CH04 Networks 1

NetworkNetwork

ModelsModels1

Chapter 4

4.1 Introduction

2

A network problem is one that can be represented by...

Nodes

Arcs

8

9

10

10

7

6

Function on Arcs

4.1 Introduction

– 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

• The importance of network models

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 flowUij = upper bound of the flowLij = 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


• Data– Unit shipping cost, supply, and demand

• 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

To From Boston Richmond Atlanta St. Louis Supply Cleveland $35 30 40 32 1200 Detroit 37 40 42 25 1000 Greensboro 40 15 20 28 800 Demand 1100 400 750 750

CARLTON PHARMACEUTICALSNetwork presentation

10

11

Boston

Richmond

Atlanta

St.Louis

Destinations

Sources

Cleveland

Detroit

Greensboro

S1=1200

S2=1000

S3= 800

D1=1100

D2=400

D3=750

D4=750

37

40

42

32

35

40

30

25

3515

20

28

CARLTON PHARMACEUTICALS – Linear Programming Model

– The structure of the model is:

Minimize Total Shipping CostST[Amount shipped from a source] [Supply at that source][Amount received at a destination] = [Demand at that destination]

– Decision variablesXij = 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

13

Boston

Richmond

Atlanta

St.Louis

D1=1100

D2=400

D3=750

D4=750

The supply constraints

Cleveland S1=1200

X11

X12

X13

X14

Supply from Cleveland X11+X12+X13+X14 = 1200

DetroitS2=1000

X21

X22

X23

X24

Supply from Detroit X21+X22+X23+X24 = 1000

GreensboroS3= 800

X31

X32

X33

X34

Supply from Greensboro X31+X32+X33+X34 = 800

CARLTON PHARMACEUTICAL – The complete mathematical model

14

Minimize 35X11+30X12+40X13+ 32X14 +37X21+40X22+42X23+25X24+ 40X31+15X32+20X33+38X34

ST

Supply constrraints:X11+ X12+ X13+ X14 1200

X21+ X22+ X23+ X24 1000X31+ X32+ X33+ X34 800

Demand constraints: X11+ X21+ X31 1000

X12+ X22+ X32 400X13+ X23+ X33 750

X14+ X24+ X34 750

All Xij are nonnegative

====

Total shipment out of a supply nodecannot exceed the supply at the node.

Total shipment received at a destinationnode, must equal the demand at that node.

CARLTON PHARMACEUTICALS Spreadsheet

15

=SUM(B7:E9)Drag to cells

C11:E11

=SUMPRODUCT(B7:E9,B15:E17) =SUM(B7:E7)Drag to cells

G8:G9

CARLTON PHARMACEUTICALS Spreadsheet

16

MINIMIZE Total Cost

SHIPMENTS

Demands are metSupplies are not exceeded

CARLTON PHARMACEUTICALS Spreadsheet - solution

17

SOLUTIONMINIMUM COST 84000

BOSTON RICHMOND ATLANTA ST. LOUIS SHIPPEDCLEVELAND 850 350 1200DETROIT 250 750 1000GREENSBORO 50 750 800

RECEIVED 1100 400 750 750

INPUTBOSTON RICHMOND ATLANTA ST. LOUIS SUPPLY

CLEVELAND 35 30 40 32 1200DETROIT 37 40 42 25 1000GREENSBORO 40 15 20 28 800

DEMAND 1100 400 750 750


COST (PER CASE)

SHIPMENTS (CASES)

CARLTON PHARMACEUTICALS Sensitivity Report

18

Adjustable CellsFinal 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$8 DETROIT BOSTON 250 0 37 5 2$C$8 DETROIT RICHMOND 0 8 40 1E+30 8$D$8 DETROIT ATLANTA 0 5 42 1E+30 5$E$8 DETROIT ST. LOUIS 750 0 25 9 1E+30$B$9 GREENSBORO BOSTON 0 20 40 1E+30 20$C$9 GREENSBORO RICHMOND 50 0 15 17 5$D$9 GREENSBORO ATLANTA 750 0 20 5 1E+30$E$9 GREENSBORO ST. LOUIS 0 20 28 1E+30 20

– Reduced costs • The unit shipment cost between Cleveland and

Atlanta must be reduced by at least $5, before it would become economically feasible to utilize it

• If this route is used, the total cost will increase by $5 for each case shipped between the two cities.


19

Adjustable CellsFinal 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$8 DETROIT BOSTON 250 0 37 5 2$C$8 DETROIT RICHMOND 0 8 40 1E+30 8$D$8 DETROIT ATLANTA 0 5 42 1E+30 5$E$8 DETROIT ST. LOUIS 750 0 25 9 1E+30$B$9 GREENSBORO BOSTON 0 20 40 1E+30 20$C$9 GREENSBORO RICHMOND 50 0 15 17 5$D$9 GREENSBORO ATLANTA 750 0 20 5 1E+30$E$9 GREENSBORO ST. LOUIS 0 20 28 1E+30 20

– Allowable Increase/Decrease• This is the range of optimality.• The unit shipment cost between Cleveland and

Boston may increase up to $2 or decrease up to $5 with no change in the current optimal transportation plan.


– Shadow prices • For the plants, shadow prices

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

ConstraintsFinal 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$11 RECEIVED RICHMOND 400 32 400 0 250$D$11 RECEIVED ATLANTA 750 37 750 0 250$E$11 RECEIVED ST. LOUIS 750 25 750 0 750


21

ConstraintsFinal 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$11 RECEIVED RICHMOND 400 32 400 0 250$D$11 RECEIVED ATLANTA 750 37 750 0 250$E$11 RECEIVED ST. LOUIS 750 25 750 0 750

– Shadow prices • 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.

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



– Assign a large objective coefficient to the route (C ij = 1,000,000)– Add a constraint to Excel solver of the form X ij = 0

23

Shipments on a Blocked Route = 0



– Assign a large objective coefficient to the route (Cij = 1,000,000)– Add a constraint to Excel solver of the form Xij = 0– Do not include the cell representing the rout in the Changing cells

24

Only Feasible Routes Included in Changing Cells

Cell C9 is NOT Included

Shipments from Greensboroto Cleveland are prohibited


– 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 COMPANYUsing 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-of-quarter 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)

27

Forecasted Production Production Costs Month Demand Capacity Regular Time OvertimeJuly 400 1000 25 30August 600 800 26 32September 1000 400 29 37

Forecasted Production Production Costs Month Demand Capacity Regular Time OvertimeJuly 400 1000 25 30August 600 800 26 32September 1000 400 29 37


• Analysis of demand:– Net demand in July = 400 - 200 = 200 pairs

– Net demand in August = 600– Net demand in September = 1000 + 1200 = 2200 pairs

• 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

Initial inventory

Forecasted demand

In house inventory


29

• 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

Network representation

30

252525.7525.7526.5026.50

303030.9030.9031.8031.80 +M+M

2626

26.7826.78

+M+M

3232

32.9632.96

+M+M

+M+M

2929

+M+M

+M+M

3737

ProductionMonth/period

Monthsold

JulyR/T

July O/T

Aug.R/T

Aug.O/T

Sept.R/T

Sept.O/T

July

Aug.

Sept.

1000

500

800

400

400

200

200

600

300

2200

Demand

Prod

uctio

n Ca

pacit

y

July R/T

MONTPELIER SKI COMPANY - Spreadsheet

31


• 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.

32

Inventory + Production -

Demand

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

36

5

6

• DATA:

-12

-13

• The stores in Falls Church (FC) and Bethesda (BA) are running low on the model 5A Arcadia workstation.

FC

BA

DEPOT MAX

37

1

2

5

6

• DATA:

-12

-13

+10

+15

• The stores in Alexandria (AA) and Chevy Chase (CC) have an access of 25 units.

FC

BA

AA

CC

DEPOT MAX

38

1

2

5

6

• DATA:

-12

-13

+10

+15

• The stores in Fairfax and Georgetown are transshipment nodes with no access supply or demand of their own.

FC

BA

AA

CC

FX

GN

• Depot Max wishes to transport the available workstations to FC and BA at minimum total cost.

4

3

DEPOT MAX

39

5

10

20

6

15

12

7

15

117

1

2

FC

BA

• DATA:

-12

-13

+10

+15

• The possible routes and the shipping unit costs are shown.

FC

BA

AA

CC

FX

GN

4

3

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

41

5

10

20

6

15

12

7

15

117

1

2 4

3 5

6

-12

-13

+10

+15

–Supply nodes: Net flow out of the node] = [Supply at the node]X12 + X13 + X15 - X21 = 10 (Node 1)X21 + X24 - X12 = 15 (Node 2)

–Intermediate transshipment nodes: [Total flow out of the node] = [Total flow into the node]X34+X35 = X13 (Node3)X46 = X24 + X34 (Node 4)

7–Demand nodes:[Net flow into the node] = [Demand for the node]X15 + X35 +X65 - X56 = 12 (Node 5)X46 +X56 - X65 = 13 (Node 6)

DEPOT MAX

• The Complete mathematical modelMin 5X12 + 10X13 + 20X15 + 6X21 + 15X24 + 12X34 + 7X35 + 15X46 + 11X56 + 7X65

S.T. X12 + X13 + X15 – X21 10

- X12 + X21 + X24 17

– X13 + X34 + X35 = 0

– X24 – X34 + X46 = 0

– X15 – X35 + X56 - X65= -12

-X46 – X56 + X65 = -13

X12 3; X15 6; X21 7; X24 10; X34 8; X35 8; X46 17; X56 7; X65 5

All variables are non-negative42

DEPOT MAX - spreadsheet

43

SOLUTIONTOTAL COST= 645

NODE NAME NODE # SUPPLY DEMAND FROM TO COST CAPACITY FROM TO FLOWSLACK 100 2

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

NODE INPUT ARC INPUT

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.

46

Inspection AreaA B C D E

1 10 4 6 10 12Assembly 2 11 7 7 9 14 Lines 3 13 8 12 14 15

4 14 16 13 17 175 19 17 11 20 19

Inspection AreaA B C D E

1 10 4 6 10 12Assembly 2 11 7 7 9 14 Lines 3 13 8 12 14 15

4 14 16 13 17 175 19 17 11 20 19

BALLSTON ELECTRONICS-NETWORK REPRESENTATION

47

48

1

2

3

4

5

Assembly Line Inspection AreasA

B

C

D

E

S1=1

S2=1

S3=1

S4=1

S5=1

D1=1

D2=1

D3=1

D4=1

D5=1

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

51

=SUM(B7:F7)Drag to cells H8:H12

=SUMPRODUCT(B7:F11,B17:F217)

=SUM(B7:B11)Drag to cells C13:F13

BALLSTON ELECTRONICS – Transportation spreadsheet

52

Each Area is ServedEach Line is Assigned

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

55

Copyright John Wiley & Sons, Inc. All rights reserved. Reproduction or translation of this work beyond that named in Section 117 of the United States Copyright Act without the express written consent of the copyright owner is unlawful. Requests for further information should be addressed to the Permissions Department, John Wiley & Sons, Inc. Adopters of the textbook are granted permission to make back-up copies for their own use only, to make copies for distribution to students of the course the textbook is used in, and to modify this material to best suit their instructional needs. Under no circumstances can copies be made for resale. The Publisher assumes no responsibility for errors, omissions, or damages, caused by the use of these programs or from the use of the information contained herein.

CH04 Networks 1

Documents

Transcript of CH04 Networks 1