SOLUTIONS MANUAL for Stochastic Modeling: Analysis and ...

29
(b)
Ȳ 1 = {0(5 − 0) + 1(6 − 5) + 2(9 − 6) + 1(11 − 9)
+2(13− 11) + 3(15 − 13) + 2(16 − 15)
+1(18 − 16) + 0(19 − 18)}÷19
= 23 ≈ 1.2 customers waiting for service
19
Ȳ 2 =
(c) N 19 = 5 arrivals
An estimate of average delay is
0(3 − 0) + 1(19 − 3)
19
= 16
19 ≈ 0.84
utilization of the copier
∫ 19
0 Y 1,t dt
= 23 =4.6 minutes
N 19 5
9. Based on the available data, the sample-average interarrival-time gap was about 17.6
seconds, and might be modeled as exponentially distributed.
Let F G be the cdf of the interarrival-time gaps, which we assume to be independent
because students act independently.
Let ∆t be the pick-up interval, in seconds.
Let S n represent the number of students waiting after the nth state change.
Define the following system events and associated clocks:
e 0 () (start of day)
S 0 ← 0
C 1 ← FG
−1 (random())
C 2 ← ∆t
e 1 () (student arrives)
S n+1 ← S n +1
C 1 ← T n+1 + FG
−1 (random())
e 2 ()
(pick up)