1 chapter 17 standing in line – at the bank, the market, the movies – is the time-waster...

1

Chapter 17

Standing in line – at the bank, the market, the movies – is the time-waster everyone loves to hate. Stand in just one 15-minute line a day, every day, and kiss goodbye to almost four days of idle time by year’s end.—Kathleen Doheny

Waiting Lines (Queues)

2

Queuing Analysis

A queue is a waiting line. A queuing system involves customers arriv-

ing for service who sometimes have to wait. Queuing analysis provides:

Summary measures for assessing a queuing system in terms of customers and time.

A way to balance the costs of providing service and costs of congestion.

Importance of Queuing Analysis: Servicing customers can be costly. Retail environments are plagued with customer

congestion. Managing that has benefits.

3

Single-Server Queuing Model(M/M/1)

Assumes exponential distribution for times: between successive arrivals—mean rate to complete service—mean rate

Provides important results: Probability for n customers in system:

P0 = 1 Pn = n P0 Mean number of customers in system:

Mean customer time spent in system:

L =

W = L = 1

4

Single-Server Queuing Model (M/M/1)

Important results (continued): Mean number of customers waiting:

Mean customer waiting time:

Server utilization factor:

Lq =

Wq = Lq =

=

5


Problem: Customers arrive at a supply room with mean rate = 25/hr. Service take a mean of 2 minutes, so that = 30/hr. Do a queuing analysis.

Solution:

= 25/30 = .833 P0 = 1 .833 = .167

P1 = (25/30)1(.167) = .139

P2 = (25/30)2(.167) = .116

167253030

251674

25303025 2

.W.L qq

hr.202530

1cust.5

253025

.WL

6


Problem (continued): Hourly cost of providing service is $10 (clerk’s wages). Lost productivity of employee customers while getting supplies is $7/hr, the congestion penalty. Evaluate total cost.

Solution (continued): The hourly system cost may be used to compare alternatives:TC = hourly queuing cost + hourly service cost = $7×W × + $10 = $7(.20)(25) + 10 = $45 The queuing cost accounts for all customers in the

system during the hour, and collective time under a congestion penalty, W × , applies above.

When the congestion penalty involves just the waiting time, Wq × is used for queuing cost. Retail customers resent the wait, not the time receiving service.

7

Multiple-Server Queuing Model (M/M/S)

Multiple servers involve greater complexity.

The model begins with computation of

1

00 1

1!!

1S

n

Sn

S/Sn/P

8

Multiple-Server Queuing Model (M/M/S)

The mean number of customers is then found:

Then the mean customer waiting time is found:

That is followed by:

02/-1!

PSS

S//L

S

q

q

q

LW

qq LLWW

1

9

Queuing SpreadsheetTemplates and Software

Queuing Templates

Palisade Decision Tools BestFit 4.0

10

Queuing Spreadsheet Templates

M/M/1 M/M/S Cost analysis for M/M/S M/M/1 non-exponential service times Exponential distribution Poisson distribution Waiting times for M/M/1 M/M/1 with a finite queue M/M/1 with a limited population M/M/1 with a constant service time

11

M/M/1 (Figure 17-4)

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10111213

1415161718

192021222324254445464748717273

A B C D E F G H I J

PROBLEM: Supply Room

Parameter Values:Mean Customer Arrival Rate: lambda = 25Mean Customer Service Rate: mu = 30

Queuing Results:Mean Number of Customers in System: L = 5Mean Customer Time Spent in System: W = 0.2Mean Number of Customers Waiting

(Length of Line): Lq = 4.166667

Mean Customer Waiting Time: Wq = 0.166667Server Utilization Factor: rho = 0.833333

Number in Probability CumulativeSystem n Pn Probability

0 0.1667 0.16671 0.1389 0.30562 0.1157 0.42133 0.0965 0.51774 0.0804 0.59815 0.0670 0.66516 0.0558 0.7209

25 0.0017 0.991326 0.0015 0.992727 0.0012 0.993928 0.0010 0.994929 0.0008 0.995852 0.0000 0.999953 0.0000 0.999954 0.0000 1.0000

BASIC QUEUING SYSTEM EVALUATION -- SINGLE SERVER

10111213

1415

G=G6/(G7-G6)=1/(G7-G6)

=G6^2/(G7*(G7-G6))

=G6/(G7*(G7-G6))=G6/G7

192021222324254445464748717273

D E=1-G15 =D19=$D$19*$G$15^C20 =E19+D20=$D$19*$G$15^C21 =E20+D21=$D$19*$G$15^C22 =E21+D22=$D$19*$G$15^C23 =E22+D23=$D$19*$G$15^C24 =E23+D24=$D$19*$G$15^C25 =E24+D25=$D$19*$G$15^C44 =E43+D44=$D$19*$G$15^C45 =E44+D45=$D$19*$G$15^C46 =E45+D46=$D$19*$G$15^C47 =E46+D47=$D$19*$G$15^C48 =E47+D48=$D$19*$G$15^C71 =E70+D71=$D$19*$G$15^C72 =E71+D72=$D$19*$G$15^C73 =E72+D73

2. Queuing summary results are here: L, W, Lq, Wq, and rho.

2. Queuing summary results are here: L, W, Lq, Wq, and rho.

3. System state probabilities are here, Pn and Pn. To fit the spreadsheet on one page some of the rows have been hidden.

3. System state probabilities are here, Pn and Pn. To fit the spreadsheet on one page some of the rows have been hidden.

1. Enter the problem parameters in G6:G7. Remember lambda must be less than mu.


12

M/M/S(Figure 17-6)

1

2

3

4

5

6

7

8

9

10

11121314

151617181920212223243839

A B C D E F G


Parameter Values:Mean Customer Arrival Rate: lambda = 0.6Mean Customer Service Rate: mu= 0.5

Number of Servers: S = 2

Queuing Results:

Mean Number of Customers in System: L = 1.8750Mean Customer Time Spent in System: W = 3.1250Mean Number of Customers Waiting



Number in Probability CumulativeSystem, n Pn Probability

0 0.2500 0.25001 0.3000 0.55002 0.1800 0.73003 0.1080 0.83804 0.0648 0.902818 0.0001 0.999919 0.0000 1.0000

BASIC QUEUING EVALUATION -- MULTIPLE SERVERS

3. The number of servers is limited to 100 or less.


Caution: For large S and n sometimes the resulting numbers are too large for Excel so error messages occur,

such as #NUM!

Caution: For large S and n sometimes the resulting numbers are too large for Excel so error messages occur,

such as #NUM!

1. Enter the problem parameters in G6:G8.


2. Remember that lambda must be less than mu times S. Otherwise the results will not be meaningful (negative probabilities).

2. Remember that lambda must be less than mu times S. Otherwise the results will not be meaningful (negative probabilities).

13

M/M/S Formulas

1

2

3

4

5

6

7

8

9

10

11121314

151617181920212223243839

A B C D E F G


Parameter Values:Mean Customer Arrival Rate: lambda = 0.6Mean Customer Service Rate: mu= 0.5

Number of Servers: S = 2

Queuing Results:

Mean Number of Customers in System: L = 1.8750Mean Customer Time Spent in System: W = 3.1250Mean Number of Customers Waiting




0 0.2500 0.25001 0.3000 0.55002 0.1800 0.73003 0.1080 0.83804 0.0648 0.902818 0.0001 0.999919 0.0000 1.0000

BASIC QUEUING EVALUATION -- MULTIPLE SERVERS

111213

14

15

16

G

=G14+(G6/G7)=G15+(1/G7)

=((G6/G7)^G8)*(G6/(G8*G7))*D20/(FACT(G8)*(1-G6/(G7*G8))^2)

=G14/G6

=G6/(G7*G8)

20

21

D

=1/(VLOOKUP(G8-1,Sum!A2:C102,3)+VLOOKUP(G8,Sum!A2:C102,2)*

(1/(1-G6/(G7*G8))))

=IF(C21>$G$8,($D$20*(($G$6/$G$7)^(C21/2))*(($G$6/$G$7)^(C21/2)))/(FACT($G$8)*($G$8^((C21-

$G$8)/2))*($G$8^((C21-$G$8)/2))),($G$6/$G$7)^C21*($D$20/FACT(C21)))

20

21

E

=D20

=E20+D21

2. The Sum !A2:C102 in cell D20 refers to the table located on the Sum worksheet, shown next.


3. The formulas in D21:E21 are copied down to the end of the table.

3. The formulas in D21:E21 are copied down to the end of the table.

1. All the calculations on this spreadsheet are based on Po found in cell D20.


14

SumWorksheet (Sum Tab)(Figure 17-19)

123456

101102

A B C

n ()n/n! Sum from 0 to n0 1.000E+00 1.000E+001 1.200E+00 2.200E+002 7.200E-01 2.920E+003 2.880E-01 3.208E+004 8.640E-02 3.294E+0099 7.395E-149 3.320E+00

100 8.874E-151 3.320E+00

23

B C=((MMS!$G$6/MMS!$G$7)^(A2))/FACT(A2) 1=((MMS!$G$6/MMS!$G$7)^(A3))/FACT(A3) =C2+B3

S

n

n

n0 !

)/( This worksheet is used to compute the sum

The sum is used in the P0 calculation.

MMS!$G$6 in cell B2 refers to cell G6 on the MMS worksheet, shown previously. The formulas in B3:C3 are copied down to B102:C102.

MMS!$G$6 in cell B2 refers to cell G6 on the MMS worksheet, shown previously. The formulas in B3:C3 are copied down to B102:C102.

15

Cost Analysis for M/M/S(Figure 17-7)

1

23456789

10111213

22232425262728293031

A B C D E F G H I J K

PROBLEM: Million Bank Tellers

Parameter Values:Mean Customer Arrival Rate: lambda = 250Mean Customer Service Rate: mu= 30Customer Cost per Unit of Time = 6Server Cost per Unit of Time = 12

Server Customer Total Customer TotalServers P0 Lq L Wq W Cost Cost(Wq) Cost(Wq) Cost(W) Cost(W)

9 0.0001 9.5049 17.8382 0.0380 0.0714 $108.00 $57.03 $165.03 $107.03 $215.0310 0.0002 2.4381 10.7714 0.0098 0.0431 $120.00 $14.63 $134.63 $64.63 $184.6311 0.0002 0.9363 9.2696 0.0037 0.0371 $132.00 $5.62 $137.62 $55.62 $187.6212 0.0002 0.3999 8.7333 0.0016 0.0349 $144.00 $2.40 $146.40 $52.40 $196.4013 0.0002 0.1761 8.5094 0.0007 0.0340 $156.00 $1.06 $157.06 $51.06 $207.0614 0.0002 0.0774 8.4107 0.0003 0.0336 $168.00 $0.46 $168.46 $50.46 $218.4615 0.0002 0.0334 8.3668 0.0001 0.0335 $180.00 $0.20 $180.20 $50.20 $230.2016 0.0002 0.0141 8.3474 0.0001 0.0334 $192.00 $0.08 $192.08 $50.08 $242.0817 0.0002 0.0057 8.3391 0.0000 0.0334 $204.00 $0.03 $204.03 $50.03 $254.0318 0.0002 0.0023 8.3356 0.0000 0.0333 $216.00 $0.01 $216.01 $50.01 $266.01

BASIC QUEUING SYSTEM EVALUATION -- MULTIPLE SERVERS

Note: In order for the spreadsheet to compute correctly, make sure automatic calculation is selected on both worksheets (Tools, Options, Calculation Tab, and click on the Automatic button under Calculation).

Note: In order for the spreadsheet to compute correctly, make sure automatic calculation is selected on both worksheets (Tools, Options, Calculation Tab, and click on the Automatic button under Calculation).





2. Remember that lambda must be less than mu times S for results to be meaningful. Consequently, rows 14 - 21 are hidden (because lambda is not less than mu times s for these rows).

2. Remember that lambda must be less than mu times S for results to be meaningful. Consequently, rows 14 - 21 are hidden (because lambda is not less than mu times s for these rows).

16

Formulas forM/M/S

Cost Analysis(Figure 17-20)

1

23456789

10111213

22232425262728293031

A B C D E F G H I J K

PROBLEM: Million Bank Tellers

Parameter Values:Mean Customer Arrival Rate: lambda = 250Mean Customer Service Rate: mu= 30Customer Cost per Unit of Time = 6Server Cost per Unit of Time = 12

Server Customer Total Customer TotalServers P0 Lq L Wq W Cost Cost(Wq) Cost(Wq) Cost(W) Cost(W)

9 0.0001 9.5049 17.8382 0.0380 0.0714 $108.00 $57.03 $165.03 $107.03 $215.0310 0.0002 2.4381 10.7714 0.0098 0.0431 $120.00 $14.63 $134.63 $64.63 $184.6311 0.0002 0.9363 9.2696 0.0037 0.0371 $132.00 $5.62 $137.62 $55.62 $187.6212 0.0002 0.3999 8.7333 0.0016 0.0349 $144.00 $2.40 $146.40 $52.40 $196.4013 0.0002 0.1761 8.5094 0.0007 0.0340 $156.00 $1.06 $157.06 $51.06 $207.0614 0.0002 0.0774 8.4107 0.0003 0.0336 $168.00 $0.46 $168.46 $50.46 $218.4615 0.0002 0.0334 8.3668 0.0001 0.0335 $180.00 $0.20 $180.20 $50.20 $230.2016 0.0002 0.0141 8.3474 0.0001 0.0334 $192.00 $0.08 $192.08 $50.08 $242.0817 0.0002 0.0057 8.3391 0.0000 0.0334 $204.00 $0.03 $204.03 $50.03 $254.0318 0.0002 0.0023 8.3356 0.0000 0.0333 $216.00 $0.01 $216.01 $50.01 $266.01

BASIC QUEUING SYSTEM EVALUATION -- MULTIPLE SERVERS

22

B C

=1/(VLOOKUP(A22-1,Sum!$A$2:$C$102,3)+VLOOKUP

(A22,Sum!$A$2:$C$102,2)*(1/(1-$G$6/($G$7*A22))))

=(($G$6/$G$7)^A22)*($G$6/(A22*$G$7))*B22/(FACT(A22)*(1-$G$6/($G$7*A22))^2)

1. The Sum !$A$2:$C$102 in cell B22 refers to the table located on the Sum worksheet. It is identical with the one for the M/M/S template shown previously.

1. The Sum !$A$2:$C$102 in cell B22 refers to the table located on the Sum worksheet. It is identical with the one for the M/M/S template shown previously.

2. The formulas in row 22 are copied down to the other rows of the table.

2. The formulas in row 22 are copied down to the other rows of the table.

17

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1011121314

1516

A B C D E F G H I J

PROBLEM: Sammy's Barbershop

Parameter Values:Mean Customer Arrival Rate: lambda = 4Mean Customer Service Rate: mu = 5Standard Deviation of Service Time: sigma = 0.05

Queuing Results:Mean Number of Customers in System: L = 2.5Mean Customer Time Spent in System: W = 0.625Mean Number of Customers Waiting



QUEUING EVALUATION FOR SINGLE SERVER WITH NON-EXPONENTIAL TIMES

M/M/1Non-exponential Service(Figure 17-8)

Enter the problem parameters in G6:G8. Remember lambda must be less than mu.


11121314

1516

G=G14+G16=G15+1/G7

=(G6^2*G8^2+G16^2)/(2*(1-G16))

=G14/G6=G6/G7

18

123456789

1011

12

13

14

15

16

17

1819

20

2122

23

A B C D E F G H I

PROBLEM: Supply Room Arrival

Mean process rate lambda = 25Lower point a = 0.05Upper point b = 0.2

Area to the left of a: Pr[X < a] = 0.713495Area between a and b: Pr[a < X < b] = 0.27977Area to the right of b: Pr[X > b] = 0.00674

Mean = 0.04 Stand. Dev. = 0.04

Interarrival Frequency Cumulative

Time Curve Distribution

t f(t) F(t)

0 25 00.05 7.16262 0.71350

0.1 2.05212 0.91792

0.2 0.16845 0.993260.3 0.01383 0.99945

0.4 0.00113 0.99995

EXPONENTIAL DISTRIBUTION

181920212223

C=EXPONDIST(B18,$D$5,FALSE)=EXPONDIST(B19,$D$5,FALSE)=EXPONDIST(B20,$D$5,FALSE)=EXPONDIST(B21,$D$5,FALSE)=EXPONDIST(B22,$D$5,FALSE)=EXPONDIST(B23,$D$5,FALSE)

91011

F=EXPONDIST(D6,D5,TRUE)=1-F11-F9=1-EXPONDIST(D7,D5,TRUE)

13B

=1/D5

13E

=B13

181920212223

D=EXPONDIST(B18,$D$5,TRUE)=EXPONDIST(B19,$D$5,TRUE)=EXPONDIST(B20,$D$5,TRUE)=EXPONDIST(B21,$D$5,TRUE)=EXPONDIST(B22,$D$5,TRUE)=EXPONDIST(B23,$D$5,TRUE)

Exponential Distribution(Figure 17-9)

1. Enter the problem parameters in D5:D7.


2. The exponential frequency curve and cumulative distribution are here:

2. The exponential frequency curve and cumulative distribution are here:

19

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101112131415161718192021222324252627282930313233

A B C D E F G

PROBLEM: Number of Arrivals at Supply Room

Mean process rate lambda = 25Duration t = 0.1Lower limit for the number of events, a = 0Upper limit for the number of events, b = 8

Pr[X < a] = 0.08208 Pr[X < a] = 0Pr[a < X < b] = 0.99886Pr[a < X < b] = 0.99575Pr[a < X < b] = 0.91677Pr[a < X < b] = 0.91367Pr[X > b] = 0.00425 Pr[X > b] = 0.00114

Number of CumulativeSuccesses ProbabilityProbability

r Pr[R = r] Pr{R < r]

0 0.08208 0.082081 0.20521 0.287302 0.25652 0.543813 0.21376 0.757584 0.13360 0.891185 0.06680 0.957986 0.02783 0.985817 0.00994 0.995758 0.00311 0.998869 0.00086 0.99972

10 0.00022 0.9999411 0.00005 0.9999912 0.00001 1.0000013 0.00000 1.00000

POISSON DISTRIBUTION

10E

=IF(E7>0,POISSON(E7-1,E5*56,TRUE),0)

15E

=1-POISSON(E8,E5*E6,TRUE)

101112131415

C=POISSON(E7,E5*E6,TRUE)=POISSON(E8,E5*E6,TRUE)-E10=POISSON(E8-1,E5*E6,TRUE)-E11=POISSON(E8,E5*E6,TRUE)-C10=POISSON(E8-1,E5*E6,TRUE)-C10=1-POISSON(E8-1,E5*E6,TRUE)

202122

B=POISSON(A20,$E$5*$E$6,FALSE)=POISSON(A21,$E$5*$E$6,FALSE)=POISSON(A22,$E$5*$E$6,FALSE)

202122

C=POISSON(A20,$E$5*$E$6,TRUE)=POISSON(A21,$E$5*$E$6,TRUE)=POISSON(A22,$E$5*$E$6,TRUE)

Poisson Distribution(Figure 17-11)



2. The Poisson probabilities and cumulative probabilities are here:

2. The Poisson probabilities and cumulative probabilities are here:

20

Waiting Times for M/M/1(Figure 17-13)

12345679

10

11121314151617181920

A B C D E F G H I J

PROBLEM: Supply Room Waiting Times


Time, t Wq(t) W(t)

0.05 0.3510 0.22120.10 0.4946 0.39350.15 0.6064 0.52760.20 0.6934 0.63210.25 0.7612 0.71350.30 0.8141 0.77690.35 0.8552 0.82620.40 0.8872 0.86470.45 0.9122 0.89460.50 0.9316 0.9179

WAITING TIME PROBABILITIES

111213

D=1-($G$6/$G$7)*EXP(-1*($G$7-$G$6)*C11)=1-($G$6/$G$7)*EXP(-1*($G$7-$G$6)*C12)=1-($G$6/$G$7)*EXP(-1*($G$7-$G$6)*C13)

111213

E=1-EXP(-1*($G$7-$G$6)*C11)=1-EXP(-1*($G$7-$G$6)*C12)=1-EXP(-1*($G$7-$G$6)*C13)


1. Enter the problem parameters in G6:G7. Remember lambda must be less than mu.2. The

cumulative distributions for the waiting times in the line and in the system are here:

2. The cumulative distributions for the waiting times in the line and in the system are here:

21

M/M/1with a Finite Queue(Figure 17-14)

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1011121314

1516171819202122232425

A B C D E F G H I J


Parameter Values:Mean Customer Arrival Rate: lambda = 25Mean Customer Service Rate: mu = 30Maximum Number of Customers in System: M = 5



Mean Customer Waiting Time: Wq = 0.0547Traffic Intensity: rho = 0.8333


0 0.2506 0.25061 0.2088 0.45942 0.1740 0.63343 0.1450 0.77844 0.1208 0.89935 0.1007 1.0000

QUEUING EVALUATION -- SINGLE SERVER WITH A FINITE QUEUE



202122232425

D E=(1-G16)/(1-G16^(G8+1)) =D20=$D$20*$G$16^C21 =E20+D21=$D$20*$G$16^C22 =E21+D22=$D$20*$G$16^C23 =E22+D23=$D$20*$G$16^C24 =E23+D24=$D$20*$G$16^C25 =E24+D25

11121314

1516

G=(G16/(1-G16))-(((G8+1)*(G16)^(G8+1))/(1-G16^(G8+1)))=G11/(G6*(1-D25))

=G11-(1-D20)

=G14/(G6*(1-D25))=G6/G7

22

M/M/1 with a Limited Population(Figure 17-15)

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1011121314

15161718192021222324252627282930

A B C D E F G H


Parameter Values:Mean Customer Arrival Rate: lambda = 2Mean Customer Service Rate: mu= 10Size of Customer Population: M = 10





0 0.0184 0.01841 0.0368 0.05522 0.0662 0.12133 0.1059 0.22724 0.1483 0.37555 0.1779 0.55346 0.1779 0.73137 0.1423 0.87368 0.0854 0.95909 0.0342 0.993210 0.0068 1.0000

QUEUING EVALUATION WITH A LIMITED POPULATION

Enter the problem parameters in G6:G8.

Enter the problem parameters in G6:G8.

NOTE: The size of the population M is limited to 100.

NOTE: The size of the population M is limited to 100.

23

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1011121314

15161718192021222324252627282930

A B C D E F G H


Parameter Values:Mean Customer Arrival Rate: lambda = 2Mean Customer Service Rate: mu= 10Size of Customer Population: M = 10





0 0.0184 0.01841 0.0368 0.05522 0.0662 0.12133 0.1059 0.22724 0.1483 0.37555 0.1779 0.55346 0.1779 0.73137 0.1423 0.87368 0.0854 0.95909 0.0342 0.993210 0.0068 1.0000

QUEUING EVALUATION WITH A LIMITED POPULATION

Formulas

M/M/1 with a Limited Population

11121314

15

16

G

=G14+1-D20=G15+(1/G7)

=G8-((G6+G7)/G6)*(1-D20)

=G14/(G6*(G8-G11))

=G6/G7



2021

E=D20=E20+D2120

21

D=1/(VLOOKUP(G8,Sum!A2:C102

,3))

=(FACT($G$8)/FACT($G$8-C21))*$D$20*($G$6/$G$7)^C21



24

SumWorksheet (Sum Tab)(Figure 17-21)

123456789

101112

A B C

n [M!/(M-n)!]()nSum from 0 to n

0 1 11 2 32 3.6 6.63 5.76 12.3600004 8.064 20.4240005 9.6768 30.1008006 9.6768 39.7776007 7.74144 47.5190408 4.644864 52.1639049 1.8579456 54.02185010 0.37158912 54.393439

4. Columns B and C will return the error function #NUM! Whenever n > M.

4. Columns B and C will return the error function #NUM! Whenever n > M.

This worksheet is used to compute the sum

nS

n NM

M

0 )!(

!

The sum is used in the P0 calculation.

2

B

=(FACT('Limited Pop'!$G$8)/FACT('Limited Pop'!$G$8-A2))*('Limited Pop'!$G$6/'Limited Pop'!$G$7)^A2

23

C

1=C2+B3

1. ‘Limited Pop’!$G$8 in cell B2 refers to cell G8 on the Limited Pop worksheet, shown previously.

1. ‘Limited Pop’!$G$8 in cell B2 refers to cell G8 on the Limited Pop worksheet, shown previously.

3. The formula in C3 is copied down to C12.

3. The formula in C3 is copied down to C12.

2. The formula in B2 is copied down to B12.

2. The formula in B2 is copied down to B12.

25

M/M/1 with Constant Service(Figure 17-16)

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10111213

141516171819

A B C D E F G H I J







0 0.1667 0.1667

SINGLE SERVER QUEUING SYSTEM EVALUATION -- CONSTANT SERVICE TIME

10111213

1415

G=G13+G15=G14+1/G7

=G15^2/(2*(1-G15))

=G13/G6=G6/G7

19D E

=1-G15 =D19



26

Palisade Decision ToolsBestFit

The BestFit 4.0 software program on the CD-ROM accompanying this book can be used to find the distribution that best fits a set of data.

It compares data with more than 30 different distributions using chi-square, Kolmogorov-Smirnov, and Anderson-Darling tests. A few of the common distributions it checks are beta, binomial, chi-square, exponential, gamma, geometric, hypergeometric, normal, Poisson, triangular, and uniform.

27

BestFit

To start BestFit, click on the Windows Start button, select Programs, Palisade Decision Tools, then BestFit 4.0

BestFit will open and the initial screen shown next will appear.

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Initial Screen

1. Enter the 25 times from Table 17-5 of the text in the Sample column.

1. Enter the 25 times from Table 17-5 of the text in the Sample column.

2. Click Fitting on the menu bar and select Run Fit. This will give the Fit Results dialog box shown next.

2. Click Fitting on the menu bar and select Run Fit. This will give the Fit Results dialog box shown next.

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BestFit Results Box(Figure 17-17)

1. This figure shows portions of the Fit Results dialog box.

1. This figure shows portions of the Fit Results dialog box.

5. Click on the GOF (Goodness of Fit) tab to see the test values of Chi-Sq, A-D, or K-S.

5. Click on the GOF (Goodness of Fit) tab to see the test values of Chi-Sq, A-D, or K-S.

2. The distributions fitted, from the best (Inverse Gaussian) to the worst (Uniform).

2. The distributions fitted, from the best (Inverse Gaussian) to the worst (Uniform).

3. The distributions are ranked by K-S values. Click on the down arrow to rank by Chi-Square or Anderson-Darling values.

3. The distributions are ranked by K-S values. Click on the down arrow to rank by Chi-Square or Anderson-Darling values.

4. Graph of the best fit and the data.

4. Graph of the best fit and the data.

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Exponential Fit(Figure 17-18)

1. The graph compares the WaySafe data with an exponential distribution.

1. The graph compares the WaySafe data with an exponential distribution.

2. The GOF tab gives the test values.

2. The GOF tab gives the test values.

3. BestFit also gives the critical values for various significance levels.

3. BestFit also gives the critical values for various significance levels.4. The Chi-Sq and K-S test values are less than all

the critical values. The A-D test value is less than the 0.01 critical value but larger than the others.

4. The Chi-Sq and K-S test values are less than all the critical values. The A-D test value is less than the 0.01 critical value but larger than the others.

1 chapter 17 standing in line – at the bank, the market, the movies – is the time-waster...

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