final assignments

D1_Answer

D.1 Customers arrive at Rich Dunn’s Styling Shop at a
rate of 3 per hour, distributed in a Poisson fashion. Rich can perform
haircuts at a rate of 5 per hour, distributed exponentially.
a) Find the average number of customers waiting for haircuts.
b) Find the average number of customers in the shop.
c) Find the average time a customer waits until it is his or her
turn.
d) Find the average time a customer spends in the shop.
e) Find the percentage of time that Rich is busy.
Arrival rate	λ = 4 customers/hour
Service rate	µ = 6 customers/hour
a)Average number of customers waiting for haircuts.
Lq =	1.3333333333
b) Average number of customers in the shop
Lq =	2
c) Average time a customer waits for his or her turn
Wq =	0.3333333333
d) Average time a customer spends in the shop
Ws =	0.5
e) Percentage of time that Rich is busy
ρ =	67%	of the times

𝐿_𝑞=〖(4)〗^2/(6(6−4))=16/(12 ) 𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟𝑠

𝐿_𝑠=4/((6−4))=4/(2 ) 𝑐𝑢𝑠𝑜𝑡𝑚𝑒𝑟𝑠

𝑊_𝑞=4/(6(6−4))= 4/(12 ) ℎ𝑜𝑢𝑟𝑠

𝑊_𝑠=1/((6−4))=1/(2 ) ℎ𝑜𝑢𝑟𝑠

𝜌=𝜆/𝜇=4/6=0.67

D3_Answer

D.3 Paul Fenster owns and manages a chili-dog and softdrink
stand near the Kean U. campus. While Paul can service 30
customers per hour on the average (m), he gets only 20 customers
per hour (l). Because Paul could wait on 50% more customers
than actually visit his stand, it doesn’t make sense to him that he
should have any waiting lines.
Paul hires you to examine the situation and to determine
some characteristics of his queue. After looking into the problem,
you find it follows the six conditions for a single-server waiting
line (as seen in Model A). What are your findings?
First, we will determine the average time a customer spends in the system
The average time a customer spends in the system is Ws
Ws =	0.1
Hence, the average time a customer spends in the system is 0.1 x 60 = 6 minutes
Next, we will determine the average number of customers waiting in the queue. Lq
Lq =	1.333
Thus, the average number of customers waiting in the queue is 1.33
The average time a customer spends waiting in the queue, Wq
Wq =	0.067
Hence, the average time a customer spends waiting in the queue is 0.067 hours
Now, we will determine the probability of the system being engaged, ρ
ρ =	67%
Hence, the probability of being engaged is 0.67

D6_Answer

D.6 Calls arrive at Lynn Ann Fish’s hotel switchboard at
a rate of 2 per minute. The average time to handle each is 20 seconds.
There is only one switchboard operator at the current time.
The Poisson and exponential distributions appear to be relevant
in this situation.
a) What is the probability that the operator is busy?
b) What is the average time that a customer must wait before
reaching the operator?
c) What is the average number of calls waiting to be answered?
Arrival rate	λ = 2 customers/minute
120 customer per hour
Service rate	µ = 20 seconds/ customer
3 customer per minute
180 customers per hour
a) Let ρ be the probability that the operator is busy
Thus, the probability that operator is busy is 0.66
b) Average time a customer must wait Wq
Wq =	0.011
Hence, the average time a customer must wait is 0.067 hours
c) To determine the average number of calls waiting to be answered is
Hence, the average number of calls waiting to be answered is 1.33

D8_Answer

D.8 Virginia’s Ron McPherson Electronics Corporation
retains a service crew to repair machine breakdowns that occur
on average l = 3 per 8-hour workday (approximately Poisson in
nature). The crew can service an average of m = 8 machines per
workday, 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 broken machine?
c) How many machines are waiting to be serviced at any given
time?
d) What is the probability that more than 1 machine is in the
system? The probability that more than 2 are broken and
waiting to be repaired or being serviced? More than 3? More
than 4?
Machine break down rate is the arrival rate
Arrival rate	λ = 3
Service rate	µ = 8
a) The utiization rate of the service system
ρ =	0.375
b) The average down time for the broken machine is Ws
Ws =	0.2	day
Thus, the average downtime for the broken machine 0.2 day
c) The number of machines waiting to be serviced at any given time is Lq
Thus, the number of machines waiting to be serviced at any given time is 0.225
d) Let ρ be the probability of machine being serviced
Probability of when more than 1 machine is serviced = 1 -ρ0
Therefore, the probability when more than 1 machine is to be serviced is 37.5%
The probability of when more than 2 machines are to be serviced = 1 -ρ0 - ρ1
Calculating ρ1
Thus, the probability of when more than 2 machines are to be serviced is 14.1%
The probability of when more than 3 machines are to be serviced = 1 -ρ0 - ρ1 - ρ2
Calculating ρ2
Thus, the probability of when more than 3 machines are to be serviced is 5.39%
The probability of when more than 4 machines are to be serviced = 1 -ρ0 - ρ1 - ρ2 - ρ3
Calculating ρ3
Thus, the probability of when more than 4 machines are to be serviced is 2.1%