Transcript No Slide Title

Chapter 4
Network
Models
1
4.1 Introduction
A network problem is one that can be
represented by...
Nodes
Arcs
10
2
Function on Arcs
4.1 Introduction
• The importance of network models
– Many business problems lend themselves to a network
formulation.
– Optimal solutions of network problems are guaranteed
integer solutions, because of special mathematical structures.
No special restrictions are needed to ensure integrality.
– Network problems can be efficiently solved by compact
algorithms due to their special mathematical structure, even
for large scale models.
3
Network Terminology
• Flow
– the amount sent from node i to node j, over an arc that connects them.
The following notation is used:
Xij = amount of flow
Uij = upper bound of the flow
Lij = lower bound of the flow
• Directed/undirected arcs
– when flow is allowed in one direction the arc is directed (marked by an
arrow). When flow is allowed in two directions, the arc is undirected (no
arrows).
• Adjacent nodes
– a node (j) is adjacent to another node (i) if an arc joins node i to node j.
4
Network Terminology
• Path / Connected nodes
– Path :a collection of arcs formed by a series of adjacent nodes.
– The nodes are said to be connected if there is a path between them.
• Cycles / Trees / Spanning Trees
– Cycle : a path starting at a certain node and returning to
the same node without using any arc twice.
– Tree : a series of nodes that contain no cycles.
– Spanning tree : a tree that connects all the nodes in a network
( it consists of n -1 arcs).
5
4.2 The Transportation Problem
Transportation problems arise when a cost-effective
pattern is needed to ship items from origins that have
limited supply to destinations that have demand for the
goods.
6
The Transportation Problem
• Problem definition
– There are m sources. Source i has a supply capacity of Si.
– There are n destinations. The demand at destination j is Dj.
– Objective:
Minimize the total shipping cost of supplying the
destinations with the required demand from the available
supplies at the sources.
7
CARLTON PHARMACEUTICALS
• Carlton Pharmaceuticals supplies drugs and other medical
supplies.
• It has three plants in: Cleveland, Detroit, Greensboro.
• It has four distribution centers in:
Boston, Richmond, Atlanta, St. Louis.
• Management at Carlton would like to ship cases of a
certain vaccine as economically as possible.
8
CARLTON PHARMACEUTICALS
• Data
– Unit shipping cost, supply, and demand
To
From
Cleveland
Detroit
Greensboro
Demand
Boston
$35
37
40
1100
Richmond
30
40
15
400
Atlanta
40
42
20
750
St. Louis
32
25
28
750
Supply
1200
1000
800
• Assumptions
–
–
–
–
Unit shipping costs are constant.
All the shipping occurs simultaneously.
The only transportation considered is between sources and destinations.
Total supply equals total demand.
9
CARLTON PHARMACEUTICALS
Network presentation
10
Destinations
Sources
D1=1100
Boston
Cleveland
S1=1200
Richmond
D2=400
Detroit
S2=1000
Atlanta
D3=750
Greensboro
S3= 800
St.Louis
D4=750 11
CARLTON PHARMACEUTICALS –
Linear Programming Model
– The structure of the model is:
Minimize Total Shipping Cost
ST
[Amount shipped from a source]  [Supply at that source]
[Amount received at a destination] = [Demand at that destination]
– Decision variables
Xij = the number of cases shipped from plant i to warehouse j.
where: i=1 (Cleveland), 2 (Detroit), 3 (Greensboro)
j=1 (Boston), 2 (Richmond), 3 (Atlanta), 4(St.Louis)
12
Supply from Cleveland X11+X12+X13+X14 = 1200
Supply from Detroit X21+X22+X23+X24 = 1000
Supply from Greensboro X31+X32+X33+X34 = 800
The supply constraints
Boston
D1=1100
X11
Cleveland
S1=1200
X12
X13
X21
X31
Richmond
X14
X22
Detroit
S2=1000
D2=400
X32
X23
X24
Atlanta
X33
St.Louis
Greensboro
S3= 800
D3=750
X34
D4=750
13
CARLTON PHARMACEUTICAL –
The complete mathematical model
Minimize 35X11+30X12+40X13+ 32X14 +37X21+40X22+42X23+25X24+
40X31+15X32+20X33+38X34
ST
Total shipment out of a supply node
cannot exceed the supply at the node.
Supply constrraints:
X11+ X12+ X13+ X14
X21+ X22+ X23+ X24
X31+ X32+ X33+ X34
Demand constraints:
X11+
X12+
X13+
X21+
X31
X22+
X32
X23+
X14+
All Xij are nonnegative
X33
X24+
X34
Total shipment received at a destination 14
node, must equal the demand at that node.
 1200
 1000
 800
= 1000
= 400
= 750
= 750
CARLTON PHARMACEUTICALS
Spreadsheet
=SUMPRODUCT(B7:E9,B15:E17)
=SUM(B7:E7)
Drag to cells
G8:G9
=SUM(B7:E9)
Drag to cells C11:E11
15
CARLTON PHARMACEUTICALS
Spreadsheet
MINIMIZE Total Cost
SHIPMENTS
Demands are met
Supplies are not
exceeded
16
CARLTON PHARMACEUTICALS
Spreadsheet - solution
CARLTON PHARMACEUTICALS
SOLUTION
MINIMUM COST
CLEVELAND
DETROIT
GREENSBORO
RECEIVED
INPUT
CLEVELAND
DETROIT
GREENSBORO
DEMAND
84000
SHIPMENTS (CASES)
BOSTON RICHMOND ATLANTA ST. LOUIS
850
350
250
750
50
750
1100
400
750
COST (PER CASE)
BOSTON RICHMOND ATLANTA ST.
35
30
40
37
40
42
40
15
20
1100
400
750
SHIPPED
1200
1000
800
750
LOUIS
32
25
28
750
SUPPLY
1200
1000
800
17
CARLTON PHARMACEUTICALS
Sensitivity Report
Adjustable Cells
Final Reduced Objective Allowable Allowable
Cell
Name
Value
Cost
Coefficient Increase Decrease
$B$7 CLEVELAND BOSTON
850
0
35
2
5
$C$7 CLEVELAND RICHMOND
350
0
30
5
17
$D$7 CLEVELAND ATLANTA
0
5
40
1E+30
5
$E$7 CLEVELAND ST. LOUIS
0
9
32
1E+30
9
$B$8 DETROIT BOSTON
0
37
5
2
– 250
Reduced costs
$C$8 DETROIT RICHMOND
0
8
40
1E+30
8
•
The
unit
shipment
cost
between
Cleveland
and
$D$8 DETROIT ATLANTA
0
5
42
1E+30
5
must be reduced
by at least
before it
$E$8 DETROIT ST. LOUIS
750 Atlanta 0
25
9 $5,1E+30
$B$9 GREENSBORO BOSTON
0 would 20
40
1E+30
20 it
become economically
feasible to utilize
$C$9 GREENSBORO RICHMOND
50
0
17
5
• If this route
is used,15the total cost
will increase
$D$9 GREENSBORO ATLANTA
750
0
20
5
1E+30
by
$5
for
each
case
shipped
between
the
two
$E$9 GREENSBORO ST. LOUIS
0
20
28
1E+30
20
cities.
18
CARLTON PHARMACEUTICALS
Sensitivity Report
Adjustable Cells
Final Reduced Objective Allowable Allowable
Cell
Name
Value
Cost
Coefficient Increase Decrease
$B$7 CLEVELAND BOSTON
850
0
35
2
5
$C$7 CLEVELAND RICHMOND
350
0
30
5
17
$D$7 CLEVELAND ATLANTA
0
5
40
1E+30
5
$E$7 CLEVELAND ST. LOUIS
0
9
32
1E+30
9
$B$8 DETROIT BOSTON
250
0
37
5
2
DETROITIncrease/Decrease
RICHMOND
0
8
40
1E+30
8
–$C$8
Allowable
$D$8 DETROIT ATLANTA
0
5
42
1E+30
5
• This is the range of optimality.
$E$8 DETROIT ST. LOUIS
750
0
25
9
1E+30
The unit shipment
cost between Cleveland
and
$B$9 • GREENSBORO
BOSTON
0
20
40
1E+30
20
Boston may increase
up to $2 or 50
decrease up0 to
$C$9 GREENSBORO
RICHMOND
15
17
5
$D$9 GREENSBORO
ATLANTA
0
20
5
1E+30
$5 with no change
in the current750
optimal
$E$9 GREENSBORO
ST. LOUIS
0
20
28
1E+30
20
transportation plan.
19
CARLTON PHARMACEUTICALS
Sensitivity Report
Constraints
Final Shadow Constraint Allowable Allowable
Cell
Name
Value
Price
R.H. Side Increase Decrease
$G$7 CLEVELAND SHIPPED
1200
-2
1200
250
0
$G$8 DETROIT SHIPPED
1000
0
1000
1E+30
0
$G$9 GREENSBORO SHIPPED
800
-17
800
250
0
$B$11 RECEIVED BOSTON
1100
37
1100
0
250
$C$11
RECEIVED
RICHMOND
400
32
400
0
250
– Shadow
prices
$D$11 RECEIVED ATLANTA
750
37
750
0
250
•
For
the
plants,
shadow
prices
$E$11 RECEIVED ST. LOUIS
750
25
750
0
750
convey the cost savings realized
for each extra case of vaccine
produced.
For each additional unit available
in Cleveland the total cost
reduces by $2.
20
CARLTON PHARMACEUTICALS
Sensitivity Report
Constraints
Cell
$G$7
$G$8
$G$9
$B$11
$C$11
$D$11
$E$11
Name
CLEVELAND SHIPPED
DETROIT SHIPPED
GREENSBORO SHIPPED
RECEIVED BOSTON
RECEIVED RICHMOND
RECEIVED ATLANTA
RECEIVED ST. LOUIS
– Shadow prices
Final Shadow Constraint Allowable Allowable
Value
Price
R.H. Side Increase Decrease
1200
-2
1200
250
0
1000
0
1000
1E+30
0
800
-17
800
250
0
1100
37
1100
0
250
400
32
400
0
250
750
37
750
0
250
750
25
750
0
750
• For the warehouses demand, shadow
prices represent the cost savings for
less cases being demanded.
For each one unit decrease in
demanded in Boston, the total cost
decreases by $37.
21
Modifications to the transportation problem
– Cases may arise that require modifications to the basic model.
• Blocked routes - shipments along certain routes are prohibited.
• Remedies:
– Assign a large objective coefficient to the route (Cij = 1,000,000)
22
Modifications to the transportation problem
– Cases may arise that require modifications to the basic model.
• Blocked routes - shipments along certain routes are prohibited.
• Remedies:
– Assign a large objective coefficient to the route (Cij = 1,000,000)
– Add a constraint to Excel solver of the form Xij = 0
Shipments on
a Blocked
Route = 0
23
Modifications to the transportation problem
– Cases may arise that require modifications to the basic model.
• Blocked routes - shipments along certain routes are prohibited.
Only Feasible Routes
• Remedies:
in Changing
Cells
– Assign a large objective Included
coefficient
to the route
(Cij = 1,000,000)
Cell C9ofis the
NOTform
Included
– Add a constraint to Excel solver
Xij = 0
– Do not include the cell representing the rout in the Changing cells
Shipments from Greensboro
to Cleveland are prohibited
24
Modifications to the transportation problem
– Cases may arise that require modifications to the basic model.
• Minimum shipment - the amount shipped along a certain route must not fall
below a pre-specified level.
– Remedy: Add a constraint to Excel of the form Xij B
• Maximum shipment - an upper limit is placed on the amount shipped along a
certain route.
– Remedy: Add a constraint to Excel of the form Xij B
25
MONTPELIER SKI COMPANY
Using a Transportation model for production scheduling
– Montpelier is planning its production of skis for the months of
July, August, and September.
– Production capacity and unit production cost will change from
month to month.
– The company can use both regular time and overtime to produce skis.
– Production levels should meet both demand forecasts and end-ofquarter inventory requirement.
– Management would like to schedule production to minimize its costs
for the quarter.
26
MONTPELIER SKI COMPANY
• Data:
– Initial inventory = 200 pairs
– Ending inventory required =1200 pairs
– Production capacity for the next quarter = 400 pairs in regular time.
= 200 pairs in overtime.
– Holding cost rate is 3% per month per ski.
– Production capacity, and forecasted demand for this quarter
(in pairs of skis), and production cost per unit (by months)
Month
July
August
September
Forecasted
Demand
400
600
1000
Production
Production Costs
Capacity Regular Time Overtime
1000
25
30
800
26
32
400
29
37
27
MONTPELIER SKI COMPANY
• Analysis of demand:
– Net demand in July = 400 - 200 = 200 pairs
Initial inventory
In house inventory
– Net demand in August = 600
– Net demand in September = 1000 + 1200 = 2200 pairs
Forecasted demand
• Analysis of Supplies:
– Production capacities are thought of as supplies.
– There are two sets of “supplies”:
• Set 1- Regular time supply (production capacity)
• Set 2 - Overtime supply
28
MONTPELIER SKI COMPANY
• Analysis of Unit costs
Unit cost = [Unit production cost] +
[Unit holding cost per month][the number of months stays in
inventory]
Example: A unit produced in July in regular time and sold in
September costs 25+ (3%)(25)(2 months) = $26.50
29
Production
Month/period
1000
800
400
July
O/T
Aug.
R/T
Aug.
O/T
Network representation
25
25.75
26.50
July
30
30.90
31.80 +M
26
26.78
200
+M
+M
+M
32
+M
32.96
29
400
Month
sold
Aug.
600
Sept.
2200
+M
37
Demand
Production Capacity
500
July
July
R/T
R/T
Sept.
R/T
300
200
Sept.
O/T
30
MONTPELIER SKI COMPANY - Spreadsheet
31
MONTPELIER SKI COMPANY
• Summary of the optimal solution
– In July produce at capacity (1000 pairs in R/T, and 500 pairs in O/T).
Store 1500-200 = 1300 at the end of July.
– In August, produce 800 pairs in R/T, and 300 in O/T. Store additional
800 + 300 - 600 = 500 pairs.
– In September, produce 400 pairs (clearly in R/T). With 1000 pairs
retail demand, there will be
(1300 + 500) + 400 - 1000 = 1200 pairs available for shipment to Ski
Chalet.
Inventory +
Production -
Demand
32
4.3 The Capacitated Transshipment
Model
• Sometimes shipments to destination nodes are
made through transshipment nodes.
• Transshipment nodes may be
– Independent intermediate nodes with no supply or
demand
– Supply or destination points themselves.
• Transportation on arcs may be bounded by given
bounds
33
The Capacitated Transshipment
Model
• The linear programming model of this problem consists
of:
– Flow on arcs decision variables
– Cost minimization objective function
– Balance constraints on each node as follows:
• Supply node – net flow out does not exceed the supply
• Intermediate node – flow into the node is equal to the flow out
• Demand node – net flow into the node is equal to the demand
– Bound constraints on each arc. Flow cannot exceed the
capacity on the arc
34
DEPOT MAX
A General Network Problem
• Depot Max has six stores located in the
Washington D.C. area.
35
DEPOT MAX
• The stores in Falls Church (FC) and Bethesda (BA)
are running low on the model 5A Arcadia workstation.
• DATA:
5
-12
FC
-13
6
BA
36
DEPOT MAX
• The stores in Alexandria (AA) and Chevy Chase (CC)
have an access of 25 units.
• DATA:
+10
AA
+15
CC
1
5
-12
FC
-13
2
6
BA
37
DEPOT MAX
• The stores in Fairfax and Georgetown are transshipment
nodes with no access supply or demand of their own.
• DATA:
+10
AA
FX
1
3
5
-12
FC
• Depot Max wishes to transport the available
workstations to FC and BA at minimum total cost.
GN
+15
CC
2
4
6
-13
BA
38
DEPOT MAX
• The possible routes and the shipping unit costs are shown.
• DATA:
+10
AA
20
10
1
6
5
FX
3
7
12
FC
11
7
GN
+15
CC
2
15
4
15
-12
FC
BA
-13
BA
39
DEPOT MAX
• Data
– There is a maximum limit for quantities shipped on
various routes.
– There are different unit transportation costs for
different routes.
40
DEPOT MAX – Types of constraints
20
+10
1
6
+15
10
5
2
3
7
–Supply nodes:
Net flow out oftransshipment
the node] = [Supply
–Intermediate
nodes: at the node]
–Demand
[Total
of the node]
flow 1)
into the node]
X12
+flow
X13 out
+ Xnodes:
10 = [Total
(Node
15 - X21 =12
7
[Net flow- into
X12 the node] ==[Demand
15(Node3)
(Nodefor2)the node]
XX3421+X+35X=24 X13
- X56 = 12
(Node
X46X=15X+24X+35X+X
(Node
4) 5)
34 65
X46 +X56 - X65 = 13 (Node 6)
15
4
-12
15
5
7
11
6
-13
41
DEPOT MAX
• The Complete mathematical model
Min 5X12 + 10X13 + 20X15 + 6X21
S.T.
10
X12 +
- X12 +
–
12
5
X13 +
X15 –
+ 15X24 + 12X34 + 7X35 + 15X46 + 11X56 + 7X65
X21
X21 +
X13 +
– X15
–
–

X24
X34 +
X24 –
 17
X35
X34 +
=0
=0
X46
X35 +
X56 -
X65
=-
-X46 –
X56 + X65 = -13
X12  3; X15  6; X21  7; X24  10; X34  8; X35  8; X46  17; X56  7; X65 
All variables are non-negative
42
DEPOT MAX - spreadsheet
NODE INPUT
NODE NAME
SLACK
ARC INPUT
NODE # SUPPLY DEMAND
100
2
FROM
TO
COST CAPACITY
SOLUTION
TOTAL
COST=
645
FROM
TO
FLOW
Alexandria
1
10
1
2
5
3
1
2
Chevy Chase
2
17
1
3
10
100000
1
3
9
Fairfax
3
1
5
20
6
1
5
6
Georgetown
4
2
1
6
7
2
1
5
Falls Church
5
12
2
4
15
10
2
4
10
Betheda
6
13
3
4
12
8
3
4
1
3
5
7
8
3
5
8
4
6
15
17
4
6
11
5
6
11
7
5
6
2
6
5
7
5
6
5
43
4.4 The Assignment Problem
• Problem definition
– m workers are to be assigned to m jobs
– A unit cost (or profit) Cij is associated with worker i performing
job j.
– Minimize the total cost (or maximize the total profit) of
assigning workers to job so that each worker is assigned a
job, and each job is performed.
44
BALLSTON ELECTRONICS
• Five different electrical devices produced on five
production lines, are needed to be inspected.
• The travel time of finished goods to inspection areas
depends on both the production line and the inspection
area.
• Management wishes to designate a separate inspection
area to inspect the products such that the total travel time
is minimized.
45
BALLSTON ELECTRONICS
• Data: Travel time in minutes from assembly
lines to inspection areas.
Assembly
Lines
1
2
3
4
5
A
10
11
13
14
19
B
4
7
8
16
17
Inspection Area
C
6
7
12
13
11
D
10
9
14
17
20
E
12
14
15
17
19
46
BALLSTON ELECTRONICSNETWORK REPRESENTATION
47
Assembly Line
S1=1
1
Inspection Areas
A D1=1
S2=1
2
B
D2=1
S3=1
3
C D3=1
S4=1
4
D
D4=1
S5=1
5
E
D5=1
48
BALLSTON ELECTRONICS –
The Linear Programming Model
Min 10X11 + 4X12 + … + 20X54 + 19X55
S.T. X11 + X12 + X13 + X14 + X15 = 1
X21 + X22 + …
+ X25 = 1
…
…
…
…
X51 + X52+ X53 + X54 + X55 = 1
All the variables are non-negative
49
BALLSTON ELECTRONICS –
Computer solutions
• A complete enumeration is not an efficient
procedure even for moderately large problems
(with m=8,
m! > 40,000 is the number of assignments to
enumerate).
• The Hungarian method provides an efficient
solution procedure.
50
BALLSTON ELECTRONICS –
Transportation spreadsheet
=SUMPRODUCT(B7:F11,B17:F217)
=SUM(B7:B11)
Drag to cells C13:F13
=SUM(B7:F7)
Drag to cells H8:H12
51
BALLSTON ELECTRONICS –
Transportation spreadsheet
Each Area is Served
Each Line is Assigned
52
BALLSTON ELECTRONICS –
Assignment spreadsheet
53
The Assignments Model - Modifications
– Unbalanced problem: The number of supply nodes
and demand nodes is unequal.
– Prohibitive assignments: A supply node should not
be assigned to serve a certain demand node.
– Multiple assignments: A certain supply node can
be assigned for more than one demand node .
– A maximization assignment problem.
54
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55