BUSINESS ANALYTICS MODULE D WA I T I N G – L I N E MO D E L S 333
D.18 Arrival rate =
= 1 computer / 85 min. = 0.706 computers/hr.
Service rate
= 1 computer / 15 min. = 4 computers/hr.
D.19 Arrival rate =
= 1 repair / 6 days = 0.1667 repairs/day
Service rate
= 1 day
334 BUSINESS ANALYTICS MODULE D WA I T I N G – L I N E MO D E L S
(b)
2.08 0.83 2.91 drivers.
LL
= + = + =
The average
waiting is 12.5 seconds.
(d) Ws = Waiting time + Service time = 12.5 seconds +
(a)
The time in the system will rise by (30-17.5)/17.5
= 71.4 percent. The employee system seems to be bet-
ter overall.
D.22 M/M/S model:
(cont. in next column)
D.22 (cont.)
Wait cost = (0.65)($100) = $65; Server cost = 5 × 15 = $75
Total Cost = $140
So M = 6 servers is optimal, at a total cost
$109=
lowest cost
Total cost = $90 + $15 +$9.50 = $114.50
For M = 7, (.06)($50) = $3 = Wait cost
Total cost = $105 + $15 + $3 = $123
two salespeople.
Ls = 9 cars in the system. Using Little’s Law:
Wq = Lq /
= 1000 / 360 = 2.78 days
D.24
The manager’s calculations are as follows:
Number of Salespeople
(g) Total cost per shift
$420
$340
$360
$380
( ) 12(12 10) 24
q
L
= = = =
−−
lowest cost
BUSINESS ANALYTICS MODULE D WA I T I N G – L I N E MO D E L S 335
D.28 = 30 loans/week
Ws = 2.4 weeks. Using Little’s Law:
Ls = Ws = 30(2.4) = 72 loans
D.29 = $800,000/month
D.31*
40 2
(a) 60 3
= = =
= = =
−−
= = =
−−
==
−−
=
−−
==
22
(40) 4
(b) ( ) 60(60 40) 3
40
(c) 2
60 40
40
(d) ( ) 60(60 40)
0.033 hours = 2 minutes
1 1 1
60 40 20
40/hour, 60/hour
q
s
q
L
L
W
D.32*
==(a) 0.833
==
−
(b) 0.20833 days = 1.667 hours
()
q
W
(b) The average time a car waits before it is washed, Wq is
given by:
20
( ) 30(30 20)
20 0.0667 hours
q
W
==
−−
(e) The probability that no cars are in the system, P0, is
given by:
01 1 1 0.667 0.33P
= − = − = − =
(f) Use of automation (with
= 60/hour) will produce the
following:
0
22
20 0.083
2 ( ) 2 60(60 20)
20 0.0042 hours
2 ( ) 2 60(60 20)
10.0042 0.0167 0.0209 hours
20 0.333
60
1 0.667
q
q
L
W
P
= = =
− −
= = =
− −
= = =
= − =
D.35*
= 4, students/minute,
= 60/12 = 5 students/minute
(a) The probability of more than two students in the
336 BUSINESS ANALYTICS MODULE D WA I T I N G – L I N E MO D E L S
(c) The average waiting time, Wq, is given by:
40.8 minutes
( ) 5(5 4)
q
W
= = =
−−
(d) The expected number of students in the queue, Lq, is
( ) 5(5 4)
−−
(e) The average number of students in the system, Ls is
(d) The average number of students in the queue for the
qs
LL
=−
where:
0
2
( 1)!( )
s
MM
−−
Then:
2
BUSINESS ANALYTICS MODULE D WA I T I N G – L I N E MO D E L S 337
(c) The average time spent in the shop, Ws, is given by:
D.38* = 10 cars/hour,
= 12 cars/hour.
338 BUSINESS ANALYTICS MODULE D WA I T I N G – L I N E MO D E L S
1
2
CASE STUDIES
NEW ENGLAND FOUNDRY
under the proposed system, will give the savings in time.
Under the present system, there are two service servers with a
The average time a person spends in the system under the present
system is 0.392 hours, or 23.5 minutes.
Adding the “system time” to the travel times involved
(6 minutes total for casting personnel and 2 minutes for molding
personnel), the total trip takes:
Under the new system, waiting lines are converted to single server,
The time spent in Bob’s department is:
1 1 1
= hour = 30 minutes
s
W
==
The reduced travel time is equal to 2 minutes, making the total
trip time equal to 32 minutes. This is an increase in time of 2
minutes and 30 seconds for the maintenance personnel.
Pete can now service 7 people per hour (
= 7). Three people
AACSB: Analytical thinking
2. To evaluate systemwide savings, the times must be monetized.
For the casting personnel who are paid $9.50 per hour, the
2.5 minutes lost per trip costs the company $0.40 per trip (2.5 60
On average there are 4 arrivals per hour from casting, and 3 arri-
vals per hour from the molding department. Therefore, the hourly
= −
+
= − + =
4 trips $0.40/trip (casting)
3 trips $1.66/trip (molding)
1.60 4.98 $3.38 per hour
S
example, the cost of changing from the old layout to the new lay-
out could completely eliminate the advantages of operating the
new layout. In addition, there may be other factors, some non–
THE WINTER PARK HOTEL
1. The current system has five clerks each with his or her own
waiting line. This can be treated as five independent queues, each
that a guest spends waiting and checking in is given by:
are diverted to a quick-serve clerk who can register them in an
average of two minutes (
= 30 per hour), their average time in
clerks ( = 63/4 = 15.75 per hour), each of whose mean service
time is 3.4 minutes (or 0.5667 hours), so that
= 1/0.5667 =
17.65 per hour. The average time in the system for these guests
queuing system. With four servers, the average time in the system
is 8.9 minutes, resulting in an overall average of:
0.2 30 + 0.8 8.9 = 13.1 minutes
With five servers, the average time is 3.9 minutes resulting in an
2. Therefore, the single waiting line with five clerks is the better
option.
BUSINESS ANALYTICS MODULE D WA I T I N G – L I N E MO D E L S 339
ADDITIONAL CASE STUDY
(available in MyOMLab)
time in system. Some students may use system time, but because
most shoppers are relieved when it is their turn, we use waiting
better system. We also assume the M/M/S model.
We begin with a rough analysis (one that is going to have a
t = 0.2 (2 minutes) + 0.8 (4 minutes) = 0.4 + 3.2 = 3.6 minutes
This means that the average service rate is 60/3.6 = 16.67 custom-
Using an arrival rate of 100 and a service rate of 16.67, the mini-
mum number of servers is 6. (This is due to round off.) In reality,
the minimum number is 7, and the average waiting time is 2.2
minutes. Trying one more server leads to a waiting time of 0.64
minutes.
Now we separate the express and regular customers. Assume
that all express customers go into the express lane (even though
they can go into any lane) and assume that all nonexpress custom-
two servers yield an average wait of 0.25 minutes.
For the regular lane, with an arrival rate of 80 and a service
lanes and 1 in an express lane. If Beth uses 8 servers, a 6–2 split
between regular lanes and express lanes yields an average wait of:
(0.2)(4) + (0.8)(0.98) = 0.8 + 0.784 = 1.584 minutes
which is better. However, the express lane would be slower than