Single channel queuing model with Poisson arrivals and exponential service times
ID: 3158377 • Letter: S
Question
Single channel queuing model with Poisson arrivals and exponential service times (M/M/1).Queuing analysis. You need to develop the formulas on excel to derive your solutions to answer questions a through w. The following table provides the information for your decisions. Check all the correct statements. Please go to chapter slides, in the last 6 pages for references.
=arrival rate=#/per hour
2
µ=service rate=#/hours
3
k=variable arrivals greater than
hours per work day
8
Cw=waiting cost of customers, $/per hour, i.e., in terms of customer dissatisfaction and lost goodwill, i.e. is $10 per hour of time spent waiting in line (after customers are actually being serviced on the rack, customers do not seem to mind waiting). Because on the average a customer has a hour waiting and there are approximately 16 customer services per day (2 per hour times 8 working hours per day), the total number of hours that customers spend on waiting for services is * 16 = 32/3, or 10
$10
m, number of service channels
1
Cs=service cost per hour per channel
$7.00
Answer these questions.
a. L= /(µ- )= the average number of customers or units in the systems, the number in line plus the number being served, is
b. W=1/(µ- )=the average time a customer spends in the system, that is, the time spent in line plus the time spent being served is
c. Lq= 2/(µ(µ- )) or the average number of customers in the queue, is
d. Wq= /(µ(µ- ))=the average time a customer spends waiting in the queue, is
e. = /µ =the utilization factor for the system, (the Greek lowercase letter rho), that is, the probability that the service facility is being used, or percentage of time mechanic is busy, or the probability that the server is busy, is
f. Po=1- /µ= the percent idle time, that is, the probability that no one is in the system or probability that there are zero customer in the system, is
Below is for your reference
Pn>k=(/µ)k+1= the probability that the number of customers in the system is greater than k. k=variable arrivals. When k between 0-7, the corresponding probabilities are (below manually calculated)
k=
prob. for k=1 through 7
=(2/3)0+1=0.666
0
=(2/3)1+1=0.444
1
=(2/3)2+1=8/27=0.296
2
=(2/3)3+1=16/81=0.198
3
=(2/3)4+1=32/243=0.132
4
=(2/3)5+1=64/729=0.088
5
=(2/3)6+1=128/2187=0.058
6
=(2/3)7+1=256/6551=0.039
7
Answer these questions.
g. (*Wq)*Cw =total waiting cost, or (hours /per day )*(*Wq)*Cw, Wq= /(µ(µ- ))=the average time a customer spends waiting in the queue, i.e. 2/(3*(3-2))=. The total daily waiting cost, i.e. =8*2*()*$10=$106.67 is
h. (m*Cs)=Total daily service cost= or = (number of channels)*(cost per channel) is
i. (m*Cs)+ (*Wq)*Cw =total costs of the queuing system, or =total service cost + total waiting cost, or total costs of the queuing system, is
if a faster mechanic can install new parts at the rate of XX/per hour, and labor rate at $XX/per hour.
2
µ
4
j. L= the average number of customers or units in the systems, the number in line plus the number being served. L= /(µ- )=2/(4-2)=1 customer in the system on the average, is
k. W=the average time a customer spends in the system, that is, the time spent in line plus the time spent being served. W=1/(µ- )=1/(4-2)=1/2, 0.5 hour in the system on the average, is
l. Lq=the average number of customers in the queue. Lq = 2/(µ(µ- ))=22/(4*(4-2))=4/8=1/2 or 0.5 customers waiting in line on the average, is
m. Wq=the average time a customer spends waiting in the queue. Wq = /(µ(µ- ))=2/(4*(4-2))=1/4 hour= 15 minutes average waiting time per customer in the queue, is
n. =the utilization factor for the system, (the Greek lowercase letter rho), that is, the probability that the service facility is is being used. = /µ=2/4=1/2, percentage of time mechanic is busy, is
o. Po= the percent idle time, that is, the probability that no one is in the system. Po =1- /µ=1-2/4=1/2, probability that there are 0 customer in the system, is
Pn>k=(/µ)k+1, when =2, µ=4
k
when =2, µ=4
when =2, µ=4
0
(2/4)0+1=0.500
1
(2/4)1+1=0.250
2
(2/4)2+1=0.125
3
(2/4)3+1=0.063
4
(2/4)4+1=0.031
5
(2/4)5+1=0.016
6
(2/4)6+1=0.008
7
(2/4)7+1=0.004
p. Wq=average time a customer spends waiting in the queue= /(µ(µ- ))=2/(4*(4-2))=1/4, is
q. the total daily waiting cost= (8 hours /per day )*(*Wq)*Cw; (8 hours /per day )*(*Wq)*Cw =8*2*1/4*$10=$40.00/per day, is
r. Notice that the total time spent waiting for the 16 customers per days is now: = 8** Wq=8*2*1/4=4 hours, is
m, number of service channels
1
Cs=service cost per hour per channel
$9
s. (*Wq)*Cw =total waiting cost, or (hours /per day )*(*Wq)*Cw, Wq= /(µ(µ- ))=the average time a customer spends waiting in the queue, is
t. (m*Cs)=Total daily service cost= or = (number of channels)*(cost per channel), is
u. (m*Cs)+ (*Wq)*Cw =total costs of the queuing system, or =total service cost + total waiting cost, or total costs of the queuing system, is
v. the lower cost is
w. the saving is
L= /(µ- )= the average number of customers or units in the systems, the number in line plus the number being served, is $2.00
W=1/(µ- )=the average time a customer spends in the system, that is, the time spent in line plus the time spent being served is 1.00
Lq= 2/(µ(µ- )) or the average number of customers in the queue, is less than 1.33
Wq= /(µ(µ- ))=the average time a customer spends waiting in the queue, is 0.67
= /µ =the utilization factor for the system, (the Greek lowercase letter rho), that is, the probability that the service facility is being used, or percentage of time mechanic is busy, or the probability that the server is busy, is 0.67
Po=1- /µ= the percent idle time, that is, the probability that no one is in the system or probability that there are zero customer in the system, is more than 0.33
(*Wq)*Cw =total waiting cost, or (hours /per day )*(*Wq)*Cw, Wq= /(µ(µ- ))=the average time a customer spends waiting in the queue, i.e. 2/(3*(3-2))=. The total daily waiting cost, i.e. =8*2*()*$10=$106.67 is $106.67
(m*Cs)=Total daily service cost= or = (number of channels)*(cost per channel) is $66.00
(m*Cs)+ (*Wq)*Cw =total costs of the queuing system, or =total service cost + total waiting cost, or total costs of the queuing system, is not $126.67
L= the average number of customers or units in the systems, the number in line plus the number being served. L= /(µ- )=2/(4-2)=1 customer in the system on the average, is greater than 1.00
W=the average time a customer spends in the system, that is, the time spent in line plus the time spent being served. W=1/(µ- )=1/(4-2)=1/2, 0.5 hour in the system on the average, is 0.4
Lq=the average number of customers in the queue. Lq = 2/(µ(µ- ))=22/(4*(4-2))=4/8=1/2 or 0.5 customers waiting in line on the average, is 0.5
Wq=the average time a customer spends waiting in the queue. Wq = /(µ(µ- ))=2/(4*(4-2))=1/4 hour= 15 minutes average waiting time per customer in the queue, is 0.25
=the utilization factor for the system, (the Greek lowercase letter rho), that is, the probability that the service facility is is being used. = /µ=2/4=1/2, percentage of time mechanic is busy, is 0.5
Po= the percent idle time, that is, the probability that no one is in the system. Po =1- /µ=1-2/4=1/2, probability that there are 0 customer in the system, is 0.6
Wq=average time a customer spends waiting in the queue= /(µ(µ- ))=2/(4*(4-2))=1/4, is 0.350
the total daily waiting cost= (8 hours /per day )*(*Wq)*Cw; (8 hours /per day )*(*Wq)*Cw is $40.00
Notice that the total time spent waiting for the 16 customers per days is now: = hrs/day** Wq is 4.00
(*Wq)*Cw =total waiting cost, or (hours /per day )*(*Wq)*Cw, Wq= /(µ(µ- ))=the average time a customer spends waiting in the queue, is not 40 hours
(m*Cs)=Total daily service cost= or = (number of channels)*(cost per channel), is more than $72
(m*Cs)+ (*Wq)*Cw =total costs of the queuing system, or =total service cost + total waiting cost, or total costs of the queuing system, is $12.00
the lower cost is $112.00
the saving is ($50.67)
I just need to know which letters to choose
=arrival rate=#/per hour
2
µ=service rate=#/hours
3
k=variable arrivals greater than
hours per work day
8
Cw=waiting cost of customers, $/per hour, i.e., in terms of customer dissatisfaction and lost goodwill, i.e. is $10 per hour of time spent waiting in line (after customers are actually being serviced on the rack, customers do not seem to mind waiting). Because on the average a customer has a hour waiting and there are approximately 16 customer services per day (2 per hour times 8 working hours per day), the total number of hours that customers spend on waiting for services is * 16 = 32/3, or 10
$10
m, number of service channels
1
Cs=service cost per hour per channel
$7.00
Explanation / Answer
a. L= /(µ- )= the average number of customers or units in the systems, the number in line plus the number being served, is
L = 2 / (3-2) = 2
b. W=1/(µ- )=the average time a customer spends in the system, that is, the time spent in line plus the time spent being served is
W = 1 / (3-2) = 1
c. Lq= 2/(µ(µ- )) or the average number of customers in the queue, is
Lq = 22 / (3(3-2)) = 1.3333
d. Wq= /(µ(µ- ))=the average time a customer spends waiting in the queue, is
Wq = 2 / (3*(3-2)) = 0.6667
e. = /µ =the utilization factor for the system, (the Greek lowercase letter rho), that is, the probability that the service facility is being used, or percentage of time mechanic is busy, or the probability that the server is busy, is
= 2/3 = 0.6667
f. Po=1- /µ= the percent idle time, that is, the probability that no one is in the system or probability that there are zero customer in the system, is
P0 = 1 - 0.6667 = 0.3333
g. (*Wq)*Cw =total waiting cost, or (hours /per day )*(*Wq)*Cw,Wq= /(µ(µ- ))=the average time a customer spends waiting in the queue, i.e. 2/(3*(3-2))=. The total daily waiting cost, i.e. =8*2*()*$10=$106.67 is
total waiting cost = (2*0.6667) * 10 = 13.3333
h. (m*Cs)=Total daily service cost= or = (number of channels)*(cost per channel) is
Total daily service cost = 1*7 = 7
i. (m*Cs)+ (*Wq)*Cw =total costs of the queuing system, or =total service cost + total waiting cost, or total costs of the queuing system, is
(1*7)+(2*0.6667)*10 = 20.3333
if a faster mechanic can install new parts at the rate of XX/per hour, and labor rate at $XX/per hour.
= 2
µ = 4
j. L= the average number of customers or units in the systems, the number in line plus the number being served. L= /(µ- )=2/(4-2)=1 customer in the system on the average, is 1
k. W=the average time a customer spends in the system, that is, the time spent in line plus the time spent being served. W=1/(µ- )=1/(4-2)=1/2, 0.5 hour in the system on the average, is 0.5
l. Lq=the average number of customers in the queue. Lq = 2/(µ(µ- ))=22/(4*(4-2))=4/8=1/2 or 0.5 customers waiting in line on the average, is 0.5
m. Wq=the average time a customer spends waiting in the queue. Wq= /(µ(µ- ))=2/(4*(4-2))=1/4 hour= 15 minutes average waiting time per customer in the queue, is 15 minutes
n. =the utilization factor for the system, (the Greek lowercase letter rho), that is, the probability that the service facility is is being used. = /µ=2/4=1/2, percentage of time mechanic is busy, is 0.5
o. Po= the percent idle time, that is, the probability that no one is in the system. Po =1- /µ=1-2/4=1/2, probability that there are 0 customer in the system, is 0.5
p. Wq=average time a customer spends waiting in the queue= /(µ(µ- ))=2/(4*(4-2))=1/4, is 0.25
q. the total daily waiting cost= (8 hours /per day )*(*Wq)*Cw; (8 hours /per day )(*Wq)*Cw =8*2*1/4*$10=$40.00/per day, is 40.00
r. Notice that the total time spent waiting for the 16 customers per days is now: = 8** Wq=8*2*1/4=4 hours, is 4 hour.
= 2
µ = 4
j. L= the average number of customers or units in the systems, the number in line plus the number being served. L= /(µ- )=2/(4-2)=1 customer in the system on the average, is 1
k. W=the average time a customer spends in the system, that is, the time spent in line plus the time spent being served. W=1/(µ- )=1/(4-2)=1/2, 0.5 hour in the system on the average, is 0.5