IEOR 4404
Assignment #1 Solutions
Simulation
January 24, 2012
Prof. Mariana OlveraCravioto
Page 1 of 4
Assignment #1 Solutions
1.
(a) To compute the departure times of the 13 customers, we proceed as follows:
A
1
= 12
and
S
1
= 40. Thus, the first customer arrives at time
t
= 12 and stays in service until
time
t
= 52.
At
t
=
A
2
= 31, another customer arrives and joins the queue (which
now consists of exactly one customer), since she cannot be served immediately.
At
t
= 52, the server becomes empty again and customer 2 starts her service which ends at
t
= 52 + 32 = 84 at which time she leaves service. In the meantime, at
t
= 63, another
customer has arrived and joined the queue. Her time of entry into service is
t
= 84 and
subsequent departure is at
t
= 84 + 55 = 139 ... etc. Proceeding as such, the departure
times are:
52
,
84
,
139
,
187
,
205
,
255
,
302
,
320
,
348
,
402
,
451
,
527
,
549
(b) Paralleling the same logic as in part (a), we get that the departure times when there are
two servers in the system are:
52
,
63
,
118
,
143
,
136
,
204
,
245
,
239
,
332
,
400
,
451
,
527
,
549
(c) We now consider a difference service mechanism:
upon completion of service of the
customer in the server, the next customer to enter service is the one that has been
waiting the least time.
To compute the departure times, we need proceed exactly as
before, except we should also keep track of the waiting times of the customers waiting in
line, and have the next entry into service be that of the customer who has been waiting
the least amount of time. The departure times are given by:
52
,
84
,
139
,
320
,
157
,
207
,
254
,
272
,
348
,
402
,
451
,
527
,
549
2. In order to determine the departure times of these customers, we note that there is no queue in
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 Spring '10
 C
 Probability theory, probability density function, Cumulative distribution function, black balls, departure times

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