570 Quantitative Business Analysis, homework 5 wk 6

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Transportation, Assignment, and Network Models

To accompany Quantitative Analysis for Management, Twelfth Edition,

by Render, Stair, Hanna and Hale

Power Point slides created by Jeff Heyl

After completing this chapter, students will be able to:

LEARNING OBJECTIVES

9 – 2

Structure LP problems for the transportation, transshipment, and assignment models.

Solve facility location and other application problems with transportation models.

Use LP to model shortest-route and maximal-flow problems.

Solve minimal-spanning tree problems.

9.1 Introduction

9.2 The Transportation Problem

9.3 The Assignment Problem

9.4 The Transshipment Problem

9.5 Maximal-Flow Problem

9.6 Shortest-Route Problem

9.7 Minimal-Spanning Tree Problem

CHAPTER OUTLINE

9 – 3

Introduction

LP problems modeled as networks

Helps visualize and understand problems

Transportation problem

Transshipment problem

Assignment problem

Maximal-flow problem

Shortest-route problem

Minimal-spanning tree problem

Specialized algorithms available

9 – 4

Introduction

Common terminology for network models

Points on the network are referred to as nodes

Typically circles

Lines on the network that connect nodes are called arcs

9 – 5

The Transportation Problem

Deals with the distribution of goods from several points of supply (sources) to a number of points of demand (destinations)

Usually given the capacity of goods at each source and the requirements at each destination

Typically objective is to minimize total transportation and production costs

9 – 6

Linear Program for Transportation

Executive Furniture Corporation transportation problem

Minimize transportation cost

Meet demand

Not exceed supply

9 – 7

Linear Program for Transportation

Let Xij = number of units shipped from source i to destination j

Where

i = 1, 2, 3, with 1 = Des Moines, 2 = Evansville, and 3 = Fort Lauderdale

j = 1, 2, 3, with 1 = Albuquerque, 2 = Boston, and 3 = Cleveland

9 – 8

Linear Program for Transportation

Minimize total cost = 5X11 + 4X12 + 3X13 + 8X21 + 4X22 + 3X23 + 9X31 +7X32 + 5X33

Subject to:

X11 + X12 + X13 ≤ 100 (Des Moines supply)

X21 + X22 + X23 ≤ 300 (Evansville supply)

X31 + X32 + X33 ≤ 300 (Fort Lauderdale supply)

X11 + X21 + X31 = 300 (Albuquerque demand)

X12 + X22 + X32 = 200 (Boston demand)

X13 + X23 + X33 = 200 (Cleveland demand)

Xij ≥ 0 for all i and j

9 – 9

Linear Program for Transportation

Optimal solution

100 units from Des Moines to Albuquerque

200 units from Evansville to Boston

100 units from Evansville to Cleveland

200 units from Ft. Lauderdale to Albuquerque

100 units from Ft. Lauderdale to Cleveland

Total cost = $3,900

9 – 10

Linear Program for Transportation

9 – 11

FIGURE 9.1 – Network Representation of a Transportation Problem

100

300

Supply

200

300

Demand

Source

Des Moines

(Source 1)

Evansville

(Source 2)

Fort Lauderdale

(Source 3)

Albuquerque

(Destination 1)

Boston

(Destination 2)

Cleveland

(Destination 3)

Destination

Using Excel QM

9 – 12

PROGRAM 9.1 – Executive Furniture Corporation Solution in Excel 2013 Using Excel QM

A General LP Model for Transportation Problems

Let

Xij = number of units shipped from source i to destination j

cij = cost of one unit from source i to destination j

si = supply at source i

dj = demand at destination j

9 – 13

A General LP Model for Transportation Problems

9 – 14

Subject to:

Facility Location Analysis

Transportation method especially useful

New location is major financial importance

Several alternative locations evaluated

Subjective factors are considered

Final decision also involves minimizing total shipping and production costs

Alternative facility locations analyzed within the framework of one overall distribution system

9 – 15

Facility Location Analysis

Hardgrave Machine Company produces computer components in Cincinnati, Salt Lake City, and Pittsburgh

Four warehouses in Detroit, Dallas, New York, and Los Angeles

Two new plant sites being considered – Seattle and Birmingham

Which of the new locations will yield the lowest cost for the firm in combination with the existing plants and warehouses?

9 – 16

Facility Location Analysis

9 – 17

WAREHOUSE	MONTHLY DEMAND (UNITS)	PRODUCTION PLANT	MONTHLY SUPPLY	COST TO PRODUCE ONE UNIT ($)
Detroit	10,000	Cincinnati	15,000	48
Dallas	12,000	Salt Lake City	6,000	50
New York	15,000	Pittsburgh	14,000	52
Los Angeles	9,000		35,000
	46,000
Supply needed from a new plant = 46,000 – 35,000 = 11,000 units per month

TABLE 9.1 – Hardgrave’s Demand and Supply Data

ESTIMATED PRODUCTION COST PER UNIT AT PROPOSED PLANTS
Seattle	$53
Birmingham	$49

Facility Location Analysis

9 – 18

TO FROM	DETROIT	DALLAS	NEW YORK	LOS ANGELES
CINCINNATI	$25	$55	$40	$60
SALT LAKE CITY	35	30	50	40
PITTSBURGH	36	45	26	66
SEATTLE	60	38	65	27
BIRMINGHAM	35	30	41	50

TABLE 9.2 – Hardgrave’s Shipping Costs

Solve two transportation problems – one for each combination

Facility Location Analysis

Xij = number of units shipped from source i to destination j

Where

i = 1, 2, 3, 4 with 1 = Cincinnati, 2 = Salt Lake City, 3 = Pittsburgh, and 4 = Seattle

j = 1, 2, 3, 4 with 1 = Detroit, 2 = Dallas, 3 = New York, and 4 = Los Angeles

9 – 19

Facility Location Analysis

Minimize total cost = 73X11 + 103X12 + 88X13 + 108X14 + 85X21 + 80X22 + 100X23 + 90X24 + 88X31 + 97X32 + 78X33 + 118X34 + 113X41 + 91X42 + 118X43 + 80X44

Subject to:

X11 + X21 + X31 + X41 = 10,000 Detroit demand

X12 + X22 + X32 + X42 = 12,000 Dallas demand

X13 + X23 + X33 + X43 = 15,000 New York demand

X14 + X24 + X34 + X44 = 9,000 Los Angeles demand

X11 + X12 + X13 + X14 ≤ 15,000 Cincinnati supply

X21 + X22 + X23 + X24 ≤ 6,000 Salt Lake City supply

X31 + X32 + X33 + X34 ≤ 14,000 Pittsburgh supply

X41 + X42 + X43 + X44 ≤ 11,000 Seattle supply

All variables Xij ≥ 0

9 – 20

Minimize total cost = 73X11 + 103X12 + 88X13 + 108X14 + 85X21 + 80X22 + 100X23 + 90X24 + 88X31 + 97X32 + 78X33 + 118X34 + 113X41 + 91X42 + 118X43 + 80X44

Subject to:

X11 + X21 + X31 + X41 = 10,000 Detroit demand

X12 + X22 + X32 + X42 = 12,000 Dallas demand

X13 + X23 + X33 + X43 = 15,000 New York demand

X14 + X24 + X34 + X44 = 9,000 Los Angeles demand

X11 + X12 + X13 + X14 ≤ 15,000 Cincinnati supply

X21 + X22 + X23 + X24 ≤ 6,000 Salt Lake City supply

X31 + X32 + X33 + X34 ≤ 14,000 Pittsburgh supply

X41 + X42 + X43 + X44 ≤ 11,000 Seattle supply

All variables Xij ≥ 0

Facility Location Analysis

9 – 21

The total cost for the Seattle alternative = $3,704,000

Minimize total cost = 73X11 + 103X12 + 88X13 + 108X14 + 85X21 + 80X22 + 100X23 + 90X24 + 88X31 + 97X32 + 78X33 + 118X34 + 113X41 + 91X42 + 118X43 + 80X44

Subject to:

X11 + X21 + X31 + X41 = 10,000 Detroit demand

X12 + X22 + X32 + X42 = 12,000 Dallas demand

X13 + X23 + X33 + X43 = 15,000 New York demand

X14 + X24 + X34 + X44 = 9,000 Los Angeles demand

X11 + X12 + X13 + X14 ≤ 15,000 Cincinnati supply

X21 + X22 + X23 + X24 ≤ 6,000 Salt Lake City supply

X31 + X32 + X33 + X34 ≤ 14,000 Pittsburgh supply

X41 + X42 + X43 + X44 ≤ 11,000 Seattle supply

All variables Xij ≥ 0

Facility Location Analysis

9 – 22

The total cost for the Seattle alternative = $3,704,000

Reformulating the problem for the Birmingham alternative and solving, the total cost = $3,741,000

Using Excel QM

9 – 23

PROGRAM 9.2 – Facility Location (Seattle) Solution in Excel 2013 Using Excel QM

Using Excel QM

9 – 24

PROGRAM 9.3 – Facility Location (Birmingham) Solution in Excel 2013 Using Excel QM

The Assignment Problem

This class of problem determines the most efficient assignment of people or equipment to particular tasks

Objective is typically to minimize total cost or total task time

9 – 25

Linear Program for Assignment Example

The Fix-it Shop has just received three new repair projects that must be repaired quickly

A radio

A toaster oven

A coffee table

Three workers with different talents are able to do the jobs

Owner estimates wage cost for workers on projects

Objective – minimize total cost

9 – 26

Linear Program for Assignment Example

9 – 27

FIGURE 9.2 – Assignment Problem in a Transportation Network Format

Supply

$11

$14

$10

$11

$12

Demand

Person

Adams

(Source 1)

Brown

(Source 2)

Cooper

(Source 3)

Project 1

(Destination 1)

Project 2

(Destination 2)

Project 3

(Destination 3)

Project

Linear Program for Assignment Example

9 – 28

Let

where

i = 1, 2, 3, with 1 = Adams, 2 = Brown, and 3 = Cooper

j = 1, 2, 3, with 1 = Project 1, 2 = Project 2, and 3 = Project 3

Linear Program for Assignment Example

9 – 29

Minimize total cost = 11X11 + 14X12 + 6X13 + 8X21 + 10X22 + 11X23 + 9X31 + 12X32 + 7X33

subject to

X11 + X12 + X13 = 1

X21 + X22 + X23 = 1

X31 + X32 + X33 = 1

X11 + X21 + X31 = 1

X12 + X22 + X32 = 1

X13 + X23 + X33 = 1

Xij = 0 or 1 for all i and j

Linear Program for Assignment Example

9 – 30

Minimize total cost = 11X11 + 14X12 + 6X13 + 8X21 + 10X22 + 11X23 + 9X31 + 12X32 + 7X33

subject to

X11 + X12 + X13 ≤ 1

X21 + X22 + X23 ≤ 1

X31 + X32 + X33 ≤ 1

X11 + X21 + X31 = 1

X12 + X22 + X32 = 1

X13 + X23 + X33 = 1

Xij = 0 or 1 for all i and j

Solution

X13 = 1, Adams assigned to Project 3

X22 = 1, Brown assigned to Project 2

X31 = 1, Cooper is assigned to Project 1

Total cost = $25

Using Excel QM

9 – 31

PROGRAM 9.4 – Mr. Fix-It Shop Assignment Solution in Excel 2013 Using Excel QM

The Transshipment Problem

Items are being moved from a source to a destination through an intermediate point (a transshipment point)

Transshipment problem

9 – 32

The Transshipment Problem

Frosty Machines manufactures snow blowers in Toronto and Detroit

Shipped to regional distribution centers in Chicago and Buffalo

Then shipped to supply houses in New York, Philadelphia, and St. Louis

Shipping costs vary by location and destination

Snow blowers cannot be shipped directly from the factories to the supply houses

9 – 33

The Transshipment Problem

9 – 34

FIGURE 9.3 – Network Representation of Transshipment Example

800

700

Supply

300

350

450

Demand

Source

Toronto

(Node 1)

Detroit

(Node 2)

New York City

(Node 5)

Philadelphia

(Node 6)

St. Louis

(Node 7)

Destination

Chicago

(Node 3)

Transshipment Point

Buffalo

(Node 4)

The Transshipment Problem

9 – 35

TABLE 9.3 – Frosty Machine Transshipment Data

	TO
FROM	CHICAGO	BUFFALO	NEW YORK CITY	PHILADELPHIA	ST. LOUIS	SUPPLY
Toronto	$4	$7	—	—	—	800
Detroit	$5	$7	—	—	—	700
Chicago	—	—	$6	$4	$5	—
Buffalo	—	—	$2	$3	$4	—
Demand	—	—	450	350	300

Minimize transportation costs associated with shipping snow blowers subject to demands and supplies

The Transshipment Problem

Minimize cost subject to

The number of units shipped from Toronto is not more than 800

The number of units shipped from Detroit is not more than 700

The number of units shipped to New York is 450

The number of units shipped to Philadelphia is 350

The number of units shipped to St. Louis is 300

The number of units shipped out of Chicago is equal to the number of units shipped into Chicago

The number of units shipped out of Buffalo is equal to the number of units shipped into Buffalo

9 – 36

The Transshipment Problem

Decision variables

Xij = number of units shipped from location (node) i to location (node) j

where

i = 1, 2, 3, 4

j = 3, 4, 5, 6, 7

9 – 37

The Transshipment Problem

Minimize cost = 4X13 + 7X14 + 5X23 + 7X24 + 6X35 + 4X36 + 5X37 + 2X45 + 3X46 + 4X47

subject to

X13 + X14 ≤ 800 (Supply at Toronto [node 1])

X23 + X24 ≤ 700 (Supply at Detroit [node 2])

X35 + X45 = 450 (Demand at New York [node 5])

X36 + X46 = 350 (Demand at Philadelphia [node 6])

X37 + X47 = 300 (Demand at St. Louis [node 7])

X13 + X23 = X35 + X36 + X37 (Shipping through Chicago [node 3])

X14 + X24 = X45 + X46 + X47 (Shipping through Buffalo [node 4])

Xij ≥ 0 for all i and j (nonnegativity)

9 – 38

Minimize cost = 4X13 + 7X14 + 5X23 + 7X24 + 6X35 + 4X36 + 5X37 + 2X45 + 3X46 + 4X47

subject to

X13 + X14 ≤ 800 (Supply at Toronto [node 1])

X23 + X24 ≤ 700 (Supply at Detroit [node 2])

X35 + X45 = 450 (Demand at New York [node 5])

X36 + X46 = 350 (Demand at Philadelphia [node 6])

X37 + X47 = 300 (Demand at St. Louis [node 7])

X13 + X23 = X35 + X36 + X37 (Shipping through Chicago [node 3])

X14 + X24 = X45 + X46 + X47 (Shipping through Buffalo [node 4])

Xij ≥ 0 for all i and j (nonnegativity)

The Transshipment Problem

9 – 39

Ship 650 units from Toronto to Chicago

Ship 150 units from Toronto to Buffalo

Ship 300 units from Detroit to Buffalo

Ship 350 units from Chicago to Philadelphia

Ship 300 units form Chicago to St. Louis

Ship 450 units from Buffalo to New York

Total cost = $9,550

Using Excel QM

9 – 40

PROGRAM 9.5 – Excel QM Solution to Frosty Machine Transshipment Problem

Maximal-Flow Problem

Determining the maximum amount of material that can flow from one point (the source) to another (the sink) in a network

Two common methods

Linear programming

Maximal-flow technique

9 – 41

Maximal-Flow Problem

Determine maximum number of cars from east to west for Waukesha WI road system

9 – 42

West Point

East Point

Capacity in Hundreds of Cars per Hour

FIGURE 9.4 – Road Network for Waukesha Maximal-Flow Example

Maximal-Flow Problem

Variables

Xij = flow from node i to node j

where

i = 1, 2, 3, 4, 5, 6

j = 1, 2, 3, 4, 5, 6

Constraints necessary for

Capacity of each arc

Equal flows into and out of each arc

9 – 43

Maximal-Flow Problem

Maximize flow = X61

subject to

9 – 44

(X21 + X61) – (X12 + X13 + X14)	=	Flows into = flows out of node 1
(X12 + X42 + X62) – (X21 + X24 + X26)	=	Flows into = flows out of node 2
(X13 + X43 + X53) – (X34 + X35)	=	Flows into = flows out of node 3
(X14 + X24 + X34 + X64) – (X42 + X43 + X46)	=	Flows into = flows out of node 4
(X35) – (X56 + X53)	=	Flows into = flows out of node 5
(X26 + X46 + X56) – (X61 + X62 + X64)	=	Flows into = flows out of node 6
Xij	≥

X12	≤	3	X13	≤	10	X14	≤	2	Capacities for arcs from node 1
X21	≤	1	X24	≤	1	X26	≤	2	Capacities for arcs from node 2
X34	≤	3	X35	≤	2				Capacities for arcs from node 3
X42	≤	1	X43	≤	1	X46	≤	1	Capacities for arcs from node 4
X53	≤	1	X56	≤	1				Capacities for arcs from node 5
X62	≤	2	X64	≤	1				Capacities for arcs from node 6

Using Excel QM

9 – 45

PROGRAM 9.6 –

Waukesha Maximal-Flow

Solution

Shortest-Route Problem

Find the shortest distance from one location to another

Can be modeled as

A linear programming problem with 0-1 variables

A special type of transshipment problem

Using specialized algorithm

9 – 46

Shortest-Route Problem

Ray Design transports beds, chairs, and other furniture items from the factory to the warehouse

Travel through several cities

No direct interstate highways

Find the route with the shortest distance

9 – 47

Plant

Warehouse

100

150

200

100

200

FIGURE 9.5 –

Roads from

Ray’s Plant

to Warehouse

Shortest-Route Problem

Variables

Xij = 1 if arc from node i to node j is selected and Xij = 0 otherwise

where

i = 1, 2, 3, 4, 5

j = 2, 3, 4, 5, 6

Constraints specify the number of units going into a node must equal the number of units going out of the node

9 – 48

Variables

Xij = 1 if arc from node i to node j is selected and Xij = 0 otherwise

where

i = 1, 2, 3, 4, 5

j = 2, 3, 4, 5, 6

Constraints specify the number of units going into a node must equal the number of units going out of the node

Shortest-Route Problem

9 – 49

Origin point must ship one unit

X12 + X13 = 1

Final destination must have one unit shipped into it

X46 + X56 = 1

Intermediate nodes must have same amounts entering and leaving

X12 + X32 = X23 + X24 + X25

X12 + X32 – X23 – X24 – X25 = 0

Shortest-Route Problem

Minimize distance = 100X12 + 200X13 + 50X23 + 50X32 + 200X24 + 200X42 + 100X25 + 100X52 + 40X35 + 40X53 + 150X45 + 150X54 + 100X46 + 100X56

subject to

9 – 50

X12 + X13	=	1	Node 1
X12 + X32 – X23 – X24 – X25	=	0	Node 2
X13 + X23 – X32 – X35	=	0	Node 3
X24 + X54 – X42 – X45 – X46	=	0	Node 4
X25 + X35 + X45 – X52 – X53 – X54 – X56	=	0	Node 5
X46 + X56	=	1	Node 6
All variables	=	0	or 1

Using Excel QM

9 – 51

PROGRAM 9.7 –

Ray Designs, Inc.

Solution

Using Excel QM

9 – 52

PROGRAM 9.7 –

Ray Designs, Inc.

Solution

X12 = X23 = X35 = X56 = 1

Route is City 1 to City 2 to City 3 to City 5 to City 6

Total distance traveled = 290 miles

Minimal-Spanning Tree Problem

Connecting all points of a network together while minimizing the total distance of the connections

Linear programming can be used but is complex

Minimal-spanning tree technique is quite easy

9 – 53

Minimal-Spanning Tree Problem

Steps for the Minimal-Spanning Tree Technique

Select any node in the network.

Connect this node to the nearest node that minimizes the total distance.

Considering all of the nodes that are now connected, find and connect the nearest node that is not connected. If there is a tie for the nearest node, select one arbitrarily. A tie suggests there may be more than one optimal solution.

Repeat the third step until all nodes are connected.

9 – 54

Minimal-Spanning Tree Problem

Lauderdale Construction

Housing project in Panama City Beach

Determine the least expensive way to provide water and power to each house

9 – 55

Gulf

FIGURE 9.6 – Network for Lauderdale Construction

Minimal-Spanning Tree Problem

Step 1 – Arbitrarily select node 1

Step 2 – Connect node 1 to node 3 (nearest)

9 – 56

Gulf

FIGURE 9.7 – First Iteration

Minimal-Spanning Tree Problem

Step 3 – Connect next nearest unconnected node, node 4

Continue for other unconnected nodes

9 – 57

FIGURE 9.8 – Second and Third Iterations

(a) Second Iteration

(b) Third Iteration

Minimal-Spanning Tree Problem

Step 4 – Repeat the process

9 – 58

FIGURE 9.9 – Last Four Iterations

(a) Fourth Iteration

(b) Fifth Iteration

Step 4 – Repeat the process

Minimal-Spanning Tree Problem

9 – 59

FIGURE 9.9 – Last Four Iterations

(d) Seventh Iteration

Minimal-Spanning Tree Problem

9 – 60

TABLE 9.4 – Summary of Steps in Lauderdale Construction Minimal-Spanning Tree Problem

STEP	CONNECTED NODES	UNCONNECTED NODES	CLOSE UNCONNECTED NODES	ARC SELECTED	ARC LENGTH	TOTAL DISTANCE
1	1	2, 3, 4, 5, 6, 7, 8	3	1–3	2	2
2	1, 3	2, 4, 5, 6, 7, 8	4	3–4	2	4
3	1, 3, 4	2, 5, 6, 7, 8	2 or 6	3–2	3	7
4	1, 2, 3, 4	5, 6, 7, 8	5 or 6	2–5	3	10
5	1, 2, 3, 4, 5	6, 7, 8	6	3–6	3	13
6	1, 2, 3, 4, 5, 6	7, 8	8	6–8	1	14
7	1, 2, 3, 4, 5, 6, 8	7	7	8–7	2	16

Using Excel QM

9 – 61

PROGRAM 9.8 –

Lauderdale Construction Minimal-Spanning Tree Example

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1.6 The Role of Computers and Spreadsheet Models in the Quantitative Analysis Approach

1.7 Possible Problems in the Quantitative Analysis Approach

1.8 Implementation—Not Just the Final Step

1.1 Introduction 1.2 What Is Quantitative Analysis? 1.3 Business Analytics 1.4 The Quantitative Analysis Approach 1.5 How to Develop a Quantitative Analysis Model

CHAPTER OUTLINE

5. Use computers and spreadsheet models to perform quantitative analysis.

6. Discuss possible problems in using quantitative analysis.

7. Perform a break-even analysis.

1. Describe the quantitative analysis approach. 2. Understand the application of quantitative analysis

in a real situation. 3. Describe the three categories of business analytics. 4. Describe the use of modeling in quantitative

analysis.

After completing this chapter, students will be able to:

Introduction to Quantitative Analysis

1CHAPTER

LEARNING OBJECTIVES

M01_REND7331_12_SE_C01_pp2.indd 1 01/10/13 9:50 AM

1.6 The Role of Computers and Spreadsheet Models in the Quantitative Analysis Approach

1.7 Possible Problems in the Quantitative Analysis Approach

1.8 Implementation—Not Just the Final Step

1.1 Introduction 1.2 What Is Quantitative Analysis? 1.3 Business Analytics 1.4 The Quantitative Analysis Approach 1.5 How to Develop a Quantitative Analysis Model

CHAPTER OUTLINE

5. Use computers and spreadsheet models to perform quantitative analysis.

6. Discuss possible problems in using quantitative analysis.

7. Perform a break-even analysis.

1. Describe the quantitative analysis approach. 2. Understand the application of quantitative analysis

in a real situation. 3. Describe the three categories of business analytics. 4. Describe the use of modeling in quantitative

analysis.

After completing this chapter, students will be able to:

Introduction to Quantitative Analysis

1CHAPTER

LEARNING OBJECTIVES

M01_REND7331_12_SE_C01_pp2.indd 1 01/10/13 9:50 AM

From Excel QM ribbon, select Menu (Alphabetical or By Chapter). Select Transportation from the drop-down menu, and then input 3 Origins (sources) and 3 Destination.

The solution is shown here.

Fill in the table with the costs, supplies, and demands.

From the Data tab, select Solver and then click Solve.

Minimize cost = cij xij i−1

∑ j=1

∑

Minimize cost= c

i-1

j=1

xij ≤ si i =1,2,...,m j=1

∑

xij i=1

∑ = dj j =1,2...,n

£s

i=1,2,...,m

j=1

i=1

j=1,2...,n

xij ≥ 0 for all i and j

³0 for all i and j

From Excel QM ribbon, select Menu (Alphabetical or By Chapter). Select Transportation from the drop- down menu, and then input 4 Origins (sources) and 4 Destination.

After entering the costs, click the Data tab and select Solver. Then click Solve.

The cost is here.

Fill in the table with the costs, supplies, and demands.

The cost is here.

Xij = 1 if person i is assigned to project j 0 otherwise

⎧ ⎨ ⎪

⎩⎪

1 if person i is assigned to project j

0 otherwise

From Excel QM ribbon, select Menu (Alphabetical or By Chapter). Select Assignment from the drop- down menu, and then input the number of assignments (3).

After entering the costs, click the Data tab and select Solver. Then click Solve.

The cost is here.

Fill in the table with the costs, supplies, and demands.