Chapter 4: Linear Programming Sensitivity Analysis

```CHAPTER 7:
TRANSPORTATION, ASSIGNMENT
AND TRANSSHIPMENT PROBLEMS
NETWORK FLOW MODEL
Consists of nodes representing a set of origins
and a set of destinations.
 An arc is used to represent the route from each
origins to each destinations.
 Each origin has a supply and each destination
has a demand.
 Objective: To determine the optimal amount to
ship from each origin to each destination

Network flow problems

Transportation Problem

Assignment Problem

Transshipment Problem
TRANSPORTATION PROBLEM






Transportation Problem
Problems of distributing goods and services
from several supply location to several
demand locations
Supply locations are called as Origin
Demand locations are called as Destination
Quantity of goods at origin are limited
Quantity of goods at destinations are known
TRANSPORTATION PROBLEM







Transportation Problem
Each origin and destinations are represented by
Circles called as nodes
Each origin and destinations are connected by
arc
Each node requires one constraint
Each arc requires one variable
The sum of variables corresponding to arcs from
an origin node must be less than or equal to
origin supply. (Rule 3)
The sum of variables corresponding to arcs into
an destination node must equal to destination ‘s
demand (Rule 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.

TRANSPORTATION PROBLEM

Network Representation
s1
s2
1
c11
c13
c21
2
c23
SOURCES
1
d1
2
d2
3
d3
c12
c22
DESTINATIONS
TRANSPORTATION PROBLEM

LP Formulation
The LP formulation in terms of the
amounts shipped from the origins to the
destinations, xij , can be written as:
Min cijxij
ij
s.t. xij &lt; si
j
xij = dj
i
xij &gt; 0
for each origin i
for each destination j
for all i and j
TRANSPORTATION PROBLEM

Distribution of goods from three plants to four
distributions
Supply
Origin
Plant
Three Months Production
Capacity (units)
1
Cleveland
5000
2
Bedford
6000
3
York
2500,Total=13,500
Demands
Destination
Distribution Centre
Three-Months Demand
Forecast (Units)
1
Boston
6000
2
Chicago
4000
3
St. Louis
2000
4
Lexington
1500,Total=13,500
TRANSPORTATION PROBLEM
Objectives
To determine the routes to be used and quantity to
be shipped from each origin to demand route
that will provide minimum total transportation
cost.
Construct a network graph
Connect each origin with the destination with arcs
representing the routes between origin and the
destinations.
12 Possible Routes
TRANSPORTATION NETWORK
Boston
3
2
Cleveland
7
2
Chicago
7
5
Bedford
5
S.Louis
6
Supplies
2
4
Demands
York
5
Transportation Cost per unit
3
Lexington
FORMULATING THE PROBLEM
X ijnumber of units shipped from origin I to destination j
X11 number of units shipped from origin (Cleveland) to
destination 1 (boston)
X12 number of units shipped from origin (Cleveland) to
destination 2 (Chicago)
Cost per unit
Origin
Boston
Chicago
St.Louis Lexington
Cleveland
3
2
7
6
Bedford
7
5
2
3
York
2
5
4
5
Objective Function=sum of cost from each source to destinations
Transportation cost shipped from Cleveland
 3x11+2x12+7x13+6x14
 Transportation cost shipped from Bedford
 7x21+5x22+2x23+3x24
 Transportation cost shipped from York
 2x31+5x32+4x33+5x34
 How many total variables and constraint??
 What are supply and demand constraints???

SUPPLY CONSTRAINT (RULE 3)
Transportation shipped from Cleveland
 X11+x12+x13+x14&lt;= 5000
 Transportation shipped from Bedford
 X21+x22+x23+x24&lt;=6000
 Transportation shipped from York
 X31+x32+x33+x34&lt;=2500

DEMAND CONSTRAINT (RULE 4)
Transportation To boston
 X11+x21+x31= 6000
 Transportation to Chicago
 X12+x22+x32=4000
 Transportation to St.Louis
 X13+x23+x33=2500
 Transportation to Lexington
 X14+x24+x34=1500

Objective function????
MODEL FORMALATION

Objective function
Min 3x11+2x12+7x13+6x14+7x21+5x22+2x23+3x24+2x31+5x32+4x33+5x34
S.T
X11+x12+x13+x14
X11
X12
X13
&lt;= 5000
X21+x22+x23+x24
&lt;=6000
X31+x32+x33+x34&lt;=2500
+x21
+x31
= 6000
+x22
+x32
=4000
+x23
+x33
=2500
X14
+x24
+x34 =1500
SOLUTION
Variable Value
Reduced Cost
X11
3500.00
0.0
X12
1500.00
0.0
X13
0.0
8.0
X14
0.0
6.0
X21
2500.00
1.0
X22
2000.00
0.0
X23
1500.00
0.0
X24
2500.00
0.0
X31
0.00
0.0
X32
0.00
4.0
X33
0.00
6.0
X34
0.00
6.0
Minimum total transportation cost?
Units shipped=3500;cost per unit=3
Total cost from Cleveland to Boston??
GENERAL LINEAR PROGRAMMING MODEL OF
TRANSPORTATION PROBLEM
m
Min 
i 1
n
c
j 1
ij
xij
s.t
n
x
j 1
m
ij
 x ij
 si , i  1,2,3.....m   sup ply
 dj , j  1,2,3....n    Demand
i 1
xij  0
Xij=number of units shipped from origin I to destination j
Cij = cost per unit shipping from origin I to destination j
Si= supply or capacity in units at origin I
Dj= demand in units at destinations j
GENERALIZED ASSIGNMENT PROBLEM
m
n
i 1
i 1
Min   cij x ij
s.t.
n
x
j 1
ij
m
x
j 1
ij
 1, i  1,2,3,.....m    Agents
 1; i ,2,3,........n      Tasks
xij  0
•Assigning jobs to machine
•Sales personnel to sales territory
•One to one assignment, i.e. one agent is assigned to one
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.

ASSIGNMENT PROBLEM

Network Representation
c
11
1
AGENTS
c13
c21
2
1
c12
c22
2
c23
c31
3
c33
c32
3
ASSIGNMENT PROBLEM

LP Formulation
Min cijxij
ij
s.t.
xij = 1
for each agent i
xij = 1
xij = 0 or 1
for all i and j
j
i

Note: A modification to the right-hand side of the first
constraint set can be made if a worker is permitted to work
more than 1 job.
Assignment Problem

LP Formulation Special Cases
•Number of agents exceeds the number of tasks:
xij &lt; 1
j
for each agent i
•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.
Estimated project completion Time
Client 1
Client 2
Client 3
Terry
10
15
9
Carle
9
18
5
Jack
6
14
3
ASSIGNMENT PROBLEM NETWORK
10
Client
1
Terry
9
15
Carle
Client 2
7
18
5
6
6
14
Supplies
Demands
Jack
Client 3
3
Number of Constraints=6;Number of variables=9
Completion Time in Days
PROBLEM FORMULATION

Days requires for Terry’s assignment


Days required for Carle ‘s assignment



9x21+18x22+5x23
Days required for Jack ‘s assignment


10x11+15x12+9x13
6x31+14x32+3x33
Objective Function
Min 10x11+15x12+9x13 +10x11+15x12+9x13+ 6x31+14x32+3x33
CONSTRAINT

Each project leader can be assigned to at most
one client.

X11+x12+x13 &lt;=1 ;Terry ‘s assignment
X21+x21+x23&lt;=1; Carle ‘s assignment
X31+X21 +x31&lt;=1 ;Jack assignment

Each Client must have at least one leader





X11+X21+X31=1; Client 1
X12+X22+X32=1; Client 2
X13+X23+X33=1 ; Client 3
SOLUTION
Variable
Value
Reduced Cost
X11
0.00
0.00
X12
1.00
0.00
X13
0.00
3.00
X21
0.00
0.00
X22
0.00
4.00
X23
1.00
0.00
X31
1.00
0.00
X32
0.00
3.0
X33
0.00
1.00
Terry is assigned to client2;x12=1
Carle is assigned to client3;x23=1
Jack is assigned to client 1 x31=1
Total completion time required is
26 days
Example: Hungry Owner
A 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.
Projects
Subcontractor A B C
Westside
50 36 16
Federated
28 30 18
Goliath
35 32 20
Universal
25 25 14
How should the contractors be assigned to minimize
total costs?
Example: Hungry Owner

Network Representation
West.
Subcontractors
50
36
16
28
Fed.
18
35
Gol.
Univ.
20
25
14
A
Projects
30
B
32
C
25
Example: Hungry Owner

Linear Programming Formulation
Min
s.t.
50x11+36x12+16x13+28x21+30x22+18x23
+35x31+32x32+20x33+25x41+25x42+14x43
x11+x12+x13 &lt; 1
x21+x22+x23 &lt; 1
Agents
x31+x32+x33 &lt; 1
x41+x42+x43 &lt; 1
x11+x21+x31+x41 = 1
x12+x22+x32+x42 = 1
x13+x23+x33+x43 = 1
xij = 0 or 1 for all i and j
HUNGARIAN METHOD




Step 1: For each row, subtract the minimum number
in that row from all numbers in that row.
Step 2: For each column, subtract the minimum
number in that column from all numbers in that
column.
Step 3: Draw the minimum number of lines to cover
all zeroes. If this number = m, STOP -- an assignment
Step 4: Subtract d (the minimum uncovered number)
from uncovered numbers. Add d to numbers covered
by two lines. Numbers covered by one line remain the
same. Then, GO TO STEP 3.
HUNGARIAN METHOD

Finding the Minimum Number of Lines and
Determining the Optimal Solution
Step 1: Find a row or column with only one unlined
zero and circle it. (If all rows/columns have two or
more unlined zeroes choose an arbitrary zero.)
 Step 2: If the circle is in a row with one zero, draw
a line through its column. If the circle is in a
column with one zero, draw a line through its row.
One approach, when all rows and columns have
two or more zeroes, is to draw a line through one
with the most zeroes, breaking ties arbitrarily.
 Step 3: Repeat step 2 until all circles are lined. If
this minimum number of lines equals m, the circles
provide the optimal assignment.

EXAMPLE: HUNGRY OWNER

Initial Tableau Setup
Since the Hungarian algorithm requires that there
be the same number of rows as columns, add a
Dummy column so that the first tableau is:
Westside
Federated
Goliath
Universal
A
50
28
35
25
B
C Dummy
36 16
0
30 18
0
32 20
0
25 14
0
EXAMPLE: HUNGRY OWNER


Step 1: Subtract minimum number in each row from all
numbers in that row. Since each row has a zero, we
would simply generate the same matrix above.
Step 2: Subtract the minimum number in each column
from all numbers in the column. For A it is 25, for B it is
25, for C it is 14, for Dummy it is 0. This yields:
Westside
Federated
Goliath
Universal
A
B
C Dummy
25 11
2
0
3
5
4
0
10
7 6
0
0
0
0
0
EXAMPLE: HUNGRY OWNER

Step 3: Draw the minimum number of lines to
cover all zeroes. Although one can &quot;eyeball&quot;
this minimum, use the following algorithm. If a
&quot;remaining&quot; row has only one zero, draw a line
through the column. If a remaining column
has only one zero in it, draw a line through the
row.
A
Westside 25
Federated 3
Goliath
10
Universal 0
B C Dummy
11
2
0
5
4
0
7
6
0
0
0
0
EXAMPLE: HUNGRY OWNER

Step 3:
Draw the minimum number of lines to cover all
zeroes.
Westside
Federated
Goliath
Universal
A
23
1
8
0
B C Dummy
9 0
0
3
2
0
5 4
0
0 0
2
EXAMPLE: HUNGRY OWNER

Step 4: The minimum uncovered number is 1.
Subtract 1 from uncovered numbers. Add 1 to
numbers covered by two lines. This gives:
A B C Dummy
Westside 23 9 0
1
Federated 0 2 1
0
Goliath
7 4 3
0
Universal 0 0 0
3
EXAMPLE: HUNGRY OWNER

Step 3: The minimum number of lines to cover all 0's is
four. Thus, there is a minimum-cost assignment of 0's
with this tableau. The optimal assignment is:
Subcontractor Project Distance
Westside
C
16
Federated
A
28
Goliath
(unassigned)
Universal
B
25
Total Distance = 69 miles
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.
TRANSSHIPMENT PROBLEM

Network Representation3
c13
s1
1
c36
c37
c14
6
d1
c46
c15
Supply
Demand
c47
4
c23
s2
2
c56
c24
c25
SOURCES
5
7
d2
c57
INTERMEDIATE
NODES
DESTINATIONS
TRANSHIPMENT PROBLEM
Contains three types of nodes: origin
,transhipment node and destination nodes
 For origin nodes sum of shipments out minus The
sum of shipment in must be less than or equal to
origin supply.
 For destination nodes sum of shipments out minus
The sum of shipment in must be equal to demand
 For transhipment nodes the sum of shipments out
must equal to sum of shipments in.

ASSIGNMENT PROBLEM NETWORK
600
400
Supplies
1
Denver
2
Atlanta
10
3
Kansas
4
Louis
ville
5
Detroit
6
Miami
150
7
Dallas
350
Distribution Routes
Number of Constraints=8;Number of variables=12
200
Demands
8
New
Orlean
s
300
CONSTRAINTS











Origin Nodes ??
X13+X14 &lt;=600 (Denver)
X23+x24 &lt;=400 (Atlanta)
For transhipment Nodes
x35+x36+x37+x38=x13+ x23 (Node 3;units in=units out)
X45+X46+X47+X48=X14+X24(Node 4;units in= units out)
For Destination nodes
X35+x45=200
X36+x46=150
X37+x47=350
X38+x48=300
TRASPORTION COST PER UNIT
Ware House
Plant
Kansas city (3)
Lousville (4)
Denver (1)
2
3
Atlanta (2)
3
1
Retail Outlet
Warehouse
Detroit (5)
Miami
(6)
Dallas (7)
New
Orleans (8)
Kansas City
2
6
3
6
Louisville
4
4
6
5
Obj function
2x13+3x14+3x23+x24+2x35+6x36+3x37+6x38+4x45+4x46+6x47+5x48+4x28+x78
SOLUTION
Variable
Value
Reduced Costs
X13
550.00
0.00
X14
50.00
0.00
X23
0.00
3.00
X24
400.00
0.00
X35
200.00
0.00
X36
0.00
1.00
X37
350.00
0.00
X38
0.00
0.00
X45
0.00
3.00
X46
150.00
0.00
X47
0.00
4.00
X48
300.00
0.00
Value of objective function?
EXAMPLE: TRANSSHIPPING
Thomas Industries and Washburn Corporation
supply three firms (Zrox, Hewes, Rockwright) 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 Rockwright. Both
Arnold and Supershelf can supply at most 75 units to
its customers.
Additional data is shown on the next slide.
Example: Transshipping
Because of long standing contracts based on past
orders, unit costs from the manufacturers to the
suppliers are:
Thomas Washburn
Arnold
5
8
Supershelf
7
4
The costs to install the shelving at the various
locations are:
Zrox
Thomas
1
Washburn
3
Hewes Rockwright
5
8
4
4
EXAMPLE: TRANSSHIPPING

Network Representation
75 ARNOLD
Arnold
5
50
Hewes
HEWES
60
RockWright
40
5
Thomas
8
3
7
Super
Zrox
1
8
75 Shelf
4
ZROX
WashWASH
BURN
Burn
4
4
EXAMPLE: TRANSSHIPPING

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 (Thomas), 4 (Washburn)
k = 5 (Zrox), 6 (Hewes), 7 (Rockwright)

Objective Function Defined
Minimize Overall Shipping Costs:
Min 5x13 + 8x14 + 7x23 + 4x24 + 1x35 + 5x36 + 8x37
+ 3x45 + 4x46 + 4x47
EXAMPLE: TRANSSHIPPING

Constraints Defined
Amount Out of Arnold:
Amount Out of Supershelf:
Amount Through Thomas:
= 0
Amount Through Washburn:
= 0
Amount Into Zrox:
Amount Into Hewes:
Amount Into Rockwright:
x13 + x14 &lt; 75
x23 + x24 &lt; 75
x13 + x23 - x35 - x36 - x37
x14 + x24 - x45 - x46 - x47
x35 + x45 = 50
x36 + x46 = 60
x37 + x47 = 40
Non-negativity of Variables: xij &gt; 0, for all i and j.
Example: Transshipping

Optimal Solution (from The Management Scientist )
Objective Function Value =
1150.000
Variable
Value
Reduced Costs
X13
X14
X23
X24
X35
X36
X37
X45
X46
X47
75.000
0.000
0.000
75.000
50.000
25.000
0.000
0.000
35.000
40.000
0.000
2.000
4.000
0.000
0.000
0.000
3.000
3.000
0.000
0.000
Example: Transshipping

Optimal Solution
ZROX
75 ARNOLD
Arnold
5
Hewes
HEWES
60
RockWright
40
5
Thomas
8
3 4
7
Super
50
1
75
8
75 Shelf
4
Zrox
WashWASH
BURN
Burn
4
```