Slides by
John
Loucks
St. Edward’s
University
Modifications by
A. Asef-Vaziri
1
Chapter 6, Part A
Distribution and Network Models



Transportation Problem
• Network Representation
• General LP Formulation
Assignment Problem
• Network Representation
• General LP Formulation
Transshipment Problem
• Network Representation
• General LP Formulation
2
Transportation, Assignment, and
Transshipment Problems


A network model is one which can be represented
by a set of nodes, a set of arcs, and functions (e.g.
costs, supplies, demands, etc.) associated with the
arcs and/or nodes.
Transportation, assignment, transshipment,
shortest-route, and maximal flow problems of this
chapter as well as the minimal spanning tree and
PERT/CPM problems (in others chapter) are all
examples of network problems.
3
Transportation, Assignment, and
Transshipment Problems



Each of the five problems of this chapter can be
formulated as linear programs and solved by
general purpose linear programming codes.
For each of the five problems, if the right-hand side
of the linear programming formulations are all
integers, the optimal solution will be in terms of
integer values for the decision variables.
However, there are many computer packages that
contain separate computer codes for these problems
which take advantage of their network structure.
4
Transportation Problem


The transportation problem seeks to minimize the
total shipping costs of transporting goods from m
origins (each with a supply si) to n destinations
(each with a demand dj), when the unit shipping
cost from an origin, i, to a destination, j, is cij.
The network representation for a transportation
problem with two sources and three destinations is
given on the next slide.
5
Transportation Problem

Network Representation
s1
s2
1
c11
Sources
c23
d1
2
d2
3
d3
c12
c13
c21
2
1
c22
Destinations
6
Transportation Problem: Example #1
Acme Block Company has orders for 80 tons of
concrete blocks at three suburban locations as follows:
Northwood -- 25 tons, Westwood -- 45 tons, and
Eastwood -- 10 tons. Acme has two plants, each of
which can produce 50 tons per week. Delivery cost per
ton from each plant to each suburban location is shown
on the next slide.
How should end of week shipments be made to fill
the above orders?
7
Transportation Problem: Example #1

Delivery Cost Per Ton
Plant 1
Plant 2
Northwood
24
30
Westwood
30
40
Eastwood
40
42
Decision Variables. the tons of concrete blocks, xij , to be
shipped from source i to destination j.
Plant 1
Plant 1
Northwood
x11
x21
Westwood
x12
x22
Eastwood
x13
x23
8
Transportation Problem: Example #2

Define the Objective Function
Minimize the total shipping cost.
Min: (shipping cost per ton for each origin to
destination) × (number of pounds shipped from
each origin to each destination).
Min: 24x11 + 30x12 + 40x13 + 30x21 + 40x22 + 42x23
9
Transportation Problem: Example #2
= Constraints

Define the Constraints
Supply constraints:
(1) x11 + x12 + x13 = 50
(2) x21 + x22 + x23 = 50
Demand constraints:
(4) x11 + x21 = 25
(5) x12 + x22 = 45
(6) x13 + x23 = 10
Non-negativity of variables:
xij > 0, i = 1, 2 and j = 1, 2, 3
10
Transportation Problem: Example #2
≤ and ≥ Constraints

Define the Constraints
Supply constraints:
(1) x11 + x12 + x13 ≤ 50
(2) x21 + x22 + x23 ≤ 50
Demand constraints:
(4) x11 + x21 ≥ 25
(5) x12 + x22 ≥ 45
(6) x13 + x23 ≥ 10
Non-negativity of variables:
xij > 0, i = 1, 2 and j = 1, 2, 3
11
Transportation Problem: Example #1

Partial Spreadsheet Showing Problem Data
A
B
C
1
2
3
4
D
E
F
G
H
X22
X23
RHS
LHSCoefficients
Constraint
X11
X12
X13
#1
1
1
1
#2
5
#3
6
#4
7
#5
8 Obj.Coefficients
X21
50
1
1
1
1
1
30
1
40
30
50
25
1
24
1
40
45
1
10
42
30
12
Transportation Problem: Example #1
Cost Table
Plant 1
Plant 2
Northwood Westwood Eastwood
24
30
40
30
40
42
`
Decision Variable Table
RHS
Northwood Westwood Eastwood LHS
Plant 1
5
45
0
50 ≤ 50
Plant 2
20
0
10
30 ≤ 50
LHS
25
45
10
2490
≥
≥
≥
RHS
25
45
10
13
Transportation Problem: Example #1
Sensitivity Report
Variable Cells
Cell
$B$9
$C$9
$D$9
$B$10
$C$10
$D$10
Name
Plant 1 Northwood
Plant 1 Westwood
Plant 1 Eastwood
Plant 2 Northwood
Plant 2 Westwood
Plant 2 Eastwood
Final Reduced Objective Allowable Allowable
Value Cost Coefficient Increase Decrease
5
0
24
4
4
45
0
30
4
36
0
4
40
1E+30
4
20
0
30
4
4
0
4
40
1E+30
4
10
0
42
4
42
Constraints
Cell
$B$11
$C$11
$D$11
$E$9
$E$10
Name
LHS Northwood
LHS Westwood
LHS Eastwood
Plant 1 LHS
Plant 2 LHS
Final Shadow Constraint Allowable Allowable
Value Price
R.H. Side Increase Decrease
25
30
25
20
20
45
36
45
5
20
10
42
10
20
10
50
-6
50
20
5
30
0
50
1E+30
20
14
Transportation Problem: Example #2
The Navy has 9,000 pounds of material in Albany,
Georgia that it wishes to ship to three installations:
San Diego, Norfolk, and Pensacola. They require 4,000,
2,500, and 2,500 pounds, respectively. Government
regulations require equal distribution of shipping
among the three carriers.
The shipping costs per pound for truck, railroad,
and airplane transit are shown on the next slide.
Formulate and solve a linear program to determine the
shipping arrangements (mode, destination, and
quantity) that will minimize the total shipping cost.
15
Transportation Problem: Example #2
Mode
Truck
Railroad
Airplane
Destination
San Diego Norfolk Pensacola
$12
20
30
$6
11
26
$5
9
28
16
Transportation Problem: Example #2

Define the Decision Variables
We want to determine the pounds of material, xij ,
to be shipped by mode i to destination j. The
following table summarizes the decision variables:
Truck
Railroad
Airplane
San Diego Norfolk Pensacola
x11
x12
x13
x21
x22
x23
x31
x32
x33
17
Transportation Problem: Example #2

Define the Objective Function
Minimize the total shipping cost.
Min: (shipping cost per pound for each mode per
destination pairing) x (number of pounds shipped
by mode per destination pairing).
Min: 12x11 + 6x12 + 5x13 + 20x21 + 11x22 + 9x23
+ 30x31 + 26x32 + 28x33
18
Transportation Problem: Example #2

Define the Constraints
Equal use of transportation modes:
(1) x11 + x12 + x13 = 3000
(2) x21 + x22 + x23 = 3000
(3) x31 + x32 + x33 = 3000
Destination material requirements:
(4) x11 + x21 + x31 = 4000
(5) x12 + x22 + x32 = 2500
(6) x13 + x23 + x33 = 2500
Non-negativity of variables:
xij > 0, i = 1, 2, 3 and j = 1, 2, 3
19
Transportation Problem: Example #2
Cost Table
Truck
Railroad
Airplane
San Siego
12
20
30
`
Decision Variable Table
San Siego
Truck
1000
Railroad
0
Airplane
3000
LHS
4000
≥
RHS
4000
Norfolk
6
11
26
Norfolk
2000
500
0
2500
≥
2500
Pensacola
5
9
28
Pensacola
0
2500
0
2500
≥
2500
LHS
RHS
3000 ≤ 3000
3000 ≤ 3000
3000 ≤ 3000
142000
20
Transportation Problem

Linear Programming Formulation
Using the notation:
xij = number of units shipped from
origin i to destination j
cij = cost per unit of shipping from
origin i to destination j
si = supply or capacity in units at origin i
dj = demand in units at destination j
continued
21
Transportation Problem

Linear Programming Formulation (continued)
Min
m
n
 c x
i 1 j 1
n
x
j 1
ij
 si
i 1, 2,
,m
Supply
ij
>
 dj
j 1, 2,
,n
Demand
m
x
i 1
ij ij
xij > 0 for all i and j
22
Transportation Problem

LP Formulation Special Cases
• Total supply exceeds total demand:
No modification of LP formulation is necessary.
• Total demand exceeds total supply:
Add a dummy origin with supply equal to
the shortage amount. Assign a zero shipping
cost per unit. The amount “shipped” from
the dummy origin (in the solution) will not
actually be shipped.
23
Transportation Problem

LP Formulation Special Cases (continued)
• The objective is maximizing profit or revenue:
Solve as a maximization problem.
• Minimum shipping guarantee from i to j:
xij > Lij
• Maximum route capacity from i to j:
xij < Lij
• Unacceptable route:
Remove the corresponding decision variable.
24
Assignment Problem




An assignment problem seeks to minimize the total
cost assignment of m workers to m jobs, given that the
cost of worker i performing job j is cij.
It assumes all workers are assigned and each job is
performed.
An assignment problem is a special case of a
transportation problem in which all supplies and all
demands are equal to 1; hence assignment problems
may be solved as linear programs.
The network representation of an assignment problem
with three workers and three jobs is shown on the
next slide.
25
Assignment Problem

Network Representation
1
Agents
c11
c13
c21
2
1
c12
Tasks
c22
2
c23
c31
3
c33
c32
3
26
Assignment Problem: Example
An electrical contractor pays his subcontractors a
fixed fee plus mileage for work performed. On a given
day the contractor is faced with three electrical jobs
associated with various projects. Given below are the
distances between the subcontractors and the projects.
Subcontractor
Westside
Federated
Goliath
Universal
Projects
A B C
50 36 16
28 30 18
35 32 20
25 25 14
How should the contractors be assigned so that total
mileage is minimized?
27
Assignment Problem: Example

Network Representation
West.
Subcontractors
50
36
16
28
Fed.
18
35
Gol.
Univ.
20
25
A
Projects
30
B
32
C
25
14
28
Assignment Problem: Example

Linear Programming Formulation
Min
s.t.
50x11+36x12+16x13+28x21+30x22+18x23
+35x31+32x32+20x33+25x41+25x42+14x43
x11+x12+x13 < 1
x21+x22+x23 < 1
Agents
x31+x32+x33 < 1
x41+x42+x43 < 1
x11+x21+x31+x41 = 1
x12+x22+x32+x42 = 1
Tasks
x13+x23+x33+x43 = 1
xij = 0 or 1 for all i and j
29
Assignment Problem: Example

Linear Programming Formulation
Min
s.t.
50x11+36x12+16x13+28x21+30x22+18x23
+35x31+32x32+20x33+25x41+25x42+14x43
x11+x12+x13 < 1
x21+x22+x23 < 1
Agents
x31+x32+x33 < 1
x41+x42+x43 < 1
x11+x21+x31+x41 ≥ 1
x12+x22+x32+x42 ≥ 1
Tasks
x13+x23+x33+x43 ≥ 1
xij = 0 or 1 for all i and j
30
Transportation Problem: Example #2
Cost Table
Project A
50
Westside
28
Federated
35
Goliath
25
Universal
`
Decision Variable Table
San Siego
0
Westside
1
Federated
0
Goliath
0
Universal
1
LHS
≥
1
RHS
Project B
36
30
32
25
Norfolk
0
0
0
1
1
≥
1
Project C
16
18
20
14
Pensacola
1
0
0
0
1
≥
1
LHS
1
1
0
1
69
≤
≤
≤
≤
RHS
1
1
1
1
31
Assignment Problem: Example

The optimal assignment is:
Subcontractor Project Distance
Westside
C
16
Federated
A
28
Goliath
(unassigned)
Universal
B
25
Total Distance = 69 miles
32
Assignment Problem

Linear Programming Formulation
Using the notation:
xij =
1 if agent i is assigned to task j
0 otherwise
cij = cost of assigning agent i to task j
continued
33
Assignment Problem

Linear Programming Formulation (continued)
Min
m
n
 c x
i 1 j 1
n
x
j 1
ij
1
i 1, 2,
,m
Agents
ij
1
j 1, 2,
,n
Tasks
m
x
i 1
ij ij
xij > 0 for all i and j
34
Assignment Problem

LP Formulation Special Cases
•Number of agents exceeds the number of tasks:
Extra agents simply remain unassigned.
•Number of tasks exceeds the number of agents:
Add enough dummy agents to equalize the
number of agents and the number of tasks.
The objective function coefficients for these
new variable would be zero.
35
Assignment Problem

LP Formulation Special Cases (continued)
•The assignment alternatives are evaluated in terms
of revenue or profit:
Solve as a maximization problem.
•An assignment is unacceptable:
Remove the corresponding decision variable.
•An agent is permitted to work t
n
x
j 1
ij
t
i 1, 2,
,m
tasks:
Agents
36
Transshipment Problem




Transshipment problems are transportation problems
in which a shipment may move through intermediate
nodes (transshipment nodes)before reaching a
particular destination node.
Transshipment problems can be converted to larger
transportation problems and solved by a special
transportation program.
Transshipment problems can also be solved by
general purpose linear programming codes.
The network representation for a transshipment
problem with two sources, three intermediate nodes,
and two destinations is shown on the next slide.
37
Transshipment Problem

Network Representation
s1
c15
Supply
s2
3
c13
1
c37
c14
Sources
c25
6
c46
c47
4
c23
2
c36
c56
c24
5
Demand
7
c57
d1
d2
Destinations
Intermediate Nodes
38
Transshipment Problem

Linear Programming Formulation
Using the notation:
xij = number of units shipped from node i to node j
cij = cost per unit of shipping from node i to node j
si = supply at origin node i
dj = demand at destination node j
continued
39
Transshipment Problem

Linear Programming Formulation (continued)
Min

c ij xij

xij 

xij 
all arcs
s.t.
arcs out
ij
 si
Origin nodes i
x
ij
0
Transhipment nodes
arcs in
arcs out
x
arcs in
x
arcs in
ij


xij  d j Destination nodes j
arcs out
xij > 0 for all i and j
continued
40
Transshipment Problem

LP Formulation Special Cases
• Total supply not equal to total demand
• Maximization objective function
• Route capacities or route minimums
• Unacceptable routes
The LP model modifications required here are
identical to those required for the special cases in
the transportation problem.
41
Transshipment Problem: Example
The Northside and Southside facilities of Zeron
Industries supply three firms (Zrox, Hewes, Rockrite)
with customized shelving for its offices. They both
order shelving from the same two manufacturers,
Arnold Manufacturers and Supershelf, Inc.
Currently weekly demands by the users are 50
for Zrox, 60 for Hewes, and 40 for Rockrite. Both
Arnold and Supershelf can supply at most 75 units to
its customers.
Additional data is shown on the next slide.
42
Transshipment Problem: Example
Because of long standing contracts based on
past orders, unit costs from the manufacturers to the
suppliers are:
Zeron N
Arnold
5
Supershelf
7
Zeron S
8
4
The costs to install the shelving at the various
locations are:
Zrox
Zeron N
1
Zeron S
3
Hewes Rockrite
5
8
4
4
43
Transshipment Problem: Example

Network Representation
ZROX
75
ARNOLD
Arnold
5
Zeron
N
8
75
4
50
Hewes
HEWES
60
RockRite
40
5
8
3
7
Super
Shelf
1
Zrox
Zeron
WASH
BURN
S
4
4
44
Transshipment Problem: Example

Linear Programming Formulation
• Decision Variables Defined
xij = amount shipped from manufacturer i to supplier j
xjk = amount shipped from supplier j to customer k
where i = 1 (Arnold), 2 (Supershelf)
j = 3 (Zeron N), 4 (Zeron S)
k = 5 (Zrox), 6 (Hewes), 7 (Rockrite)
• Objective Function Defined
Minimize Overall Shipping Costs:
Min 5x13 + 8x14 + 7x23 + 4x24 + 1x35 + 5x36 + 8x37
+ 3x45 + 4x46 + 4x47
45
Transshipment Problem: Example

Constraints Defined
Amount Out of Arnold:
Amount Out of Supershelf:
Amount Through Zeron N:
Amount Through Zeron S:
Amount Into Zrox:
Amount Into Hewes:
Amount Into Rockrite:
x13 + x14 < 75
x23 + x24 < 75
x13 + x23 - x35 - x36 - x37 = 0
x14 + x24 - x45 - x46 - x47 = 0
x35 + x45 = 50
x36 + x46 = 60
x37 + x47 = 40
Non-negativity of Variables: xij > 0, for all i and j.
46
Transshipment Problem: Example
Cost Table
Zeron N
Arnold
5
Supershelf
7
Zeron S
8
4
Decision Variable Table
Zeron N
Zeron S
Arnold
75
0
Supershelf
0
75
75
75
Zeron N
Zeron S
75
75
≤
≤
75
75
Hewes
5
4
Rockrite
8
4
Zrox
Hewes
25
35
60
≥
60
Rockrite
50
0
50
≥
50
`
Zeron N
Zeron S
Zeron N
Zeron S
Zrox
1
3
0
0
=
=
0
40
40
≥
40
75
75
1150
0
0
47
Transshipment Problem: Example

Solution
ZROX
75
ARNOLD
Arnold
5
75
Zeron
N
8
75
4
50
Hewes
HEWES
60
RockRite
40
5
8
3 4
7
Super
Shelf
1
Zrox
Zeron
WASH
BURN
S
4
48
Transshipment Problem: Example
Variable Cells
Cell
$B$8
$C$8
$B$9
$C$9
$I$8
$J$8
$K$8
$I$9
$J$9
$K$9
Final Reduced Objective Allowable Allowable
Name
Value Cost Coefficient Increase Decrease
Arnold Zeron N
75
0
5
2
2
Arnold Zeron S
0
2
8
1E+30
2
Supershelf Zeron N
0
4
7
1E+30
4
Supershelf Zeron S
75
0
4
2
1E+30
Zeron N Zrox
50
0
1
3
6
Zeron N Hewes
25
0
5
2
2
Zeron N Rockrite
0
3
8
1E+30
3
Zeron S Zrox
0
3
3
1E+30
3
Zeron S Hewes
35
0
4
2
2
Zeron S Rockrite
40
0
4
3
10
Constraints
Cell
$D$14
$D$15
$D$8
$D$9
$I$10
$J$10
$K$10
Name
Zeron N
Zeron S
Arnold
Supershelf
Zrox
Hewes
Rockrite
Final Shadow Constraint Allowable Allowable
Value Price
R.H. Side Increase Decrease
0
5
0
0
75
0
6
0
0
25
75
0
75
1E+30
0
75
-2
75
25
0
50
6
50
0
50
60
10
60
0
25
40
10
40
0
25
49
Transshipment Problem: Example

Computer Output (continued)
Constraint
1
2
3
4
5
6
7
Slack/Surplus
0.000
0.000
0.000
0.000
0.000
0.000
0.000
Dual Values
0.000
2.000
-5.000
-6.000
-6.000
-10.000
-10.000
50
Transshipment Problem: Example

Computer Output (continued)
OBJECTIVE COEFFICIENT RANGES
Variable Lower Limit Current Value Upper Limit
X13
X14
X23
X24
X35
X36
X37
X45
X46
X47
3.000
6.000
3.000
No Limit
No Limit
3.000
5.000
0.000
2.000
No Limit
5.000
8.000
7.000
4.000
1.000
5.000
8.000
3.000
4.000
4.000
7.000
No Limit
No Limit
6.000
4.000
7.000
No Limit
No Limit
6.000
7.000
51
Transshipment Problem: Example

Computer Output (continued)
RIGHT HAND SIDE RANGES
Constraint
1
2
3
4
5
6
7
Lower Limit
75.000
75.000
-75.000
-25.000
0.000
35.000
15.000
Current Value Upper Limit
75.000
No Limit
75.000
100.000
0.000
0.000
0.000
0.000
50.000
50.000
60.000
60.000
40.000
40.000
52
Transshipment Transformed into
Transportation Problem
Cost Table
Zeron N
Arnold
5
Supershelf
7
Zeron N
0
Zeron S
1000
`
Zeron S
8
4
1000
0
Zrox
1000
1000
1
3
Hewes
1000
1000
5
4
Rockrite
1000
1000
8
4
Decision Variable Table
Zeron N
Zeron S
Arnold
75
0
Supershelf
0
75
Zeron N
75
0
Zeron S
0
75
LHS
150
150
=
=
RHS
150
150
Zrox
0
0
50
0
50
=
50
Hewes
0
0
25
35
60
=
60
Rockrite
0
0
0
40
40
=
40
LHS
75
75
150
150
1150
=
=
=
=
RHS
75
75
150
150
53
End of Chapter 6, Part A
54