University of Illinois Spring 2007
ECE 413: Solutions to Problem Set 7
1. The reliability of multiple connections
The probability of total outage on a given day is pk, so the probability of total outage at least one
day during a year is 1 −(1 −pk)365. For p= 0.001, this gives values 0.3059 for k= 1, 0.000365 for
k= 2, and 3.65 e-07 for k= 3. We find that k= 2 suffices (even for a leap year).
2. Failure rate function for discrete random variables
(a) Since {T=k}⊂{T≥k}, the definitions of conditional probability and pmf yield λT(k) =
P(T=k|T≥k) = P(T=k)
P(T≥k)=pT(k)
P∞
l=kpT(l). Or, equivalently, λT(k) = pT(k)
1−Pk−1
l=1 pT(l). (In particular,
λT(1) = pT(1). Note that λT(k) is only well defined if P(T≥k)>0.)
(b) By definition, λT(kf) = 1 if and only P(T=kf) = P(T≥kf)>0. So λT(kf) = 1 if and only
if P(T > kf) = P(T≥kf)−P(T=kf) = 0 and P(T=kf)>0. Thus, the condition λT(kf)=1
means that Thas a largest possible value, and that value is kf.
(c) For 1 ≤k≤n,λT(k) = pT(k)
1−Pk−1
l=1 pT(l)=
1
n
1−k−1
n
=1
n−k+1 . (That is, the first nvalues of λTare
1
n,1
n−1, . . . , 1
1.)
(d) One solution is to first derive a general formula. Note that
P(T≥k+ 1) = P({T≥k+ 1}∩{T≥k})
=P(T≥k+ 1|T≥k)P(T≥k)
= (1 −P(T=k|T≥k))P(T≥k)
= (1 −λT(k))P(T≥k) (1)
Starting with the initial condition P(T≥1) = 1, (1) can be used to solve for P(T≥k) for all k.
The result is P(T≥k) = (1 −λT(1))(1 −λ2(T)) · · · (1 −λT(k−1)) for k≥1. Therefore,
P(T=k) = P({T≥k}∩{T=k})
=P(T≥k)P(T=k|T≥k) = P(T≥k)λT(k)
= (1 −λT(1)) · · · (1 −λT(k−1))λT(k).
In particular, if λT(k) = pfor all k≥1, then pT(k) = (1 −p)k−1pfor k≥1. In other words, Thas
the geometric distribution with parameter p.
(e) Fix k≥0. Consider conditional probabilities given the event {T≥k}. Equivalently, consider
conditional probabilities given that both links are still working at time k. Then the conditional
probability that the network fails at time k, is the conditional probability that the first link fails at
time k, plus the conditional probability that the second link fails at time k, minus the conditional
probability that both links fail at time k. That is λT(k) = λ1(k) + λ2(k)−λ1(k)λ2(k). (Note: If
λ1(k) and λ2(k) are both very small, then λT(k)≈λ1(k) + λ2(k).)
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