University of Southern California
Daniel J. Epstein Department of Industrial and Systems Engineering
ISE 330: Introduction to Operations Research
Lecture 21 Outline
Read:
H & L chapter 8.1-8.2
EXAMPLE – P & T COMPANY
SOURCES
[ supply ]
DESTINATIONS
[ demand ]
Bellingham
[–80]
Sacramento
[75]
[–65]
Salt Lake City
Eugene
[125]
Albert Lea
[–70]
Rapid City
[100]
[–85]
Albuquerque
Shipping Data
Cannery
Allocation
1
2
3
Shipping Cost ($) per Truckload
Warehouse
1
2
3
4
464
513
654
867
352
416
690
791
995
682
388
685
80
65
70
85
Output
75
125
100
See also: Jan-Feb 1997 isse of Interfaces, Proctor & Gamblecase study.
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ISE 330: Lecture 21 Notes
http://www-classes.usc.edu/engr/ise/330
A Transportation Problem
Requirements Assumption: Each source has a fixed supply of units, si, where the
entire supply must be distributed to he destinations. Each destination has a
fixed demand for units, dj, and the entire demand must be received from the
sources.
Feasible Solutions Property: A transportation problem will have feasible solutions
if and only if…
Cost Assumption: The cost of distributing units from any particular source to any
particular destination, cij, is directly proportional to the number of units
distributed, xij.
The model: Any transportation model can be described completely by a parameter
table…
Source
1
2
3
Demand
1
c11
c21
:
cm1
d1
Cost per Unit Distributed
Destination
2
…
c12
…
c22
…
:
cm2
…
d2
…
n
c1n
c2n
:
cmn
dn
Supply
s1
s2
:
sm
The LP Formulation of the Problem…
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ISE 330: Lecture 21 Notes
http://www-classes.usc.edu/engr/ise/330
Do you detect a pattern in the coefficients? If so, what is it?
Integer Solutions Property: For transportation problems where every si and dj have
integer values, all the basic variables in every BFS also have integer values.
Dummy Sources
Water Resources data for Metro Water District
Colombo River
Sacron River
Calorie River
Min needed
Requested
Cost (Tens of Dollars) per Acre Foot
Berdoo
Los Devils
San Go Hollyglass
16
13
22
17
14
13
19
15
19
20
23
–
30
70
0
10
50
70
30

Supply
50
60
50
(in 1M
acre feet)
Formulation…
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ISE 330: Lecture 21 Notes
http://www-classes.usc.edu/engr/ise/330
The Transportation Simplex Method
Row 0 of simplex tableaux after standardization…
ui
= multiple of original row i that has been subtracted (directly or indirectly)
from original row 0 by the simlex method during all iterations leading to the
current simplex tableaux.
vj
= multiple of original row m+j that has been subtracted (directly or indirectly)
from original row 0 by the simplex method during all iterations leading to the
current simplex tableaux.
Observe that:
[1] no artificial variables are needed
[2] the current row 0 can be obtained without using any other row simply by
calculating the current values of ui and vj directly.
[3] the leaving basic variable can be identified in a simple way without (explicitly)
using the coefficients of the entering basic variable. Hence, the new BFS can
be identified immediately without any algebraic manipulations on the rows of
the simplex tableaux.
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ISE 330: Lecture 21 Notes
http://www-classes.usc.edu/engr/ise/330
The only information needed by the transportation simplex method is:
Compare: A transportation problem with m sources and n destinations, the simplex
tableau would have _________ rows and ____________ columns, and the
transportation simplex tableau would have ________ rows and ________
columns.
Format of a transportation simplex tableau:
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ISE 330: Lecture 21 Notes
http://www-classes.usc.edu/engr/ise/330
Constructing an Initial BFS:
[1] From the rows and columns still under consideration, select the next basic
variable (allocation) according to some criterion.
[2] Make the allocation large enough to exactly use up the remaining supply in its
row or the remaining demand in its column (whichever is smaller).
[3] Eliminate that row or column from further consideration. (If the row and column
have the same remaining supply and demand, then arbitrarily select the row
as the one to be eliminated. The column will be used later to provide a
degenerate basic variable, circled.)
[4] If only one row or only one column remains under consideration, then the
procedure is completed by selecting every remaining variable associated
with that row or column to be basic with the only feasible allocation.
Otherwise, return to [1].
For step [1]:
Northwest corner rule: Begin by selected x11. Thereafter, if xij was the last basic
variable selected, then next select xi,j+1 (one column to the right) if source i
has any supply remaining. Otherwise, next select xi+1, j (one row down).
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ISE 330: Lecture 21 Notes
http://www-classes.usc.edu/engr/ise/330
Vogel’s approximation method: For each row and column remaining under
consideration, calculate its difference, the arithmetic difference between the
smallest and next-to-the-smallest unit cost cij still reamining in that row or
column. In that row or column having the largest difference, select the
variable having the smallest remaining unit cost.
Russel’s approximation method: For each source row i remaining under
consideration, determine its ui, the largest unit cost cij still remaining in that
row. For each destination column j remaining under consideration,
determine its vj, which is the largest unit cost cij still remaining in that column.
For each variable xij not previously selected in these rows and columns,
calculate ij = cij – ui – vj. Select the variable having the largest (in absolute
terms) negative value of ij.
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ISE 330: Lecture 21 Notes
http://www-classes.usc.edu/engr/ise/330
The Transportation Simplex Method
Initialization: Construct an initial BFS by the procedure outlined above.
Optimality Test: Derive ui and vj by selecting the row having the largest number of
allocations, setting its ui = 0, then solving the set of equations cij = ui + vj for
each (i,j) such that xij is basic. If cij – ui – vj  0 for every (i,j) such that xij is
nonbasic, then the current solution is optimal.
Iteration:
1. Determine the entering basic variable: Select the nonbasic variable xij having
the largest (in absolute terms) negative value of cij – ui – vj.
2. Determine the leaving basic variable: Identify the chain reaction required to
retain feasibility when the entering basic variable is increased. Fromt eh
donor cells, select the basic variable having the smallest value.
3. Determine the new BFS: Add the value of the leaving basic variable to the
allocation for each recipient cell. Subtract this value from the allocation for
each donor cell.
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