Iterative computations of the Transportation algorithm

1
Iterative computations of the Transportation
algorithm
2
Iterative computations of the Transportation
algorithm After determining the starting BFS by
any one of the three methods discussed earlier,
we use the following algorithm to determine the
optimum solution Step1 Use the Simplex
optimality condition to determine the entering
variable as a current non-basic variable that can
improve the solution. If the optimality condition
is satisfied by all non-basic variables, the
current solution is optimal and we stop.
Otherwise we go to Step 2.
3
Step 2. Determine the leaving variable using the
Simplex feasibility condition. Change the basis
and go to Step 1.
The determination of the entering variable from
among the current non-basic variables is done by
the method of multipliers.
In the method of multipliers, we associate with
each row a dual variable (also called a
multiplier) ui and with each column we associate
a dual variable (also called a multiplier) vj.
4
Noting that each row corresponds to a constraint
and each column corresponds to a constraint we
recall from duality theory that
At any simplex iteration , Primal z-equation
Left hand side Right hand side
coefficient of of corresponding -
of corresponding
variable xj dual constraint
dual constraint
That is
(Verify this by taking m3 and n4 !)
5
Since there are mn-1 basic variables and since
for all such basic variables, we have mn-1
equations
to determine the mn variables
We arbitrarily choose one of them and equate to
zero and determine the remaining mn-1 of them.
Then we calculate
for all non-basic variables xij. Then the
entering variable is that one for which
is most positive.
6
We do all this on the transportation tableau
itself (and NOT separately) as the following
example shows.
7
Destination
Starting Tableau
Total Cost 48

Supply
v13
v27
v36
v43
Source
u10
3
2
0
-1
1
1
u2 -3
-2
-2
u3 2
1
2
1
6
Demand
Thus x32 enters the basis.
8
Determining the leaving variable We first
construct a closed loop that starts and ends at
the entering variable cell. The loop consists of
connected horizontal and vertical segments only
(no diagonals are allowed). Except for the
entering variable cell, each vertex (or corner)
of the closed loop must correspond to a basic
variable cell. The loop can cross itself and
bypass one or more basic variables. The amount ?
to be allocated to the entering variable cell is
such that it satisfies all the demand and supply
restrictions and must be non-negative. Usually
9
? is the minimum of the amounts allocated to the
basic cells adjacent to the entering variable
cell. Having decided about the amount ? to be
allocated to the entering cell, for the supply
and demand limits to remain satisfied, we must
alternate between subtracting and adding the
amount ? at the successive corners of the loop.
In this process one of the basic variables will
drop to zero. In simplex language, we say it
leaves the basis. We repeat this process till
optimality is reached. We illustrate with a
numerical example.
10
Destination
Starting Tableau
Total Cost 48

Supply
v13
v27
v36
v43
Source
u10
3
2
0
-1
1
1
u2 -3
-2
-2
u3 2
1
2
?
1
6
Demand
Thus x32 enters the basis.
11
Thus ? will become 1 and in the process both the
basic variables x22 and x33 will become
simultaneously zero. Since only one of them
should leave the basis we make x22 leave the
basis and keep x33 in the basis but with value
zero. Thus the transportation cost reduces by 6
(as x23 increases by 1) and we say one iteration
is over. The resulting new tableau is on the next
slide.
12
Destination
Total Cost 42
Start of Iteration 2

Supply
v13
v27
v312
v49
Source
?
3
2
u10
5
6
2
u2 -9
-2
-8
-6
u3 -4
0
2
1
-5
Demand
Thus x13 enters the basis.
13
Thus ? will become 0 and x32 leaves the basis.
Again the BFS is degenerate . But the
transportation cost remains the same and we say
the second iteration is over. The resulting new
tableau is on the next slide.
14
Destination
Total Cost 42
Start of Iteration 3
v36

Supply
v13
v27
v49
Source
?
3
2
0
u10
5
2
u2 -3
4
-2
0
2
u3 -4
1
-6
-5
Demand
Thus x14 enters the basis.
15
Sometimes it might be difficult to find the
closed loop from the entering cell by inspection.
In that case the following method can be used to
find the closed loop. We sketch the flowchart of
the sequence in which the variables ui and vj
were determined. For example in the above case
the flowchart is on the next slide. Now to find
the loop emanating from the non-basic cell (1,4),
join u1 and v4 by a dotted line (as shown). Then
the closed loop is
(1,4) (1,2) (2,3) (3,4)
(1,4)
16
v13
u10
v27
u3 -4
v4 9
u2 -9
v36
Thus the closed loop is
(1,4) (1,2) (2,3)
(3,4) (1,4)
17
Thus ? will become 2 and in the process both the
basic variables x12 and x32 will become
simultaneously zero. Since only one of them
should leave the basis we make x32 leave the
basis and keep x12 in the basis but with value
zero. Also x32 becomes 3. Thus the
transportation cost reduces by 124210 and we
say third iteration is over. The resulting new
tableau is on the next slide.
18
Destination
Total Cost 32
Start of Iteration 4

Supply
v13
v27
v36
v44
Source
2
3
0
0
u10
2
u2 -3
-1
-2
0
u3 -4
3
-6
-5
-5
Demand
Thus this is the optimal tableau. Alt Opt
solutions exist.
19
Problem 4 Problem Set 5.3B Page 193 In
the unbalanced transportation problem given in
the table below, if a unit from a source is not
shipped out (to any of the destinations) a
storage cost is incurred at the rate of 5, 4,
and 3 per unit for sources 1,2, and 3
respectively. Additionally all the supply at
source 2 must be shipped out completely to make
room for a new product. Use VAM to determine the
starting solution and determine the optimum
solution.
20
1 2 3
20
1 1 2 1
2 3 4 5
40
3 2 3 3
30
30 20 20
To balance the problem, we introduce a dummy
destination with transportation costs
5, M, 3 respectively.
(Solution in the next slide)
21
Destination
1 2 3 Dummy
Supply Row Penalties
1
0
-
-
-
20
Source
10
2
1
1
30
1
1
10
10
1
3
1
1
0
20
10
0
Demand
0
10
1
1
2
2
Column Penalties
1
1
-
M-3
-
-
1
1
-
-
-
1
Total
shipping cost
240
22
Destination
Total Cost 240
Starting Tableau

Supply
v12
v23
v33
v43
Source
20
u1-2
-4
-1
-1
30
10
u2 1
-1
4-M
u3 0
0
20
10
0
Demand
Thus this is the optimal tableau. Alt Opt
solutions exist.
23
Problem 5 Problem Set 5.3 B Page 193
In a 3?3 transportation problem, let xij be the
amount shipped from source i to destination j and
cij be the corresponding transportation cost per
unit. The amounts of supply at sources 1, 2, and
3 are 15, 30, and 85 units, respectively and the
demands at destinations 1, 2, and 3 are 15, 30,
and 85 units, respectively. Assume that the
northwest corner solution is optimal and that the
associated values of the multipliers are
24

• given by u1 -2, u2 3, u3 5, v1 2, v2 5,
  and v3 10.
• Find the associated optimal cost
• Determine the smallest values of cij associated
  with each non-basic variable that will maintain
  the optimality of the northwest corner solution.

25
Supply
v25
v12
v310
0
3
8
15
u1-2
5
13
8
u23
5
25
25
10
15
7
5
80
u35
Demand
5
5
Associated cost
1475
26
Problem 8.1-6 Page 393 Hillier and Lieberman
(Operations Research 7th Edition) The Onenote
Co. produces a single product at three plants for
four customers. The three plants will produce 60,
80, and 40 units respectively. The firm has made
a commitment to sell 40 units to customer 1, 60
units to customer 2, and at least 20 units to
customer 3. Both customers 3 and 4 also want to
buy as many of the remaining units as possible.
The net profit associated with shipping a unit
from plant i to customer j is given by the
following table.
27
Customer
1 2
3 4
1 800 700 500
200
Plant 2 500 200 100
300
3 600 400 300
500
Management wants to know how many units to sell
to customers 3 and 4 and how many units to ship
from each of the plant to each of the customers
to maximize profits. Formulate the problem as a
transportation model and solve it.
28
There are 3 sources, viz. Plants 1, 2 and 3.
Right now there are 4 destinations, viz.
customers 1, 2, 3, and 4. The supplies ai at the
three sources are 60, 80, and 40 respectively.
The demands at the three destinations are b1
40, b2 60, b3 ? 20, b4 ?
Since in a transportation model, all constraints
are equalities, we shall put b3 80 ( since
customer 3 must get at least 20 units) and b4
60 as the supply remaining after satisfying the
three customers 1, 2, and 3 is 60 and since
customers 3 and 4 will buy as much as possible.
29
But now the demand has become 240 and so we
introduce a dummy source SF with supply 60. Since
the customers 1, 2 must definitely get 40 and 60
units respectively, the dummy source cannot send
any amount to these destinations. This is
achieved by putting the cost from the dummy to
these destinations as big M. Now the cost from
dummy to the destinations 3 and 4 are put as
zero. Also since these are actually profits, and
since the transportation model is a minimization
problem, to maximize the total profit we take cij
as negative of the profits given. The starting
tableau is given below.
30
Destination
1 2 3 4 Supply
1
0
1
-
-
-
60
0
Source
2
2
2
40
2
-
40
40
3
1
1
20
2
2
20
20
SF
0
0
0
0
60
20
Demand
2
3
2
2
1
-
2
2
-
-
2
2
-
-
3
5
31
Destination
v2 -7
v3 -5
v4 -7
v1 -9
u1 0
60
0
Source
-5
-1
u2 4
40
40

0
-5
u3 2
20
20
-9
-1
60
u4 5
-2
-2-M
-4-M
This is the optimal tableau.
Max Profit - min z - (-900) 900.
32
Compre Question I Semester 2003-2004 The
table below gives the times taken by 3 persons to
complete 4 tasks( i.e. cell (i,j) is the time
taken by person i to complete the task j).
Tasks
1 2
3 4
1 4 1
2 6
Person 2 6 4 3
5
3 5 2
6 4
33
If each task is to be allocated to a person (i.e.
no splitting of the task between 2 or more
persons is allowed) and if each person can be
assigned at most two tasks, find the optimum
allocation of the jobs to the persons to minimize
the total time taken to complete all the 4 tasks.
This can be formulated as a transportation model
with three sources (persons) and 4 destinations
(tasks). The demands at the three destinations
are bj 1 for j1,2,3,4. But the availabilities
are ai 2 for i1, 2, 3 as each person can be
assigned a maximum of two tasks. Thus to balance
the problem, we introduce a dummy task with
demand 2 and time 0. Thus we get the starting
tableau
34
Destination
Starting Tableau
Dummy Supply
1
2
3
4
Source
1
1
3
1
1
1
1
2
3
2
-
-
3
2
2
2
1
0
0
1
0
Demand
0
1 1 1 1 0
1 1 4 2 -
1 1 - 2 -
35
Destination
Starting Tableau
v33
v44
v50
Supply
v15
v22
Source
u1 -1
1
1
0
-3
-1
u2 0
2
-1
-2
0
-1
0
1
1
0
u3 0
-3
Demand
Thus this is the optimal tableau. Optimal cost
12