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Unformatted text preview: Transportation problems In D1 you studied linear programming problems and their
solutions using the simplex algorithm. In this chapter and the next
one the solutions of two special types of linear programming
problems are considered by methods other than the simplex
algorithm. These are transportation problems and assignment
problems. Both of these problem types could be solved using the
simplex algorithm, but the process would result in very large
simplex tableaux and numerous simplex iterations.
Because of the special characteristics of each problem, however,
alternative solution methods requiring significantly less mathematical
manipulation than the simplex method have been developed. 1.1 The transportation problem
The transportation problem deals with the distribution of goods
from several points of supply, such as factories, often known as
sources, to a number of points of demand, such as warehouses,
often known as destinations.
Each source is able to supply a fixed number of units of the
product, usually called the capacity or availability, and each
destination has a fixed demand, usually known as the requirement.
The objective is to schedule shipments from sources to destinations
so that the total transport cost is a minimum.
Example 1
A concrete company transports concrete from three plants, 1, 2
and 3, to three construction sites, A, B and C.
The plants are able to supply the following numbers of tons per week:
Plant Supply (capacity) 1 300 2 300 3 100 1 2 Transportation problems The requirements of the sites, in numbers of tons per week, are:
Construction site Demand (requirement) A
B
C 200
200
300 The cost of transporting 1 ton of concrete from each plant to each
site is shown in the figure below in pounds per ton.
A 4
7 1 4
3
5 Plants B
Sites 5 2 8
9 C 5 3 For computational purposes it is convenient to put all the above
information into a table, as in the simplex method. In this table
each row represents a source and each column represents a
destination.
Sites
To A B C Supply
(availability) 1 4 3 8 300 2 7 5 9 300 3 4 5 5 100 Demand
(requirement) 200 200 300 From
Plants 1.2 Formulation as a linear
programming problem
Before proceeding with the solution of the transportation problem,
using the method developed especially for it, we will show that it
can be formulated as a linear programming problem.
The decision variables xij are the numbers of tons transported
from plant i (where i 1, 2, 3) to each site j (where j A, B, C). Transportation problems The objective function represents the total transportation cost £Z.
Each term in the objective function Z represents the cost of
tonnage transported on one route. For example, for the route
2 3 C the term is 9x2C , that is:
(cost per ton 9) Â (number of tons transported x2C )
Hence the objective function is:
Z 4x1A 3x1B 8x1C
7x2A 5x2B 9x2C
4x3A 5x3B 5x3C (1) Notice that in this problem the total supply is 300 300
100 700 and the total demand is 200 200 300 700.
so (total supply) (total demand).
This is called a balanced problem. Balanced problems are
considered throughout this chapter unless otherwise stated.
In a balanced problem all the products that can be supplied are
used to meet the demand. There are no slacks and so all
constraints are equalities rather than inequalities.
The constraints for the problem are obtained by considering the
rows and columns of the above table.
Plants: The total number of tons transported from plant 1 to
sites A, B and C must be equal to the supply, so:
x1A x1B x1C 300 (2a) Similarly, for plants 2 and 3:
x2A x2B x2C 300 (2b) x3A x3B x3C 100 (2c) Sites: The total number of tons received by site A from plants 1,
2 and 3 must be equal to the demand, so:
x1A x2A x3A 200 (3a) Similarly, for sites B and C:
x1B x2B x3B 200 (3b) x1C x2C x3C 300 (3c) All decision variables must be nonnegative, so:
xij 5 0 (4) for i 1, 2, 3 and j A, B, C.
The linear programming problem is then:
Minimise Z subject to the constraints given by equations (2), (3)
and (4). 3 4 Transportation problems Exercise 1A
1 A steel company has three mills, M1 , M2 and M3 , which can
produce 40, 10 and 20 kilotonnes of steel each year. Three
customers, C1 , C2 and C3 , have requirements of 12, 18 and 40
kilotonnes respectively in the same period. The cost, in units
of £1000, of transporting a kilotonne of steel from each mill to
each customer is shown in the figure below.
11 M1 C1 7
9 3 M2 9
C2
3 5
M3 3
3 C3 (a) Summarise all the above information in the form of a
table.
(b) Formulate the problem of carrying out the transportation
at minimum cost as a linear programming problem.
2 A transportation problem involves the following costs, supply
and demand:
W1 W2 W3 Supply F1 7 8 6 4 F2 9 2 4 3 F3 5 6 3 8 Demand 2 9 4 Formulate this problem as a linear programming problem.
3 Downside Mills produces carpet at plants in Abbeyville (A)
and Bridgeway (B). The carpet is shipped to two outlets in
Courtney (C) and Dove Valley (D). The cost, in £ per ton, of
shipping carpet from each of the two plants to each of the two
outlets is as follows: Transportation problems To C D A 20 32 B 35 15 From The plant at Abbeyville can supply 250 tons of carpet per
week and the plant at Bridgeway can supply 300 tons of carpet
per week. The Courtney outlet has a demand for 320 tons per
week and the outlet at Dove Valley has a demand for 230 tons
per week. The company wants to know the number of tons of
carpet to ship from each plant to each outlet in order to
minimise the total shipping cost. Formulate this transportation
problem as a linear programming problem. 1.3 Solution of the transportation
problem
Setting up a transportation tableau
The purpose of the transportation tableau is to summarise
conveniently and concisely all relevant data and keep track of
algorithm computations. In this respect it serves the same role as
the simplex tableau did for linear programming problems.
The transportation tableau for Example 1 is:
To
From A Supply
(availability) C 4 3
200 8 5 9 4 2 3 7 1 Demand
(requirement) B 5 5 200 300 300
300
100
ÀÀÀ
700 2 À À Total demand
and supply Each cell in the tableau represents the amount transported from
one source to one destination. The amount placed in each cell is
therefore the value of a decision variable for the cell. For example,
the cell at the intersection of row 2 and column C represents the
decision variable x2C . The smaller box within each cell contains the
unit transportation cost for that route. The final column and the
final row are the capacities of the plants and the requirements of 5 6 Transportation problems the sites respectively. The total demand and total supply has been
placed in the bottom righthand corner. It is suggested that you
should always do this to check that you have a balanced problem.
Later we will see how to deal with unbalanced problems where
total destination requirements are greater than or less than total
source capacities. The transportation algorithm
The transportation algorithm consists of three stages.
Stage 1
Stage 2
Stage 3 Find a transportation pattern that uses all the products
available and satisfies all the requirements. This is called
developing an initial solution.
Test the solution for optimality. If the solution is optimal
stop. If the solution is not optimal move to stage 3.
Use the steppingstone method to obtain an improved
solution and then return to stage 2. The algorithm stops when no further improvement is possible. Finding an initial solution
An initial feasible solution can be found by several alternative
methods. We will give just one, known as the northwest corner
method.
You will recall that prior to applying the simplex algorithm an
initial solution had to be established in the initial simplex tableau.
The initial solution in the transportation problem is not the origin.
The northwest corner method requires that we start in the upper
lefthand cell or northwest corner of the table and allocate units to
shipping routes as follows:
(i) Exhaust the supply at each row before moving down to the
next row.
(ii) Exhaust the requirements of each column before moving to
the right to the next column.
Let us implement this rule for Example 1.
From
1 To A B 200 100 2 100 Supply
300 200 200 200 300 100 3
Demand C 100 300 The figures in the above table were obtained in the following way: Transportation problems (i) Site A requires 200. All of these can be supplied by plant 1.
Plant 1 still has 100 left. These are sent to site B.
(ii) We now move to line 2. Site A is satisfied. Site B still requires
100. These can be sent from plant 2. Plant 2 still has 200 left.
These are sent to site C.
(iii) Site C still requires 100. These can be sent from plant 3.
This solution is feasible since supply and demand constraints are
all satisfied.
This solution corresponds to x1A 200, x1B 100, x2B 100,
x2C 200, x3C 100 and all other variables zero.
The total cost of this solution is:
4 Â (200) 3 Â (100) 5 Â (100) 9 Â (200) 5 Â (100)
800 300 500 1800 500
£3900
Example 2
Use the northwest corner rule to obtain an initial solution for the
transportation tableau below.
Site
To
From A 1
Plant 40 2 30 10
4
11 3
Demand B 70 C Supply 12 20
30
50 9 5 7 8 30
30 6 40
50
60
150 (i) Not all of site A's demand (70) can be met by plant 1. Site A
can receive 40 from plant 1 and plant 2 can provide the
remaining 30.
(It is a good idea to fill in the cells as you go along.)
(ii) Plant 2 still has 20 left and these can be sent to site B.
(iii) Site B still requires 30 and these can be provided by plant 3.
Plant 3 still has 30 left and these can be sent to site C to
exactly meet its requirement.
The total cost of this solution is:
(40 Â 10) (30 Â 4) (20 Â 5) (30 Â 8) (30 Â 6)
400 120 100 240 180
1040 7 8 Transportation problems Notice that at each step a row or column constraint is satisfied
except at the last stage when both a row and a column constraint
are satisfied.
An initial solution obtained using the northwest corner rule is
often called the northwest corner solution. As illustrated in the
example below, we can obtain initial solutions by starting at other
corners of the tableau.
Example 3
For the transportation tableau below write down:
(a) the northwest corner solution
(b) the northeast corner solution
(c) the southwest corner solution
(d) the southeast corner solution.
Assuming one of these gives an optimal solution, obtain the
optimal solution.
A B
40 2
300 65 70 1 30 250
400 350 (a) The northwest corner solution obtained using the rule given
earlier is:
A B 1 250 250 2 50 350 300 350 400 (b) The northeast corner solution is obtained by starting in the
top righthand corner and is:
A
1
2 B
250 250 300 100 400 300 350 Transportation problems (c) The southwest corner solution is obtained by starting in the
bottom lefthand corner and is:
A B 1 250
300 100 400 300 2 250 350 This is the same as (b).
(d) The southeast corner solution is obtained by starting in the
bottom righthand corner and is:
A B 1 250 250 2 50 350 300 350 400 This is the same as (a).
The cost of (a) is:
(250 Â 40) (50 Â 70) (350 Â 30) 24 000
The cost of (b) is:
(250 Â 65) (300 Â 70) (100 Â 30) 40 250
So the optimal solution is:
250 from 1 3 A
50 from 2 3 A
350 from 2 3 B
The minimum cost is 24 000. Exercise 1B
1 Use the northwest corner rule to obtain an initial solution for
the following transportation tableaux:
Supply
(a)
100
40
50
Demand 60 60 70 9 10 Transportation problems Supply (b) 60
40
140
Demand 120 80 40
Supply (c) 60
100
80
Demand 60 70 110
Supply (d) 90
40
50
Demand 60 70 50 2 For the transportation tableau below write down:
(a) the northwest corner solution
(b) the northeast corner solution
(c) the southwest corner solution
(d) the southeast corner solution.
Given that one of these gives an optimal solution, obtain the
optimal solution.
A Supply 8 2
90 6 10 1 Demand B 12 160 100
150 Transportation problems Testing the solution for optimality
An important rule
The method we shall now describe for testing a solution for
optimality can only be applied if one essential condition is satisfied.
This condition is:
The number of occupied cells (routes used) must be equal to
one less than the sum of the number of rows and the number
of columns.
In both Example 1 and Example 2 we have three rows and three
columns. Since the number of occupied cells is 5 in each case and
this is equal to (3 3 À 1), the condition is satisfied.
In the general case, when we have m sources and n destinations the
number of occupied cells must be (m n À 1).
When the number of occupied cells is less than this the solution is
said to be degenerate. Later in this chapter we will consider what
to do when we have degeneracy. Calculation of improvement indices
To test a solution for optimality we need to calculate an
improvement index for each unused cell. As a first step in this
process we must compute:
(i) a value for each row, denoted by Ri
(ii) a value for each column, denoted by Kj .
If Cij is the unit cost in the cell in the ith row and jth column of
the transportation tableau, then we can obtain the above values by
using:
Ri Kj Cij & for the occupied (used) cells. The Ri and Kj are called shadow costs.
The northwest corner solution for Example 1 is repeated below.
To
From
1 A
200 4
7 2 100
100 4 3
Demand B 200 C
3
5
5 200 Supply
8 200
100
300 9
5 300
300
100
700 11 12 Transportation problems So for this solution we have:
cell (1, 1):
cell (1, 2):
cell (2, 2):
cell (2, 3):
cell (3, 3): R1 K1
R1 K2
R2 K2
R2 K3
R3 K3 4
3
5
9
5 (1)
(2)
(3)
(4)
(5) Here we have five equations in six unknowns: R1 , R2 , R3 , K1 , K2 ,
and K3 . We may therefore choose one of the unknowns and solve
for the others. It is usual to choose R1 0, but you do not have
to, you may choose any one of the unknowns.
Then from:
(1)
(2)
(3)
(4)
(5) 0 K1 4 A K1 4
0 K2 3 A K2 3
R2 K2 R2 3 5 A R2 2
R2 K3 2 K3 9 A K3 7
R3 K3 R3 7 5 A R3 À2 It is always possible to solve the resulting set of equations in this
stepbystep way but the calculations may not proceed from
equation (1) 3 equation (2) 3 equation (3) 3 equation (4)
3 equation (5) as above. Notice that it is possible that the Ri and
Kj values may be positive, negative or zero. After solving for the R
and K values a few times you may become so proficient that you
will be able to do the calculations in your head. The following
table summarises the results and may be written down straight
away if you do the calculations in your head. This table is also
useful in calculating improvement indices.
K1 4
R1 0
R2 2
R3 À2 X 4
7
4 K2 3
X
X K3 7 3
5
5 8 X
X 9
5 Notice that we have indicated the used cells by X and only
included the costs. Do not get confused: it is the costs that are
involved in the calculations of the R and K values and only the
costs. Some students do get confused on this matter.
Having calculated the R and K values we can now calculate, for
each unused cell, an improvement index Iij using the formula:
& Iij Cij À Ri À Kj By definition of the Ri and Kj values the improvement index for
the occupied (used) cells is zero. Transportation problems For the northwest corner solution for Example 1 we have, using
the table above:
cell (1, 3):
cell (2, 1):
cell (3, 1):
cell (3, 2): I13 C13 À R1 À K3 8 À 0 À 7 1
I21 C21 À R2 À K1 7 À 2 À 4 1
I31 C31 À R3 À K1 4 À (À2) À 4 2
I32 C32 À R3 À K2 5 À (À2) À 3 4 A warning: Do be careful when you are subtracting negative
numbers. For example, 2 À (À2) 4.
Again, with a little practice these calculations can be done in your
head and the results written into a table such as that below.
Improvement indices
K1 4
R1 0 0 R2 2 1 R3 À2 2 4
7
4 K2 3
0
0
4 3
5
5 K3 7
1
0
0 8
9
5 Notice that the occupied cells have been indicated with ringed zeros.
& If all the improvement indices are greater than or equal to zero, an
optimal solution has been reached. If there are any negative improvement indices then it is possible to
improve the current solution and decrease the total shipping costs.
Each negative index computed represents the amount by which the
total transportation costs could be decreased if 1 unit were shipped
on that route.
In the light of the above, the northwest corner solution to
Example 1 is optimal. The solution to the transportation problem
posed in Example 1 is:
send 200 tons from plant 1 to site A
100 tons from plant 1 to site B
100 tons from plant 2 to site B
200 tons from plant 2 to site C
100 tons from plant 3 to site C The cost of the transportation pattern, which is the minimum total
cost, is £3900, as shown earlier.
Example 4
Due to roadworks the cost of shipping 1 ton from plant 1 to site A
in Example 1 increases to £6. All the remaining data remain the
same. Obtain the improvement indices for the northwest corner
solution now. 13 14 Transportation problems The northwest corner solution is:
To
From
1 A B
6 200 7 2 100
100 4 3
Demand 200 C
3
5
5 200 Supply
8 200
100
300 9
5 300
300
100
700 The equations to be solved for the R and K values are now:
R1 K1
R1 K2
R2 K2
R2 K3
R3 K3 6
3
5
9
5 Taking R1 0, we obtain R2 2, R3 À2, K1 6, K2 3 and
K3 7.
The improvement indices for the nonoccupied cells are then:
I13
I21
I31
I32 8À0À71
7 À 2 À 6 À1
4 À (À2) À 6 0
5 À (À2) À 3 4 As there is now a negative improvement index the solution is no
longer optimal and can be improved. Exercise 1C
1 In Example 2 we considered the transportation problem given
by the following table of supply, demand and unit costs:
To A B C Supply 1 10 12 9 40 2 4 5 7 50 3 11 8 6 60 Demand 70 50 30 From Transportation problems We showed in Example 2 that the northwest corner rule gave
the following initial solution:
To
From A 1 40 2 30 B
10
4
11 3
Demand 70 C Supply 12 20
30
50 9 5 7 8 30
30 6 40
50
60
150 By calculating improvement indices for the unoccupied cells,
show that this solution is optimal. Give the transportation
pattern and its cost.
2 A company has three factories, F1 , F2 and F3 , and three
warehouses, W1 , W2 and W3 . The table below shows the costs
Cij of sending one unit of product from factory Fi to
warehouse Wj . Also shown are the availabilities at each
factory and the requirements of each warehouse.
To W1 W2 W3 Availability F1 8 6 7 4 F2 2 4 9 3 F3 6 3 5 8 Requirement 9 4 2 From (a) Use the northwest corner rule to write down a possible
pattern of distribution and find its cost.
(b) Calculate the improvement indices for the unused cells
and hence show that this distribution pattern is optimal.
3 A company has three warehouses, W1 , W2 and W3 , which are
supplied by three suppliers, S1 , S2 and S3 . The table below
shows the cost Cij of sending one case of goods from supplier Si
to warehouse Wj , in appropriate units. Also shown in the table
are the number of cases available at each supplier and the 15 16 Transportation problems number of cases required at each warehouse. The total number
of cases available is equal to the total number of cases required.
W1 W2 W3 Availability S1 10 4 11 14 S2 12 5 8 10 S3 9 6 7 6 Requirement 8 10 12 (a) Use the northwest corner rule to obtain a possible pattern
of distribution.
(b) Calculate the improvement indices for the unused cells
and hence show that this distribution pattern is not optimal. Obtaining an improved solution
(the steppingstone method)
Now that we have decided that the northwest corner solution in
Example 4 can be improved by using route (2, 1), we must decide
which route is to be removed from the present solution. This
procedure is similar to that used in the simplex algorithm, where we
first choose an entering variable and then choose a departing variable.
In order to determine the cell to be used we need to draw a closed
path or loop in the transportation tableau.
& A closed path or loop is a sequence of cells in the transportation
tableau such that:
(i) each pair of consecutive cells lies in either the same row or
the same column
(ii) no three consecutive cells lie in the same row or column
(iii) the first and last cells of a sequence lie in the same row or
column
(iv) no cell appears more than once in the sequence. The loops we are interested in have the following property:
the first cell is unused and all the other cells are used.
For Example 4 above the used cells are shown by X in the table below:
X X
X X X X Transportation problems The possible loops with the above property are:
X X
X X X X X X X X X (i) starting at (2, 1)
X (ii) starting at (1, 3) X
X X X X X X X
X (iii) starting at (3, 2) (iv) starting at (3, 1) It should be noted that in developing the loop it is possible to skip
over both unused and used cells.
Since we wish to use cell (2, 1), we now concentrate our attention on
diagram (i) above. To ensure that the constraints remain satisfied, we
add y to the number in (2, 1) and (Ày) and (y) to the numbers in
the other cells, alternately, as we move round the loop. We then have:
A B C 1 200 À y 100 y 2 y 100 À y 300
200
100 3
200 200 300
100 300 You may check each row and column in turn to see that the
constraints are satisfied.
To satisfy the nonnegativity condition and improve the solution
as much as possible we make y as large as possible without making
any entry negative. So choose y min(100, 200) 100.
Taking y 100 gives us the new improved solution:
100 6 200 3 7 5 4 100 5 8 200
100 Notice that route (2, 2) is not now used.
We will now test this solution for optimality. 9
5 17 18 Transportation problems The equations to be solved for the R and K values are:
R1 K1
R1 K2
R2 K1
R2 K3
R3 K3 6
3
7
9
5 K1 6
R1 0 X R2 1 X K2 3 6 X 3 7 5 4 R 3 À3 K3 8 5 8 X
X 9
5 Taking R1 0, we obtain R2 1, R3 À3, K1 6, K2 3 and
K3 8 (see table).
The improvement indices for the unused cells are:
I13
I22
I31
I32 8À0À80
5À1À31
4 À (À3) À 6 1
5 À (À3) À 3 5 All improvement indices are nonnegative and so this solution is
optimal:
send 100 tons from plant 1 to site A
200 tons from plant 1 to site B
100 tons from plant 2 to site A
200 tons from plant 2 to site C
100 tons from plant 3 to site C Take cost of this transportation pattern is:
(100 Â 6) (200 Â 3) (100 Â 7) (200 Â 9) (100 Â 5)
600 600 700 1800 500
£4200 Alternative optimal solutions
You may have noticed that one of the improvement indices
calculated above (I13 ) was zero. This indicates that the route (1, 3)
could be used without changing the total overall cost:
X
X X
X
X The routes used in the optimal solution obtained are shown by X
in the above diagram. Also shown is the closed path or loop
starting at (1, 3). Proceeding as above we have: Transportation problems 100 À y 200 y 100 y 200 À y
100 In this case y min(100, 200) 100, so we have the alternative
solution:
6 200 3 send 7 5 4 200 100 5 100
100 8
9
5 200 tons from plant 1 to site B
100 tons from plant 1 to site C
200 tons from plant 2 to site A
100 tons from plant 2 to site C
100 tons from plant 3 to site C The total cost of this transportation pattern is:
(200 Â 3) (100 Â 8) (200 Â 7) (100 Â 9) (100 Â 5)
600 800 1400 900 500
£4200, as before Exercise 1D
1 In question 3 of Exercise 1C you showed that the solution
obtained by the northwest corner rule was not optimal.
Obtain an improved solution using the steppingstone method
and show that this solution is optimal.
2 A lumber company ships pine flooring from its three mills, A1 ,
A2 and A3 , to three building suppliers, B1 , B2 and B3 . The
table below shows the demand, availabilities and unit costs of
transportation. Starting with the northwest corner solution
and using the steppingstone method, determine the
transportation pattern that minimises the total cost.
B1 B2 B3 Availability A1 3 3 2 25 A2 4 2 3 40 A3 3 4 3 31 Demand 30 30 36 19 20 Transportation problems 1.4 Some special situations
Unbalanced transportation problems
So far in this chapter we have considered only balanced
transportation problems in which (total supply) (total demand).
In many realworld situations, however, supply exceeds demand, or
vice versa. When we have an unbalanced transportation problem we
need to convert it into a balanced transportation problem in order
to use the solution procedure described earlier in this chapter.
Example 5
Consider the problem given in Example 3 but with source 1 now
having 350 products available rather than 250. The transportation
tableau for this problem is now obtained by adding a dummy
column, that is a dummy destination.
This destination is assigned a dummy requirement equal to the
surplus available, in this case (400 350) À (300 350) 100. The
transportation costs in the dummy column are zero, as these
surplus products will not, of course, be transported.
The transportation tableau is now:
To
From A B Dummy Supply 40 2
Demand 65 0 70 1 30 0 300 350 350
400 100 750 We can now apply the transportation algorithm described earlier.
The northwest corner solution is:
300 40
70 300 50
300 65
30 350 0 100 0 100 350
400
750 The R and K values are:
K1 40 K2 65 K3 35
R1 0
R2 À35 X 40
70 X
X 65
30 0 X 0 Transportation problems The improvement indices for the unused cells are:
I13 0 À 0 À 35 À35
I21 70 À (À35) À 40 65
Since I13 À35, that is it is negative, we can improve the solution
by using the route (1, 3).
The required loop is:
X X
X X Adding and subtracting y as required gives:
300 50 À y y 300 y 100 À y In this case y min(50, 100) 50 and so we obtain the improved
pattern:
300 50
350 50 The R and K values for this pattern are:
K1 40 K2 30
R1 0
R2 0 X 40
70 K3 0 65 X 30 X
X 0
0 and the improvement indices for the unused cells are:
I12 65 À 0 À 30 35
I21 70 À 0 À 40 30
Since all improvement indices are positive, this pattern is optimal.
Send 300 tons from 1 to A
350 tons from 2 to B There will be 50 products left at 1 and 50 products left at 2.
The cost of this transportation pattern is:
(300 Â 40) (350 Â 30)
22 500
This is to be compared with the cost of the northwest corner
solution, which is:
(300 Â 40) (50 Â 65) (300 Â 30)
24 250 21 22 Transportation problems If the total demand exceeds the total supply, then we add a
dummy source with supply (total demand) À (total supply). A
dummy row is added to the tableau and as before the
transportation costs in this row will be zero. The transportation
algorithm can then be used to solve the problem. Exercise 1E
1 A manufacturing company has three factories, F1 , F2 and F3 ,
and two retail outlets, R1 and R2 . It wishes to transport its
products from its factories to its outlets at minimum total
cost. The table below gives details of demand and supply, and
also the unit costs of transportation.
To R1 R2 Supply F1 2 6 30 F2 2 4 60 F3 6 9 20 Demand 60 20 From (a) Write down the northwest corner solution to the
corresponding balanced problem.
(b) Show that this solution is optimal.
(c) State the optimal transportation pattern and gives its cost.
2 A transportation problem involves the following costs, supply
and demand:
To D1 D2 D3 Supply S1 7 8 10 50 S2 9 7 8 60 Demand 70 30 40 From (a) Show that the problem is unbalanced and draw a
transportation tableau for the corresponding balanced
problem. Transportation problems (b) Write down the northwest corner solution and then use
the steppingstone method to obtain an optimal solution.
(c) Interpret your optimal solution and obtain the minimum
cost. Degeneracy
& A feasible solution to a transportation problem is degenerate if the
number of used cells is less than the magic number (m n À 1). Degeneracy can occur during the determination of the initial
feasible solution or it can occur during subsequent iterations when
the steppingstone method is used.
Degeneracy requires a special procedure to deal with the problem. To
handle degenerate problems we create an artificially occupied cell,
that is we place a zero in one of the unused cells and then treat that
cell as if it were occupied. The cell must be chosen so that it is
possible to calculate all the R and K values. There is usually some
flexibility in selecting the unused cell that will receive the zero. Degeneracy in an initial solution
Consider the following transportation tableau:
To
From A B
5 1 C
4 Supply
12 8 3
Demand 6 10 11 2 7 11 100 200 200 200
100
200
500 Using the northwest corner method gives the initial solution:
To
From
1 A
100 5
8 2 100
100 11 3
Demand B 100 C
4 12 6 10 7 200 Supply 200
200 11 200
100
200
500 23 24 Transportation problems This initial solution is degenerate since the number of used cells is
4, which is less than the magic number (3 3 À 1) 5. The
degeneracy arises because when 100 is placed in (2, 2) a row and a
column requirement are satisfied simultaneously. To deal with this
problem we place a zero in an unused cell. In this case we may
choose either cell (2, 3) or cell (3, 2).
If we choose to place a zero in (2, 3), then the equations to be
solved for R and K are:
R1 K1
R1 K2
R2 K2
R2 K3
R3 K3 5
4
6
10
11 K1 5
R1 0 X R2 2
R3 3 K2 4 5
8
11 X
X K3 8 4
6
7 12 X
X 10
11 Taking R1 0, we obtain R2 2, R3 3, K1 5, K2 4 and
K3 8.
The improvement indices for the unused cells are:
I13
I21
I31
I32 12 À 0 À 8 4
8À2À51
11 À 3 À 5 3
7À3À40 Since all these indices are nonnegative, this solution is optimal.
Since also I32 0 there is an alternative solution in which route (3,
2) is used. This corresponds to making the alternative choice for
the cell to receive the zero.
The optimal solution to this problem is:
send 100 units from 1 to A
100 units from 1 to B
100 units from 2 to B
200 units from 3 to C The total cost of this pattern is:
(100 Â 5) (100 Â 4) (100 Â 6) (200 Â 11)
3700 Transportation problems Degeneracy during later solution stages
Consider the transportation tableau:
To A From B
2 1 C
4 3 2
Demand 5 8 100 Supply
150 6 150 200 100 350 The initial solution obtained by the northwest corner rule is:
To A From
1 100 B
2 50 3 2
Demand C
4 100 5 8 100 Supply 100 150 150 6 200 100 350 The R and K values are:
K1 2
R1 0 X K2 4 2 X 3 R2 4 X K3 2 4
8 5 X 6 The improvement indices for the unused cells are:
I13 5 À 0 À 2 3
I21 3 À 4 À 2 À3
There is a negative improvement index I21 and so the solution may
be improved:
A
1
2 B 100 À y
X 50 y
X y X
100 À y C X 25 26 Transportation problems The relevant loop is shown above together with the y adjustment.
In this case we choose y 100 and obtain the new solution:
To A From B
2 1
2 100 Demand 150 3 C
4 5 8 100 Supply 6 100 150 150
200 100 350 The number of used cells is 3 and since this is less than the magic
number (3 2 À 1) 4 the solution is degenerate. This is because
two formerly occupied cells have dropped to zero. To proceed we
have to add a zero to one of the unoccupied cells and then treat it
as occupied. It is usual to choose the cell with the lowest shipping
cost, in this case (1, 1).
The R and K values are:
K1 2
R1 0 X R2 1 X K2 4
2
3 X K3 5
4
8 5 X 6 The improvement indices of the unused cells are:
I13 5 À 0 À 5 0
I22 8 À 1 À 4 3
Again there is an alternative solution obtained by using (1, 3)
instead of (1, 1) as the cell in which to place the zero.
The optimal solution is:
send 150 units from 1 to B
100 units from 2 to A
100 units from 2 to C Total cost of this pattern is:
(150 Â 4) (100 Â 3) (100 Â 6)
1500 Transportation problems Exercise 1F
1 A manufacturing company produces diesel engines in three
cities, C1 , C2 and C3 , and they are purchased by three trucking
companies, T1 , T2 and T3 . The table below shows the number
of engines available at C1 , C2 and C3 and the number of
engines required by T1 , T2 and T3 . It also shows the
transportation cost per engine (in £100s) from sources to
destinations. The company wishes to keep the total
transportation costs to a minimum.
To T1 T2 T3 Supply C1 3 2 3 25 C2 4 2 3 35 C3 3 2 6 20 Demand 30 30 20 From (a) Write down the northwest corner solution and state why
it is degenerate.
(b) Use the steppingstone method to obtain the optimal
solution.
(c) Give the transportation pattern and its total cost.
2 A builders' merchant has 13 tons of sand at site X, 11 tons at site
Y and 10 tons at site Z. He has orders for 9 tons from customer
A, 13 tons from customer B and 12 tons from customer C. The
cost per ton (in £10s) of moving the sand between depots and
customers is given in the table below.
A B C X 1 2 4 Y 1 3 4 Z 5 7 5 (a) Write down the northwest corner solution.
(b) Use the steppingstone method to obtain an optimal
solution. State the minimum cost of transportation. 27 28 Transportation problems 3 A transportation problem involves the following costs, supply
and demand:
D1 D2 D3 Supply S1 17 8 14 30 S2 15 10 20 20 S3 20 5 10 10 Demand 10 20 30 (a) Write down the northwest corner solution.
(b) Use the steppingstone method to obtain an optimal
solution and give its cost. SUMMARY OF KEY POINTS
1 The shadow costs Ri , for the ith row, and Kj , for the jth
column, are obtained by solving Ri Kj Ci j for
occupied cells, taking R1 0 arbitrarily.
2 The improvement index Ii j for an unoccupied cell is
defined by
Ii j Ci j À Ri Kj .
3 If all improvement indices are greater than or equal to
zero then an optimal solution has been reached.
4 A closed path or loop is a sequence of cells in the
transportation tableau such that
(i) each pair of consecutive cells lies in either the same
row or the same column
(ii) no three consecutive cells lie in the same row, or
column
(iii) the first and last cells of a sequence lie in the same
row or column
(iv) no cell appears more than once in the sequence.
5 A feasible solution to a transportation problem is
degenerate of the number of used cells is less than the
magic number (m n À 1), where m is the number of
rows and n is the number of columns. ...
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 Spring '12
 AbhinavDhar

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