Problem 1 R 6 customers per hour Rp 15 customer per minute or 6015 12hour U R Rp 612 05 a How long does a customer stay in the processor with the server Tp 5 minutes ID: 131623
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Slide1
Example: The average arrival rate to a GAP store is 6 customers per hour. The average service time is 5 min per customer.
Problem 1
R = 6 customers per hourRp =1/5 customer per minute, or 60(1/5) = 12/hour U= R/Rp = 6/12 = 0.5
a) How long
does a
customer stay in the processor (with the server)?
Tp
= 5 minutes
b) On average how many customers are
there
with the server?
RTp
=
Ip
=
6(5/60)
= 0.5
Alternatively;
Ip
=
cU
=1(0.5) = 0.5Slide2
Problem 2
What if the arrival rate is 11 per hour? Processing rate is still
Rp=12U= R/Rp U=11/12a) How long does a customer stay in the processor (with the server)?Tp = 5 minutesb) On average how many customers are there with the server?RTp = Ip = (11/60)(5) = 11/12AlternativelyIp
=
cU
= 1(11/12) = 11/12Slide3
A local GAP store on average has 10 customers per hour
for the checkout line. The store has two cashiers
. The average service time for checkout is 5 minutes . Problem 3Arrival rate: R = 10 per hourAverage inter-arrival time: Ta = 1/R = 1/10 hr = 6 min
Average service time:
Tp
= 5 min
Number of servers:
c =2
Rp
= c/
Tp
= 2/5 per min or 24 per hour
U= R/
Rp
= 10/24 = 0.42Slide4
Problem 3
a
) How long does a customer stay in the processors (with the servers)?Average service time: Tp = 5 minb) On average how many customers are there with the servers?Ip =?RTp = Ip = (1/12)(10) = 0.84 Alternatively
Ip
=
cU
= 2(0.42) = 0.84Slide5
A call center has 11 operators. The average arrival rate of calls is 200 calls per hour. Each of the operators on average can serve 20 customers per hour. U = 200/220 =0.91.
Problem 4
a) How long does a customer stay in the processors (with the servers)?Average service time: Tp = 60/20 = 3 min
b) On average how many customers are there
with the servers?
Ip
=?
RTp
=
Ip
=
(200/60)(3) = 10
Alternatively
Ip
=
cU
= 11(200/220) = 10