Transcript decision analysis

Slides by
John
Loucks
St. Edward’s
University
© 2009 South-Western, a part of Cengage Learning
Slide 1
Chapter 10, Part B
Distribution and Network Models



Shortest-Route Problem
Maximal Flow Problem
A Production and Inventory Application
© 2009 South-Western, a part of Cengage Learning
Slide 2
Shortest-Route Problem



The shortest-route problem is concerned with
finding the shortest path in a network from one
node (or set of nodes) to another node (or set of
nodes).
If all arcs in the network have nonnegative values
then a labeling algorithm can be used to find the
shortest paths from a particular node to all other
nodes in the network.
The criterion to be minimized in the shortest-route
problem is not limited to distance even though the
term "shortest" is used in describing the procedure.
Other criteria include time and cost. (Neither time
nor cost are necessarily linearly related to distance.)
© 2009 South-Western, a part of Cengage Learning
Slide 3
Shortest-Route Problem

Linear Programming Formulation
Using the notation:
xij =
1 if the arc from node i to node j
is on the shortest route
0 otherwise
cij = distance, time, or cost associated
with the arc from node i to node j
continued
© 2009 South-Western, a part of Cengage Learning
Slide 4
Shortest-Route Problem

Linear Programming Formulation (continued)
Min

c ij xij

xij

xij 
all arcs
s.t.
 1 Origin node i
arcs out
arcs out
x
ij
0
x
ij
 1 Destination node j
Transhipment nodes
arcs in
arcs in
© 2009 South-Western, a part of Cengage Learning
Slide 5
Example: Shortest Route
Susan Winslow has an important business meeting
in Paducah this evening. She has a number of alternate
routes by which she can travel from the company
headquarters in Lewisburg to Paducah. The network of
alternate routes and their respective travel time, ticket
cost, and transport mode appear on the next two slides.
If Susan earns a wage of $15 per hour, what route
should she take to minimize the total travel cost?
© 2009 South-Western, a part of Cengage Learning
Slide 6
Example: Shortest Route

Network Model
F
2
A
K
B
C
1
Lewisburg
5
D
E
L
G
3
J
6
I
H
4
© 2009 South-Western, a part of Cengage Learning
M
Paducah
Slide 7
Example: Shortest Route
Route
A
B
C
D
E
F
G
H
I
J
K
L
M
Transport
Mode
Train
Plane
Bus
Taxi
Train
Bus
Bus
Taxi
Train
Bus
Taxi
Train
Bus
Time
(hours)
4
1
2
6
3 1/3
3
4 2/3
1
2 1/3
6 1/3
3 1/3
1 1/3
4 2/3
© 2009 South-Western, a part of Cengage Learning
Ticket
Cost
$ 20
$115
$ 10
$ 90
$ 30
$ 15
$ 20
$ 15
$ 15
$ 25
$ 50
$ 10
$ 20
Slide 8
Example: Shortest Route
Route
A
B
C
D
E
F
G
H
I
J
K
L
M
Transport
Mode
Train
Plane
Bus
Taxi
Train
Bus
Bus
Taxi
Train
Bus
Taxi
Train
Bus
Time
(hours)
4
1
2
6
3 1/3
3
4 2/3
1
2 1/3
6 1/3
3 1/3
1 1/3
4 2/3
Time
Cost
$60
$15
$30
$90
$50
$45
$70
$15
$35
$95
$50
$20
$70
© 2009 South-Western, a part of Cengage Learning
Ticket
Cost
$ 20
$115
$ 10
$ 90
$ 30
$ 15
$ 20
$ 15
$ 15
$ 25
$ 50
$ 10
$ 20
Total
Cost
$ 80
$130
$ 40
$180
$ 80
$ 60
$ 90
$ 30
$ 50
$120
$100
$ 30
$ 90
Slide 9
Example: Shortest Route

LP Formulation
• Objective Function
Min 80x12 + 40x13 + 80x14 + 130x15 + 180x16 + 60x25
+ 100x26 + 30x34 + 90x35 + 120x36 + 30x43 + 50x45
+ 90x46 + 60x52 + 90x53 + 50x54 + 30x56
• Node Flow-Conservation Constraints
x12 + x13 + x14 + x15 + x16 = 1 (origin)
– x12 + x25 + x26 – x52 = 0 (node 2)
– x13 + x34 + x35 + x36 – x43 – x53 = 0 (node 3)
– x14 – x34 + x43 + x45 + x46 – x54 = 0 (node 4)
– x15 – x25 – x35 – x45 + x52 + x53 + x54 + x56 = 0 (node 5)
x16 + x26 + x36 + x46 + x56 = 1 (destination)
© 2009 South-Western, a part of Cengage Learning
Slide 10
Example: Shortest Route

Solution Summary
Minimum total cost = $150
x12 = 0
x13 = 1
x14 = 0
x15 = 0
x16 = 0
x25 = 0
x26 = 0
x34 = 1
x35 = 0
x36 = 0
x43 = 0
x45 = 1
x46 = 0
© 2009 South-Western, a part of Cengage Learning
x52 = 0
x53 = 0
x54 = 0
x56 = 1
Slide 11
Maximal Flow Problem


The maximal flow problem is concerned with
determining the maximal volume of flow from one
node (called the source) to another node (called the
sink).
In the maximal flow problem, each arc has a
maximum arc flow capacity which limits the flow
through the arc.
© 2009 South-Western, a part of Cengage Learning
Slide 12
Example: Maximal Flow




A capacitated transshipment model can be developed
for the maximal flow problem.
We will add an arc from the sink node back to the
source node to represent the total flow through the
network.
There is no capacity on the newly added sink-tosource arc.
We want to maximize the flow over the sink-to-source
arc.
© 2009 South-Western, a part of Cengage Learning
Slide 13
Maximal Flow Problem

LP Formulation
(as Capacitated Transshipment Problem)
• There is a variable for every arc.
• There is a constraint for every node; the flow out
must equal the flow in.
• There is a constraint for every arc (except the
added sink-to-source arc); arc capacity cannot be
exceeded.
• The objective is to maximize the flow over the
added, sink-to-source arc.
© 2009 South-Western, a part of Cengage Learning
Slide 14
Maximal Flow Problem

LP Formulation
(as Capacitated Transshipment Problem)
Max xk1
s.t.
(k is sink node, 1 is source node)
xij - xji = 0
(conservation of flow)
xij < cij
(cij is capacity of ij arc)
i
j
xij > 0, for all i and j
(non-negativity)
(xij represents the flow from node i to node j)
© 2009 South-Western, a part of Cengage Learning
Slide 15
Example: Maximal Flow
National Express operates a fleet of cargo planes and
is in the package delivery business. NatEx is interested
in knowing what is the maximum it could transport in
one day indirectly from San Diego to Tampa (via Denver,
St. Louis, Dallas, Houston and/or Atlanta) if its direct
flight was out of service.
NatEx's indirect routes from San Diego to Tampa,
along with their respective estimated excess shipping
capacities (measured in hundreds of cubic feet per day),
are shown on the next slide. Is there sufficient excess
capacity to indirectly ship 5000 cubic feet of packages in
one day?
© 2009 South-Western, a part of Cengage Learning
Slide 16
Example: Maximal Flow

Network Model
Denver
3
2
4
San
Diego
2
4
1
3
4
3
7
Tampa
1
1
5
3
St. Louis
2
3
4
3
3
Houston
3
5
6
5
6
© 2009 South-Western, a part of Cengage Learning
Dallas
Atlanta
Slide 17
Example: Maximal Flow

Modified Network Model
3
2
4
Source
2
4
1
3
3
4
Sink
3
7
1
1
5
3
2
3
4
3
3
5
6
5
Added
arc
6
© 2009 South-Western, a part of Cengage Learning
Slide 18
Example: Maximal Flow

LP Formulation
• 18 variables (for 17 original arcs and 1 added arc)
• 24 constraints
• 7 node flow-conservation constraints
• 17 arc capacity constraints (for original arcs)
© 2009 South-Western, a part of Cengage Learning
Slide 19
Example: Maximal Flow

LP Formulation
• Objective Function
Max x71
• Node Flow-Conservation Constraints
x12 + x13 + x14 – x71 = 0
(node 1)
– x12 + x24 + x25 – x42 – x52 = 0
(node 2)
– x13 + x34 + x36 – x43 = 0
(and so on)
– x14 – x24 – x34 + x42 + x43 + x45 + x46 + x47 – x54 – x64 = 0
– x25 – x45 + x52 + x54 + x57 = 0
– x36 – x46 + x64 + x67 = 0
– x47 – x57 – x67 + x71 = 0
© 2009 South-Western, a part of Cengage Learning
Slide 20
Example: Maximal Flow

LP Formulation (continued)
• Arc Capacity Constraints
x12 < 4 x13 < 3 x14 < 4
x24 < 2
x25 < 3
x34 < 3
x42 < 3
x36 < 6
x43 < 5
x45 < 3
x52 < 3
x54 < 4
x57 < 2
x64 < 1
x67 < 5
© 2009 South-Western, a part of Cengage Learning
x46 < 1
x47 < 3
Slide 21
Example: Maximal Flow

Alternative Optimal Solution #1
Objective Function Value =
10.000
Variable
Value
Variable
Value
x12
x13
x14
x24
x25
x34
x36
x42
x43
3.000
3.000
4.000
1.000
2.000
0.000
5.000
0.000
2.000
x45
x46
x47
x52
x54
x57
x64
x67
x71
0.000
0.000
3.000
0.000
0.000
2.000
0.000
5.000
10.000
© 2009 South-Western, a part of Cengage Learning
Slide 22
Example: Maximal Flow

Alternative Optimal Solution #1
2
2
3
Source
3
10
2
1
Sink
4
1
5
4
3
5
2
3
7
5
6
© 2009 South-Western, a part of Cengage Learning
Slide 23
Example: Maximal Flow

Alternative Optimal Solution #2
Objective Function Value =
10.000
Variable
Value
Variable
Value
x12
x13
x14
x24
x25
x34
x36
x42
x43
3.000
3.000
4.000
1.000
2.000
0.000
4.000
0.000
1.000
x45
x46
x47
x52
x54
x57
x64
x67
x71
0.000
1.000
3.000
0.000
0.000
2.000
0.000
5.000
10.000
© 2009 South-Western, a part of Cengage Learning
Slide 24
Example: Maximal Flow

Alternative Optimal Solution #2
2
2
3
Source
2
1
Sink
4
1
5
3
4
7
1
3
10
5
1
3
4
6
© 2009 South-Western, a part of Cengage Learning
Slide 25
A Production and Inventory Application


Transportation or transshipment models can be
developed for applications that have nothing to do
with the physical shipment of goods from origins to
destinations.
We now show how to use a transshipment model to
solve a production and inventory problem.
© 2009 South-Western, a part of Cengage Learning
Slide 26
Example: Production & Inventory Problem
Contour Carpets is a small manufacturer of carpeting
for home and office installations. Production capacity,
demand, production cost per square yard, and inventory
holding cost per square yard for the next four quarters are
shown below .
Quarter
1
2
3
4
Production
Capacity
(sq. yds.)
600
300
500
400
Production
Demand
Cost
(sq. yds.) ($/sq.yd.)
400
2
500
5
400
3
400
3
© 2009 South-Western, a part of Cengage Learning
Inventory
Cost
($/sq.yd.)
0.25
0.25
0.25
0.25
Slide 27
Example: Production & Inventory Problem
Note that production capacity, demand, and
production costs vary by quarter, whereas the cost of
carrying inventory from one quarter to the next is
constant at $0.25 per yard. Contour wants to determine
how many yards of carpeting to manufacture each
quarter to minimize the total production and inventory
cost for the four-quarter period.
The objective is to determine a production scheduling
and inventory policy that will minimize the total
production and inventory cost for the four quarters.
Constraints involve production capacity and demand in
each quarter.
© 2009 South-Western, a part of Cengage Learning
Slide 28
Example: Production & Inventory Problem

LP Formulation
Let x15 denote the number of square yards of carpet
manufactured in quarter 1. The capacity of the
facility is 600 square yards in quarter 1, so the
production capacity constraint is:
x15 ≤ 300
Using similar decision variables, we obtain the
production capacities for quarters 2–4:
x26 ≤ 300
x37 ≤ 500
x48 ≤ 400
© 2009 South-Western, a part of Cengage Learning
Slide 29
Example: Production & Inventory Problem

LP Formulation
In general, for each quarter the beginning inventory
plus the production minus the ending inventory
must equal demand. However, because quarter 1 has
no beginning inventory, the constraint for node 5 is:
x15 – x56 = 400
The constraints associated with the demand nodes in
quarters 2, 3, and 4 are:
x56 + x26 – x67 = 500
x67 + x37 – x78 = 400
x78 + x48 = 400
© 2009 South-Western, a part of Cengage Learning
Slide 30
Example: Production & Inventory Problem

LP Formulation
The objective is to minimize total production and
inventory cost, so we write the objective function as:
Min 2x15 + 5x26 + 3x37 + 3x48 + 0.25x56 + 0.25x67 + 0.25x78
© 2009 South-Western, a part of Cengage Learning
Slide 31
Example: Production & Inventory Problem

LP Solution
Contour Carpets should manufacture:
600 square yards of carpet in quarter 1
300 square yards in quarter 2,
400 square yards in quarter 3, and
400 square yards in quarter 4.
Note also that 200 square yards will be carried over
from quarter 1 to quarter 2.
The total production and inventory cost is $5150.
© 2009 South-Western, a part of Cengage Learning
Slide 32
End of Chapter 10, Part B
© 2009 South-Western, a part of Cengage Learning
Slide 33