Copy of Mod 4 Practice Problems (with solutions)
xlsx
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School
University of Wisconsin, Platteville *
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Course
4120
Subject
Industrial Engineering
Date
Jan 9, 2024
Type
xlsx
Pages
7
Uploaded by SuperHumanBook14421
Customers send emails to a help desk of an online retailer every two minutes (on average). The standard deviation of the inter arrival time is also 2 min. There are three employees answering emails and it takes an average of 4 min to write a response. The standard deviation of processing time is 2 min. How long does each customer wait to be served? Flow unit:
Email
Resource:
Help Desk employees
Process:
Customer service
Given data:
Ave. interarrival time (a)
2 mins
2 mins
Ave. processing time (p):
4 mins
2 mins
Number of servers (m):
3 emps
Solution:
Coef. of variations
1
0.5
exponent = sqrt ( 2 x ( m+1) ) -1 =
1.8284271
utilization = p/(a x m) =
66.67% 1.19 min
std dev of interarrival time (σ
a)
:
std dev of proc time (σ
p)
:
CVa=σ
a/
a
CV
p
=σ
p/
p
T
q =processing time/m x [ (utilization ^
exponent / (1-utilization)
] x
[(Cva^2 +
CV
p
^2) / 2] =
?_𝑞=((Processing time)/𝑚)×(Utilization^(√(2(𝑚+1) )−1)/
(1−Utilization))×((CV_𝑎^2+CV_𝑝^2)/2)
arrival rate
DVD Copy
1
st dev of inter
P
7 days
m
CVP
1
process time
Interarrival
10 days
stdev P utilization
70%
Utilization
CVA
CVP
16.3333333
exponent
1.82842712
Q6
Tq
22.3909687
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6 per/hr
0.1 per min
5 min
process time
3 people
interarrival
25 min
m
25 min
Cva
CVp
83%
0.625
0.5
exponent
1
2.16227766
tq
Q7
3.7702026
1 min
10 sec
6 per min
7 people
1.3
0.8
utilization
0.85748585
3
0.7362932
0.7362932
44.1775922 sec
You need to get your drivers license renewed. You must wait in line with your documentation at the DMV. You made the mistake of arriving on a Friday afternoon – one of the busiest times! On average there are 20 drivers arriving per hour, however there are 6 stations open serving drivers. The actual renewal process is relatively short at 15 minutes. (Note CVa and CVp are given at 1 each). Flow unit:
driver
Resource:
License renewal stations Process:
Driver License Renewal Process
What is the processing time (p)?
p =
15 minutes
What is the inter arrival time (a)?
Given: arrival rate (1/a) =
20 drivers/hr
interarrival time (a) =
0.05 hour
or
3 minutes
What is the flow rate (or throughput rate)?
Flow rate is same as arrival rate
20 drivers/hr
What is the average person’s time in queue? Given Data:
Coef. of variations
1
1
number of servers (m) =
6 stations
processing time (p) =
15 mins/driver
exponent = sqrt ( 2 x ( m+1) ) -1 =
2.74165739
2.741657
utilization = p/(a x m) =
83.33%
83.33%
9.10 min
9.099204
Cv
a
=
CV
p
=
T
q =[processing time/m] x [ (utilization ^
exponent / (1-utilization)
] x
[(Cva^2 +
CV
p
^2) / 2] =
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On average, how many people are waiting in line with their documents? Flow rate = 20 drivers/hr
or
0.3333 driver/min
Wait time (Tq)=
9.10 min
Littles law: I=RT
3.03 drivers
On average, how many total people are in the system? Flow rate = 20 drivers/hr
or
0.3333 driver/min
Total time in system = Tq + Tp =
24.10 minutes
This is the sum of waiting and processing times.
Littles law: I=RT
I =
8.03 drivers
I
q
=