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Answers to the exam ISP of December 19, 2005 1. Counting time in hours
 

Summary: Answers to the exam ISP of December 19, 2005
1. Counting time in hours:
a. Taking t = 2 in P(N(t) = 20) = e-10t
(10t)2
0/20! yields 0.089
b. P(N(0.01) 1) = 1 - e-0.1
, or P(X 0.01) where X is an interarrival time,
hence exponentially distributed with parameter 10.
c. For the remaining interarrival time X we have EX = 1/10 hours, so 6 minutes
after 12.00 hours. Hence at 12.06 hours.
d. Let p = P(job has > 3 pages) = 1-e-2
(1+2+2+4/3) 0.143. Then {M(t)}
is a Poisson process with rate 10p, so P(M(t) = m) = e10pt
(10pt)m
/m!.
2. a. Given the present, say Xn = i, the process will jump to 0 or i + 1 with probabi-
lities that do not depend on Xn-1, Xn-2, . . .. Hence the process is a DTMC. It
is irreducible (the path 0, 1, 2, 3, 4, 5, 6, 0 has positive probability so all states
communicate), aperiodic (GCD(2,3,4,. . . )=1, so period of state 0, and hence all
other states, is 1), and not transient but recurrent (finite closed class).

  

Source: Al Hanbali, Ahmad - Department of Applied Mathematics, Universiteit Twente
Litvak, Nelly - Department of Applied Mathematics, Universiteit Twente

 

Collections: Engineering