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Excel con formulario para resolver problemas de colas
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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
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