Developing the Binomial Distribution
3. Occurrences are independent; that is, an occurrence in one interval does not influence the probability of an occurrence in another interval
We can derive the equation for computing Poisson probabilities directly from the bi-nomial probability distribution by taking the mathematical limits as P S 0 and n S`.
With these limits, the parameter l = nP is a constant that specifies the average number of occurrences (successes) for a particular time and/or space. We can see intuitively that the Poisson is a special case of the binomial obtained by extending these limits. However, the mathematical derivation is beyond the scope of this book. The interested reader is referred to page 244 of Hogg and Craig (1995). The Poisson probability distribution function is given in Equation 4.21.
168 Chapter 4 Discrete Probability Distributions
The sum of Poisson random variables is also a Poisson random variable. Thus, the sum of K Poisson random variables, each with mean l, is a Poisson random variable with mean Kl.
Two important applications of the Poisson distribution in the modern global economy are the probability of failures in complex systems and the probability of defective products in large production runs of several hundred thousand to a million units. A large world-wide shipping company such as Federal Express has a complex and extensive pickup, classification, shipping, and delivery system for millions of packages each day. There is a very small probability of handling failure at each step for each of the millions of packages handled every day. The company is interested in the probability of various numbers of failed deliveries each day when the system is operating properly. If the number of actual failed deliveries observed on a particular day has a small probability of occurring, given proper targeted operations, then the management begins a systematic checking process to identify and correct the reason for excessive failures.
The Poisson Distribution Function, Mean, and Variance The random variable X is said to follow the Poisson distribution if it has the probability distribution
P1x2 = e-llx
x! , for x = 0, 1, 2, . . . (4.21) where
P1x2 = the probability of x successes over a given time or space, given l l = the expected number of successes per time or space unit, l 7 0 e > 2.71828 1the base for natural logarithms2
The mean and variance of the Poisson distribution are mx = E3X4 = l and s2x = E 31X - mx224 = l
Example 4.10 System Component Failure (Poisson Probabilities)
Andrew Whittaker, computer center manager, reports that his computer system expe-rienced three component failures during the past 100 days.
a. What is the probability of no failures in a given day?
b. What is the probability of one or more component failures in a given day?
c. What is the probability of at least two failures in a 3-day period?
Solution A modern computer system has a very large number of components, each of which could fail and thus result in a computer system failure. To compute the probability of failures using the Poisson distribution, assume that each of the millions of components has the same very small probability of failure. Also assume that the first failure does not affect the probability of a second failure (in some cases, these assumptions may not hold, and more complex distributions would be used). In par-ticular, for this problem we assume that the past 100 days have been a good standard performance for the computer system and that this standard will continue into the future.
From past experience the expected number of failures per day is 3/100, or l = 0.03.
a. P1no failures in a given day2 = P1X = 0 u l = 0.032 = e-0.03l0
0! = 0.970446
4.5 Poisson Distribution 169 The Poisson distribution has been found to be particularly useful in waiting line, or queuing, problems. These important applications include the probability of various num-bers of customers waiting for a phone line or waiting to check out of a large retail store.
These queuing problems are an important management issue for firms that draw custom-ers from large populations. If the queue becomes too long, customcustom-ers might quit the line or might not return for a future shopping visit. If a store has too many checkout lines, then there will be personnel idle waiting for customers, resulting in lower productivity.
By knowing the probability of various numbers of customers in the line, management can balance the trade-off between long lines and idle customer service associates. In this way the firm can implement its strategy for the desired customer service level—shorter wait times imply higher customer-service levels but have a cost of more idle time for checkout workers.
b. The probability of at least one failure is the complement of the probability of 0 failures:
P1X Ú 12 = 1 - P1X = 02 = 1 - ce-llx
x! d = 1 - ce-0.03l0 0! d
= 1 - e-0.03 = 1 - 0.970446 = 0.029554
c. P1at least two failures in a 3@day period2 = P1X Ú 2 u l = 0.092, where the average over a 3-day period is l = 310.032 = 0.09 :
P1X Ú 2 u l = 0.092 = 1 - P1X … 12 = 1 - 3P1X = 02 + P1X = 124
= 1 - 30.913931 + 0.0822544 and, thus,
P1X Ú 2 u l = 0.092 = 1 - 0.996185 = 0.003815
Example 4.11 Customers at a Photocopying Machine (Poisson Probability)
Customers arrive at a photocopying machine at an average rate of 2 every five minutes.
Assume that these arrivals are independent, with a constant arrival rate, and that this problem follows a Poisson model, with X denoting the number of arriving customers in a 5-minute period and mean l = 2. Find the probability that more than two custom-ers arrive in a 5-minute period.
Solution Since the mean number of arrivals in five minutes is 2, then l = 2. To find the probability that more than 2 customers arrive, first compute the probability of at most 2 arrivals in a five-minute period, and then use the complement rule.
These probabilities can be found in Table 5 in the appendix or by using a computer:
P1X = 02 = e-220
0! = e-2 = 0.135335 P1X = 12 = e-221
1! = 2e-2 = 0.27067 P1X = 22 = e-222
2! = 2e-2 = 0.27067
Thus, the probability of more than 2 arrivals in a five-minute period is as follows:
P1X 7 22 = 1 - P1X … 22 = 1 - 30.135335 + 0.27067 + 0.270674 = 0.323325
170 Chapter 4 Discrete Probability Distributions
Example 4.12 Ship Arrivals at a Dock
The Canadian government has built a large grain-shipping port at Churchill, Manitoba, on the Hudson Bay. Grain grown in southern Manitoba is carried by rail to Churchill during the open-water shipping season. Unfortunately the port is open only 50 days per year during July and August. This leads to some critical crew staffing decisions by management. The port has the capacity to load up to 7 ships simultaneously, provided that each loading bay has an assigned crew. The remote location and short shipping season results in a very high labor cost for each crew assigned, and management would like to minimize the number of crews. Ships arrive in a random pattern that can be modeled using the Poisson probability model. If a ship arrives and all available loading bays are filled, the ship will be delayed, resulting in a large cost that must be paid to the owner of the ship. This penalty was negotiated to encourage ship owners to send their ships to Churchill.
Results of an initial analysis indicate that each ship requires six hours for loading by a single crew. The port can remain open only 50 days per year, and 500 ships must be loaded during this time. Each additional crew costs $180,000, and each boat delay costs $10,000. How many crews should be scheduled?
Solution The final decision is based on the probability of ship arrivals during a 6-hour period and the cost of additional crews versus the penalty cost for delayed ships. The first step is to compute the probabilities of various numbers of ships arriving during a 6-hour period and then the cost of ship delays. Then, we compute the cost of crews and the cost of ship delays for various levels of crew assignment.
Ship arrivals can be modeled by assuming that there are thousands of ships in the world and each has a small probability of arriving during a 6-hour loading period.
An alternative model assumption is that during six hours there are a large number of small time intervals—say, 0.1 second—in this case, 216,000 such intervals. We also need to assume that ships do not travel in convoys. With 500 ships arriving over 50 days, we have a mean of 10 ships per day, or l = 2.5 ship arrivals during a 6-hour period. The probability of x arrivals during a 6-hour period is computed using the following:
P1X = x u l = 2.52 = e-2.52.5x x!
If four crews are scheduled, the probabilities of delaying ships are as follows:
P1delay 1 ship2 = P15 ships arrive2 = e-2.52.55
5! = 0.0668 P1delay 2 ships2 = P16 ships arrive2 = e-2.52.56
6! = 0.0278 P1delay 3 ships2 = P17 ships arrive2 = e-2.52.57
7! = 0.0099 The probabilities of idle crews are as follows:
P11 crew idle2 = P13 ships arrive2 = e-2.52.53
3! = 0.2138 P12 crews idle2 = P12 ships arrive2 = e-2.52.52
2! = 0.2565 P13 crews idle2 = P11 ship arrive2 = e-2.52.51
1! = 0.2052 P14 crews idle2 = P10 ship arrive2 = e-2.52.50
0! = 0.0821
4.5 Poisson Distribution 171