# Computer Science

Solution
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Question:
14-11
The Rockwell Electronics Corporation retains a service crew to repair machine breakdowns that occur on an average of 饾渾=3 per day (approximately Poisson in nature), the crew can service an average of 碌 = 8 machines per day, with a repair time distribution that resembles the exponential distribution.
(a) What is the utilization rate of this service system?
(b) What is the average downtime for a machine that is broken?
(c) How many machines are waiting to be serviced at any given time?
(d) What is the probability that more than one machine is in the system? Probability that more than
two are broken and waiting to be repaired or being serviced? More than three? More than four?
Arrival rate (位) 3 Per day
Service rate (渭) 8 Per day
(a) What is the utilization rate of this service system?
Solution: computation of the following
Utilization rate(U)=位/渭
server utilization (U) 37.50%
(b) What is the average downtime for a machine that is broken?
Solution: computation of the following
The average down time is the time that the machine waits to be serviced plus the time taken to repair the machine.
The average down time is given by W
W=1/1(渭-位)
W 0.2 Day
assuming 8 hrs/day 1.6 Hours
(c) How many machines are waiting to be serviced at any given time?
Solution: computation of the following
Lq=位^2/渭(渭-位)
Lq 0.225 Machines
(d) What is the probability that more than one machine is in the system? Probability that more than two are broken and waiting to be repaired or being serviced? More than three? More than four?
Solution: computation of the following
Pn>k=(位/渭)^(k+1)
Pn>1 0.141
Pn>2 0.053
Pn>3 0.020
Pn>4 0.007
Solution 2
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Question:
From historical data, Harry鈥檚 car wash estimates that dirty cars arrive at the rate of 10 per hour all day Saturday. With a crew working the wash line, Harry figures that cars can be cleaned at the rate of one every five minutes. One car at a time is cleaned in this example of single-server waiting line. Assuming Poisson arrivals and exponential service times, find the
A) average number of cars in line
B) average time a car waits before it is washed
C) average time a car spends in the service system
D) utilization rate of the car wash
E) probability that no cars are in the system
Arrival rate 10 cars Per hour
Service rate One car at every 5 minutes
Service rate 12 cars per hour
Number of servers (s) 1
Entering above values in the Excelmodules Queuing models—->M/M/s, we get following results:
A) average number of cars in line
Avg no of cars in line(Lq) 4.1666666667
B) average time a car waits before it is washed
Avg waiting time in queue(Wq) 0.4166666667 hours
Avg waiting time in queue(Wq) 25 Mins
C) average time a car spends in the service system
Avg time in service system(W) 0.5 hours
Avg time in service system(W) 30 Mins
D) utilization rate of the car wash
Average utilization of service system 0.8333333333 83.33 Percent
E) probability that no cars are in the system
Probability of no car in the system(P(0)) 0.1666666667
Solution_Excel modules
Harry’s car wash
Queuing Model M/M/s (Exponential Service Times)
Input Data Operating Characteristics
Arrival rate (l) 10 Average server utilization (r) 0.8333
Service rate (m) 12 Average number of customers in the queue (Lq) 4.1667
Number of servers (s) 1 Average number of customers in the system (L) 5.0000
Average waiting time in the queue (Wq) 0.4167
Average time in the system (W) 0.5000
Probability (% of time) system is empty (P0) 0.1667
0
Probabilities
Number of Units Probability Cumulative Probability
0 0.1667 0.1667
1 0.1389 0.3056
2 0.1157 0.4213
3 0.0965 0.5177
4 0.0804 0.5981
5 0.0670 0.6651
6 0.0558 0.7209
7 0.0465 0.7674
8 0.0388 0.8062
9 0.0323 0.8385
10 0.0269 0.8654
11 0.0224 0.8878
12 0.0187 0.9065
13 0.0156 0.9221
14 0.0130 0.9351
15 0.0108 0.9459
16 0.0090 0.9549
17 0.0075 0.9624
18 0.0063 0.9687
19 0.0052 0.9739
20 0.0043 0.9783
Computations
n or s (lam/mu)^n/n! Cumsum(n-1) term2 P0(s) Rho(s) Lq(s) L(s) Wq(s) W(S)
0 1
1 0.8333333333 1 5 0.1666666667 0.8333333333 4.1666666667 5 0.4166666667 0.5
2 0.3472222222 1.8333333333 0.5952380952 0.4117647059 0.4166666667 0.175070028 1.0084033613 0.0175070028 0.1008403361
3 0.0964506173 2.1805555556 0.1335470085 0.432132964 0.2777777778 0.0221961787 0.855529512 0.0022196179 0.0855529512
4 0.0200938786 2.2770061728 0.0253817414 0.4343316753 0.2083333333 0.0029010774 0.8362344108 0.0002901077 0.0836234411
5 0.0033489798 2.2971000514 0.0040187757 0.434571213 0.1666666667 0.0003492888 0.8336826222 0.0000349289 0.0833682622
6 0.0004651361 2.3004490312 0.000540158 0.4345956968 0.1388888889 0.000037863 0.8333711963 0.0000037863 0.0833371196
7 0.0000553733 2.3009141673 0.0000628562 0.4345979946 0.119047619 0.0000036915 0.8333370248 0.0000003692 0.0833337025
8 0.0000057681 2.3009695406 0.0000064388 0.4345981918 0.1041666667 0.0000003254 0.8333336587 0.0000000325 0.0833333659
9 0.0000005341 2.3009753087 0.0000005886 0.4345982073 0.0925925926 0.0000000261 0.8333333594 0.0000000026 0.0833333359
10 0.0000000445 2.3009758428 0.0000000486 0.4345982084 0.0833333333 0.0000000019 0.8333333353 0.0000000002 0.0833333335
11 0.0000000034 2.3009758873 0.0000000036 0.4345982085 0.0757575758 0.0000000001 0.8333333335 0 0.0833333333
12 0.0000000002 2.3009758906 0.0000000003 0.4345982085 0.0694444444 0 0.8333333333 0 0.0833333333
13 0 2.3009758909 0 0.4345982085 0.0641025641 0 0.8333333333 0 0.0833333333
14 0 2.3009758909 0 0.4345982085 0.0595238095 0 0.8333333333 0 0.0833333333
15 0 2.3009758909 0 0.4345982085 0.0555555556 0 0.8333333333 1.34352048387924E-16 0.0833333333
16 0 2.3009758909 0 0.4345982085 0.0520833333 6.51218686419213E-17 0.8333333333 6.51218686419213E-18 0.0833333333
17 1.26721499130873E-16 2.3009758909 1.33253535168546E-16 0.4345982085 0.0490196078 2.98514163203532E-18 0.8333333333 2.98514163203532E-19 0.0833333333
18 5.86673607087373E-18 2.3009758909 6.15152908402294E-18 0.4345982085 0.0462962963 1.29778811625998E-19 0.8333333333 1.29778811625998E-20 0.0833333333
19 2.57312985564637E-19 2.3009758909 2.69116333526318E-19 0.4345982085 0.0438596491 5.36502185461152E-21 0.8333333333 5.36502185461152E-22 0.0833333333
20 1.07213743985266E-20 2.3009758909 1.1187521111506E-20 0.4345982085 0.0416666667 2.11394636204157E-22 0.8333333333 2.11394636204157E-23 0.0833333333
21 4.25451365020895E-22 2.3009758909 4.43031999939114E-22 0.4345982085 0.0396825397 7.95623609441516E-24 0.8333333333 7.95623609441516E-25 0.0833333333
22 1.61155820083672E-23 2.3009758909 1.67500537409801E-23 0.4345982085 0.0378787879 2.86596194812096E-25 0.8333333333 2.86596194812096E-26 0.0833333333
23 5.83897898853885E-25 2.3009758909 6.05848947682979E-25 0.4345982085 0.0362318841 9.89852884544816E-27 0.8333333333 9.89852884544816E-28 0.0833333333
24 2.02742325990932E-26 2.3009758909 2.10035215415067E-26 0.4345982085 0.0347222222 3.28348663103548E-28 0.8333333333 3.28348663103548E-29 0.0833333333
25 6.75807753303108E-28 2.3009758909 6.9911146893425E-28 0.4345982085 0.0333333333 1.04769859291578E-29 0.8333333333 1.04769859291578E-30 0.0833333333
26 2.16605049135612E-29 2.3009758909 2.23777401755996E-29 0.4345982085 0.0320512821 3.22030655322929E-31 0.8333333333 3.22030655322929E-32 0.0833333333
27 6.68534102270406E-31 2.3009758909 6.89824997247171E-31 0.4345982085 0.0308641975 9.54766585945925E-33 0.8333333333 9.54766585945925E-34 0.0833333333
28 1.98968482818573E-32 2.3009758909 2.05071810512395E-32 0.4345982085 0.0297619048 2.73386016760705E-34 0.8333333333 2.73386016760705E-35 0.0833333333
29 5.71748513846475E-34 2.3009758909 5.88664150350809E-34 0.4345982085 0.0287356322 7.56900547795275E-36 0.8333333333 7.56900547795275E-37 0.0833333333
30 1.58819031624021E-35 2.3009758909 1.6335671824185E-35 0.4345982085 0.0277777778 2.02841534558582E-37 0.8333333333 2.02841534558582E-38 0.0833333333
1. Both l and m must be RATES, and use the same time unit. For example, given a service time such as 10 minutes per customer, convert it to a service rate such as 6 per hour. 2. The total service rate (rate x servers) must be greater than the arrival rate.

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