session 4a

Session 4a

Decision Models -- Prof. Juran

2

OverviewMore Network Models

– Assignment Model (Contract Bidding)• “Big Cost” trick

– Project Management (House Building)• More binary / integer tricks• Critical Path / Slack Time• Excel trick: Conditional Formatting• Cost Crashing

– Changing an objective to a constraint– Issues with Integers– Location Analysis (Hospital Location)


3

Contract Bidding ExampleA company is taking bids on four construction jobs. Three contractors have placed bids on the jobs. Their bids (in thousands of dollars) are given in the table below. (A dash indicates that the contractor did not bid on the given job.) Contractor 1 can do only one job, but contractors 2 and 3 can each do up to two jobs.

Job 1 Job 2 Job 3 Job 4 Contractor 1 50 46 42 40 Contractor 2 51 48 44 — Contractor 3 — 47 45 45


4

Formulation Decision VariablesWhich contractor gets which job(s).ObjectiveMinimize the total cost of the four jobs.ConstraintsContractor 1 can do no more than one job.Contractors 2 and 3 can do no more than two jobs each.Contractor 2 can’t do job 4.Contractor 3 can’t do job 1.Every job needs one contractor.


5

Formulation Decision VariablesDefine Xij to be a binary variable representing the assignment of contractor i to job j. If contractor i ends up doing job j, then Xij = 1. If contractor i does not end up with job j, then Xij = 0.

Define Cij to be the cost; i.e. the amount bid by contractor i for job j.

ObjectiveMinimize Z =

3

1

4

1i jijijCX


6

Formulation Constraints for all j.

for i = 1.

for i = 2, 3.

13

1

iijX

14

1

jijX

24

1

jijX

01,34,2 XX


7

Solution Methodology123456789101112131415

A B C D E F G HJob 1 Job 2 Job 3 Job 4

Contractor 1 50 46 42 40Contractor 2 51 48 44 10000Contractor 3 10000 47 45 45

Assignment of contractors to jobsJob 1 Job 2 Job 3 Job 4 Total Max

Contractor 1 0 0 0 0 0 <= 1Contractor 2 0 0 0 0 0 <= 2Contractor 3 0 0 0 0 0 <= 2Total 0 0 0 0

= = = =Required 1 1 1 1

Total cost ($1000s) 0=SUMPRODUCT(B2:E4,B8:E10)

=SUM(E8:E10)

=SUM(B8:E8)


8

Solution MethodologyNotice the very large values in cells B4 and E3. These specific values (10,000) aren’t important; the main thing is to assign these particular contractor-job combinations costs so large that they will never be in any optimal solution.


9

Solution Methodology


10

Optimal Solution123456789101112131415

A B C D E F G HJob 1 Job 2 Job 3 Job 4

Contractor 1 50 46 42 40Contractor 2 51 48 44 10000Contractor 3 10000 47 45 45

Assignment of contractors to jobsJob 1 Job 2 Job 3 Job 4 Total Max

Contractor 1 0 0 0 1 1 <= 1Contractor 2 1 0 1 0 2 <= 2Contractor 3 0 1 0 0 1 <= 2Total 1 1 1 1

= = = =Required 1 1 1 1

Total cost ($1000s) 182


11

Conclusions

The optimal solution is to award Job 4 to Contractor 1, Jobs 1 and 3 to Contractor 2, and Job 2 to Contractor 3. The total cost is $182,000.


12

Sensitivity Analysis1. What is the “cost” of restricting

Contractor 1 to only one job?2. How much more can Contractor 1 bid

for Job 4 and still get the job?


13

1234567891011121314151617181920212223242526272829

A B C D E F G HMicrosoft Excel 15.0 Sensitivity Report

Variable CellsFinal Reduced Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease$B$8 Contractor 1 Job 1 0 1 50 1E+30 1$C$8 Contractor 1 Job 2 0 1 46 1E+30 1$D$8 Contractor 1 Job 3 0 0 42 1 3$E$8 Contractor 1 Job 4 1 0 40 3 1E+30$B$9 Contractor 2 Job 1 1 0 51 1 1E+30$C$9 Contractor 2 Job 2 0 1 48 1E+30 1$D$9 Contractor 2 Job 3 1 0 44 1 1$E$9 Contractor 2 Job 4 0 9958 10000 1E+30 9958$B$10 Contractor 3 Job 1 0 9949 10000 1E+30 9949$C$10 Contractor 3 Job 2 1 0 47 1 1E+30$D$10 Contractor 3 Job 3 0 1 45 1E+30 1$E$10 Contractor 3 Job 4 0 3 45 1E+30 3

ConstraintsFinal Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease$B$11 Total Job 1 1 51 1 0 1$C$11 Total Job 2 1 47 1 1 1$D$11 Total Job 3 1 44 1 0 1$E$11 Total Job 4 1 42 1 0 1$F$8 Contractor 1 Total 1 -2 1 1 0$F$9 Contractor 2 Total 2 0 2 1E+30 0$F$10 Contractor 3 Total 1 0 2 1E+30 1


14

ConclusionsThe sensitivity report indicates a shadow price of –2 (cell E29).

(Allowing Contractor 1 to perform one additional job would reduce the total cost by 2,000.)

The allowable increase in the bid for Job 4 by Contractor 1 is 3. (He could have bid any amount up to $43,000 and still have won that job.)


15

Con. 3Con. 2Con. 1

Network Representation

Job 3Job 2 Job 4Job 1


16

Optimal Solution

Con. 1

Job 3

Con. 2 Con. 3

Job 2 Job 4Job 1

4047

4451


17

House-Building ExampleActivity Description Predecessors Duration (days)

Activity A Build foundation — 5 Activity B Build walls and ceilings A 8 Activity C Build roof B 10 Activity D Do electrical wiring B 5 Activity E Put in windows B 4 Activity F Put on siding E 6 Activity G Paint house C, F 3


18

Managerial Problem Definition

Find the critical path and the minimum number of days needed to build the house.


19

Network Representation1

0Start

43

2

A5

B8

C10

E4

D5

G3

F6

5End


20

FormulationDecision VariablesWe are trying to decide when to begin and end each of the activities.ObjectiveMinimize the total time to complete the project.ConstraintsEach activity has a fixed duration.There are precedence relationships among the activities.We cannot go backwards in time.


21

FormulationDecision VariablesDefine the nodes to be discrete events. In other words, they occur at one exact point in time. Our decision variables will be these points in time.Define ti to be the time at which node i occurs, and at which time all activities preceding node i have been completed.Define t0 to be zero.

ObjectiveMinimize t5.


22

FormulationConstraintsThere is really one basic type of constraint. For each activity x, let the time of its starting node be represented by tjx and the time of its ending node be represented by tkx. Let the duration of activity x be represented as dx.

For every activity x,

For every node i,

xjxkx dtt

0it


23


A B C D E F G H I J K L M N O P

1

t0 t1 t2 t3 t4 t50 1 1 1 1 1

t0 t1 t2 t3 t4 t5A -1 1 0 0 0 0 1 >= 5B 0 -1 1 0 0 0 0 >= 8C 0 0 -1 0 1 0 0 >= 10D 0 0 -1 0 0 1 0 >= 5E 0 0 -1 1 0 0 0 >= 4F 0 0 0 -1 1 0 0 >= 6G 0 0 0 0 -1 1 0 >= 3

1

0Start

43

2

A5

B8

C10

E4

D5

G3

F6

5End


24

Solution MethodologyThe matrix of zeros, ones, and negative ones (B12:G18) is a means for setting up the constraints.

The sumproduct functions in H12:H18 calculate the elapsed time between relevant pairs of nodes, corresponding to the various activities.

The duration times of the activities are in J12:J18.


25


26



26

t0 t1 t2 t3 t4 t50 5 13 17 23 26

t0 t1 t2 t3 t4 t5A -1 1 0 0 0 0 5 >= 5B 0 -1 1 0 0 0 8 >= 8C 0 0 -1 0 1 0 10 >= 10D 0 0 -1 0 0 1 13 >= 5E 0 0 -1 1 0 0 4 >= 4F 0 0 0 -1 1 0 6 >= 6G 0 0 0 0 -1 1 3 >= 3

1

0Start

43

2

A5

B8

C10

E4

D5

G3

F6

5End


27

Conclusions

The project will take 26 days to complete.

The only activity that is not critical is the electrical wiring.


28

CPM JargonAny activity for which

is said to have slack time, the amount of time by which that activity could be delayed without affecting the overall completion time of the whole project. In this example, only activity D has any slack time (13 – 5 = 8 units of slack time).

xjxkx dtt


29

CPM Jargon

Any activity x for which

is defined to be a “critical” activity, with zero slack time.

xjxkx dtt


30

CPM JargonEvery network of this type has at least one critical path, consisting of a set of critical activities. In this example, there are two critical paths: A-B-C-G and A-B-E-F-G.

1

0Start

43

2

A5

B8

C10

E4

D5

G3

F6

5End


31

Excel Tricks: Conditional Formatting1

23456789101112131415161718


26

t0 t1 t2 t3 t4 t50 5 13 17 23 26

t0 t1 t2 t3 t4 t5A -1 1 0 0 0 0 5 >= 5B 0 -1 1 0 0 0 8 >= 8C 0 0 -1 0 1 0 10 >= 10D 0 0 -1 0 0 1 13 >= 5E 0 0 -1 1 0 0 4 >= 4F 0 0 0 -1 1 0 6 >= 6G 0 0 0 0 -1 1 3 >= 3

1

0Start

43

2

A5

B8

C10

E4

D5

G3

F6

5End


32

Critical Activities: Using the Solver Answer Report

ConstraintsCell Name Cell Value Formula Status Slack

$H$12 A 5 $H$12>=$J$12 Binding 0$H$13 B 8 $H$13>=$J$13 Binding 0$H$14 C 10 $H$14>=$J$14 Binding 0$H$15 D 13 $H$15>=$J$15 Not Binding 8$H$16 E 4 $H$16>=$J$16 Binding 0$H$17 F 6 $H$17>=$J$17 Binding 0$H$18 G 3 $H$18>=$J$18 Binding 0


33

House-Building Example, Continued

Suppose that by hiring additional workers, the duration of each activity can be reduced. Use LP to find the strategy that minimizes the cost of completing the project within 20 days.


34

Crashing Parameters

Activity Cost per Day to

Reduce Duration Maximum Possible Reduction (Days)

Build foundation $30 2 Walls and ceilings $15 3 Build roof $20 1 Electrical wiring $40 2 Put in windows $20 2 Put on siding $30 3 Paint house $40 1


35


Find a way to build the house in 20 days.


36

FormulationDecision VariablesNow the problem is not only when to schedule the activities, but also which activities to accelerate. (In CPM jargon, accelerating an activity at an additional cost is called “crashing”.)

ObjectiveMinimize the total cost of crashing.


37

FormulationConstraintsThe project must be finished in 20 days.Each activity has a maximum amount of crash time.Each activity has a “basic” duration. (These durations were considered to have been fixed in Part a; now they can be reduced.)There are precedence relationships among the activities.We cannot go backwards in time.


38

FormulationDecision VariablesDefine the number of days that activity x is crashed to be Rx.

For each activity there is a maximum number of crash days Rmax, x

Define the crash cost per day for activity x to be Cx

ObjectiveMinimize Z =

7

1xxxCR


39

FormulationConstraints



For every node i,

xxjxkx Rdtt

xx RR max,

0it


40


A B C D E F G H I J K L M N O

30

t0 t1 t2 t3 t4 t50 1 1 1 1 1

20 <-- Max Completion Time

t0 t1 t2 t3 t4 t5 Improved Duration Basic Duration Crash Time Max Crash Cost/TimeA -1 1 0 0 0 0 1 >= 4 5 1 2 30$ B 0 -1 1 0 0 0 0 >= 8 8 0 3 15$ C 0 0 -1 0 1 0 0 >= 10 10 0 1 20$ D 0 0 -1 0 0 1 0 >= 5 5 0 2 40$ E 0 0 -1 1 0 0 0 >= 4 4 0 2 20$ F 0 0 0 -1 1 0 0 >= 6 6 0 3 30$ G 0 0 0 0 -1 1 0 >= 3 3 0 1 40$

1

0Start

43

2

A5

B8

C10

E4

D5

G3

F6

5End


41

Solution MethodologyG3 now contains a formula to calculate the total crash cost. The new decision variables (how long to crash each activity x, represented by Rx) are in M12:M18.

G8 contains the required completion time, and we will constrain the value in G6 to be less than or equal to G8.

The range J12:J18 calculates the revised duration of each activity, taking into account how much time is saved by crashing.


42

Solution Methodology


43


A B C D E F G H I J K L M N O

145

t0 t1 t2 t3 t4 t50 3 8 11 17 20

20 <-- Max Completion Time

t0 t1 t2 t3 t4 t5 Improved Duration Basic Duration Crash Time Max Crash Cost/TimeA -1 1 0 0 0 0 3 >= 3 5 2 2 30$ B 0 -1 1 0 0 0 5 >= 5 8 3 3 15$ C 0 0 -1 0 1 0 9 >= 9 10 1 1 20$ D 0 0 -1 0 0 1 12 >= 5 5 0 2 40$ E 0 0 -1 1 0 0 3 >= 3 4 1 2 20$ F 0 0 0 -1 1 0 6 >= 6 6 0 3 30$ G 0 0 0 0 -1 1 3 >= 3 3 0 1 40$

1

0Start

43

2

A5

B8

C10

E4

D5

G3

F6

5End


44

ConclusionsIt is feasible to complete the project in 20 days, at a cost of $145.

1

0Start

43

2

A3

B5

C9

E3

D5

G3

F6

5End


45

ConclusionsBasic Duration Crash Time Improved Duration Cost/Time Cost

A 5 2 3 30$ 60$ B 8 3 5 15$ 45$ C 10 1 9 20$ 20$ D 5 0 5 40$ -$ E 4 1 3 20$ 20$ F 6 0 6 30$ -$ G 3 0 3 40$ -$

145$


46

An Alternative SolutionBasic Duration Crash Time Improved Duration Cost/Time Cost

A 5 2 3 30$ 60$ B 8 3 5 15$ 45$ C 10 0 10 20$ -$ D 5 0 5 40$ -$ E 4 0 4 20$ -$ F 6 0 6 30$ -$ G 3 1 2 40$ 40$

145$


47

Excel Tricks: VLOOKUP

Looks for a specific value in the left column of a table and finds the row where that value appears, then returns the value corresponding to another specified column in that row.


48

12345678910111213141516171819202122232425262728293031

A B C D E F G H I JSchool CostNYU 31,746$

School Accept % Enroll % GMAT GPA Cost Minority pct Non-U.S. pct Female % SalaryStanford 8.8% 78.4% 718 3.58 31,002$ 24.6% 30.9% 38.7% 124,740$ Harvard 11.6% 87.3% 703 3.58 30,050$ 18.6% 37.2% 34.2% 121,979$ Penn 15.5% 70.2% 703 3.57 31,218$ 19.0% 43.1% 29.3% 121,200$ MIT 18.1% 69.9% 703 3.50 31,200$ 19.0% 35.5% 26.5% 118,381$ Northwestern 16.6% 61.8% 700 3.45 30,255$ 17.5% 30.8% 27.5% 114,664$ Duke 20.1% 52.8% 690 3.43 30,323$ 20.6% 32.7% 34.8% 115,444$ Chicago 28.4% 59.7% 695 3.33 30,596$ 18.2% 33.2% 23.6% 115,331$ Columbia 13.5% 68.7% 705 3.40 31,912$ 19.3% 26.5% 36.5% 117,989$ Dartmouth 18.8% 53.3% 690 3.40 30,250$ 17.2% 29.9% 27.6% 121,692$ Berkeley 14.5% 50.3% 688 3.43 21,242$ 19.4% 30.2% 30.4% 111,321$ Michigan 19.6% 58.5% 676 3.34 30,686$ 20.5% 31.2% 27.6% 117,498$ Virginia 16.6% 51.0% 681 3.40 27,283$ 15.1% 30.1% 29.1% 112,706$ NYU 22.3% 48.2% 689 3.40 31,746$ 19.0% 30.2% 34.1% 112,900$ Yale 20.0% 56.7% 686 3.50 29,055$ 17.3% 38.9% 27.5% 112,514$ UCLA 17.4% 48.5% 691 3.60 22,490$ 21.6% 24.0% 28.6% 103,364$ Cornell 25.8% 51.2% 669 3.32 30,455$ 22.6% 25.6% 27.1% 116,588$ North Carolina 23.8% 44.2% 674 3.30 25,525$ 16.2% 30.4% 30.6% 109,420$ Carnegie-Mellon 28.2% 60.1% 652 3.25 28,452$ 21.3% 38.3% 26.8% 111,211$ Texas 30.0% 52.4% 680 3.34 21,200$ 12.3% 26.7% 22.7% 106,859$ USC 29.0% 43.4% 670 3.30 30,082$ 32.3% 23.8% 28.7% 95,213$ Indiana 24.7% 49.8% 651 3.35 20,696$ 14.3% 34.5% 20.7% 99,732$ Emory 31.5% 46.7% 651 3.30 28,012$ 10.7% 29.9% 23.4% 104,417$ Rochester 32.7% 42.2% 637 3.33 28,620$ 10.3% 54.2% 25.2% 103,466$ Georgetown 20.8% 46.2% 662 3.35 28,440$ 10.8% 38.7% 28.1% 99,528$ Michigan State 22.3% 55.0% 641 3.36 16,955$ 13.6% 37.9% 25.2% 88,746$ Ohio State 29.0% 41.4% 645 3.38 22,151$ 15.5% 29.2% 36.7% 99,598$ Minnesota 34.4% 40.6% 645 3.33 20,352$ 8.6% 29.1% 27.9% 91,140$

=VLOOKUP(A2,A5:J31,6,0)


49

Integer Programs• We have seen models where the

decision variables had to be integers, but we didn’t have to impose a Solver constraint to make that happen (a special attribute of some transportation models)

• In general, you do need to impose a Solver constraint to force an integer solution

• No more Simplex algorithm!• Binary = Special case of integer


50

Difficulties with Integer Programs

• Linear approximations are not necessarily feasible or optimal

• Integer optimal solutions require much more complicated algorithms

• Integer algorithms do not yield a sensitivity report


51

Maximize z = YX 1121 Subject to: YX 47 13

X, Y 0 Linear Program

0

1

2

3

4

0 1 2 3 4

Optimal SolutionX1=1.857, X2=0


52

Optimal linear-programming solution: X = 1.857, Y = 0Rounded to X = 2, Y = 0 is infeasibleRounded to X = 1, Y = 0 is not optimalOptimal integer-programming solution: X = 0, Y = 3

Integer Program

0

1

2

3

4

0 1 2 3 4

Optimal SolutionX1=0, X2=3


53

Hospital Location ExampleA county is going to build two hospitals. There are nine cities in which the hospitals can be built. The number of hospital visits per year made by people in each city and the x-y coordinates of each city are listed in the table below. The county's goal is to minimize the total distance that patients must travel to hospitals. Where should it locate the hospitals?


54

Hospital Location ExampleCity x y Hospital visits per Year

1. Kimsbury 0 0 3000 2. Jozefacki Junction 10 3 4000 3. Sapras Falls 12 15 5000 4. Lowthersburg 14 13 6000 5. Leesville 16 9 4000 6. Sanjay Beach 18 6 3000 7. Patel Point 8 12 2000 8. Rothsboro 6 10 4000 9. Nikolova City 4 8 1200


55

Hospital Location ExampleCounty Map

0

2

4

6

8

10

12

14

16

0 2 4 6 8 10 12 14 16 18 20

X

Y

1. Kimsbury 2. Jozefacki Junction

9. Nikolova City

8. Rothsboro7. Patel Point

3. Sapras Falls

6. Sanjay Beach

5. Leesville

4. Lowthersburg


56


Decision VariablesWe need to decide whether or not to build a hospital in each of nine cities.We also need to decide how many visits there will be from each city to each hospital.ObjectiveWe want to minimize the total distance traveled to the hospital by all county residents.


57

Managerial Problem DefinitionConstraintsThe cities can’t be moved.Exactly two hospitals will be built.All of the planned hospital visits must be accounted for and included in the total distance calculation.No hospital visits are allowed to a city that has no hospital.


58


98

7

6

5

4

3

2

1


59


98

7

6

5

4

3

2

1 One Hospital


60

FormulationDecision Variables Define Vij to be the number of visits from people in city i to the hospital in city j. Define Xj to be a binary variable. If a hospital is built in city j, then Xj = 1; otherwise, Xj = 0. These Vij and Xj are the decision variables. There are 81 + 9 = 90 decision variables here. Objective Define Dij to be the distance from city i to city j.

Minimize Z = iji j

ijDV

9

1

9

1


61

Formulation


62


A B C D E F G H I J K L M N O P Q R S T U V WXj

1 2 3 4 5 6 7 8 9 Sum Required Total miles (x 1000)1 1 1 1 1 1 1 1 1 9 = 2 0.00

Vij to City coordinates1 2 3 4 5 6 7 8 9 Sum Ri x y

1 3.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 3 = 3 1 0 02 0.0 4.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 4 = 4 2 10 33 0.0 0.0 5.0 0.0 0.0 0.0 0.0 0.0 0.0 5 = 5 3 12 154 0.0 0.0 0.0 6.0 0.0 0.0 0.0 0.0 0.0 6 = 6 4 14 13

from 5 0.0 0.0 0.0 0.0 4.0 0.0 0.0 0.0 0.0 4 = 4 5 16 96 0.0 0.0 0.0 0.0 0.0 3.0 0.0 0.0 0.0 3 = 3 6 18 67 0.0 0.0 0.0 0.0 0.0 0.0 2.0 0.0 0.0 2 = 2 7 8 128 0.0 0.0 0.0 0.0 0.0 0.0 0.0 4.0 0.0 4 = 4 8 6 109 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 1.2 1.2 = 1.2 9 4 8

3.0 4.0 5.0 6.0 4.0 3.0 2.0 4.0 1.2Logical upper bounds <= <= <= <= <= <= <= <= <= Distances 1 2 3 4 5 6 7 8 9

32.2 32.2 32.2 32.2 32.2 32.2 32.2 32.2 32.2 1 0.0 10.4 19.2 19.1 18.4 19.0 14.4 11.7 8.92 10.4 0.0 12.2 10.8 8.5 8.5 9.2 8.1 7.83 19.2 12.2 0.0 2.8 7.2 10.8 5.0 7.8 10.64 19.1 10.8 2.8 0.0 4.5 8.1 6.1 8.5 11.25 18.4 8.5 7.2 4.5 0.0 3.6 8.5 10.0 12.06 19.0 8.5 10.8 8.1 3.6 0.0 11.7 12.6 14.17 14.4 9.2 5.0 6.1 8.5 11.7 0.0 2.8 5.78 11.7 8.1 7.8 8.5 10.0 12.6 2.8 0.0 2.89 8.9 7.8 10.6 11.2 12.0 14.1 5.7 2.8 0.0

=SUM($N$7:$N$15)*C3

=SUMPRODUCT(C7:K15,N18:V26)

=SUM(C10:K10)

=SUM(K7:K15)

=SQRT((VLOOKUP($M22,$P$6:$R$15,2)-VLOOKUP(O$17,$P$6:$R$15,2))^2+(VLOOKUP($M22,$P$6:$R$15,3)-VLOOKUP(O$17,$P$6:$R$15,3))^2)


63

Solution MethodologyThe 9 Xj decision variables are in the range C3:K3.The 81 Vij decision variables are in the range C7:K15.The objective function is in cell P3.The matrix in the range N18:V26 calculates the distance between each pair of cities using a long and ugly Excel function based on the famous Pythagorean theorem.Cell N3 is used to keep track of constraint (1).Cells L7:L15 and N7:N15 are used to keep track of constraint (2).Cells C16:K16 and C18:K18 are used to keep track of constraint (3).


64


65


1011121314151617181920212223242526

A B C D E F G H I J K L M N O P Q RXj

1 2 3 4 5 6 7 8 9 Sum Required Total miles (x 1000)0 0 0 1 0 0 0 1 0 2 = 2 132.50

Vij to City coordinates1 2 3 4 5 6 7 8 9 Sum Ri x y

1 0.0 0.0 0.0 0.0 0.0 0.0 0.0 3.0 0.0 3 = 3 1 0 02 0.0 0.0 0.0 0.0 0.0 0.0 0.0 4.0 0.0 4 = 4 2 10 33 0.0 0.0 0.0 5.0 0.0 0.0 0.0 0.0 0.0 5 = 5 3 12 154 0.0 0.0 0.0 6.0 0.0 0.0 0.0 0.0 0.0 6 = 6 4 14 13

from 5 0.0 0.0 0.0 4.0 0.0 0.0 0.0 0.0 0.0 4 = 4 5 16 96 0.0 0.0 0.0 3.0 0.0 0.0 0.0 0.0 0.0 3 = 3 6 18 67 0.0 0.0 0.0 0.0 0.0 0.0 0.0 2.0 0.0 2 = 2 7 8 128 0.0 0.0 0.0 0.0 0.0 0.0 0.0 4.0 0.0 4 = 4 8 6 109 0.0 0.0 0.0 0.0 0.0 0.0 0.0 1.2 0.0 1.2 = 1.2 9 4 8

0.0 0.0 0.0 18.0 0.0 0.0 0.0 14.2 0.0Logical upper bounds <= <= <= <= <= <= <= <= <= Distances 1 2 3 4 5

0.0 0.0 0.0 32.2 0.0 0.0 0.0 32.2 0.0 1 0.0 10.4 19.2 19.1 18.42 10.4 0.0 12.2 10.8 8.53 19.2 12.2 0.0 2.8 7.24 19.1 10.8 2.8 0.0 4.55 18.4 8.5 7.2 4.5 0.06 19.0 8.5 10.8 8.1 3.67 14.4 9.2 5.0 6.1 8.58 11.7 8.1 7.8 8.5 10.09 8.9 7.8 10.6 11.2 12.0


66

Optimal SolutionIf the county wants to build two hospitals, then the optimal locations are in City 4 (Lowthersburg) and City 8 (Rothsboro). The total miles traveled would be 132,500.


67


98

7

6

5

4

3

2

1 Two Hospitals


68


98

7

6

5

4

3

2

1 Three Hospitals


69


98

7

6

5

4

3

2

1 Four Hospitals


70


98

7

6

5

4

3

2

1 Five Hospitals


71


98

7

6

5

4

3

2

1 Six Hospitals


72


98

7

6

5

4

3

2

1 Seven Hospitals


73


98

7

6

5

4

3

2

1 Eight Hospitals


74


98

7

6

5

4

3

2

1 Nine Hospitals


75

SummaryMore Network Models

– Assignment Model (Contract Bidding)• “Big Cost” trick

– Project Management (House Building)• More binary / integer tricks• Critical Path / Slack Time• Excel trick: Conditional Formatting• Cost Crashing

– Changing an objective to a constraint– Issues with Integers– Location Analysis (Hospital Location)

session 4a

Documents

adecision models

juran2decision models

juran13decision models

juran7decision models

juran10decision models

juran9decision models

juran1decision models

juran5decision models