤ 1000Y34;
Xij, Zij ≥ 0 and integer, Yij = 0,1
The optimum solution is given by Z11 = 100, Z12 =20, Z22 = 60, Z23 = 70, Z24 = 40, Z34 = 160, X11
= 40, X12 =40, X22 = 20, X23 = 30, X24 = 20, X34 = 50, Y11 = Y12 = Y22 = Y23 = Y24 = Y34 = 1 and the
total cost is 13183. The gain in combining the items and solving a single problem is 1060.
Multi stage with fixed charge and node Charge
The basic transportation problem considers transportation between a set of supply points
and a set of demand points. This can be thought of as transportation in a single stage
involving supply and demand points. In practice multi stage transportation exists where
items are transported from factories to warehouses and from warehouses to retailers. In
practice multiple items are transported taking into consideration multiple time periods.
There could be fixed costs to transport between arcs (factories to warehouses and between
warehouses to retailers). There could also be node charges which would be the rent for
warehouse space.
We consider m factories, n warehouses and p retailers and q items. Let X ijk be the quantity
of item k transported between factory i and warehouse j. Let Y ij = 1 if we transport between
i and j. Let Zjlk be the quantity of item k transported between warehouse j and retailer l. Let
Rjl = 1 if we transport between warehouse j and retailer l. Let Wj = 1 if warehouse j is chosen
and items go through warehouse j.
Cij = cost of unit transportation between factory i and warehouse j (assumed to be the same
for all items)
Tjl = cost of unit transportation between warehouse j and retailer l (assumed to be the same
for all items)
aik = availability of item k in factory i
blk = requirement if item k in retailer l
Gj = fixed cost at warehouse j
Mjk = capacity available in warehouse j for item k
The problem is to
m
n
m
n
q
 f Y   C X
Minimize
ij ij
i 1 j 1
i 1 j 1 k 1
ij
ijk
p
n
n
j 1
j 1 l 1
m
n
q
  G jW j   s jl R jl   Tjl Z jlk
i 1 j 1 k 1
Subject to
n
X
j 1
ijk
 aik i, k
jlk
 b jk j , k
ijk
 MYij i, j
jlk
 MR jl , j , l
p
Z
l 1
q
X
k 1
q
Z
k 1
m
q
p
q
 X ijk   Z jlk  MW j i, k
i 1 k 1
X
i 1
l 1 k 1
q
m
ijk
  X jlk j , k
l 1
Xij integer and Yij binary.
The objective function minimizes the sum of the fixed charges between the factories and
warehouses and between warehouses and retailers. The cost of transportation between the
factories and warehouses as well as between warehouses and retailers are included in the
objective function. The fixed costs of using the warehouses are also included.
The first set of constraints ensures that the quantities transported from the factories are
less than the availability of each item. The second set of constraints ensures that the
quantities transported to the retailers meet the demand of each item. The third and fourth
set of constraints ensures that the items are transported using routes for which the fixed
costs are incurred. The fifth and sixth set of constraints ensures that the transportations into
and out of the warehouses happen using those for which the fixed costs are incurred. The
seventh set of constraints ensures that the items that come into the warehouses leave
them.
The fixed costs of setting up the facilities are not included in the model. It is assumed that
transportation happens among the facilities that have already been created
Illustration 2.14
Consider 3 factories, 3 warehouses and 4 retailers. Two items are transported. The
availability in the factories are 1000, 2000, 1000 of item 1 and 1500, 1000 and 1200 of item
2. The requirements in the four retailers are 200, 300, 500 and 800 of item 1 and 400, 600,
700 and 200 of item 2. The fixed costs associated with the three warehouses are 500, 800
and 1000. The unit costs and fixed costs of transportation among the facilities are given in
Tables 2.32 and 2.33
Table 2.32 – Cost of unit transportation
R1
R2
R3
R4
W1
18
24
22
17
W2
21
22
21
15
W3
24
26
25
16
F1
9
11
10
F2
11
12
10
F3
8
9
6
Table 2.33 – fixed cost of transportation
R1
R2
R3
R4
W1
240
180
265
188
W2
147
158
170
180
W3
187
198
165
175
F1
160
185
170
F2
220
170
140
F3
180
185
155
The formulation is as follows:
Minimize 160Y11 + 220Y12 + 180Y13 + 185Y21 + 170Y22 +185Y23 + 170Y31 + 140Y32 + 155Y33 +
9X111 + 11X121 + 8X131 + 11X211 + 12X221 + 9X231 + 10X311 + 10X321 + 6X331 + 9X112 + 11X122 +
8X132 + 11X212 + 12X222 + 9X232 + 10X312 + 10X322 + 6X332 + 500W1 + 800W2 + 1000W3 + 240R11
+ 180R12 + 265R13 + 188R14 + 147R21 + 158R22 + 170R23 + 180R24 + 187R31 + 198R32 + 165R33 +
175R34 + 18Z111 + 24Z121 + 22Z131 + 17Z141 + 21Z211 + 22Z221 + 21Z231 + 15Z241 + 24Z311 +
26Z321 + 25Z231 + 16Z241 + 18Z112 + 24Z122 + 22Z132 + 17Z142 + 21Z212 + 22Z222 + 21Z232 +
15Z242 + 24Z312 + 26Z322 + 25Z232 + 16Z242
Subject to
X111 + X121 + X131 ≤ 1000
X211 + X221 + X231 ≤ 2000
X311 + X321 + X331 ≤ 1000
X112 + X122 + X132 ≤ 1500
X212 + X222 + X232 ≤ 1000
X312 + X322 + X332 ≤ 1200
Z111 + Z211 + Z311 ≥ 200
Z121 + Z221 + Z321 ≥ 300
Z131 + Z231 + Z331 ≥ 500
Z141 + Z241 + Z341 ≥ 800
Z112 + Z212 + Z312 ≥ 400
Z122 + Z222 + Z322 ≥ 600
Z132 + Z232 + Z332 ≥ 700
Z142 + Z242 + Z342 ≥ 200
X111 + X112 ≤ 10000Y11
X121 + X122 ≤ 10000Y21
X131 + X132 ≤ 10000Y31
X211 + X212 ≤ 10000Y21
X221 + X222 ≤ 10000Y22
X231 + X232 ≤ 10000Y23
X311 + X 312 ≤ 10000Y31
X321 + X322 ≤ 10000Y32
X331 + X332 ≤ 10000Y33
Z111 + Z112 ≤ 10000R11
Z121 + Z122 ≤ 10000R12
Z131 + Z132 ≤ 10000R13
Z141 + Z142 ≤ 10000R14
Z211 + Z212 ≤ 10000R21
Z221 + Z222 ≤ 10000R22
Z231 + Z232 ≤ 10000R23
Z241 + Z242 ≤ 10000R24
Z311 + Z312 ≤ 10000R31
Z321 + Z322 ≤ 10000R32
Z331 + Z332 ≤ 10000R33
Z341 + Z342 ≤ 10000R34
X111 + X112 + X211 + X212 + X311 + X312 + Z111 + Z112 + Z121 + Z122 + Z131 + Z132 + Z141 + Z142 ≤
10000W1
X121 + X122 + X221 + X222 + X321 + X322 + Z211 + Z212 + Z221 + Z222 + Z231 + Z232 + Z241 + Z242 ≤
10000W2
X131 + X132 + X231 + X232 + X331 + X332 + Z311 + Z312 + Z321 + Z322 + Z331 + Z332 + Z341 + Z342 ≤
10000W3
X111 + X211 + X311 = Z111 + Z121 + Z131 + Z141
X112 + X212 + X312 = Z112 + Z122 + Z132 + Z142
X121 + X221 + X321 = Z211 + Z221 + Z231 + Z241
X122 + X222 + X322 = Z212 + Z222 + Z232 + Z242
X131 + X231 + X331 = Z311 + Z321 + Z331 + Z341
X132 + X232 + X332 = Z312 + Z322 + Z332 + Z342
Xijk, Zjlk ≥ 0 and Wj, Yij, Rjl = 0,1
The above binary IP has 66 variables (including 24 binaries) and 44 constraints. The
optimum solution is given by
Y11 = Y33 = W1 = W3 = R11 = R12 = R13 = R32 = R34 = 1; X111 = 800, X112 = 1100, X331 = 1000, X332 =
800, Z111 = 200, Z121 = 100, Z112 = 400, Z132 = 700, Z322 = 600, Z131 = 500, Z341 = 800, Z321 = 200,
Z342 = 200 with minimum total cost = 107173.
Multi stage capacitated problems
In the earlier formulation, we assumed that the warehouses have unlimited capacities while
there were capacity restrictions on the factories. The availability of the items in the factories
represents a capacity restriction. When we include constraints on the capacities, the
problem becomes a multistage capacitated transportation problem. We could also add fixed
costs of transporting using a particular arc or through a particular warehouse or node.
Consider a two stage transportation problem where we have m factories, n warehouses and
p retailers. We consider a single item in this formulation though this can easily be extended
to multiple items. Let Cij be the unit transportation cost between factory i and warehouse j.
Let tjk be the cost of transportation between warehouse j and retailer k. Let ai and bk be the
availability and requirement of the item at factory i and retailer k respectively. Let d j be the
maximum capacity of warehouse j. Let Xij be the quantity transported between factory i and
warehouse j. Let Zjk be the quantity transported between warehouse j and retailer k.
The objective is to
m
n
n
p
 C X   t Z
Minimize
i 1 j 1
ij
ij
j 1 k 1
ij
ij
Subject to
n
X
j 1
ij
 ai
ij
 dj
jk
 bk
m
X
i 1
n
Z
j 1
p
m
k 1
i 1
 Z jk   X ij
Xij, Zjk ≥ 0
Illustration 2.15
Consider a two stage transportation problem with 3 factories, 3 warehouses and 4 retailers.
The availability in the factories is 1000, 2000, 1000 and the requirements in the four
retailers are 200, 300, 500 and 800. The unit costs of transportation among the facilities are
given in Table 2.26. The capacities in the warehouses are 800, 600 and 700.
Using the variables given we formulate a problem to minimize 9X11 + 11X12 + 8X13 + 11X21 +
12X22 + 9X23 + 10X31 + 10X32 + 6X33 + 18Z11 + 24Z12 + 22Z13 + 17Z14 + 21Z21 + 22Z22 + 21Z23 +
15Z24 + 24Z31 + 26Z32 + 25Z33 + 16Z34
Subject to
X11 + X12 + X13 ≤ 1000
X21 + X22 + X23 ≤ 2000
X31 + X32 + X33 ≤ 1000
Z11 + Z21 + Z31 ≥ 200
Z12 + Z22 + Z32 ≥ 300
Z13 + Z23 + Z33 ≥ 500
Z14 + Z24 + Z34 ≥ 800
X11+X21+X31 ≤ 800
X12+X22+X32 ≤ 600
X13+X23+X33 ≤ 700
X11+X21+X31-Z11-Z12-Z13-Z14=0
X12+X22+X32 - Z21- Z22 -Z23 - Z24=0
X13+X23+X33 - Z31 - Z32 - Z33 - Z34=0
Xij, Zjk ≥ 0
The optimum solution to the problem is given by X11 = 800, X32 = 300, X33 = 700, Z11 = 200,
Z12 = 100, Z13 = 500, Z22 = 200 with cost = 4850.
If we have a very large supply in each of the factories and a very large capacity, each retailer
would get it from the factory with the least cost of transportation. The solution changes
since we have restrictions on the availability of material in the factories as well as in the
warehouses.
In the example, we obtained integer solutions though the variables were defined as
continuous. This is guaranteed if there are no capacity restrictions on the warehouses.
Illustration 2.16 - Caterer problem
Consider the caterer problem discussed in Illustration 1.13. Formulate the caterer problem
as a transportation problem. The caterer problem is as follows:
Consider a caterer who has to provide food for several dinners happening in the next 8 days.
The demand for cloth napkins that are used in the dinners is 80, 120, 66, 75, 100, 120 135
and 150. New napkins cost Rs 20. Napkins can be put to laundry and washed napkins can be
used on subsequent days. Two types of laundry are available. The fast laundry that charges
Rs 6 per unit and will deliver for use on the second day and the slow laundry that costs Rs
4/unit and can deliver for use on the third day. Find the least cost purchase and use plan for
the caterer?
The caterer problem can be formulated as a transportation problem. We assume that new
napkins can be bought from the market and create market as a supply with 846 napkins
(which is the total napkins required for all 8 days). A maximum of 80, 120, 66, 75, 100 and
120 napkins can be sent to laundries (fast or slow) and can be used to meet the demands of
days 3 to 8. There are 7 supplies with quantities 846, 80, 120, 66, 75, 100 and 120. The
demand for 8 days is 80, 120, 66, 75, 100, 120 135 and 150. The cost of a new napkin is 8,
the cost of sending to the laundries are 6 and 4 for the fast and slow type. The total supply is
1407 and the total demand is 846. We can balance the transportation table by creating a
dummy demand of 561. Table 2.34 shows the transportation costs and the supply and
demand quantities.
Table 2.34– Transportation costs, supply and demand quantities.
20
20
20
20
20
20
20
M
M
6
4
4
4
4
M
M
M
6
4
4
4
M
M
M
M
6
4
4
M
M
M
M
M
6
4
M
M
M
M
M
M
6
M
M
M
M
M
M
M
80
120
66
75
100
120
135
20
4
4
4
4
4
6
150
0
0
0
0
0
0
0
561
846
80
120
66
75
100
120
The big M denotes a very large cost of transportation between the source and the demand.
This actually means that we cannot transport from the source to destination. In our
example, we cannot use the 80 napkins that can be sent to the laundry to meet the demand
of day 1 or day 2. We can use these napkins that return from the laundry to meet the
demands of days 3 to 8. We give a value of M = 1000 indicating that the cost is large enough
for the solution to not consider the possibility of transporting from a source to a destination
that has cost = M.
Let Xij represent the quantity transported from source i to destination j. The objective is to
Minimize 20X11 + 20X12 +20X13 + 20X14 +20X15 + 20X16 +20X17 + 20X18 + 6X23 + 6X34 + 6X45 +
6X56 + 6X67 + 6X78 + 4X24 + 4X25 +4X26 +4X27 + 4X28 +4X34 +4X35 + 4X36 +4X37 + 4X38 + 4X45 +4X46
+4X47 + 4X48 +4X56 +4X57 + 4X58 +4X67 + 4X68 + 4X78
Subject to
X11 + X12 + X13 + X14 +X15 + X16 +X17 + X18 ≤ 846
X23 + X24 + X25 + X26 + X27 + X28 ≤ 80
X34 + X35 + X36 + X37 + X38 ≤ 120
X45 + X46 + X47 + X48 ≤ 66
X56 + X57 + X58 ≤ 75
X67 + X68 ≤ 100
X78 ≤ 120
X11 ≥ 80
X12 ≥ 120
X13 + X23 ≥ 66
X14 + X24 +X34 ≥ 75
X15 + X25 +X35 + X45 ≥ 100
X16 + X26 +X36 + X46 + X56 ≥ 120
X17 + X27 +X37 + X47 + X57 + X67 ≥ 135
X18 + X28 +X38 + X48 + X58 + X68 + X78 ≥ 135
Xij ≥ 0
The optimum solution is given by X11 = 80, X12 = 120, X13 = 66, X15 = 19, X56 = 10, X67 = 70, X78
= 120, X24 = 75, X25 = 5, X35 = 76, X36 = 44, X46 = 66, X57 = 65, X68 = 30 with Z = 8344.
The optimum solution given by the transportation model is the same given in Chapter 1. The
LP formulation has 19 variables and 14 constraints. The transportation formulation has 29
variables and 14 constraints.
Illustration 2.17 - Production planning problem – revisited.
Consider a company making a single product. The demand for the next four days is 60, 200,
160 and 100 units. The company employs 15 people and each person works for 8 hours
daily. Each person makes one unit of the product/hour. The employees can work for 1 extra
hour per day overtime. The costs of regular and over time production are Rs 80 and 120 per
product. The company has to meet the daily demand but can produce more and store them
for further use at Rs 4/unit/day. The company can get the product made outside and buy it
at Rs 125/unit. If the company does not use an employee, it costs Rs 60/hour. Find the least
cost production plan?
Minimize 125X11 + 125X12 + 125X13 + 125X14 + 80X21 + 84X22 + 88X23 + 92X24 + 60X25 + 120X31
+ 124X32 + 128X33 + 132X34 + 80X42 + 84X43 + 88X44 + 60X45 + 120X52 + 124X53 + 128X54 + 80X63
+ 84X64 + 60X65 + 120X73 + 124X74 + 80X84 + 60X85 + 120X94
Subject to
X11 + X12 + X13 + X14 ≤ 520
X21 + X22 + X23 + X24 + X25 = 120
X31 + X32 + X33 + X34 ≤ 15
X42 + X43 + X44 + X45 = 120
X52 + X53 + X54 ≤ 15
X63 + X64 + X65 = 120
X73 + X74 ≤ 15
X84 + X85 = 120
X94 ≤ 15
X11 + X21 + X31 = 60
X12 + X22 + X32 + X42 + X52 = 200
X13 + X23 + X33 + X43 + X53 + X63 + X63 = 160
X14 + X24 + X34 + X44 + X54 + X64 + X74 + X84 + X94 = 100
The optimum solution is given by X13 = 25, X21 = 55, X22 = 65, X31 = 5, X42 = 120, X52 = 15, X63 =
120, X73 = 15, X84 = 100, X85 = 20 with Z = 45585.
The LP formulation had 19 variables and 12 constraints. The transportation formulation has
28 variables and 13 constraints.
Application 2.1
Consider the transportation of two items from two supply points to three destination
points. The availability in the two supply points are 8000, 6000, of item 1 and 10000 and
9000 of item 2. The requirements in the three destinations are 2000, 3000, 5000 of item 1
and 4000, 6000, 7000 of item 2. The distances among the facilities are given in Table 2.35
Table 2.35 – Distances (in km)
D1
D2
D3
S1
240
180
265
S2
147
158
170
The transportation is carried out using two types of trucks with capacity 9 and 16 tonnes.
There is a fixed cost of Rs 5000 and Rs 10000 respectively for each truck of each type hired.
There is an additional cost of Rs 3/km travelled. Find the number of trucks of each type that
are used to transport the items at minimum total cost?
Let Xijk be the quantity of item k transported between supply point i and destination j.
Let nijl the number of trucks of type l used to transport between supply point i and
destination j.
The objective is to minimize the sum of fixed cost of hiring the trucks and the costs
associated with the distances. The objective is to
Minimize 5720n111 + 10720n112 + 5540n121 + 10540n122 + 5795n131 + 10795n132 + 5441n211 +
10441n212 + 5474n221 + 10474n222 + 5510n231 + 10510n232
Subject to
X111 + X121 + X131 ≤ 8000
X112 + X122 + X132 ≤ 10000
X211 + X221 + X231 ≤ 6000
X212 + X222 + X232 ≤ 9000
X111 + X211 = 2000
X112 + X212 = 4000
X121 + X221 = 3000
X122 + X222 = 6000
X131 + X231 = 5000
X132 + X232 = 7000
X111 + X112 ≤ 9n111 + 16n112
X121 + X122 ≤ 9n121 + 16n122
X131 + X132 ≤ 9n131 + 16n132
X211 + X212 ≤ 9n211 + 16n212
X221 + X222 ≤ 9n221 + 16n222
X231 + X232 ≤ 9n231 + 16n232
nijl ≥ 0 and integer; Xijk ≥ 0
The objective function minimizes the total cost of engaging the trucks. The coefficient 5720
for n111 is the sum of the fixed cost of 5000 per truck and the operating cost of 3 x 240 = 720
for the distance travelled. The first ten constraints are the supply and demand constraints
for the items. The next six constraints relate the number of trucks of each type to the
quantities transported between the supply and demand points.
The optimum solution to the mixed integer linear program is given by n 111 = n121 = n232 = 1;
X111 =2000, X121 = 3000, X112 = 4000, X122 = 6000, X231 = 5000, X232 = 7000. In the optimal
solution two 9 tonne trucks and one 16 tonne truck are used.