# 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

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```					                   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)

1
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.
2
Single-Server Queuing Model
(M/M/1)
 Assumes exponential distribution for times:
 between successive arrivals—mean rate l
 to complete service—mean rate m
 Provides important results:
 Probability for n customers in system:
P0 = 1- l/m         Pn = ( l/m)n P0
 Mean number of customers in system:
L= l
m-l
 Mean customer time spent in system:
W = L = 1
3
l    m-l
Single-Server Queuing Model
(M/M/1)
 Important results (continued):
 Mean number of customers waiting:
Lq =    l2
m(m- l)
 Mean customer waiting time:

Wq =
Lq
=     l
l m(m - l)
 Server utilization factor:
r = l
m

4
Single-Server Queuing Model
(M/M/1)
 Problem: Customers arrive at a supply room with
mean rate l = 25/hr. Service take a mean of 2
minutes, so that m = 30/hr. Do a queuing analysis.
 Solution:
25                            1
L           5 cust.       W             .20 hr.
30 - 25                      30 - 25
(25)2                          25
Lq               4.167 Wq                    .167
30(30 - 25)                  30(30 - 25)
r = 25/30 = .833        P0 = 1 - .833 = .167
P1 = (25/30)1(.167) = .139

5
P2 = (25/30)2(.167) = .116
Single-Server Queuing Model
(M/M/1)
 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 ×l + \$10 = \$7(.20)(25) + 10 = \$45
 The queuing cost accounts for all l customers in the
system during the hour, and collective time under a
congestion penalty, W × l, applies above.
 When the congestion penalty involves just the waiting
time, Wq × l is used for queuing cost. Retail customers
6        resent the wait, not the time receiving service.
Multiple-Server Queuing Model
(M/M/S)
 Multiple servers involve greater complexity.

 The model begins with computation of
  l n  l  S                   
 S -1  m   m  
              1     
P0  1 /                                
n  0 n!       S !  1 - l / Sm  
7                                                 
Multiple-Server Queuing Model
(M/M/S)
 The mean number of customers is then found:
(l / m )S (l / Sm )
Lq                      P0
S!(1 - l/Sm ) 2

 Then the mean customer waiting time is found:
Lq
Wq 
l
 That is followed by:
1                l
W  Wq             L  Lq 
m                m
8
Templates and Software
 Queuing Templates

 Palisade Decision Tools BestFit 4.0

9

   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
10
1. Enter the problem
parameters in G6:G7.                                M/M/1
Remember lambda
must be less than mu.                               (Figure 17-4)
A      B      C        D         E           F            G      H         I        J
1        BASIC QUEUING SYSTEM EVALUATION -- SINGLE SERVER
2
3 PROBLEM: Supply Room
4
5        Parameter Values:
6             Mean Customer Arrival Rate: lambda =                   25
7             Mean Customer Service Rate: mu =                       30
8                                                                                   G
9        Queuing Results:                                               10 =G6/(G7-G 6)
2. Queuing           10             Mean Number of Customers in System: L =                 5 11 =1/(G7-G 6)
11             Mean Customer Time Spent in System: W =               0.2 12
summary              12             Mean Number of Customers Waiting                          13 =G6^2/(G7*(G 7-G6))
results are          13                                   (Length of Line): Lq =     4.166667 14 =G6/(G7*(G 7-G6))
14             Mean Customer Waiting Time: Wq =                 0.166667
here: L, W, Lq,                                                                               15 =G6/G7
15             Server Utilization Factor: rho =                 0.833333
Wq, and rho.         16
17             Number in Probability Cumulative                        D              E
18             System n        Pn     Probability          19 =1-G15            =D19
3. System state      19                        0   0.1667    0.1667          20   =\$D\$19*\$G\$15^C20   =E19+D20
20                        1   0.1389    0.3056          21   =\$D\$19*\$G\$15^C21   =E20+D21
probabilities        21                        2   0.1157    0.4213          22   =\$D\$19*\$G\$15^C22   =E21+D22
are here, Pn         22                        3   0.0965    0.5177          23   =\$D\$19*\$G\$15^C23   =E22+D23
and S Pn. To
23                        4   0.0804    0.5981          24   =\$D\$19*\$G\$15^C24   =E23+D24
24                        5   0.0670    0.6651          25   =\$D\$19*\$G\$15^C25   =E24+D25
fit the              25                        6   0.0558    0.7209          44   =\$D\$19*\$G\$15^C44   =E43+D44
44                       25   0.0017    0.9913
45   =\$D\$19*\$G\$15^C45   =E44+D45
46   =\$D\$19*\$G\$15^C46   =E45+D46
on one page          46                       27   0.0012    0.9939
47   =\$D\$19*\$G\$15^C47   =E46+D47
47                       28   0.0010    0.9949
some of the          48                       29   0.0008    0.9958          48   =\$D\$19*\$G\$15^C48   =E47+D48
71   =\$D\$19*\$G\$15^C71   =E70+D71
rows have            71                       52   0.0000    0.9999
11       72                       53   0.0000    0.9999          72   =\$D\$19*\$G\$15^C72   =E71+D72
been hidden.         73                       54   0.0000    1.0000          73   =\$D\$19*\$G\$15^C73   =E72+D73
1. Enter the problem           M/M/S
parameters in G6:G8.            (Figure 17-6)
A       B      C        D          E           F         G
1 BASIC QUEUING EVALUATION -- MULTIPLE SERVERS
2
3 PROBLEM: Supply Room
2. Remember               4
that lambda               5        Parameter Values:
must be less              6             Mean Customer Arrival Rate: lambda =                   0.6
than mu times S.          7             Mean Customer Service Rate: mu =                       0.5
Otherwise the             8             Number of Servers: S =                                   2
results will not          9
10        Queuing Results:
be meaningful
11             Mean Number of Customers in System: L =             1.8750
(negative
12             Mean Customer Time Spent in System: W =             3.1250
probabilities).          13             Mean Number of Customers Waiting
14                                   (Length of Line): Lq =        0.6750
3. The number of         15             Mean Customer Waiting Time: Wq =                    1.1250
servers is limited       16             Server Utilization Factor: rho =                    0.6000
17
to 100 or less.
18             Number in Probability Cumulative
Caution: For large S
19             System, n       Pn     Probability
20                  0       0.2500       0.2500
and n sometimes the
21                  1       0.3000       0.5500             resulting numbers are
22                  2       0.1800       0.7300             too large for Excel so
23                  3       0.1080       0.8380             error messages occur,
24                  4       0.0648       0.9028
12            38                 18       0.0001       0.9999             such as #NUM!
39                 19       0.0000       1.0000
D                                    2. The Sum !A2:C102
=1/(VLOOKUP(G8-                    M/M/S         in cell D20 refers to
1,Sum!A2:C102,3)+VLOOKUP(G8,Sum!A2:C102,2)*                    the table located on
20               (1/(1-G6/(G7*G8))))
A     B
=IF(C21>\$G\$8,(\$D\$20*((\$G\$6/\$G\$7)^(C21/2))*((\$G  C
Formulas
D
the Sum worksheet,
E shown next.
F         G
1 BASIC QUEUING EVALUATION -- MULTIPLE SERVERS
\$6/\$G\$7)^(C21/2)))/(FACT(\$G\$8)*(\$G\$8^((C21-
2
\$G\$8)/2))*(\$G\$8^((C21-
3 PROBLEM: Supply Room
21 \$G\$8)/2))),(\$G\$6/\$G\$7)^C21*(\$D\$20/FACT(C21)))
4
1. All the calculations on
5       Parameter Values:
this spreadsheet are                               Mean Customer Arrival Rate: lambda =
6                                                               0.6
based on Po found in cell             7            Mean Customer Service Rate: mu =                   0.5
D20.                                  8            Number of Servers: S =                               2
G                   9
10       Queuing Results:
11 =G14+(G6/G7)
11            Mean Number of Customers in System: L =        1.8750
12 =G15+(1/G7)
12            Mean Customer Time Spent in System: W =        3.1250
13                                   13            Mean Number of Customers Waiting
=((G6/G7)^G8)*(G6/(G8*G7))*D20/   14                                  (Length of Line): Lq =   0.6750
15            Mean Customer Waiting Time: W q =              1.1250
14 (FACT(G8)*(1-G6/(G7*G8))^2)
16            Server Utilization Factor: rho =               0.6000
15 =G14/G6                           17
16 =G6/(G7*G8)                       18            Number in Probability Cumulative
19            System, n     Pn      Probability
20                0      0.2500       0.2500              E
3. The formulas in               21                1      0.3000       0.5500
D21:E21 are copied                                                                  20 =D20
22                2      0.1800       0.7300
down to the end of the           23                3      0.1080       0.8380       21 =E20+D21
table.                           24                4      0.0648       0.9028
13                        38               18      0.0001       0.9999
39               19      0.0000       1.0000
SumWorksheet (Sum Tab)
(Figure 17-19)

This worksheet is used to compute the sum
(l / m ) n
S

 n!
n 0
The sum is used in the P0 calculation.
A             B             C
1     n         (l/m)n/n!   Sum from 0 to n
2     0        1.000E+00      1.000E+00
MMS!\$G\$6 in cell B2       3     1        1.200E+00      2.200E+00
refers to cell G6 on      4     2        7.200E-01      2.920E+00
the MMS worksheet,        5     3        2.880E-01      3.208E+00
shown previously.         6     4        8.640E-02      3.294E+00
The formulas in          101    99       7.395E-149     3.320E+00
B3:C3 are copied         102   100       8.874E-151     3.320E+00
down to B102:C102.
B                                   C
14
2 =((MMS!\$G\$6/MMS!\$G\$7)^(A2))/FACT(A2)               1
3 =((MMS!\$G\$6/MMS!\$G\$7)^(A3))/FACT(A3)               =C2+B3
1. Enter the
problem
parameters in
G6:G9.
Cost Analysis for M/M/S
(Figure 17-7)

A       B       C       D        E        F         G        H          I        J         K
1        BASIC QUEUING SYSTEM EVALUATION -- MULTIPLE SERVERS
2
3 PROBLEM: Million Bank Tellers
2. Remember         4
3. The number of
that lambda         5         Parameter Values:                                              servers is
must be less        6                 Mean Customer Arrival Rate: lambda =    250            limited to 100 or
7                 Mean Customer Service Rate: mu =          30
than mu times       8                 Customer Cost per Unit of Time =           6           less.
S for results to    9                 Server Cost per Unit of Time =            12
10
be meaningful.     11
Consequently,      12                                                     Server Customer     Total  Customer Total
rows 14 - 21       13 Servers    P0      Lq       L        Wq        W     Cost    Cost(Wq) Cost(Wq) Cost(W) Cost(W)

are hidden         22    9     0.0001   9.5049 17.8382   0.0380   0.0714   \$108.00   \$57.03   \$165.03   \$107.03   \$215.03
23   10     0.0002   2.4381 10.7714   0.0098   0.0431   \$120.00   \$14.63   \$134.63    \$64.63   \$184.63
(because           24   11     0.0002   0.9363 9.2696    0.0037   0.0371   \$132.00    \$5.62   \$137.62    \$55.62   \$187.62
lambda is not      25   12     0.0002   0.3999 8.7333    0.0016   0.0349   \$144.00    \$2.40   \$146.40    \$52.40   \$196.40
26   13     0.0002   0.1761 8.5094    0.0007   0.0340   \$156.00    \$1.06   \$157.06    \$51.06   \$207.06
less than mu       27   14     0.0002   0.0774 8.4107    0.0003   0.0336   \$168.00    \$0.46   \$168.46    \$50.46   \$218.46
times s for        28   15     0.0002   0.0334 8.3668    0.0001   0.0335   \$180.00    \$0.20   \$180.20    \$50.20   \$230.20
these rows).       29   16     0.0002   0.0141 8.3474    0.0001   0.0334   \$192.00    \$0.08   \$192.08    \$50.08   \$242.08
30   17     0.0002   0.0057 8.3391    0.0000   0.0334   \$204.00    \$0.03   \$204.03    \$50.03   \$254.03
31   18     0.0002   0.0023 8.3356    0.0000   0.0333   \$216.00    \$0.01   \$216.01    \$50.01   \$266.01

Note: In order for the spreadsheet to compute correctly, make sure
automatic calculation is selected on both worksheets (Tools, Options,
15
Calculation Tab, and click on the Automatic button under Calculation).
B                                  C
=1/(VLOOKUP(A22-
1,Sum!\$A\$2:\$C\$102,3)+VLOOKUP =((\$G\$6/\$G\$7)^A22)*(\$G\$6/(A22*\$
(A22,Sum!\$A\$2:\$C\$102,2)*(1/(1- G\$7))*B22/(FACT(A22)*(1-
Formulas for
22        \$G\$6/(\$G\$7*A22))))

D                E
\$G\$6/(\$G\$7*A22))^2)

F               G
M/M/S
22 =C22+(\$G\$6/\$G\$7)   =C22/\$G\$6         =E22+(1/\$G\$7)    =\$G\$9*A22

H               I            J                    K                  Cost Analysis
22 =\$G\$6*E22*\$G\$8     =G22+H22 =\$G\$6*F22*\$G\$8           =G22+J22
(Figure 17-20)
1. The Sum                  A       B        C      D        E        F         G        H          I        J         K
!\$A\$2:\$C\$102           1        BASIC QUEUING SYSTEM EVALUATION -- MULTIPLE SERVERS
in cell B22            2
3 PROBLEM: Million Bank Tellers
refers to the          4
2. The formulas
table located          5         Parameter Values:                                       in row 22 are
on the Sum             6                 Mean Customer Arrival Rate: lambda =    250     copied down to
7                 Mean Customer Service Rate: mu =          30
worksheet. It          8                 Customer Cost per Unit of Time =           6    the other rows
is identical           9                 Server Cost per Unit of Time =            12    of the table.
10
with the one          11
for the M/M/S         12                                                     Server Customer    Total   Customer Total
13 Servers    P0      Lq       L        Wq        W     Cost    Cost(Wq) Cost(Wq) Cost(W) Cost(W)
template
22    9     0.0001   9.5049 17.8382   0.0380   0.0714   \$108.00   \$57.03   \$165.03   \$107.03   \$215.03
shown                 23   10     0.0002   2.4381 10.7714   0.0098   0.0431   \$120.00   \$14.63   \$134.63    \$64.63   \$184.63
previously.           24   11     0.0002   0.9363 9.2696    0.0037   0.0371   \$132.00    \$5.62   \$137.62    \$55.62   \$187.62
25   12     0.0002   0.3999 8.7333    0.0016   0.0349   \$144.00    \$2.40   \$146.40    \$52.40   \$196.40
26   13     0.0002   0.1761 8.5094    0.0007   0.0340   \$156.00    \$1.06   \$157.06    \$51.06   \$207.06
27   14     0.0002   0.0774 8.4107    0.0003   0.0336   \$168.00    \$0.46   \$168.46    \$50.46   \$218.46
28   15     0.0002   0.0334 8.3668    0.0001   0.0335   \$180.00    \$0.20   \$180.20    \$50.20   \$230.20
29   16     0.0002   0.0141 8.3474    0.0001   0.0334   \$192.00    \$0.08   \$192.08    \$50.08   \$242.08
16        30   17     0.0002   0.0057 8.3391    0.0000   0.0334   \$204.00    \$0.03   \$204.03    \$50.03   \$254.03
31   18     0.0002   0.0023 8.3356    0.0000   0.0333   \$216.00    \$0.01   \$216.01    \$50.01   \$266.01
M/M/1Non-exponential Service
(Figure 17-8)
Enter the problem parameters in G6:G8. Remember
lambda must be less than mu.
A        B     C        D         E           F           G        H         I         J
1   QUEUING EVALUATION FOR SINGLE SERVER WITH NON-EXPONENTIAL TIMES
2
3 PROBLEM: Sammy's Barbershop
4
5        Parameter Values:
6             Mean Customer Arrival Rate: lambda =             4
7             Mean Customer Service Rate: mu =                 5
8             Standard Deviation of Service Time: sigma =   0.05
9
10        Queuing Results:                                                        G
11             Mean Number of Customers in System: L =        2.5 11 =G14+G16
12             Mean Customer Time Spent in System: W =      0.625 12 =G15+1/G7
13             Mean Number of Customers Waiting                   13
14                                   (Length of Line): Lq =   1.7 14 =(G6^2*G8^2+G16^2)/(2*(1-G16))
15 =G14/G6
15             Mean Customer Waiting Time: Wq =             0.425 16 =G6/G7
16             Server Utilization Factor: rho =               0.8

17
1. Enter the
problem
parameters          Exponential Distribution
in D5:D7.
(Figure 17-9)
A         B           C              D          E               F       G       H        I
1               EXPONENTIAL DISTRIBUTION
2
3   PROBLEM: Supply Room Arrival
4                                                                     F
5   Mean process rate lambda =                     25    9 =EXPONDIST(D6,D5,TRUE)
6   Lower point a =                              0.05   10 =1-F11-F9
7   Upper point b =                               0.2   11 =1-EXPONDIST(D7,D5,TRUE)
8
9   Area to the left of a: Pr[X < a] =                             0.713495
10   Area between a and b: Pr[a < X < b] =                           0.27977
11   Area to the right of b: Pr[X > b] =                E            0.00674
13 =B13                           C
12                              B
18   =EXPONDIST(B18,\$D\$5,FALSE)
13   Mean =          0.04 13 =1/D5      Stand. Dev. =        0.04
19   =EXPONDIST(B19,\$D\$5,FALSE)
14                                                                  20   =EXPONDIST(B20,\$D\$5,FALSE)
2. The          15             Interarrival Frequency    Cumulative                 21   =EXPONDIST(B21,\$D\$5,FALSE)
exponential     16               Time        Curve       Distribution               22   =EXPONDIST(B22,\$D\$5,FALSE)
frequency       17                 t           f(t)          F(t)                   23   =EXPONDIST(B23,\$D\$5,FALSE)
curve and                                                                                           D
18                 0           25              0
18   =EXPONDIST(B18,\$D\$5,TRUE)
cumulative      19               0.05       7.16262       0.71350
19   =EXPONDIST(B19,\$D\$5,TRUE)
distribution    20                0.1       2.05212       0.91792                   20   =EXPONDIST(B20,\$D\$5,TRUE)
are here:       21                0.2       0.16845       0.99326                   21   =EXPONDIST(B21,\$D\$5,TRUE)
22                0.3       0.01383       0.99945                   22   =EXPONDIST(B22,\$D\$5,TRUE)
23                 0.4      0.00113       0.99995                   23   =EXPONDIST(B23,\$D\$5,TRUE)
18
Poisson Distribution
(Figure 17-11)

A            B          C               D        E         F       G
1                          POISSON DISTRIBUTION
2
3   PROBLEM: Number of Arrivals at Supply Room
4                                                                                1. Enter the
5   Mean process rate lambda =                                 25
6   Duration t =                                              0.1
problem
7   Lower limit for the number of events, a =                   0                parameters
8   Upper limit for the number of events, b =                   8
9                                                                                in D5:D8.
10   Pr[X < a] =             0.08208      Pr[X < a] = 0
11   Pr[a < X < b] =         0.99886                         E
12   Pr[a < X < b] =         0.99575 10 =IF(E7>0,POISSON(E7-1,E5*56,TRUE),0)
13   Pr[a < X < b] =         0.91677                         E
2. The          14   Pr[a < X < b] =         0.91367      15 =1-POISSON(E8,E5*E6,TRUE)
Poisson         15   Pr[X > b] =             0.00425      Pr[X > b] = 0.00114
16
probabilities   17   Number of             Cumulative
C
18   Successes Probability Probability
and             19       r      Pr[R = r] Pr{R < r]          10   =POISSON(E7,E5*E6,TRUE)
11   =POISSON(E8,E5*E6,TRUE)-E10
cumulative      20       0      0.08208     0.08208
12   =POISSON(E8-1,E5*E6,TRUE)-E11
21       1      0.20521     0.28730
probabilities   22       2      0.25652     0.54381
13   =POISSON(E8,E5*E6,TRUE)-C10
14   =POISSON(E8-1,E5*E6,TRUE)-C10
are here:       23       3      0.21376     0.75758
15   =1-POISSON(E8-1,E5*E6,TRUE)
24       4      0.13360     0.89118
B
25       5      0.06680     0.95798
20 =POISSON(A20,\$E\$5*\$E\$6,FALSE)
26       6      0.02783     0.98581
21 =POISSON(A21,\$E\$5*\$E\$6,FALSE)
27       7      0.00994     0.99575
22 =POISSON(A22,\$E\$5*\$E\$6,FALSE)
28       8      0.00311     0.99886
29       9      0.00086     0.99972                           C
30      10      0.00022     0.99994          20 =POISSON(A20,\$E\$5*\$E\$6,TRUE)
31      11      0.00005     0.99999          21 =POISSON(A21,\$E\$5*\$E\$6,TRUE)
19   32      12      0.00001     1.00000          22 =POISSON(A22,\$E\$5*\$E\$6,TRUE)
33      13      0.00000     1.00000
Waiting Times for M/M/1
(Figure 17-13)

1. Enter the
A     B     C       D        E       F       G         H         I
problem      J
1                   WAITING TIME PROBABILITIES                  parameters in
2
3 PROBLEM: Supply Room Waiting Times
G6:G7.
4                                                               Remember
2. The
5         Parameter Values:                                     lambda must be
6             Mean Customer Arrival Rate: lambda = 25
cumulative       7             Mean Customer Service Rate: mu =     30
less than mu.
distributions    9
10              Time, t     W q(t)   W(t)
for the                                                                        D
11               0.05   0.3510   0.2212
waiting         12               0.10   0.4946   0.3935   11 =1-(\$G\$6/\$G\$7)*EXP(-1*(\$G\$7-\$G\$6)*C11)
12 =1-(\$G\$6/\$G\$7)*EXP(-1*(\$G\$7-\$G\$6)*C12)
times in the    13               0.15   0.6064   0.5276
13 =1-(\$G\$6/\$G\$7)*EXP(-1*(\$G\$7-\$G\$6)*C13)
line and in     14               0.20   0.6934   0.6321
15               0.25   0.7612   0.7135                     E
the system      16               0.30   0.8141   0.7769     11 =1-EXP(-1*(\$G \$7-\$ G\$6)*C11)
are here:       17               0.35   0.8552   0.8262     12 =1-EXP(-1*(\$G \$7-\$ G\$6)*C12)
18               0.40   0.8872   0.8647     13 =1-EXP(-1*(\$G \$7-\$ G\$6)*C13)
19               0.45   0.9122   0.8946
20               0.50   0.9316   0.9179

20
M/M/1with a Finite Queue
(Figure 17-14)
Enter the problem parameters in G6:G8.
Remember lambda must be less than mu.
A      B       C         D         E             F           G         H         I         J
1     QUEUING EVALUATION -- SINGLE SERVER WITH A FINITE QUEUE
2
3 PROBLEM: Supply Room
4
5        Parameter Values:                                                                G
6             Mean Customer Arrival Rate: lambda =                      25     =(G16/(1-G16))-
7             Mean Customer Service Rate: mu =                          30     (((G8+1)*(G16)^(G8+1))/
8             Maximum Number of Customers in System: M =                 5 11 (1-G16^(G8+1)))
9                                                                           12 =G11/(G6*(1-D25))
10        Queuing Results:
13
11             Mean Number of Customers in System: L =               1.9788
14 =G11-(1-D20)
12             Mean Customer Time Spent in System: W =               0.0880
13             Mean Number of Customers Waiting                              15 =G14/(G6*(1-D25))
14                                    (Length of Line): Lq =         1.2294 16 =G6/G7
15             Mean Customer Waiting Time: Wq =                      0.0547
16             Traffic Intensity: rho =                              0.8333
17
18             Number in Probability Cumulative
19             System n        Pn      Probability                      D                 E
20                       0    0.2506      0.2506         20 =(1-G16)/(1-G16^(G8+1)) =D20
21                       1    0.2088      0.4594         21 =\$D\$20*\$G\$16^C21        =E20+D21
22                       2    0.1740      0.6334         22 =\$D\$20*\$G\$16^C22        =E21+D22
23                       3    0.1450      0.7784         23 =\$D\$20*\$G\$16^C23        =E22+D23
21   24                       4    0.1208      0.8993         24 =\$D\$20*\$G\$16^C24        =E23+D24
25                       5    0.1007      1.0000         25 =\$D\$20*\$G\$16^C25        =E24+D25
M/M/1 with a Limited Population
(Figure 17-15)

A       B       C          D          E           F         G        H
1   QUEUING EVALUATION WITH A LIMITED                            POPULATION
2
3 PROBLEM:     Supply Room
4                                                                               Enter the
5          Parameter Values:
6              Mean Customer Arrival Rate: lambda =                     2       problem
7              Mean Customer Service Rate: mu =                        10       parameters in
8              Size of Customer Population: M =                        10
9                                                                               G6:G8.
10          Queuing Results:
11              Mean Number of Customers in System: L =             5.0919
12              Mean Customer Time Spent in System: W =             0.5187
13              Mean Number of Customers Waiting
14                                       (Length of Line): Lq =     4.1103
15              Mean Customer Waiting Time: Wq =                    0.4187
16              Traffic Intensity: rho =                            0.2000
17
18               Number in Probability Cumulative
19               System, n        Pn      Probability
20                   0         0.0184      0.0184
21                   1         0.0368      0.0552
NOTE: The size    22                   2         0.0662      0.1213
of the            23                   3         0.1059      0.2272
24                   4         0.1483      0.3755
population M is   25                   5         0.1779      0.5534
limited to 100.   26                   6         0.1779      0.7313
27                   7         0.1423      0.8736
28                   8         0.0854      0.9590
22      29                   9         0.0342      0.9932
30                   10        0.0068      1.0000
1. All the calculations on this
Formulas
spreadsheet are based on Po found in
cell D20.

M/M/1 with a Limited Population
A      B      C        D         E         F        G        H
G                 1   QUEUING EVALUATION WITH A LIMITED POPULATION
2
11 =G14+1-D20                    3 PROBLEM:     Supply Room
12 =G15+(1/G7)                   4
5          Parameter Values:
13                               6              Mean Customer Arrival Rate: lambda =                 2
14 =G8-((G6+G7)/G6)*(1-D20)      7              Mean Customer Service Rate: mu =                    10
8              Size of Customer Population: M =                    10
15 =G14/(G6*(G8-G11))            9
10          Queuing Results:
16 =G6/G7                       11              Mean Number of Customers in System: L =         5.0919
12              Mean Customer Time Spent in System: W =         0.5187
2. The Sum !A2:C102 in cell D20      13              Mean Number of Customers Waiting
14                                       (Length of Line): Lq = 4.1103
refers to the table located on the
15              Mean Customer Waiting Time: Wq =                0.4187
Sum worksheet, shown next.           16              Traffic Intensity: rho =                        0.2000
17
18               Number in Probability Cumulative
19               System, n        Pn      Probability
D                     20                   0         0.0184      0.0184                E
21                   1         0.0368      0.0552
=1/(VLOOKUP(G8,Sum!A2:C102                                                                  20 =D20
22                   2         0.0662      0.1213
20            ,3))                   23                   3         0.1059      0.2272         21 =E20+D21
24                   4         0.1483      0.3755
=(FACT(\$G\$8)/FACT(\$G\$8-          25                   5         0.1779      0.5534
26                   6         0.1779      0.7313
21 C21))*\$D\$20*(\$G\$6/\$G\$7)^C21       27                   7         0.1423      0.8736
28                   8         0.0854      0.9590
23                        29                   9         0.0342      0.9932
30                   10        0.0068      1.0000
SumWorksheet (Sum Tab)
(Figure 17-21)

This worksheet is used to compute the sum
1. ‘Limited Pop’!\$G\$8
n                 3. The
in cell B2 refers to                  S
M!  l 
 (M - N )! m 
cell G8 on the                                                                    formula in
Limited Pop                                                                     C3 is copied
worksheet, shown                     n 0                                       down to
C12.
previously.
The sum is used in the P0 calculation.
B                      A             B               C                C
n
=(FACT('Limited                     1   n     [M!/(M-n)!](l/m) Sum from 0 to n   2 1
Pop'!\$G\$8)/FACT('Limited            2   0                      1              1 3 =C2+B3
Pop'!\$G\$8-A2))*('Limited            3   1                      2              3
4   2                    3.6            6.6 4. Columns B
2 Pop'!\$G\$6/'Limited Pop'!\$G\$7)^A2
5   3                   5.76     12.360000 and C will
6   4                 8.064      20.424000 return the
2. The formula in
7   5                9.6768      30.100800 error function
B2 is copied                   8   6                9.6768      39.777600
down to B12.                                                               #NUM!
9   7              7.74144       47.519040
10   8             4.644864       52.163904
Whenever n >
24                        11   9           1.8579456        54.021850 M.
12   10         0.37158912        54.393439
M/M/1 with Constant Service
(Figure 17-16)

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

A      B      C         D         E           F          G        H         I         J
1   SINGLE SERVER QUEUING SYSTEM EVALUATION -- CONSTANT SERVICE TIME
2
3 PROBLEM: Supply Room
4
5        Parameter Values:
6             Mean Customer Arrival Rate: lambda =                   25
7             Mean Customer Service Rate: mu =                       30
8
9        Queuing Results:
10             Mean Number of Customers in System: L =            2.9167              G
11             Mean Customer Time Spent in System: W =            0.1167  10 = G13+ G15
12             Mean Number of Customers Waiting                           11 =G 14+1/G 7
13                                   (Length of Line): Lq =       2.0833  12
13 = G15^2/(2*(1-G15))
14             Mean Customer Waiting Time: Wq =                   0.0833
14 =G13/G 6
15             Server Utilization Factor: rho =                   0.8333
16                                                                        15 = G6/G 7
17             Number in Probability Cumulative
18             System n        Pn     Probability               D       E
19 =1-G15 =D19
19                     0      0.1667      0.1667

25
BestFit

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.
26
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.
27
Initial Screen

1. Enter      2. Click Fitting on
the 25        the menu bar and
times from    select Run Fit. This
Table 17-5    will give the Fit
of the text   Results dialog box
in the        shown next.
Sample
column.

28
1. This figure
shows
portions of the
Fit Results
BestFit Results Box
2. The                                 (Figure 17-17)
dialog box.
distributions
fitted, from
the best
5. Click on the
(Inverse
GOF (Goodness of
Gaussian) to
Fit) tab to see the
the worst
test values of Chi-
(Uniform).
Sq, A-D, or K-S.
3. The
distributions
are ranked
by K-S
values.
Click on the
down arrow
to rank by
Chi-Square
4. Graph of the
or
best fit and the
Anderson-
data.
Darling
values.
29
Exponential Fit                   2. The GOF tab
gives the test
(Figure 17-18)           values.

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

3. BestFit also
gives the critical
values for
various
significance
4. The Chi-Sq and K-S test values are less than all    levels.
the critical values. The A-D test value is less than
the 0.01 critical value but larger than the others.
30

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