transportation winston 2
TRANSCRIPT
-
7/30/2019 Transportation Winston 2
1/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.1
Transportation, Assignment andTransshipment Problems
fromOperations Research: Applications & Algorithms,
4th edition, by Wayne L. Winston
-
7/30/2019 Transportation Winston 2
2/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.2
Description
A transportation problem basically deals with the
problem, which aims to find the best way to fulfill
the demand of n demand points using the
capacities of m supply points. While trying to findthe best way, generally a variable cost of shipping
the product from one supply point to a demand
point or a similar constraint should be taken into
consideration.
-
7/30/2019 Transportation Winston 2
3/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.3
7.1 Formulating Transportation
ProblemsExample 1: Powerco has three electric
power plants that supply the electric needs
of four cities.
The associated supply of each plant and
demand of each city is given in the table 1.
The cost of sending 1 million kwh ofelectricity from a plant to a city depends on
the distance the electricity must travel.
-
7/30/2019 Transportation Winston 2
4/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.4
Transportation tableau
A transportation problem is specified by
the supply, the demand, and the shipping
costs. So the relevant data can be
summarized in a transportation tableau.The transportation tableau implicitly
expresses the supply and demand
constraints and the shipping cost between
each demand and supply point.
-
7/30/2019 Transportation Winston 2
5/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.5
Table 1. Shipping costs, Supply, and Demand
for Powerco Example
From To
City 1 City 2 City 3 City 4 Supply
(Million kwh)
Plant 1 $8 $6 $10 $9 35
Plant 2 $9 $12 $13 $7 50
Plant 3 $14 $9 $16 $5 40Demand
(Million kwh)
45 20 30 30
Transportation Tableau
-
7/30/2019 Transportation Winston 2
6/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.6
Solution
1. Decision Variable:
Since we have to determine how much electricity
is sent from each plant to each city;
Xij = Amount of electricity produced at plant i
and sent to city j
X14 = Amount of electricity produced at plant 1
and sent to city 4
-
7/30/2019 Transportation Winston 2
7/86Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.7
2. Objective function
Since we want to minimize the total cost of shipping
from plants to cities;
Minimize Z = 8X11+6X12+10X13+9X14
+9X21+12X22+13X23+7X24
+14X31+9X32+16X33+5X34
-
7/30/2019 Transportation Winston 2
8/86Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.8
3. Supply Constraints
Since each supply point has a limited productioncapacity;
X11+X12+X13+X14
-
7/30/2019 Transportation Winston 2
9/86Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.9
4. Demand Constraints
Since each supply point has a limited productioncapacity;
X11+X21+X31 >= 45X12+X22+X32 >= 20
X13+X23+X33 >= 30
X14+X24+X34 >= 30
-
7/30/2019 Transportation Winston 2
10/86Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.10
5. Sign Constraints
Since a negative amount of electricity can not be
shipped all Xijs must be non negative;
Xij >= 0 (i= 1,2,3; j= 1,2,3,4)
-
7/30/2019 Transportation Winston 2
11/86Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.11
LP Formulation of Powercos Problem
Min Z = 8X11+6X12+10X13+9X14+9X21+12X22+13X23+7X24
+14X31+9X32+16X33+5X34
S.T.: X11+X12+X13+X14 = 30
X14+X24+X34 >= 30
Xij >= 0 (i= 1,2,3; j= 1,2,3,4)
-
7/30/2019 Transportation Winston 2
12/86Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
12
General Description of a Transportation
Problem
1. A set of m supply points from which a good is
shipped. Supply point i can supply at most si
units.
2. A set of n demand points to which the good is
shipped. Demand pointj must receive at least di
units of the shipped good.
3. Each unit produced at supply point i and shippedto demand pointj incurs a variable costofcij.
-
7/30/2019 Transportation Winston 2
13/86Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
13
Xij = number of units shipped from supply point i to
demand point j
),...,2,1;,...,2,1(0
),...,2,1(
),...,2,1(..
min
1
1
1 1
njmiX
njdX
misXts
Xc
ij
mi
i
jij
nj
j
iij
mi
i
nj
j
ijij
-
7/30/2019 Transportation Winston 2
14/86Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
14
Balanced Transportation Problem
If Total supply equals to total demand, theproblem is said to be a balanced
transportation problem:
nj
j
j
mi
i
i ds11
-
7/30/2019 Transportation Winston 2
15/86Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
15
Balancing a TP if total supply exceeds total
demand
If total supply exceeds total demand, we
can balance the problem by adding dummy
demand point. Since shipments to thedummy demand point are not real, they are
assigned a cost of zero.
-
7/30/2019 Transportation Winston 2
16/86Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
16
Balancing a transportation problem if total
supply is less than total demand
If a transportation problem has a total
supply that is strictly less than total
demand the problem has no feasiblesolution. There is no doubt that in such a
case one or more of the demand will be left
unmet. Generally in such situations a
penalty cost is often associated with unmetdemand and as one can guess this time the
total penalty cost is desired to be minimum
-
7/30/2019 Transportation Winston 2
17/86Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
17
7.2 Finding Basic Feasible
Solution for TPUnlike other Linear Programming
problems, a balancedTP with m supply
points and n demand points is easier to
solve, although it has m + n equality
constraints. The reason for that is, if a set
of decision variables (xijs) satisfy all but
one constraint, the values for xijs willsatisfy that remaining constraint
automatically.
-
7/30/2019 Transportation Winston 2
18/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.18
Methods to find the bfs for a balanced TP
There are three basic methods:
1. Northwest Corner Method
2. Minimum Cost Method
3. Vogels Method
-
7/30/2019 Transportation Winston 2
19/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.19
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 demandpoint 1 and the supply of supply point 1. Your
x11 value can not be greater than minimum of
this 2 values).
-
7/30/2019 Transportation Winston 2
20/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.20
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
-
7/30/2019 Transportation Winston 2
21/86
-
7/30/2019 Transportation Winston 2
22/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.22
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 X
3 2 1
2
X X X 3
3 2 X
3 2 1 X
2
X X X 2
-
7/30/2019 Transportation Winston 2
23/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.23
Finally, we will have the following bfs, which is:
x11
=3, x12
=2, x22
=3, x23
=2, x24
=1, x34
=2
3 2 X
3 2 1 X
2 X
X X X X
-
7/30/2019 Transportation Winston 2
24/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.24
2. Minimum Cost Method
The Northwest Corner Method dos not utilize shippingcosts. 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 costmethod, first we find the decision variable with the
smallest shipping cost (Xij). Then assignXijits largest
possible value, which is the minimum ofsi and dj
-
7/30/2019 Transportation Winston 2
25/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.25
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 orcolumn 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.
-
7/30/2019 Transportation Winston 2
26/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.26
An example for Minimum Cost Method
Step 1: Select the cell with minimum cost.
2 3 5 6
2 1 3 5
3 8 4 6
5
10
15
12 8 4 6
-
7/30/2019 Transportation Winston 2
27/86
-
7/30/2019 Transportation Winston 2
28/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.28
Step 3: Find the new cell with minimum shipping
cost and cross-out row 2
2 3 5 6
2 1 3 5
2 8
3 8 4 6
5
X
15
10 X 4 6
-
7/30/2019 Transportation Winston 2
29/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.29
Step 4: Find the new cell with minimum shipping
cost and cross-out row 1
2 3 5 6
5
2 1 3 5
2 8
3 8 4 6
X
X
15
5 X 4 6
-
7/30/2019 Transportation Winston 2
30/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.30
Step 5: Find the new cell with minimum shipping
cost and cross-out column 1
2 3 5 6
5
2 1 3 5
2 8
3 8 4 6
5
X
X
10
X X 4 6
-
7/30/2019 Transportation Winston 2
31/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.31
Step 6: Find the new cell with minimum shipping
cost and cross-out column 3
2 3 5 6
5
2 1 3 5
2 8
3 8 4 6
5 4
X
X
6
X X X 6
-
7/30/2019 Transportation Winston 2
32/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.32
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
5
2 1 3 5
2 8
3 8 4 6
5 4 6
X
X
X
X X X X
-
7/30/2019 Transportation Winston 2
33/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.33
3. Vogels Method
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.
Identify the row or column with the largest penalty.
Find the first basic variable which has the smallestshipping cost in that row or column. Then assign the
highest possible value to that variable, and cross-out
the row or column as in the previous methods.
Compute new penalties and use the same procedure.
-
7/30/2019 Transportation Winston 2
34/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.34
An example for Vogels Method
Step 1: Compute the penalties.
Supply Row Penalty
6 7 8
15 80 78
Demand
Column Penalty 15-6=9 80-7=73 78-8=70
7-6=1
78-15=63
15 5 5
10
15
-
7/30/2019 Transportation Winston 2
35/86
-
7/30/2019 Transportation Winston 2
36/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.36
Step 3: Identify the largest penalty and assign the
highest possible value to the variable.
Supply Row Penalty
6 7 8
5 5
15 80 78
Demand
Column Penalty 15-6=9 _ _
_
_
15 X X
0
15
-
7/30/2019 Transportation Winston 2
37/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.37
Step 4: Identify the largest penalty and assign the
highest possible value to the variable.
Supply Row Penalty
6 7 8
0 5 5
15 80 78
Demand
Column Penalty _ _ _
_
_
15 X X
X
15
-
7/30/2019 Transportation Winston 2
38/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.38
Step 5: Finally the bfs is found as X11=0, X12=5,
X13=5, and X21=15
Supply Row Penalty
6 7 8
0 5 5
15 80 78
15
Demand
Column Penalty _ _ _
_
_
X X X
X
X
-
7/30/2019 Transportation Winston 2
39/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.39
7.3 The Transportation Simplex
Method
In this section we will explain how the simplex
algorithm is used to solve a transportation problem.
-
7/30/2019 Transportation Winston 2
40/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.40
How to Pivot a Transportation Problem
Based on the transportation tableau, the followingsteps should be performed.
Step 1. Determine (by a criterion to be developed
shortly, for example northwest corner method) the
variable that should enter the basis.
Step 2. Find the loop (it can be shown that there is
only one loop) involving the entering variable and
some of the basic variables.Step 3. Counting the cells in the loop, label them as
even cells or odd cells.
-
7/30/2019 Transportation Winston 2
41/86
-
7/30/2019 Transportation Winston 2
42/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.42
Illustration of pivoting procedure on the Powerco
example. We want to find the bfs that would result if
X14 were entered into the basis.
Nothwest Corner bfs and loop for Powerco
35 35
10 20 20 50
10 30 40
45 20 30 30
E O E O E O
(1, 4) (3, 4) (3, 3) (2, 3) (2, 1) (1, 1)
0
12
34
5
-
7/30/2019 Transportation Winston 2
43/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.43
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 0+20 35
10+20 20
20-20
(nonbasic) 50
10+20 30-20 40
45 20 30 30
After the pivot the new bfs is X11=15, X14=20,
X21=30, X22=20, X33=30 and X34=10.
-
7/30/2019 Transportation Winston 2
44/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
44
Two important points!
In the pivoting procedure:
1. Since each row has as many +20s as20s, the
new solution will satisfy each supply and demand
constraint.
2. By choosing the smallest odd variable (X23) to
leave the basis, we ensured that all variables willremain nonnegative.
-
7/30/2019 Transportation Winston 2
45/86
-
7/30/2019 Transportation Winston 2
46/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
46
Since the example is a minimization problem, the
current bfs will be optimal if all the ijs are
nonpositive; otherwise, we enter into the basis withthe most positive ij.
After determining cBVB-1 we can easily determine
ij. Since the first constraint has been dropped,
cBVB-1 will have m+n-1 elements.cBVB
-1= [u2 u3um v1 v2vn]
Where u2 u3um are elements ofcBVB-1
corresponding to the m-1 supply constraints, and v1v2vn are elements ofcBVB-1 corresponding to the n
demand constraints.
-
7/30/2019 Transportation Winston 2
47/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
47
To determine cBVB-1 we use the fact that in any
tableau, each basic variableXij must have ij=0.
Thus for each of the m+n-1 variables in BV,
cBVB-1aijcij=0
For the Northwest corner bfs of Powerco problem,BV={X11, X21, X22, X23, X33, X34}. Applying the
equation above we obtain:
11= [u2 u3 v1 v2 v3 v4] -8 = v1-8=0
0
0
0
1
0
0
1
-
7/30/2019 Transportation Winston 2
48/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
48
21= [u2 u3 v1 v2 v3 v4] -9 = u2+v1-9=0
22= [u2 u3 v1 v2 v3 v4] -12 = u2+v2-12=0
23= [u2 u3 v1 v2 v3 v4] -13 = u2+v3-13=0
0
0
0
1
0
1
0
0
1
0
0
1
0
1
0
00
1
-
7/30/2019 Transportation Winston 2
49/86
-
7/30/2019 Transportation Winston 2
50/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
50
u1=0
u1+v1=8
u2+v1=9
u2+v2=12
u2+v3=13
u3+v3=16
u3+v4=5
By solving the system above we obtain:v1=8, u2=1, v2=11,v3=12, u3=4, v4=1
-
7/30/2019 Transportation Winston 2
51/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
51
For each nonbasic variable, we now compute
ij = ui+vjcij
We obtain:
12 = 0+116 = 5 13 = 0+1210 = 2
14 = 0+19 = -8 24 = 1+17 = -531 = 4+814 = -2 32 = 4+119 = 6
Since 32 is the greatest positive ij, we would nextenter X32 into the basis. Each unit that is entered
into the basis will decrease Powercos cost by $6.
-
7/30/2019 Transportation Winston 2
52/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
52
We have determined that X32 should enter the basis. As
shown in the table below the loop involving X32 and
some of the basic variables is (3,2), (3,3), (2,3), (2,2).
The odd cells in the loop are (2,2) and (3,3). Since the
smallest value of these two is 10 the pivot is 10.
35 35
10 20 20 50
10 30 40
45 20 30 30
-
7/30/2019 Transportation Winston 2
53/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
53
The resulting bfs will be:
X11=35, X32=10, X21=10, X22=10, X23=30 and X34=30
The uis and vjs for the new bfs were obtained by
solving
u1=0
u2+v2=12
u3+v4=5
u1+v1=8
u2+v3=13
u2+v1=9
u3+v2=9
-
7/30/2019 Transportation Winston 2
54/86
-
7/30/2019 Transportation Winston 2
55/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
55
7.4 Sensitivity AnalysisIn this section we discuss the following three aspects
of sensitivity analysis for the transportation problem:
1. Changing the objective function coefficient of anonbasic variable.
2. Changing the objective function coefficient of a
basic variable.
3. Increasing a single supply by and a single demand
by .
-
7/30/2019 Transportation Winston 2
56/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
56
1. Changing the objective function coefficient
of a nonbasic variable.
Changing the objective function coefficient of a
nonbasic variableXij will leave the right hand side of
the optimal tableau unchanged. Thus the current
basis will still be feasible. Since we are not changingcBVB
-1, the uis and vjs remain unchanged. In row 0
only the coefficient ofXij will change. Thus as long
as the coefficient ofXij in the optimal row 0 is
nonpositive, the current basis remains optimal.
L t t t th f ll i ti b t
-
7/30/2019 Transportation Winston 2
57/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
57
Lets try to answer the following question about
Powerco as an example:
For what range of values of the cost of shipping 1million kwh of electricity from plant 1 to city 1 will the
current basis remain optimal?
Suppose we change c11from 8 to 8+ .
Now 11 = u1+v1-c11=0+6-(8+ )=2- .
Thus the current basis remains optimal for2- =-2, and c11>=8-2=6.
-
7/30/2019 Transportation Winston 2
58/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
58
2. Changing the objective function coefficient of a
basic variable.
Since we are changing cBVB-1, the coefficient of each
nonbasic variable in row 0 may change, and to determine
whether the current basis remain optimal, we must find
the new uis and vjs and use these values to price out all
nonbasic variables. The current basis remains optimal as
long as all nonbasic variables price out nonpositive.
-
7/30/2019 Transportation Winston 2
59/86
-
7/30/2019 Transportation Winston 2
60/86
-
7/30/2019 Transportation Winston 2
61/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
61
Thus, the current basis remains optimal for2
-
7/30/2019 Transportation Winston 2
62/86
-
7/30/2019 Transportation Winston 2
63/86
-
7/30/2019 Transportation Winston 2
64/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
64
7.5. Assignment Problems
Example: Machineco has four jobs to be completed.
Each machine must be assigned to complete one job.
The time required to setup each machine for
completingeach job is shown in the table below.Machinco wants to minimize the total setup time needed
to complete the four jobs.
-
7/30/2019 Transportation Winston 2
65/86
-
7/30/2019 Transportation Winston 2
66/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
66
The Model
According to the setup table Machincos problem can be
formulated as follows (for i,j=1,2,3,4):
10
1
1
1
1
1
1
1
1..
10629387
5612278514min
44342414
43332313
42322212
41312111
44434241
34333231
24232221
14131211
4443424134333231
2423222114131211
ijij orXX
XXXX
XXXX
XXXX
XXXX
XXXX
XXXX
XXXX
XXXXts
XXXXXXXX
XXXXXXXXZ
-
7/30/2019 Transportation Winston 2
67/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
67
For the model on the previous page note that:
Xij=1 if machine i is assigned to meet the demands of
jobj
Xij=0 if machine i is assigned to meet the demands ofjobj
In general an assignment problem is balancedtransportation problem in which all supplies and
demands are equal to 1.
Although the transportation simplex appears to be very
-
7/30/2019 Transportation Winston 2
68/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
68
Although the transportation simplex appears to be very
efficient, there is a certain class of transportation
problems, called assignment problems, for which the
transportation simplex is often very inefficient. For thatreason there is an other method called The Hungarian
Method. The steps of The Hungarian Method are as
listed below:
Step1. Find a bfs. Find the minimum element in each row
of the mxm cost matrix. Construct a new matrix by
subtracting from each cost the minimum cost in its row.
For this new matrix, find the minimum cost in eachcolumn. Construct a new matrix (reduced cost matrix) by
subtracting from each cost the minimum cost in its
column.
-
7/30/2019 Transportation Winston 2
69/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
69
Step2. Draw the minimum number of lines (horizontal
and/or vertical) that are needed to cover all zeros in the
reduced cost matrix. If m lines are required , an optimalsolution is available among the covered zeros in the
matrix. If fewer than m lines are required, proceed to step
3.
Step3. Find the smallest nonzero element (call its value
k) in the reduced cost matrix that is uncovered by the
lines drawn in step 2. Now subtract k from eachuncovered element of the reduced cost matrix and add k
to each element that is covered by two lines. Return to
step2.
-
7/30/2019 Transportation Winston 2
70/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
70
7.6 Transshipment Problems
A transportation problem allows only shipments that godirectly from supply points to demand points. In many
situations, shipments are allowed between supply points
or between demand points. Sometimes there may also
be points (called transshipment points) through whichgoods can be transshipped on their journey from a
supply point to a demand point. Fortunately, the optimal
solution to a transshipment problem can be found by
solving a transportation problem.
-
7/30/2019 Transportation Winston 2
71/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
71
The following steps describe how the optimal solution to
a transshipment problem can be found by solving a
transportation problem.
Step1. If necessary, add a dummy demand point (with a
supply of 0 and a demand equal to the problems excess
supply) to balance the problem. Shipments to the dummyand from a point to itself will be zero. Let s= total
available supply.
Step2. Construct a transportation tableau as follows: A
row in the tableau will be needed for each supply pointand transshipment point, and a column will be needed for
each demand point and transshipment point.
-
7/30/2019 Transportation Winston 2
72/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
72
Each supply point will have a supply equal to its
original supply, and each demand point will have a
demand to its original demand. Let s= total available
supply. Then each transshipment point will have a supply
equal to (points original supply)+s and a demand equal
to (points original demand)+s. This ensures that anytransshipment point that is a net supplier will have a net
outflow equal to points original supply and a net
demander will have a net inflow equal to points original
demand. Although we dont know how much will beshipped through each transshipment point, we can be
sure that the total amount will not exceed s.
Transportation Problem
-
7/30/2019 Transportation Winston 2
73/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
73
Transportation Problem
Powerco Power Plant: Formulation
Define Variables:Let Xij = number of (million kwh) produced at plant i and
sent to city j.
Objective Function:Min z = 8*X11 + 6*X12 + 10*X13 + 9*X14
+ 9*X21 + 12*X22 + 13*X23 + 7*X24
+ 14*X31 + 9*X32 + 16*X33 + 5*X34
Supply Constraints:X11 + X12 + X13 + X14 < = 35 (Plant 1)
X21 + X22 + X23 + X24 < = 50 (Plant 2)
X31 + X32 + X33 + X34 < = 40 (Plant 3)
Transportation
-
7/30/2019 Transportation Winston 2
74/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
74
pProblem
Powerco Power Plant: Formulation (Contd.)
Demand Constraints:X11 + X21 +X31 > = 45 (City 1)
X12 + X22 +X32 > = 20 (City 2)
X13 + X23 +X33 > = 30 (City 3)
X14 + X24 +X34 > = 30 (City 4)
Nonnegativity Constraints:Xij > = 0 (i=1,..,3; j=1,..,4)
Balanced?Total Demand = 45 + 20 + 30 + 30 = 125
Total Supply = 35 + 50 + 40 = 125
Yes, Balanced Transportation Problem! (If problem is
unbalanced, add dummy supply or demand as required.)
Transportation Problem
-
7/30/2019 Transportation Winston 2
75/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
75
Transportation Problem
Powerco Power Plant: LINGO Formulation
model:
! A 3 Plant, 4 Customer
Transportation Problem;
SETS:
PLANT / P1, P2, P3/ : CAPACITY;CUSTOMER / C1, C2, C3, C4/ : DEMAND;
ROUTES( PLANT, CUSTOMER) : COST, QUANTITY;
ENDSETS
! The objective;
[OBJ] MIN = @SUM( ROUTES: COST * QUANTITY);
Transportation Problem
-
7/30/2019 Transportation Winston 2
76/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
76
Transportation ProblemPowerco Power Plant: LINGO Formulation (Contd.)
! The demand constraints;
@FOR( CUSTOMER( J): [DEM]
@SUM( PLANT( I): QUANTITY( I, J))>= DEMAND( J));
! The supply constraints;
@FOR( PLANT( I): [SUP]
@SUM( CUSTOMER(J): QUANTITY(I,J))
-
7/30/2019 Transportation Winston 2
77/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
77
Transportation ProblemPowerco Power Plant: LINGO Solution
Optimal solution found at step: 7Objective value: 1020.000
QUANTITY( P1, C1) 0.0000000
QUANTITY( P1, C2) 10.00000
QUANTITY( P1, C3) 25.00000QUANTITY( P1, C4) 0.0000000
QUANTITY( P2, C1) 45.00000
QUANTITY( P2, C2) 0.0000000
QUANTITY( P2, C3) 5.000000
QUANTITY( P2, C4) 0.0000000QUANTITY( P3, C1) 0.0000000
QUANTITY( P3, C2) 10.00000
QUANTITY( P3, C3) 0.0000000
QUANTITY( P3, C4) 30.00000
All Quantities
are Integers!
Transportation Problem
-
7/30/2019 Transportation Winston 2
78/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
78
Transportation ProblemPowerco Power Plant: LINGO Solution (Contd.)
Row Slack or Surplus Dual Price
OBJ 1020.000 1.000000
DEM( C1) 0.0000000 -9.000000
DEM( C2) 0.0000000 -9.000000
DEM( C3) 0.0000000 -13.00000DEM( C4) 0.0000000 -5.000000
SUP( P1) 0.0000000 3.000000
SUP( P2) 0.0000000 0.0000000
SUP( P3) 0.0000000 0.0000000
All constraints are binding!(Typical of Balanced Transportation Problems; results in
simple algorithms)
-
7/30/2019 Transportation Winston 2
79/86
-
7/30/2019 Transportation Winston 2
80/86
Powerco Power Plant: Sensitivity Analysis (Contd.)V i bl V l R d d C t
-
7/30/2019 Transportation Winston 2
81/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
81
Variable Value Reduced Cost
QUANTITY( P1, C1) 0.00 2.00 (c11)
QUANTITY( P1, C2) 10.00 0.00 (c12)
QUANTITY( P1, C3) 25.00 0.00 (c13)
QUANTITY( P1, C4) 0.00 7.00 (c14)
QUANTITY( P2, C1) 45.00 0.00 (c21)
QUANTITY( P2, C2) 0.00 3.00 (c22)
QUANTITY( P2, C3) 5.00 0.00 (c23)QUANTITY( P2, C4) 0.00 2.00 (c24)
QUANTITY( P3, C1) 0.00 5.00 (c31)
QUANTITY( P3, C2) 10.00 0.00 (c32)
QUANTITY( P3, C3) 0.00 3.00 (c33)QUANTITY( P3, C4) 30.00 0.00 (c34)
z = c11* Q11 + c12* Q12 + c13* Q13 + c14* Q14
+ c21* Q21 + c22* Q22 + c23* Q23 + c24* Q24
+ c31* Q31 + c32* Q32 + c33* Q33 + c34* Q34
Nonbasic
Basic
Basic
Nonbasic
Basic
Nonbasic
BasicNonbasic
Nonbasic
Basic
NonbasicBasic
A i t P bl
-
7/30/2019 Transportation Winston 2
82/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
82
Assignment ProblemsPersonnel Assignment:
Time (hours)
Job 1 Job 2 Job 3 Job 4
Person 1 14 5 8 7
Person 2 2 12 6 5Person 3 7 8 3 9
Person 4 2 4 6 10
Assignment Problem Formulation
-
7/30/2019 Transportation Winston 2
83/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
83
Assignment Problem Formulation
Define Variables:
Let Xij = 1 if ith person is assigned to jth job
Xij = 0 if ith person is not assigned to jth job
Objective Function:
Min z = 14*X11 + 5*X12 + + 10*X44Personnel Constraints:
X11 + X12 + X13 + X14 = 1
X21 + X22 + X23 + X24 = 1
X31 + X32 + X33 + X34 = 1X41 + X42 + X43 + X44 = 1
Demand Constraints:
X11 + X21 + X31 + X41 = 1
X12 + X22 + X32 + X42 = 1
X13 + X23 + X33 + X43 = 1X14 + X24 + X34 + X44 = 1
Binary Constraints:
Xij = 0 or Xij = 1
-
7/30/2019 Transportation Winston 2
84/86
Assignment Problem Algorithm
-
7/30/2019 Transportation Winston 2
85/86
Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.
85
Assignment Problem Algorithm
9 0 3 00 10 4 1
4 5 0 4
0 2 4 6
Subtract Column Minimum from Each Column:
10 0 3 0
0 9 3 05 5 0 4
0 1 3 5
Subtract Minimum uncrossed value from uncrossed valuesand add to twice-crossed values:
Draw lines to cross out zeros and read solution from zeros
0
00
0
Solution:
Hungarian Method
-
7/30/2019 Transportation Winston 2
86/86
gStep 1: Find the minimum element in each row of the m x m cost
matrix. Construct a new matrix by subtracting from each cost the
minimum cost in its row. For this new matrix, find the minimum cost ineach column. Construct a new matrix (called the reduced cost matrix)
by subtracting from each cost the minimum cost in its column.
Step 2: Draw the minimum number of lines (horizontal and/or vertical)
that are needed to cover all the zeros in the reduced cost matrix. If mlines are required, an optimal solution is available among the covered
zeros in the matrix. If fewer than m lines are needed, proceed to Step
3.
Step 3: Find the smallest nonzero element (call its value k) in the
reduced cost matrix that is uncovered by the lines drawn in Step 2.Now subtract k from each uncovered element of the reduced cost
matrix and add k to each element that is covered by two lines. Return
to Step 2