Assignment #3: Queuing Systems |
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Course: |
CEG 4183 |
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Name: |
Jim Sellers |
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Student #: |
2030251 |
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Due: |
2003.02.04 |
1) [2 marks] What is the utilization of an
M/M/1 queue that has four (4) people waiting on average?
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ω |
= |
ρ2 / (1 ρ) |
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4 |
= |
ρ2 / (1 ρ) |
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4 * (1 ρ) |
= |
ρ2 |
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4 4ρ |
= |
ρ2 |
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ρ2 + 4ρ - 4 |
= |
0 |
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Now using the Quadratic Equation, I solved for p to get values of aprox. 0.828 and 4.828. Since you cant have a negative utilization, the answer is aprox. 0.828
2) [6 marks] At
an ATM machine in a supermarket, the average length of a transition is 2
minutes, and on average, customers arrive to use the machine once every 5
minutes.
a) How long is the average time that a person must spend waiting and using the machine?
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ρ |
= |
λ / μ |
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= |
2 / 5 |
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= |
0.4 |
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Tw |
= |
ρ Ts / (1 - ρ) |
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Tw |
= |
(0.4) Ts / (1 (0.4)) |
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Tw |
= |
(0.4) * (5 min) / (0.6) |
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Tw |
= |
3.333 min |
b)
What is the 90th percentile of residence time?
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mq(r) |
= |
Tw / p * ln(10p) |
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mq(90) |
= |
3.333 / 0.4 * ln(10 * 0.4) |
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mq(90) |
= |
11.55 minutes |
c)
On average, how many people are waiting to use the
machine? Assume M/M/1.
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ω |
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ρ2 / (1 ρ) |
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(2/5) / (1 2/5) |
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2/5 / 3/5 |
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2/3 |
On average, there is 2/3 of a person waiting to use the machine.
3) [4 marks] Messages
arrive at a switching center for a particular outgoing communication line in a
Poisson manner with a
mean arrival rate of 180 messages per hour. Message length in distributed
exponentially with a mean length of 14,400 characters. Line speed is 9600 bps.
a) What is the mean waiting time in the switching center?
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λ |
= |
180 messages / hour * 1 hour / 3600 seconds |
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= |
0.05 messages / second |
Assume 8 bits per character.
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Ts |
= |
(14 400 chars) * (8 bits / char) / 9600 bps |
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12 seconds |
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ρ |
= |
λ Ts |
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(0.05 msg / s) * (12 s) |
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= |
0.6 messages |
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Tq |
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Ts / (1 ρ) |
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= |
12 s / (1 0.6) |
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30 seconds |
b)
How many messages will be waiting in the switching center
for transmission on the average?
Assume in switching centre means both the queue and the message being processed.
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q |
= |
p / (1 p) |
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= |
0.6 / 0.4 |
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= |
1.5 |
4) [8 marks]
Refer to the attached pages. Calculate
p, Tw, and mtw(90) and mtw(95)
for M/M/3, and 5 M/M/1s. Correct steps are needed.
For 5 M / M / 1s:
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p |
= |
λ Ts / number of workstations-queues |
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5 (1/8) / 5 |
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0.125 |
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Ts |
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½ hour use / 8 hours |
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1/16 |
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Tw |
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(p Ts ) / (1 - p) |
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(1/8 * 1/16) / (1 0.125) * (8 hours / day * 60 min / hour) |
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= |
4.29 min / day |
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mtw(y) |
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Tw / p * ln(100p / 100 y) |
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mtw(90) |
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4.29 / 0.125 * ln(10 * 0.125) |
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= |
7.650 (aprox) |
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mtw(y) |
= |
Tw / p * ln(100p / 100 y) |
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mtw(95) |
= |
4.29 / 0.125 * ln(20 * 0.125) |
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= |
31.416 (aprox) |
For M / M / 3:
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p |
= |
λ Ts / N |
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= |
5 (1/8) / 3 |
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5 / 24 |
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K |
= |
Sum( (Np)i / i! ) from i=0 to n-1 Sum( (Np)i / i! ) from i=0 to n |
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= |
1 + (5/24) + (5/24)^2 / 2 1 + (5/24) + (5/24)^2 / 2 + (5/24)^3 / 3! |
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0.998 776 297 |
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C |
= |
(1 K) / (1 pK) |
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= |
(1 - 0.998 776 297) / (1 (5/24) * 0.998 776 297) |
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= |
0.001 545 232 |
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Tw |
= |
C / N * Ts / (1 - p) |
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= |
0.001 545 232 / 3 * 5 / (1 5 / 24) |
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= |
0.003 253 12 * (8 hours / day * 60 min / hour) |
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= |
1.56 min / day |
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mtw(y) |
= |
Ts / (N p) * ln(100C / 100 y) |
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mtw(90) |
= |
5 / (3 5 / 24) * ln(10 * 0.001 545 232) |
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= |
- 7.469 (aprox) |
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mtw(y) |
= |
Ts / (N p) * ln(100C / 100 y) |
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mtw(95) |
= |
5 / (3 5 / 24) * ln(20 * 0.001 545 232) |
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= |
- 6.227 (aprox) |
The negative numbers mean that 90% of the time, you will not wait at all.