1

In this chapter you will:

• learn the terminology used to describe and model
transportation problems
• learn about the ‘north-west corner method’
• learn about unbalanced transportation problems
• understand what is meant by a degenerate solution
• learn to find shadow costs
• find improvement indices
• use the stepping-stone method
• formulate a transportation problem as a linear
programming problem.

Transportation
problems

In industry, people are concerned with efficiency and

cost-effectiveness at all stages. In this chapter you
will look at the costs due to the transportation of
goods – to factories and from factories to warehouses
and customers. The problems considered are usually
concerned with minimising distribution costs where
there are multiple sources and multiple destinations.
1
CHAPTER 1

1.1 You should be familiar with the terminology used in describing and modelling
the transportation problem.

In order to solve transportation problems you need to consider:

䊏 The capacity of each of the supply points (or sources) – the quantity of goods that can be
produced at each factory or held at each warehouse. This is called the supply or stock.

䊏 The amount required at each of the demand points – the quantity of goods that are
needed at each shop or by each customer. This is called the demand (or destination).

䊏 The unit cost of transporting goods from the supply points to the demand points.

The unit cost is the cost of transporting one item. If one item costs
c pounds to transport from A to X then two items will cost 2c pounds
to transport along that route, and n items nc pounds.

Example 1

Three suppliers A, B and C, each produce road grit which has to be delivered to council depots W,
X, Y and Z. The stock held at each supplier and the demand from each depot is known. The cost,
in pounds, of transporting one lorry load of grit from each supplier to each depot is also known.
This information is given in the table.

Depot Depot Depot Depot Stock This table is often

W X Y Z (lorry loads) referred to as the
cost matrix.
Supplier A 180 110 130 290 14
Supplier B 190 250 150 280 16 This is the cost of
Supplier C 240 270 190 120 20 transporting one
lorry load from B to Y
Demand (in £s).
11 15 14 10 50
(lorry loads)

Notice that the total supply is equal to the total demand. If this is
not the case we simply introduce a dummy destination to absorb the
excess supply, with transportation costs all zero (see Section 1.3).

Use the information in the table to write down:

a the number of lorry loads of grit that each supplier can supply
b the number of lorry loads of grit required at each depot
c the cost of transporting a lorry load of grit from A to W
d the cost of transporting a lorry load of grit from C to Z.
e Which is the cheapest route to use?
f Which is the most expensive route to use?

2
 Transportation problems

a Suppliers A, B and C can provide 14, 16 and 20 lorry loads respectively.

b Depots W, X Y and Z require 11, 15, 14 and 10 lorry loads respectively.
c The cost of transporting one lorry load from A to W is £180.
d The cost of transporting one lorry load from C to Z is £120.
e The cheapest route is A to X at £110 per load.
f The most expensive route is A to Z at £290 per load

Solving the transportation problem

The method is as follows.
1 First find an initial solution that uses all the stock and meets all the demands.
2 Calculate the total cost of this solution and see if it can be reduced by transporting some
goods along a route not currently in the solution.
(If this is not possible then the solution is optimal.)
3 If the cost can be reduced by using a new route, as many units as possible are allocated to
this new route to create a new solution.
4 The new solution is checked in the same way as the initial solution to see if it is optimal.
If not, any new routes found are included.
5 When no further savings are possible, an optimal solution has been found.

1.2 You can find an initial solution to the transportation problem.

A method often called the ‘north-west corner method’ is used.

1 Create a table, with one row for every source and one column for every destination. Each cell
represents a route from a source to a destination. Each destination’s demand is given at the
foot of each column and each source’s stock is given at the end of each row. Enter numbers in
each cell to show how many units are to be sent along that route.
2 Begin with the top left-hand corner (the north-west corner). Allocate the maximum available
quantity to meet the demand at this destination (whilst not exceeding the stock at this source!).
3 As each stock is emptied, move one square down and allocate as many units as possible from
the next source until the demand of the destination is met.
As each demand is met, move one square to the right and again allocate as many units as
possible.
4 When all the stock is assigned, and all the demands met, stop.

In order to avoid degenerate For a problem involving In the examination,

solutions, movements are made m source vertices and n problems will be
between squares either vertically destination vertices you must restricted to a
or horizontally but never enter n  m  1 transportation maximum of 4 supply
diagonally. See example 4 quantities  0. This will points and 4 demand
not reduce throughout the points.
problem.

3
CHAPTER 1

Example 2

Depot W Depot X Depot Y Depot Z Stock

Supplier A 180 110 130 290 14
Supplier B 190 250 150 280 16
Supplier C 240 270 190 120 20
Demand 11 15 14 10 50

Use the north-west corner method to find an initial solution to the problem described in
example 1 and shown in the table.

1 W X Y Z Stock
A 14
B 16 1 Set up the table.
C 20
Demand 11 15 14 10 50

2 Beginning with the north-west corner, start to fill in

the number of units you wish to send along each route.
W X Y Z Stock 2 Start to fill in the number
of units you wish to send
A 11 14 along each route.
B 16
C 20
Depot W requires 11 lorry
Demand 11 15 14 10 50 loads. This does not exhaust
the stock of supplier A.

W X Y Z Stock 3 The demand at W has

been met so move one
A 11 3 14 square to the right and
B 16 allocate 14  11  3 units.
C 20
The stock at A is now
Demand 11 15 14 10 50 exhausted. The demand at
X has not been met.

W X Y Z Stock
A 11 3 14
B 12 16 As the stock at A has been
exhausted, move one square
C 20 down and allocate the
Demand 11 15 14 10 50 maximum possible number
of units from supplier B to
depot X. In this case,
15  3  12.

4
 Transportation problems

W X Y Z Stock
Now that the demand at
A 11 3 14 X has also been met, move
B 12 4 16 one square to the right and
use the remaining stock at B
C 20 to start to meet the demand
Demand 11 15 14 10 50 at Y.

W X Y Z Stock
The stock at B is now
A 11 3 14 exhausted (12  4  16)
B 12 4 16 and so move one square
down and use the stock at
C 10 20 C to fulfill the remaining
Demand 11 15 14 10 50 demand at Y.

W X Y Z Stock
A 11 3 14 Finally, move one square
to the right and use the
B 12 4 16 remaining stock at C to
C 10 10 20 meet the demand at Z.
Demand 11 15 14 10 50
This is the final table. All of the stock has been used and
all of the demands met.

The number of occupied cells (routes used) in the table  number of supply points  number of
demand points 1.
In this case the
number of occupied cells (routes used)  6,
number of supply points  3,
number of demand points  4
and 6  3  4  1.

Use this table, together with the table showing costs, to

work out the total cost of the solution.
W X Y Z Stock The cells shaded indicate the
costs of the routes currently
A 180 110 130 290 14 being used. It is these,
B 190 250 150 280 16 together with the number
of units being transported
C 240 270 190 120 20 along each route, that give
Demand 11 15 14 10 50 the cost of the solution.

The total cost of this solution is

(11  180)  (3  110)  (12  250)  (4  150)
 (10  190)  (10  120)  £9 010.

5
CHAPTER 1

1.3 You can adapt the algorithm to deal with unbalanced transportation problems.

䊏 When the total supply  total demand, we say the problem is unbalanced.

䊏 If the problem is unbalanced we simply add a dummy demand point with a demand chosen
so that total supply ⴝ total demand, with transportation costs of zero.

Example 3

A B C Supply
X 9 11 10 40
Y 10 8 12 60
Z 12 7 8 50
Demand 50 40 30

Three outlets A, B and C are supplied by three suppliers X, Y and Z. The table shows the cost, in
pounds, of transporting each unit, the number of units required at each outlet and the number
of units available at each supplier.
a Explain why it is necessary to add a dummy demand point in order to solve this problem.
b Add a dummy demand point and appropriate costs to the table.
c Use the north-west corner method to obtain an initial solution.

a The total supply is 150, but the total demand is 120.

A dummy is needed to absorb this excess, so that
total supply equals total demand.

b A B C D Supply We add a dummy column, D,

where the demand is 30 (the
X 9 11 10 0 40 amount by which the supply
Y 10 8 12 0 60 exceeds the demand), and
the transportation costs are
Z 12 7 8 0 50 zero (since there is no actual
Demand 50 40 30 30 150 transporting done)! The problem
is now balanced, the total
supply  the total demand.
c A B C D Supply
X 40 40
Y 10 40 10 60
Z 20 30 50
Demand 50 40 30 30 150

6
 Transportation problems

1.4 You understand what is meant by a degenerate solution and know how to
manage such solutions.

䊏 In a feasible solution to a transportation problem with m rows and n columns, if the number
of cells used is less than n ⴙ m ⴚ 1, then the solution is degenerate.
䊏 This will happen when an entry, other than the last, is made that satisfies the supply for a
given row, and at the same time, satisfies the demand for a given column.
䊏 The algorithm requires that n ⴙ m ⴚ 1 cells are used in every solution, so a zero needs to be
placed in a currently unused cell.

Example 4

A B C Supply
W 10 11 6 30
X 4 5 9 20
Y 3 8 7 35
Z 11 10 9 35
Demand 30 40 50 120
a Demonstrate that the north-west corner method gives a degenerate solution and explain why
it is degenerate.
b Adapt your solution to give a non-degenerate initial solution and state its cost.

a A B C Supply
W 30 30
X 20 20 Notice that there has been
a diagonal ‘move’ from cell
Y 20 15 35
WA to cell XB. Degenerate
Z 35 35 solutions can be avoided by
not allowing diagonal moves.
Demand 30 40 50 120
This solution is degenerate since it fulfils all the supply
and demand needs but only uses 5 cells WA, XB, YB,
YC and ZC. There are 4 rows and 3 columns so a non-
degenerate solution will use 4  3  1  6 cells.
b Start by placing the largest possible number in the
north-west corner
Having placed the 30 in the
A B C Supply NW corner, you now need to
place the next number in the
W 30 30 square to its right or in the
X 20 square underneath.

Y 35 Since both the supply and

the demand are satisfied, this
Z 35 means that you will have to
Demand 30 40 50 120 place a zero in either the cell to
its right or the cell underneath.

7
CHAPTER 1

There are two possible initial solutions, depending on

where you chose to place the zero.
Either A B C Supply
W 30 0 30
X 20 20
Y 20 15 35
Z 35 35
Demand 30 40 50 120

Or A B C Supply
W 30 30
X 0 20 20
Y 20 15 35 In fact the zero can be placed
Z 35 35 anywhere in the table, but
it is convenient to ‘stick to
Demand 30 40 50 120 the rule’ about restricting
the movement to one square
Both have a cost of (30  10)  0  (20  5) down or one square right.
 (20  8)  (15  7)  (35  9)  £980

Exercise 1A

Photocopy masters are available for the questions in this exercise.

In Questions 1 to 4, the tables show the unit costs of transporting goods from supply points to
demand points. In each case:
a use the north-west corner method to find the initial solution,
b verify that, for each solution, the number of occupied cells  number of supply points
 number of demand points  1.
c determine the cost of each initial solution.

1 P Q R Supply
A 150 213 222 32
B 175 204 218 44
C 188 198 246 34
Demand 28 45 37 110

2 P Q R S Supply
A 27 33 34 41 54
B 31 29 37 30 67
C 40 32 28 35 29
Demand 21 32 51 46 150

8
 Transportation problems

3 P Q R Supply
A 17 24 19 123
B 15 21 25 143
C 19 22 18 84
D 20 27 16 150
Demand 200 100 200 500

4 P Q R S Supply
A 56 86 80 61 134
B 59 76 78 65 203
C 62 70 57 67 176
D 60 68 75 71 187
Demand 175 175 175 175 700

5 A B C D Supply
X 27 33 34 41 60
Y 31 29 37 30 60
Z 40 32 28 35 80
Demand 40 70 50 20

Four sandwich shops A, B, C and D can be supplied with bread from three bakeries, X, Y,
and Z. The table shows the cost, in pence, of transporting one tray of bread from each
supplier to each shop, the number of trays of bread required by each shop and the number
of trays of bread that can be supplied by each bakery.
a Explain why it is necessary to add a dummy demand point in order to solve this problem,
and what this dummy point means in practical terms.
b Use the north-west corner method to determine an initial solution to this problem and
the cost of this solution.

6 K L M N Supply
A 35 46 62 80 20
B 24 53 73 52 15
C 67 61 50 65 20
D 92 81 41 42 20
Demand 25 10 18 22

A company needs to supply ready-mixed concrete from four depots A, B, C and D to four
work sites K, L, M and N. The number of loads that can be supplied from each depot and
the number of loads required at each site are shown in the table above, as well as the
transportation cost per load from each depot to each work site.
a Explain what is meant by a degenerate solution.
b Demonstrate that the north-west corner method gives a degenerate solution.
c Adapt your solution to give a non-degenerate intial solution.

9
CHAPTER 1

7 L M N Supply
P 3 5 9 22
Q 4 3 7 a
R 6 4 8 11
S 8 2 5 b
Demand 15 17 20

The table shows a balanced transportation problem. The initial solution, given by the north-
west corner method, is degenerate.
a Use this information to determine the values of a and b.
b Hence write down the initial, degenerate solution given by the north west-corner method.

Finding an improved solution

䊏 To find an improved solution, you need to:
1 use the non-empty cells to find the shadow costs (see Section 1.5)
2 use the shadow costs and the empty cells to find improvement indices (see Section
1.6)
3 use the improvement indices and the stepping stone algorithm to find an improved
solution (see Section 1.7).

1.5 You can find shadow costs.

䊏 Transportation costs are made up of two components, one associated with the source and
one with the destination. These costs of using that route, are called shadow costs.

Depot W Depot X Depot Y Depot Z Stock

Supplier A 180 110 14
Supplier B 250 150 16
Supplier C 190 120 20
Demand 11 15 14 10 50

In example 2, the cost of £250 in transporting one unit from supplier B to depot X must be
dependent on the features  location, toll costs etc, of both B and X.
Using the routes currently in use you can build up equations, showing the cost of transporting one
unit, such as
S(A)  D(X)  110 and S(C)  D(Z)  120 etc.
where S(A), D(X) are the costs due to supply point A and demand point X and so on,
respectively.
You need a value for each of the source components and each
You are only looking at the
of the destination components. You do not have sufficient costs of the routes used in
equations for a solution (five equations and six unknowns) but your current solution.
relative costs will do.
10
 Transportation problems

To find the shadow costs, follow these steps.

1 Start with the north-west corner, set the cost Put the first source component
linked with its source to zero. to zero, and set the destination
component to carry the cost of the
2 Move along the row to any other non-empty squares transportation in row 1.
and find any other destination costs in the same way.
3 When all possible destination costs for that row have been established, go to the start of the
next row.
4 Move along this row to any non-empty squares and use the destination costs found earlier,
to establish the source cost for the row. Once that has been done, find any further unknown
destination costs.
5 Repeat steps 3 and 4 until all source and destination costs have been found.

Example 5

Depot W Depot X Depot Y Depot Z Stock

Supplier A 180 110 130 290 14
Supplier B 190 250 150 280 16
Supplier C 240 270 190 120 20
Demand 11 15 14 10 50
Calculate the shadow costs given by the initial solution of the problem given in example 2 and
shown in the table.

Initial solution (see page 5) was

W X Y Z Stock
A 11 3 14
B 12 4 16
C 10 10 20
Demand 11 15 14 10 50

Focus on the costs of the routes being used  the non-

Remember to use the cost
empty squares values not the number
of items currently being
Depot Depot Depot Depot transported along that route.
Stock
W X Y Z
Supplier A 180 110 14
Supplier B 250 150 16
Supplier C 190 120 20
Demand 11 15 14 10 50

11
CHAPTER 1

Putting S(A) to zero, from row 1 we get D(W)  180 Put S(A)  0 arbitrarily and
and D(X)  110 then solve the equations
S(A)  D(W)  180 and
S(A)  D(X)  110.
Find the remaining shadow costs by ‘walking round’ the current
solution, noting the shadow costs you find round the edge
of the table, and using shadow costs found earlier to find the
remaining ones.

Shadow costs 180 110

Depot W Depot X Depot Y Depot Z Stock
0 Supplier A 180 110 14
Supplier B 250 150 16
Supplier C 190 120 20
Demand 11 15 14 10 50

Knowing that D(X)  110, you solve

Now move to Row 2. S(B)  D(X)  250 to get S(B)  140,
You know that D(X)  110, so you find S(B)  140 and use this together with the equation
hence D(Y)  10 S(B)  D(Y)  150 to get D(Y)  10.

Shadow costs 180 110 10

Depot W Depot X Depot Y Depot Z Stock
0 Supplier A 180 110 14
140 Supplier B 250 150 16
Supplier C 190 120 20
Demand 11 15 14 10 50
Move to Row 3. Knowing that D(Y)  10, you solve
You know that D(Y)  10, so we find S(C)  D(Y)  190 to get S(C)  180, and
S(C)  180 and hence that D(Z)  60. then solve S(C)  D(Z)  120 to find D(Z).

Shadow costs 180 110 10 ⴚ60

Depot W Depot X Depot Y Depot Z Stock
0 Supplier A 180 110 14
140 Supplier B 250 150 16
180 Supplier C 190 120 20
Demand 11 15 14 10 50

You have now found all source and all destination shadow costs.
S(A)  0 S(B)  140 S(C)  180
Do not be alarmed at the negative shadow
D(W)  180 D(X)  110 D(Y)  10 cost found for D(Z). It is a feature of
D(Z)  60 arbitrarily putting S(A)  0. You are simply
finding costs relative to the cost S(A).

12
 Transportation problems

Example 6

A B C D Supply
X 9 11 10 0 40
Y 10 8 12 0 60
Z 12 7 8 0 50
Demand 50 40 30 30 150
Calculate the shadow costs given by the initial solution of the problem given in example 3 and
shown in the table.

The north-west corner method gave the following initial solution

(see page 6)
A B C D Supply
X 40 40
Y 10 40 10 60
Z 20 30 50
Demand 50 40 30 30
We need to use the costs rather than the number of items being
transported. So we use the following numbers.
Shadow costs
A B C D Supply
X 9 40
Y 10 8 12 60
Z 8 0 50
Demand 50 40 30 30 150
Arbitrarily assign S(X)  0.
Shadow costs
A B C D Supply
0 X 9 40
Y 10 8 12 60
Z 8 0 50
Demand 50 40 30 30 150
Use this to work out the shadow cost for D(A).
Shadow costs 9
A B C D Supply
0 X 9 40
Y 10 8 12 60
Z 8 0 50
Demand 50 40 30 30 150

13
CHAPTER 1

Use this to work out the shadow cost for S(Y).

Shadow costs 9
A B C D Supply
0 X 9 40
1 Y 10 8 12 60
Z 8 0 50
Demand 50 40 30 30 150
We use this to work out the shadow costs for D(B) and D(C).
Shadow costs 9 7 11
A B C D Supply
0 X 9 40
1 Y 10 8 12 60
Z 8 0 50
Demand 50 40 30 30 150
Use these to work out the shadow cost for S(Z).
Shadow costs 9 7 11
A B C D Supply
0 X 9 40
1 Y 10 8 12 60
ⴚ3 Z 8 0 50
Demand 50 40 30 30 150
Use this to work out the shadow cost for D(D).
Shadow costs 9 7 11 3
A B C D Supply
0 X 9 40
1 Y 10 8 12 60
ⴚ3 Z 8 0 50
Demand 50 40 30 30 150

You do not have to show each stage of the table in the examination.
Just this final list of shadow costs is sufficient.

14
 Transportation problems

1.6 You can find improvement indices and use these to find entering cells.

It may be possible to reduce the cost of the initial solution by introducing a route that is not
currently in use. You consider each unused route in turn and calculate the reduction in cost which
would be made by sending one unit along that route. This is called the improvement index.
䊏 The improvement index in sending a unit from a source P to a demand point Q is found
by subtracting the source cost S(P) and destination cost D(Q) from the stated cost of
transporting one unit along that route C(PQ). i.e.
Improvement index for PQ  IPQ C(PQ)  S(P)  D(Q)
䊏 The route with the most negative improvement index will be introduced into the solution.
䊏 The cell corresponding to the value with the most negative improvement index becomes the
entering cell (or entering square or entering route) and the route it replaces is referred
to as the exiting cell (or exiting square or exiting route).
䊏 If there are two equal potential entering cells you may choose either. Similarly, if there are
two equal exiting cells, you may select either
䊏 If there are no negative improvement indices the solution is optimal.

Example 7

Shadow costs 180 110 10 ⴚ60

Depot W Depot X Depot Y Depot Z Stock
0 Supplier A 180 110 130 290 14
140 Supplier B 190 250 150 280 16
180 Supplier C 240 270 190 120 20
Demand 11 15 14 10 50

Use the shadow costs found in example 5, and shown in the table above, to calculate
improvement indices, and use these to identify the entering cell.

Focus on the routes not currently being used, BW, CW, CX, AY, AZ and BZ.
You already know that
S(A)  0 S(B)  140 S(C)  180 D(W)  180 D(X)  110 D(Y)  10 D(Z)  60
Improvement index for BW  IBW  C(BW)  S(B)  D(W)  190  140  180  130
Improvement index for CW  ICW  240  180  180  120
Improvement index for CX  ICX  270  180  110  20
Improvement index for AY  IAY  130  0  10  120
Improvement index for AZ  IAZ  290  0  (ⴚ60)  350
Improvement index for BZ  IBZ  280  140  (ⴚ60)  200
The entering cell is therefore BW, since this is the most negative.

15
CHAPTER 1

Example 8

X Y Z Supply
A 11 12 17 11
B 13 10 13 15
C 15 18 9 14
Demand 10 15 15
a Use the north-west corner method to find an initial solution to the transportation problem
shown in the table.
b Find the shadow costs and improvement indices.
c Hence determine if the solution is optimal.

a X Y Z Supply
A 10 1 11
B 14 1 15
C 14 14
Demand 10 15 15

b Shadow costs 11 12 15
X Y Z Supply
0 A 11 12 17 11
ⴚ2 B 13 10 13 15
ⴚ6 C 15 18 9 14
Demand 10 15 15
Improvement indices for cells:
BX  13  2  11  4
CX  15  6  11  10
CY  18  6  12  12
AZ  17  0  15  2
c There are no negative improvement indices, so the solution is
optimal.

Exercise 1B

Photocopy masters are available for the questions in this exercise.

Questions 1 to 4
Start with the initial, north-west corner, solutions found in questions 1 to 4 of exercise 1A.
In each case use the initial solution, and the original cost matrix, shown below, to find
a the shadow costs,
b the improvement indices
c the entering cell, if appropriate.
16
 Transportation problems

1 P Q R Supply
A 150 213 222 32
B 175 204 218 44
C 188 198 246 34
Demand 28 45 37

2 P Q R S Supply
A 27 33 34 41 54
B 31 29 37 30 67
C 40 32 28 35 29
Demand 21 32 51 46

3 P Q R Supply
A 17 24 19 123
B 15 21 25 143
C 19 22 18 84
D 20 27 16 150
Demand 200 100 200

4 P Q R S Supply
A 56 86 80 61 134
B 59 76 78 65 203
C 62 70 57 67 176
D 60 68 75 71 187
Demand 175 175 175 175

1.7 You can use the stepping-stone method to obtain an improved solution.

In example 7 you discovered that the most negative improvement index was BW with a value
of 130 (see page 15). This means that every time you send a unit along BW you save a cost of
130. Therefore you want to send as many units as possible along this new route. You have to
be careful, however, not to exceed the stock or the demand. To ensure this, you go through a
sequence of adjustments, called the stepping-stone method.
You are looking therefore for a cycle of adjustments, where you increase the value in one cell and
then decrease the value in the next cell, then increase the value in the next, and so on.

A popular mind picture is that you are using the cells as ‘stepping-stones’, placing one foot on each,
and alternately putting down your left foot (increasing) then right foot (decreasing) as you journey
around the table – hence the method’s nickname.

17
CHAPTER 1

The stepping-stone method

1 Create the cycle of adjustments. The two basic rules are:
a within any row and any column there can only be one increasing cell and one decreasing
cell.
b apart from the entering cell, adjustments are only made to non-empty cells.
2 Once the cycle of adjustments has been found you transfer the maximum number of units
through this cycle. This will be equal to the smallest number in the decreasing cells (since
you may not have negative units being transported).
3 You then adjust the solution to incorporate this improvement.

Example 9

In example 2, the table of costs was

Depot W Depot X Depot Y Depot Z Stock

Supplier A 180 110 130 290 14
Supplier B 190 250 150 280 16
Supplier C 240 270 190 120 20
Demand 11 15 14 10 50

and the initial solution was

W X Y Z Stock
A 11 3 14
B 12 4 16
C 10 10 20
Demand 11 15 14 10 50

at a cost of £9 010 (see page 5).

Obtain an improved solution and find the improved cost.

Use BW as the entering cell, since this gave the most

negative improvement index, 130 (see Example 7, page 15).
So BW will be an increasing cell. We enter a value of  into
this cell.

W X Y Z Stock
A 11 3 14
B ␪ 12 4 16
C 10 10 20
Demand 11 15 14 10 50

18
 Transportation problems

In order to keep the demand at W correct, you must

therefore decrease the entry at AW, so AW will be a
decreasing cell.
W X Y Z Stock
A 11 ⴚ ␪ 3 14
B ␪ 12 4 16
C 10 10 20
Demand 11 15 14 10 50

In order to keep the stock at A correct, you must therefore

increase the entry at AX, so AX will be an increasing cell.
W X Y Z Stock
A 11 ⴚ ␪ 3ⴙ␪ 14
B ␪ 12 4 16
C 10 10 20
Demand 11 15 14 10 50

In order to keep the demand at X correct, you must

therefore decrease the entry at BX, so BX will be a
decreasing cell.
W X Y Z Stock
This is as far as you can
A 11 ⴚ ␪ 3ⴙ␪ 14 go with adjustments since
B ␪ 12 ⴚ ␪ 4 16 the top two rows both
have an increasing and
C 10 10 20 decreasing cell.
Demand 11 15 14 10 50

Now choose a value for , the greatest value you can,

without introducing negative entries into the table. Look at
the decreasing cells and see that the greatest value of  is
11 (since 11  11  0).
Replace  by 11 in the table:
W X Y Z Stock
A 11 ⴚ 11 3 ⴙ 11 14
B 11 12 ⴚ 11 4 16
C 10 10 20
Demand 11 15 14 10 50

19
CHAPTER 1

This gives the improved solution:

W X Y Z Stock
A 14 14
B 11 1 4 16
C 10 10 20
Demand 11 15 14 10 50
Looking at the table of transportation costs, this solution
has a cost of £7 580 As a double check, it is always true that
New cost  cost of former solution  improvement index  .
In this case 7 580  9 010  (130)  11

You will notice that AW has become empty. AW is therefore the exiting cell.
Remember that the number of cells used in a feasible solution must equal the number of rows
plus the number of columns minus 1. So if you put a number into an entering cell, it must be
balanced by a number being removed from an exiting cell.

䊏 At each iteration we create one entering cell and one exiting cell.

䊏 To find an optimal solution, continue to calculate new shadow costs and improvement
indices and then apply the stepping-stone method. Repeat this iteration until all the
improvement indices are non-negative.

Example 10

Find an optimal solution for example 9.

This second iteration indicates the amount of working you need to

show in the examination.

Second iteration
Find the new shadow costs:
Shadow costs 50 110 10 ⴚ60
W X Y Z Stock
0 A 180 110 130 290 14
140 B 190 250 150 280 16
180 C 240 270 190 120 20
Demand 11 15 14 10 50

20
 Transportation problems

Finding the new improvement indices for the non-used cells:

AW  180  0  50  130
CW  240  180  50  10
CX  270  180  110  20
AY  130  0  10  120
AZ  290  0  60  350
BZ  280  140  60  200

So the new entering cell is CX, since this has the most negative
improvement index.
Applying the stepping-stone method gives
W X Y Z Stock
A 14 14
B 11 1 4 16
C  10   10 20
Demand 11 15 14 10 50

Looking at cells BX and CY we see that the greatest value for 

is 1
The new exiting cell will be BX,   1 and we get
W X Y Z Stock
A 14 14
B 11 5 16
C 1 9 10 20
Demand 11 15 14 10 50

The new cost is £7 560

Checking, 7 580  (20)  1  7560

Third iteration
New shadow costs
Shadow costs 70 110 30 ⴚ40
W X Y Z Stock
0 A 180 110 130 290 14
120 B 190 250 150 280 16
160 C 240 270 190 120 20
Demand 11 15 14 10 50

21
CHAPTER 1

New improvement indices for the non-used cells:

AW  180  0  70  110
CW  240  160  70  10
BX  250  120  110  20
AY  130  0  30  100
AZ  290  0  40  330
BZ  280  120  40  200
There are no negative improvement indices so this solution is optimal.
The solution is 110 units A to X
190 units B to W At this point, if there is an improvement index of
150 units B to Y 0, this would indicate that there is an alternative
optimal solution. To find it, simply use the cell
270 units C to X with the zero improvement index as the entering
190 units C to Y cell. (See question 2 in Mixed Exercise 1E.)
120 units C to Z

Some stepping-stone routes are not rectangles and some  values are not immediately apparent.

Example 11

Supermarket X Supermarket Y Supermarket Z Stock

Warehouse A 24 22 28 13
Warehouse B 26 26 14 11
Warehouse C 20 22 20 12
Demand 10 13 13
The table shows the unit cost, in pounds, of transporting goods from each of three warehouses,
A, B and C to each of three supermarkets X, Y and Z. It also shows the stock at each warehouse
and the demand at each supermarket.
Solve the transportation problem shown in the table. Use the north-west corner method to
obtain an initial solution. You must state your shadow costs, improvement indices, stepping-
stone routes,  values, entering cells and exiting cells. You must state the initial cost and the
improved cost after each iteration.

Check the problem is balanced, Supply  Demand  36, so we do not need to add a
dummy.
The north-west corner method gives the following initial solution
X Y Z Stock
A 10 3 13
B 10 1 11
C 12 12
Demand 10 13 13

22
 Transportation problems

The cost of the initial solution is £820

Calculate shadow costs
Shadow costs 24 22 10
X Y Z Stock
0 A 24 22 28 13
4 B 26 26 14 11
10 C 20 22 20 12
Demand 10 13 13
Calculating the improvement indices for the empty cells:
BX  26  4  24  2
CX  20  10  24  14
CY  22  10  22  10
AZ  28  0  10  18

Use CX as the entering cell, since it has the most negative improvement index.
X Y Z Stock
A 10   3 13
B 10   1 11
C  12   12
Demand 10 13 13

This stepping stone route is quite complicated. Take a minute or two to check how it has been
created. Start by putting  in CX, then to correct the demand of X, subtract  from cell AX, then to
correct the supply in A, add  to cell AY and so on, finishing at cell CZ.

The maximum value of  is 10

Either cell AX or cell BY can be the exiting cell, since both of these will go to zero. We simply
choose one of them, AX to be the exiting cell, the other will have a numerical value of zero.

Many candidates fail to understand the difference between an empty cell and one with a zero entry.
Zero is a number, just like 4 and ‘counts’ towards our m  n  1 entries in the table. An empty cell
has no number in it.

Improved solution is
X Y Z Stock
A 13 13
B 0 11 11
C 10 2 12
Demand 10 13 13

23
CHAPTER 1

The cost is now £680

We need to check for optimality, by calculating improvement indices.
The second set of shadow costs are:
Shadow costs 10 22 10
X Y Z Stock
0 A 24 22 28 13
4 B 26 26 14 11
10 C 20 22 20 12
Demand 10 13 13
Improvement indices
AX BX CY AZ
14 12 10 18
So the solution is not yet optimal and the next entering cell is CY.
X Y Z Stock
A 13 13
B 0 11   11
C 10  2 12
Demand 10 13 13
We can see from cell BY that the maximum value This looks odd, but the algorithm must
of  is 0. be followed.  can not be negative but
it can be any non-negative number,
The entering cell is CY and will have an entry of 0. and, as has been already mentioned, 0
The exiting cell is BY and it will now be empty. is a number.
The improved solution is:
X Y Z Stock
A 13 13
B 11 11
C 10 0 2 12
Demand 10 13 13
The cost is unchanged at £680
We again need to check for optimality, by calculating improvement indices.
Shadow costs:
Shadow costs 20 22 20
X Y Z Stock
0 A 24 22 28 13
ⴚ6 B 26 26 14 11
0 C 20 22 20 12
Demand 10 13 13

24
 Transportation problems

Improvement indices are

AX BX BY AZ
4 12 10 8
All the improvement indices are non-negative and so the solution is optimal.
X Y Z Stock
A 13 13
B 11 11
C 10 0 2 12
Demand 10 13 13

So the optimal solution is to send 13 units from A to Y

11 units from B to Z
10 units from C to X
2 units from C to Z
at a cost of £680

Exercise 1C

Questions 1 to 3
Complete your solutions to the transportation problems from questions 1, 2 and 4 in exercises
1A and 1B. You should demonstrate that your solution is optimal.

1 P Q R Supply
A 150 213 222 32
B 175 204 218 44
C 188 198 246 34
Demand 28 45 37

2 P Q R S Supply
A 27 33 34 41 54
B 31 29 37 30 67
C 40 32 28 35 29
Demand 21 32 51 46

3 P Q R S Supply
A 56 86 80 61 134
B 59 76 78 65 203
C 62 70 57 67 176
D 60 68 75 71 187
Demand 175 175 175 175

25
CHAPTER 1

The solution to question 3 requires a number of iterations, plus the optimality check – you will
certainly get lots of practise in implementing the algorithms!

P Q Stock The table shows the unit cost, in pounds,

4
of transporting goods from each of three
A 2 6 3
warehouses A, B and C to each of two
B 2 7 5 supermarkets P and Q. It also shows the
C 6 9 2 stock at each warehouse and the demand at
Demand 6 4 each supermarket.
Solve the transportation problem shown in the table. Use the north-west corner method to
obtain an initial solution. You must state your shadow costs, improvement indices, stepping-
stone routes,  values, entering cells and exiting cells. You must state the initial cost and the
improved cost after each iteration.

1.8 You can formulate a transportation problem as a linear programming problem.

In book D1 you met linear programming problems in two variables. In chapter 4 of this book you will
study linear programming problems in more than two variables.

Consider our first example

Depot W Depot X Depot Y Depot Z Stock

Supplier A 180 110 130 290 14
Supplier B 190 250 150 280 16
Supplier C 240 270 190 120 20
Demand 11 15 14 10 50

Let x11 (the entry in the 1st row, 1st column) be the number of x11, x23, x34 and so on
units transported from A to W, and x24 (the entry in the 2nd row are called the decision
and 4th column) be the number of units transported from B to Z, variables.
and so on, then you have the following solution.

Depot W Depot X Depot Y Depot Z Stock

Supplier A x11 x12 x13 x14 14
Supplier B x21 x22 x23 x24 16
Supplier C x31 x32 x33 x34 20
Demand 11 15 14 10 50

In this case a lot of these entries will be empty  there will be only 3  4  1  6 non-empty
cells, and all the rest will be blank. However, since you do not yet know which will be empty you
allow for any of them to be non-empty as you formulate the problem.

The objective is to minimise the total cost, which will be calculated by finding the sum of the
product of number of units transported along each route and the cost of using that route.
26
 Transportation problems

In this case the objective is

Minimise C  180x11  110x12  130x13  290x14
 190x21  250x22  150x23  280x24
 240x31  270x32  190x33  120x34
Looking at the stock, you know that the four entries in the first row must not exceed 14, so you
have our first constraint:
x11  x12  x13  x14 14

We can write similar constraints for each supplier and each depot, and include non-negativity
constraints. This enables you to formulate the transportation problem as a linear programming
problem.

䊏 The standard way of presenting a transportation problem as a linear programming problem:

• first define your decision variables,
• next write down the objective function,
• finally write down the constraints.

Example 12

R S T Supply
A 3 3 2 25
B 4 2 3 40
C 3 4 3 31
Demand 30 30 36

Formulate the transportation problem as a linear programming problem. You must state your
decision variables, objective and constraints.

Let xij be the number of units transported from i to j. FIRST define your decision
where i ∈ {A, B, C} variables
j ∈ {R, S, T }
and xij  0

Minimise C  3x11  3x12  2x13 NEXT write down the

 4x21  2x22  3x23 objective function.
 3x31  4x32  3x33

Subject to: x11  x12  x13 25 FINALLY write down the

x21  x22  x23 40 constraints.
x31  x32  x33 31
x11  x21  x31 30
x12  x22  x32 30
x13  x23  x33 36

27
CHAPTER 1

Exercise 1D

Formulate the following transportation problems as linear programming problems.

1 P Q R Supply
A 150 213 222 32
B 175 204 218 44
C 188 198 246 34
Demand 28 45 37

2 P Q R S Supply
A 27 33 34 41 54
B 31 29 37 30 67
C 40 32 28 35 29
Demand 21 32 51 46

3 P Q R Supply
A 17 24 19 123
B 15 21 25 143
C 19 22 18 84
D 20 27 16 150
Demand 200 100 200

4 P Q R S Supply
A 56 86 80 61 134
B 59 76 78 65 203
C 62 70 57 67 176
D 60 68 75 71 187
Demand 175 175 175 175

Mixed exercise 1E

1 L M Supply
A 20 70 15
B 40 30 5
C 60 90 8
Demand 16 12

The table shows the cost, in pounds, of transporting a car from each of three factories A, B
and C to each of two showrooms L and M. It also shows the number of cars available for
delivery at each factory and the number required at each showroom.

28
 Transportation problems

a Use the north-west corner method to find an initial solution.

b Solve the transportation problem, stating shadow costs, improvement indices, entering
cells, stepping-stone routes,  values and exiting cells.
c Demonstrate that your solution is optimal and find the cost of your optimal solution.
d Formulate this problem as a linear programming problem, making your decision variables,
objective function and constraints clear.
e Verify that your optimal solution lies in the feasible region of the linear programming
problem.

2 P Q R Supply
F 23 21 22 15
G 21 23 24 35
H 22 21 23 10
Demand 10 30 20

The table shows the cost of transporting one unit of stock from each of three supply points
F, G and H to each of three sales points P, Q and R. It also shows the stock held at each
supply point and the amount required at each sales point.
a Use the north-west corner method to obtain an initial solution.
b Taking the most negative improvement index to indicate the entering square, perform
two complete iterations of the stepping-stone method. You must state your shadow costs,
improvement indices, stepping-stone routes and exiting cells.
c Explain how you can tell that your current solution is optimal.
d State the cost of your optimal solution.
e Taking the zero improvement index to indicate the entering square, perform one futher
iteration to obtain a second optimal solution.

3 X Y Z Supply
J 8 5 7 30
K 5 5 9 40
L 7 2 10 50
M 6 3 15 50
Demand 25 45 100

The transportation problem represented by the table above is to be solved.

A possible north-west corner solution is

X Y Z Supply
J 25 5 30
K 40 40
L 0 50 50
M 50 50
Demand 25 45 100

29
CHAPTER 1

a Explain why it is was necessary to add a zero entry (in cell LY) to the solution.
b State the cost of this initial solution.
c Choosing cell MX as the entering cell, perform one iteration of the stepping-stone
method to obtain an improved solution. You must make your route clear, state your
exiting cell and the cost of the improved solution.
d Determine whether your current solution is optimal. Give a reason for your answer.
After two more iterations the following solution was found.

X Y Z Supply
J 30 30
K 20 20 40
L 50 50
M 25 25 50
Demand 25 45 100
e Taking the most negative improvement index to indicate the entering square, perform
one further complete iteration of the stepping-stone method to obtain an optimal
solution. You must state your shadow costs, improvement indices, stepping-stone route
and exiting cell.

4 S T U Supply
A 6 10 7 50
B 7 5 8 70
C 6 7 7 50
Demand 100 30 20
a Explain why a dummy demand point might be needed when solving a transportation
problem.

The table shows the cost, in pounds, of transporting one van load of fruit tree seedlings from
each of three greenhouses A, B and C to three garden centres S, T and U. It also shows the
stock held at each greenhouse and the amount required at each garden centre.
b Use the north-west corner method to obtain an initial solution.
c Taking the most negative improvement index in each case to indicate the entering square,
use the stepping-stone method to obtain an optimal solution. You must state your shadow
costs, improvement indices, stepping-stone routes, entering squares and exiting cells.
d State the cost of your optimal solution.
e Formulate this problem as a linear programming problem. Make your decision variables,
objective function and constraints clear.

30
 Transportation problems

Summary of key points

1 In solving transportation problems, you need to know
• the supply or stock
• the demand
• the unit cost of transporting goods.

2 When total supply  total demand, the problem is unbalanced.

3 A dummy destination is needed if total supply does not equal total demand.
Transport costs to this dummy destination are zero.

4 To solve the transportation problem, the north-west corner method is used.

5 The north-west corner method:

䊏 Create a table, with one row for every source and one column for every destination.
Each destination’s demand is given at the foot of each column and each source’s stock is
given at the end of each row. Enter numbers in each cell to show how many units are to
be sent along that route.
䊏 Begin with the top left-hand corner. Allocate the maximum available quantity to meet
the demand at this destination (but do not exceed the stock at this source).
䊏 As each stock is emptied, move one square down and allocate as many units as possible
from the next source until the demand of the destination is met.
䊏 As each demand is met, move one square to the right and again allocate as many units
as possible.
䊏 Stop when all the stock is assigned and all the demands are met.

6 In a feasible solution to a transportation problem with m rows and n columns, if the

number of cells used is less than n  m  1 then the solution is degenerate.

7 The algorithm requires n  m  1 cells to be used in every solution, so a zero must be

placed in an unused cell in a degenerate solution.

8 To find an improved solution, you need to:

䊏 use the non-empty cells to find the shadow costs
䊏 use the shadow costs and the empty cells to find improvement indices
䊏 use the improvement indices and the stepping-stones method to find an improved
solution.

9 Transportation costs are made up of two components, one associated with the source and
one with the destination. The costs of using a route are called shadow costs.
(See example 5 for how to work out shadow costs.)

10 The improvement index of a route is the reduction in cost which would be made by
sending one unit along that route.
Improvement index for route PQ  IPQ  C(PQ)  S(P)  D(Q) (see example 7).

11 The stepping-stone method is used to find an improved solution (see example 9).

31