Computational Methods for
Management and Economics
Carla Gomes
Module 8b
The transportation simplex method
The transportation and assignment
problems
• Special types of linear programming
problems.
• The structure of these problems leads to
algorithms – streamlined versions of the
simplex method - more efficient than the
standard simplex method.
The transportation problem
Setting up the tableau for the general
simplex method
• Convert minimization problem into maximization
problem
• Big M or 2 phase method to introduce m+ n
artificial variables (equality constraints)
• Find an initial BF solution
Insert table 8.13
Note:all entries not shown are 0.
Iterations in the general simplex
method
• Optimality test and step 1 of iteration:
– Select an entering basic variable  knowing current
row 0 (which is obtained by subtracting a certain
multiple of another row from the preceding row 0).
• Step 2 (leaving basic variable)  identifies which basic
variable reaches zero as the entering variable is increased
(which is done by comparing the current coefficients of the
entering basic variable and the corresponding rhs).
• Step 3 (determine new BF solution)  subtracting certain
multiples of one row from other rows
Iterations in the general simplex method
applied to transportation problem
• General form of row 0
– ui – multiple of the original row i that has been
subtracted (directly or indirectly) from original row 0.
– vj – multiple of the original row i that has been
subtracted (directly or indirectly) from original row 0.
– cij – ui – vj – 0 if it is a basic variable; otherwise it is
interpreted as the rate at which Z will change as xij is
increased.
Insert table – 8:14
Transportation simplex method
Does the transportation simplex method
do all the general simplex method operations?
•
•
•
•
NO!
No artificial variables are needed  simple procedure is used to
identify the initial BF solution.
Row 0 can be obtained without using any other row –a simple
procedure allows us to calculate the values of ui and vi directly.
Basically by solving a simple procedure that allows us to find the
solution to the set of equations for the basic variables that have a
coefficient 0 in row 0, i.e., cij – ui – vj=0.
The leaving variable can also be identified in a simple way due to the
special structure of the problem, without explicitly using the
coefficients of the new entering basic variable.
The new BF solution can also be identified very easily.
Almost the entire simplex tableau can be eliminated!
Transportation simplex method
• Much faster than standard simplex method
– standard simplex method
m+n+1 rows
(m+1)(n+1) = mn + m + n + 1 columns
– transportation simplex method
m + 2 rows
n + 2 columns
Assume m = 10 n = 100 and see the difference!
Tableau for Transportation simplex
method
Insert table 8.15
(*) Because there are m+n equations a BF solution in this case only has m+n-1
basic variables – the other variable can be obtained as a function of the m+n-1 variables
Transportation simplex method
• Initialization: finding the initial BF solution
• Optimality test
• Iteration
General procedure for construction of
an initial basic feasible solution for TP
1.
2.
3.
4.
From the rows and columns still under consideration select the next
basic variable (allocation) according to some criterion
Make that allocation the largest possible (the smallest of the
remaining supply in its row vs. remaining demand in its column)
Eliminate the row or column that zeroed the corresponding
supply/demand (in case of tie, randomly pick between the row vs.
column; this is a case of a degenerate basic solution, i.e, a basic
variable with value 0).
If only one row or column remains under consideration, then the
procedure is complete by selecting every remaining variable
associated with that row or column to be basic with the only feasible
allocation. Otherwise return to step 1.
Criteria for selecting next basic variable
There are several methods. Here are three:
1. Northwest Corner Method
2. Minimum Cost Method
3. Vogel’s Method
1. Northwest Corner Method
To find the bfs by the NWC method:
Begin in the upper left (northwest) corner of the
transportation tableau and set x11 as large as
possible (here the limitations for setting x11 to a
larger number, will be the demand of demand
point 1 and the supply of supply point 1. Your
x11 value can not be greater than minimum of
this 2 values).
According to the explanations in the previous slide
we can set x11=3 (meaning demand of demand
point 1 is satisfied by supply point 1).
5
6
2
3
5
2
3
3
2
6
2
X
5
2
3
After we check the east and south cells, we saw that
we can go east (meaning supply point 1 still has
capacity to fulfill some demand).
3
2
X
6
2
X
3
2
3
3
2
X
3
3
2
X
X
2
3
After applying the same procedure, we saw that we
can go south this time (meaning demand point 2
needs more supply by supply point 2).
3
2
3
X
2
1
2
X
X
3
2
3
X
3
X
2
1
X
2
X
X
X
2
Finally, we will have the following bfs, which is:
x11=3, x12=2, x22=3, x23=2, x24=1, x34=2
3
2
3
X
X
X
2
X
1
X
2
X
X
2. Minimum Cost Method
The Northwest Corner Method dos not utilize shipping
costs. It can yield an initial bfs easily but the total
shipping cost may be very high. The minimum cost
method uses shipping costs in order come up with a
bfs that has a lower cost. To begin the minimum cost
method, first we find the decision variable with the
smallest shipping cost (Xij). Then assign Xij its largest
possible value, which is the minimum of si and dj
After that, as in the Northwest Corner Method we
should cross out row i and column j and reduce the
supply or demand of the noncrossed-out row or
column by the value of Xij. Then we will choose the
cell with the minimum cost of shipping from the
cells that do not lie in a crossed-out row or column
and we will repeat the procedure.
An example for Minimum Cost Method
Step 1: Select the cell with minimum cost.
2
3
5
6
5
2
1
3
5
10
3
12
8
8
4
4
6
6
15
Step 2: Cross-out column 2
6
5
3
2
5
5
3
1
2
2
8
12
X
6
4
8
3
4
6
15
Step 3: Find the new cell with minimum shipping
cost and cross-out row 2
2
3
5
6
5
2
1
3
5
X
2
8
3
10
8
X
4
4
6
6
15
Step 4: Find the new cell with minimum shipping
cost and cross-out row 1
2
3
5
6
X
5
2
1
3
5
X
2
8
3
5
8
X
4
4
6
6
15
Step 5: Find the new cell with minimum shipping
cost and cross-out column 1
2
3
5
6
X
5
2
1
3
5
X
2
8
3
8
4
6
5
X
X
4
6
10
Step 6: Find the new cell with minimum shipping
cost and cross-out column 3
2
3
5
6
X
5
2
1
3
5
X
2
8
3
8
5
4
6
4
X
X
X
6
6
Step 7: Finally assign 6 to last cell. The bfs is found
as: X11=5, X21=2, X22=8, X31=5, X33=4 and X34=6
2
3
5
6
X
5
2
1
3
5
X
2
8
3
8
5
4
4
X
X
6
6
X
X
X
3. Vogel’s Method
1. Begin with computing each row and column a
penalty. The penalty will be equal to the difference
between the two smallest shipping costs in the row
or column.
2. Identify the row or column with the largest penalty.
Find the first basic variable which has the smallest
shipping cost in that row or column. Assign the
highest possible value to that variable, and cross-out
the row or column as in the previous methods.
3. Compute new penalties and use the same procedure.
An example for Vogel’s Method
Step 1: Compute the penalties.
6
7
15
Demand
Column Penalty
Supply
Row Penalty
10
7-6=1
15
78-15=63
8
80
78
15
5
5
15-6=9
80-7=73
78-8=70
Step 2: Identify the largest penalty and assign the
highest possible value to the variable.
6
7
Supply
Row Penalty
5
8-6=2
15
78-15=63
8
5
15
Demand
Column Penalty
80
78
15
X
5
15-6=9
_
78-8=70
Step 3: Identify the largest penalty and assign the
highest possible value to the variable.
6
7
5
Column Penalty
Row Penalty
0
_
15
_
8
5
15
Demand
Supply
80
78
15
X
X
15-6=9
_
_
Step 4: Identify the largest penalty and assign the
highest possible value to the variable.
6
0
7
5
Supply
Row Penalty
X
_
15
_
8
5
15
80
78
Demand
15
X
X
Column Penalty
_
_
_
Step 5: Finally the bfs is found as X11=0, X12=5,
X13=5, and X21=15
6
0
7
5
Supply
Row Penalty
X
_
X
_
8
5
15
80
78
15
Demand
X
X
X
Column Penalty
_
_
_
Russell’s method
1. For each row and column remaining under
consideration determine its ui – the largest cij still in
that row, and its vj -- the largest cij still in that column.
2. For each variable xij not previous selected in these
rows and columns calculate ij = cij - ui - vj.
3. Select the variable xij with the largest (in absolute
terms) negative ij (ties my be broken randomly).
An example for Russell’s Method
Step 1: Compute ui’s and vj’s for each row and column .
Supply
u
16
16
13
22
17
50
22
14
14
13
19
15
60
19
19
19
20
23
M
50
M
M
0
M
0
0
50
M
Demand
30
20
70
30
60
v
M
19
M
23
M
Step 2: For each variable xij not previous selected in these rows and
columns calculate ij = cij - ui - vj.
Supply
16
-25
16
–9 - M
13
-23
22
-5-M
17
50
22
-5–M
14
-24
14
-6-M
13
-23
19
-4-M
15
60
19
19-2M
19
-M
19
20-2M
20
-M
23
-M
M
50
M
M
-M-19
0
-M
M
-M-23
0
-2M
0
0
M
-6-M
–M
50
Demand
30
20
70
30
v
M
19
M
23
11 = 16 - 22 – M = - 6 – M
21 = 14 - 19 – M = - 5 – M
31 = 19 - M– M = 19 – 2 M
41 = M - M– M = – M
u
12 = 16 - 22 –19= – 25
22 = 14 - 19 – 19 = – 24
32 = 19 - M– 19 = - M
42 = 0 - M– 19= – M - 19
13 = 13 - 22 –M= –9 - M
23 = 13 - 19 – M = – 6 - M
33 = 20 - M– M = 20 – 2 M
43 = M - M– M = – M
50
10
50
M
13 = 22 - 22 –23= –23
23 = 19 - 19 – 23 = – 23
33 = 23 - M– 23 = – M
43 = 0 - M– 23 = – M -23
14 = 17 - 22 –M= –5-M
24 = 15 - 19 – M = –4-M
34 = M - M– M = – M
44 = 0 - M– M = – 2M
Step 2: For each variable xij not previous selected in these rows and
columns calculate ij = cij - ui - vj.
Supply
-25
16
-25
16
-29
13
-23
22
-5-M
17
10
-24
-M
–M
u
40
22
50
14
-24
14
-25
13
-23
19
-4-M
15
60
19
19
-M
19
-M
20
-M
23
-M
M
50
M
M
-M-19
0
-M
M
-M-23
0
-2M
0
0
50
Demand
30
20
70
30
v
19
19
20
23
11 = 16 - 22 –19 = - 25
21 = 14 - 19 – 19= - 24
31 = 19 - M– 19 = – M
12 = 16 - 22 –19= – 25
22 = 14 - 19 – 19 = – 24
32 = 19 - M– 19 = - M
Note: v1 and v3 changed
13 = 13 - 22 –20= – 29
23 = 13 - 19 – 20= –26
33 = 20 - M– 20 = – M
0
60
M
14 = 22 - 22 –23= –23
24 = 19 - 19 – 23 = – 23
34 = 23 - M– 23 = – M
15 = 17 - 22 –M= –5-M
25 = 15 - 19 – M = –4-M
35 = M - M– M = – M
Step 2: For each variable xij not previous selected in these rows and
columns calculate ij = cij - ui - vj.
Supply
-25
16
-25
–29
16
13
-23
22
40
-24
-M
–M
-5-M
17
10
0
22
50
14
-24
14
-24
13
-23
19
-4-M
15
60
19
19
-M
19
-M
20
-M
23
-M
M
50
M
M
-M-19
0
-M
M
-M-23
0
-2M
0
0
50
Demand
30
20
v
19
19
11 = 16 - 22 –19 = - 25
21 = 14 - 19 – 19= - 24
31 = 19 - M– 19 = – M
u
12 = 16 - 22 –19= – 25
22 = 14 - 19 – 19 = – 24
32 = 19 - M– 19 = - M
30
70
20
13 = 13 - 22 –20= – 29
23 = 13 - 19 – 20= –26
33 = 20 - M– 20 = – M
30
23
0
60
M
14 = 22 - 22 –23= –23
24 = 19 - 19 – 23 = – 23
34 = 23 - M– 23 = – M
Russell’s method – Interactive Solver
Initial Basic Feasible Solution
Comparison of different methods for finding
initial basic feasible solution
• NW corner rule – very simple to implement and therefore fast.
However it doesn’t factor in the costs.
• The minimum cost method and Vogel’s method take into consideration
the costs and they are simple to compute
• Russell’s method – also takes into consideration the costs and it closely
patterns the decision criterion used by the transportation simplex
method to pick the entering variable.(*)
• Nevertheless it is not clear which method performs better on average,
which may depend on the particular class of problem
(*) In particular ij = cij - ui – vj estimate the relative values of cij - ui – vj
Transportation simplex method
• Initialization: finding the initial BF solution
• Optimality test – how do we check if it is
the optimal solution?
• Iteration
Transportation simplex method
Optimality Test
• Very similar to the standard simplex method
– we check if all the cij - ui – vj are nonnegative.
Optimality test – A BF solution is optimal iff
cij - ui – vj 0 for every (i,j) such that xij is non-basic .
Note: the cij - ui – vj correspond to the coefficients in row 0.
Transportation simplex method
Optimality Test
So what do we have to do?
1. calculate the ui’s and vj’s for the current BF
solution
2. Use the values obtained in 1 to calculate the cij
- ui – vj for the non-basic variables.
Optimality test – A BF solution is optimal iff
cij - ui – vj 0 for every (i,j) such that xij is non-basic .
Note: the cij - ui – vj correspond to the coefficients in row 0.
Calculation of the ui’s and vj’s
for the current BF solution
• We know that when a variable is basic its
corresponding row 0 coefficient is zero
So cij - ui – vj= 0  cij = ui – vj for all the basic variables
 Given that there are m + n – 1 variables, there are m + n – 1 of
these equations
 Since the number of unknowns (ui’s and vj’s ) is m+n, we can
assign an arbitrary value to one of them; a convenient
choice is to assign to the ui that has the largest number of
allocations the value 0.
 It is then very easy to solve the remaining equations.
Calculation of the ui’s and vj’s for the current BF solution
x31: 19 = u3 + v1; set u3=0
x32: 19 = u3 + v2;
x34: 19 = u3 + v4;
x21: 14 = u2 + v1;
v1 = 19;
x23: 13 = u2 + v3;
u2 = -5;
x13: 13 = u1 + v3;
v3 = 18;
x15: 17 = u1 + v5;
u1 = -5;
x45: 0 = u4 + v5;
v5 = 22;
 v1 = 19;
 v2 = 19;
 v4 = 23;
 u2 = -5;
 v3 = 18;
 u1 = -5;
 v5 = 22;
 u4 = -22;
Now we can compute the cij - ui – vj for the non-basic variables.
x11: 16 +5 - 19 = 2;
x12: 16 + 5 – 19 = 2;
x14: 22 + 5 – 23 = 4;
x22: 14 + 5 –19 = 0; v1 = 19;
x24: 19 + 5 –23 = 1; + v3;
x25: 15 + 5 –22 = -2;
x33: 20 –0 –18 = 2;
x41: M + 22 – 19 = M – 3;
x42: 0 + 22 – 19 = 3;
x43: M + 22 – 18 = M + 4;
x43: 0 + 22 – 23 = -1;
Is it optimal?
Which non-basic variable will enter the basis?
The one with the largest cij - ui – vj in absolute value
Transportation simplex method
• Initialization: finding the initial BF solution
• Optimality test – how do we check if it is
the optimal solution?
• Iteration
How to Pivot a Transportation Problem
1.
Select as the entering variable the variable with the largest
(in absolute value) negative cij - ui – vj value.
2. Select as the leaving variable. Increasing the entering
variable from zero sets off a chain reaction of
compensating changes in other basic variables. The first
variable to be decreased to zero then becomes the leaving
basic variable.
1. Find the chain involving the entering variable and some
of the basic variables.
2. Counting the cells in the chain, label them alternating
as (+) and (–) cells, starting with the entering variable
with label (+);
3. Find the (-) cells whose variable assumes the smallest
value. Call this value . The variable corresponding to
this (-) cell will leave the basis.
4. To perform the pivot, decrease the value of each (-) cell
by  and increase the value of each + cell by . The
variables that are not in the loop remain unchanged. The
pivot is now complete.
( If  =0, the entering variable will equal 0, and a - variable that has a
current value of 0 will leave the basis. In this case a degenerate bfs
existed before and will result after the pivot. If more than one - cell in
the loop equals , you may arbitrarily choose one of these odd cells to
leave the basis; again a degenerate bfs will result)
Illustration of pivoting procedure. We want to find the bfs that
would result if x14 were entered into the basis.
Example of the chain
-
Ent.
Variable
35
35
+
+ 10
20
+
45
20
20
50
10
30
30
30
What is the value of the entering variable?
-
40
New bfs after x14 is pivoted into basis. Since There is no loop involving
the cells (1,1), (1,4), (2,1), (2,2), (3,3) and (3, 4) the new solution is a bfs.
35-20
10+20
45
0+20
20
20
20-20
(nonbasic)
35
50
10+20
30-20
30
30
40
After the pivot the new bfs is x11=15, x14=20, x21=30, X22=20,
X33=30 and X34=10.
Two important points!
In the pivoting procedure:
1. Since each row has as many +20s as –20s, the
new solution will satisfy each supply and demand
constraint.
2. By choosing the smallest (-) variable (X23) to
leave the basis, we ensured that all variables will
remain nonnegative.
• The 1st initial BF solution is degenerate because x31=0. But
it is not a problem since this cell becomes a (+) cell.
• Another degenerate solution arises in the 3rd tableau (two
(-) cells tie for the smallest value - 30). We selected (2,1)
to be the leaving variable; if we selected (3,4), (2,1) would
become the degenerate basic variable. Cell (3,4) is a (-)
cell but at the level 0. So the entering variable has the
value 0.
• Because none of the cij - ui – vj is negative in the 4th
tableau, it means that the 3rd tableau is also optimal.