Example 10.1 - Customers in Line

Arrival rate 1

Service rate 2

Interarrival Time 1.0000Service time 0.5000

System Utilization 0.5000

Probability system is empty 0.5000

Average number in line 0.5000

Average number in system 1.0000

Average time in line 0.5000

Average time in system 1.0000

n = 10

P(10 units in system) = 0.0005

P(n < 10 units in system) = 0.9990

n P(n) P(n < n)0 0.50001 0.2500 0.50002 0.1250 0.75003 0.0625 0.87504 0.0313 0.93755 0.0156 0.96886 0.0078 0.98447 0.0039 0.99228 0.0020 0.99619 0.0010 0.9980

10 0.0005 0.999011 0.0002 0.999512 0.0001 0.999813 0.0001 0.999914 0.0000 0.999915 0.0000 1.000016 0.0000 1.000017 0.0000 1.000018 0.0000 1.000019 0.0000 1.0000

=

=

1/ = 1/ =

=P

0 =

Lq =

Ls =

Wq =

Ws =

20 0.0000 1.000021 0.0000 1.000022 0.0000 1.000023 0.0000 1.000024 0.0000 1.000025 0.0000 1.000026 0.0000 1.000027 0.0000 1.000028 0.0000 1.000029 0.0000 1.000030 0.0000 1.000031 0.0000 1.000032 0.0000 1.000033 0.0000 1.000034 0.0000 1.000035 0.0000 1.000036 0.0000 1.000037 0.0000 1.000038 0.0000 1.000039 0.0000 1.000040 0.0000 1.000041 0.0000 1.000042 0.0000 1.000043 0.0000 1.000044 0.0000 1.000045 0.0000 1.000046 0.0000 1.000047 0.0000 1.000048 0.0000 1.000049 0.0000 1.000050 0.0000 1.000051 0.0000 1.000052 0.0000 1.000053 0.0000 1.0000

M M s

Page 3

Example 10.3 - Determining the Number of Servers

Arrival Rate 0.1 0.1 0.1 0.1

Service Rate 0.2 0.2 0.2 0.2

Number of servers S = 3 4 5 6

Average number being served r = 0.500 0.500 0.500 0.500

Average number in line 0.003 0.000 0.000 0.000

Average number in system 0.503 0.500 0.500 0.500

Average time in line 0.030 0.003 0.000 0.000

Average time in system 5.030 5.003 5.000 5.000 System Utilization rho = 0.167 0.125 0.100 0.083

P(zero units in system) 0.606 0.606 0.607 0.607

Average waiting time 2.000 1.429 1.111 0.909

P(wait) 0.015 0.002 0.000 0.000

Note: The Lq value in Exhibit 10.9 may differ slightly from the Lq

on this template. The value of Lq in this template is

based on a queue formula.

=

=

Lq =

Ls =

Wq =

Ws =

P0 =

Wa =

Pw =

M M s

Page 4

0.1 0.1

0.2 0.2

7 8

0.500 0.500

0.000 0.000

0.500 0.500

0.000 0.000

5.000 5.000 0.071 0.063

0.607 0.607

0.769 0.667

0.000 0.000

M M s

Page 5

M P0 1 0.5002 0.6003 0.606

4 #N/A

5 #N/A

6 #N/A

7 #N/A8 #N/A

9 #N/A

10 #N/A

11 #N/A12 #N/A

M M s

Page 6

Calculations:

M P0 1 1.000 1.000 0.500 2 1.500 0.167 0.600 3 1.625 0.025 0.606

4 1.646 0.003 0.606

5 1.648 0.000 0.607

6 1.649 0.000 0.607

7 1.649 0.000 0.607 8 1.649 0.000 0.607

9 1.649 0.000 0.607

10 1.649 0.000 0.607

11 1.649 0.000 0.607 12 1.649 0.000 0.607

Example 10.5 - Waiting Line Approximation Spreadsheet

Mean (average) Time Between Customer Arrivals (minutes) =

Standard Deviation of the Time Between Customer Arrivals =

Mean (average) Service Time (minutes)=

Standard Deviation of the Service Time =

Number of Servers =

Customer Arrival Rate (customers per minute) λ = #DIV/0!Service Rate (customers per minute) μ = #DIV/0!

Expected Server Utilization ρ = #DIV/0!

Coefficient of Variation of Customer Arrival Time #DIV/0!

Coefficient of Variation of Service Time #DIV/0!

Expected Number of Customers Waiting in Line #DIV/0!Expected Number of Customers in the System #DIV/0!

Expected Time Waiting to be Served #DIV/0! ====>

Expected Time Waiting to be Served #DIV/0!

Ca =

Cs =

Lq =

Ls = W

q =

Ws =

#DIV/0!

Example 10.2 - Equipment Selection

Arrival rate 10

Service rate 12

Interarrival Time 0.1000Service time 0.0833

System Utilization 0.8333

Probability system is empty 0.1667

Average number in line 2.0833

Average number in system 2.9167

Average time in line 0.2083

Average time in system 0.2917

=

=

1/ = 1/ =

=P

0 =

Lq =

Ls =

Wq =

Ws =

Note: This problem uses Model 2 (Constant Service time) equations

n P(n) P(n < n)0 0.16671 0.1389 0.16672 0.1157 0.30563 0.0965 0.42134 0.0804 0.51775 0.0670 0.59816 0.0558 0.66517 0.0465 0.72098 0.0388 0.76749 0.0323 0.8062

10 0.0269 0.838511 0.0224 0.865412 0.0187 0.887813 0.0156 0.906514 0.0130 0.922115 0.0108 0.935116 0.0090 0.945917 0.0075 0.954918 0.0063 0.962419 0.0052 0.968720 0.0043 0.973921 0.0036 0.978322 0.0030 0.981923 0.0025 0.984924 0.0021 0.987425 0.0017 0.9895

26 0.0015 0.991327 0.0012 0.992728 0.0010 0.993929 0.0008 0.994930 0.0007 0.995831 0.0006 0.996532 0.0005 0.997133 0.0004 0.997634 0.0003 0.998035 0.0003 0.998336 0.0002 0.998637 0.0002 0.998838 0.0002 0.999039 0.0001 0.999240 0.0001 0.999341 0.0001 0.999442 0.0001 0.999543 0.0001 0.999644 0.0001 0.999745 0.0000 0.999746 0.0000 0.999847 0.0000 0.999848 0.0000 0.999849 0.0000 0.999950 0.0000 0.999951 0.0000 0.999952 0.0000 0.999953 0.0000 0.9999

Example 10.4 - Finite Population Source

Population Size N = 4 4Number of servers S = 1 2Average service time T = 7.5 7.5Average time between service calls U = 60 60P(wait) - from table D = 0.3210 0.0370 Efficiency factor - from table F = 0.9570 0.9980 Service factor 0.111 0.111 Average number waiting L = #NAME? #NAME?Average waiting time W = #NAME? #NAME?Average number running J = #NAME? #NAME?Average number being serviced H = #NAME? #NAME?

Per TimeUnit

Service cost = $7.00Downtime cost = $40.00

Comparison of Downtime Cost

Number of Servers 1 2

Number of Machines Down (H+L) #NAME? #NAME?

Cost per Hour of Machines Down #NAME? #NAME?[ (H+L) x $40 per unit time ]

Cost of Service $7.00 $14.00($7 per unit time )

Total Cost per Unit Time #NAME? #NAME?

=

Note: You must enter D and F (based on N, , and M) from the table in the text

Comparison of Downtime Cost

1 2

$0$2$4$6$8

$10$12$14$16

Comparison of Dow ntime Cost for Different Num ber of Servers Cost of Server

Cost of Machine

Number of Servers