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Section 4-9/Erlang and Gamma Distributions 139

pp. 1290–1300)] presented a screening analysis model of breast of healthcare workers to reduce the hazard rate of inl uenza virus
cancer based on data from India. In this analysis, the time that infection for patients in regular hospital departments. In this anal-
a breast cancer case stays in a preclinical state is modeled to be ysis, each patient’s length of stay in the department is taken as
exponentially distributed with a mean depending on the state. exponentially distributed with a mean of 7.0 days.
For example, the time that a cancer case stays in the state of T1C (a) What is the probability that a patient stays in hospital for
(tumor size of 11–20 mm) is exponentially distributed with a less than 5.5 days?
mean of 1.48 years. (b) What is the probability that a patient stays in hospital for
(a) What is the probability that a breast cancer case in India more than 10.0 days if the patient has currently stayed for
stays in the state of T1C for more than 2.0 years? 7.0 days?
(b) What is the proportion of breast cancer cases in India that (c) Determine the mean length of stay such that the probability
spend at least 1.0 year in the state of T1C? is 0.9 that a patient stays in the hospital less than 6.0 days.
(c) Assume that a person in India is diagnosed to be in the state 4-136. An article in Ad Hoc Networks [“Underwater Acous-
of T1C. What is the probability that the patient is in the tic Sensor Networks: Target Size Detection and Performance
same state six months later? Analysis” (2009, Vol.7(4), pp. 803–808)] discussed an under-
4-134. Requests for service in a queuing model follow a Pois- water acoustic sensor network to monitor a given area in an
son distribution with a mean of i ve per unit time. ocean. The network does not use cables and does not interfere
(a) What is the probability that the time until the i rst request is with shipping activities. The arrival of clusters of signals gen-
less than 4 minutes? erated by the same pulse is taken as a Poisson arrival process
(b) What is the probability that the time between the second with a mean of λ per unit time. Suppose that for a specii c
and third requests is greater than 7.5 time units? underwater acoustic sensor network, this Poisson process has a
(c) Determine the mean rate of requests such that the probabil- rate of 2.5 arrivals per unit time.
ity is 0.9 that there are no requests in 0.5 time units. (a) What is the mean time between 2.0 consecutive arrivals?
(d) If the service times are independent and exponentially distrib- (b) What is the probability that there are no arrivals within 0.3
uted with a mean of 0.4 time units, what can you conclude time units?
about the long-term response of this system to requests? (c) What is the probability that the time until the i rst arrival
4-135. An article in Vaccine [“Modeling the Effects of Inl uenza exceeds 1.0 unit of time?
Vaccination of Health Care Workers in Hospital Departments” (d) Determine the mean arrival rate such that the probability is
(2009, Vol.27(44), pp. 6261–6267)] considered the immunization 0.9 that there are no arrivals in 0.3 time units.

4-9 Erlang and Gamma Distributions

An exponential random variable describes the length until the i rst count is obtained in a Pois-
son process. A generalization of the exponential distribution is the length until r events occur
in a Poisson process. Consider Example 4-23.

Example 4-23 Processor Failure The failures of the central processor units of large computer systems are often
modeled as a Poisson process. Typically, failures are not caused by components wearing out but
by more random failures of the large number of semiconductor circuits in the units. Assume that the units that fail are
immediately repaired, and assume that the mean number of failures per hour is 0.0001. Let X denote the time until four
failures occur in a system. Determine the probability that X exceeds 40,000 hours.
Let the random variable N denote the number of failures in 40,000 hours of operation. The time until four failures
occur exceeds 40,000 hours if and only if the number of failures in 40,000 hours is three or less. Therefore,
( ,000) = ( 3)
P N ≤
P X > 40
The assumption that the failures follow a Poisson process implies that N has a Poisson distribution with
E N ( ) = 40 ,000 . (0 0001 ) = 4 failures per 40 ,000 hours
Therefore, − k
4
( ,000) = ( 3) = ∑ e 4 = .
3
P N ≤
P X > 40 0 433
k = 0 k!
The previous example can be generalized to show that if X is the time until the rth event in a
Poisson process, then r −1 − x λ x λ ( ) k
(
P X > x) = ∑ e (4-17)
k = 0 k!