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IEOR 4106: Introduction to Operations Research: Stochastic Models
Spring 2011, Professor Whitt, March 10
- Money Withdrawn from an ATM Machine
Customers arrive at an automated teller machine (ATM) at the times of a Poisson process with a rate of λ = 10 per hour. Suppose that the amount of money withdrawn on each transaction has a mean of $30 and a standard deviation of $20.
(a) Find the mean and variance of the total amount of dollars withdrawn in 8 hours.
(b) What is the approximate probability that the total amount of money withdrawn in the first 8 hours exceeds $3, 400?
(c) How do the answers change if the Poisson arrival process is a nonhomogeneous Poisson process witharrival rate function λ(t) = 4t, t ≥ 0?
———————————————————————- (a) Assuming that successive withdrawals are IID, this is a compound Poisson process; see Section 5.4.2. Let X(t) be the total amount withdrawn in the time interval [0, t]. Let N (t) be the number of customers to come to the ATM in the interval [0, t]. Let Yn be the amount of the nth^ withdrawal. Then X(t) can be represented as the following random sum of random variables
X(t) =
N ∑ (t)
i=
Yi.
Hence, from §5.4 (p. 346 in the last edition),
E[X(t)] = λtE[Y 1 ] and var(X(t)) = λtE[Y 12 ] , (1)
so that
E[X(8)] = 10 × 8 × 30 = 2400 and var(X(t)) = 10 × 8 × ((30^2 + (20)^2 ) = 104, 000.
The standard deviation is
But why are those the correct formulas? To see why, look at Examples 3.10 and 3.17 in Chapter 3. We discuss the harder variance formula. Let
Y =
∑^ N
i=
Xi ,
where N is a nonnegative-integer-valued random variable and Xi are IID random variables. (We are using new notation here.) Then, by the conditional variance formula in Proposition 3.1, V ar(Y ) = E[V ar(Y |N )] + V ar(E[Y |N ]) = E[N ]V ar(X 1 ) + E[X 1 ]^2 V ar(N ).
In the Poisson-process case, with rate λ,
E[N (t)] = V ar(N (t)) = λt.
That yields formula (1) above.
(b) Use a normal approximation. It can be justified by applying the central limit theorem, because the stochastic process {X(t) : t ≥ 0 } has stationary and independent increments and the summands all have finite mean and variance. We can think of
X(t) =
∑^ n
i=
[X(k/n) − X((k − 1)/n)],
which is the sum of i.i.d. random variables. Hence,
P (X(t) > 3400) = P
X(t) − E[X(t)] std(X(t))
3400 − E[X(t)] std(X(t))
≈ P
N (0, 1) >
3400 − E[X(t)] std(X(t))
≈ P
N (0, 1) >
3400 − 2400]
≈ P (N (0, 1) > 3) ≈ 0. 0013. (2)
(c) The random variable N (t) is still a Poisson random variable, but we need to calculate the new mean. In formulas (1), replace λt = 10 × 8 = 80 by
∫ (^8)
0
λ(s) ds =
0
4 s ds = 128.
Then (1) is modified to become
E[X(t)] =
0
λ(s) dsE[Y 1 ] and var(X(t)) =
0
λ(s) dsE[Y 12 ] , (3)
so that E[X(8)] = 128E[Y 1 ] and var(X(t)) = 128E[Y 12 ] ,
The method from here is the same as in parts (a) and (b).
E[X(8)] = 128 × 30 = 3840 and var(X(t)) = 128 × ((30^2 + (20)^2 ) = 166, 400.
The standard deviation is 407.92.
P (X(t) > 3400) = P
X(t) − E[X(t)] std(X(t))
3400 − E[X(t)] std(X(t))
≈ P
N (0, 1) >
3400 − E[X(t)] std(X(t))
≈ P
N (0, 1) >
3400 − 3840]
≈ P (N (0, 1) > − 1 .1) = P (N (0, 1) < 1 .1) = 0. 8643 (4)
- Columbia Space Company, from 2005 midterm exam
———————————————————————-
- Typographical Errors, Exercise 5.62 on p. 363.
First, by (b), p 1 =
1 + E E[[XX^23 ]]
E[X 3 ]
(E[X 2 ] + E[X 3 ]
Hence we can estimate p 1 by X 3 /(X 2 +X 3 ). Thus p 1 is estimated by the fraction of error found by proofreader 2 that are also found by proofreader 1. Similarly (just change the labels!), we can estimate p 2 by X 3 /(X 1 + X 3 ).
The total number of errors found has mean
E[X 1 + X 2 + X 3 ] = E[X 1 ] + E[X 2 ] + E[X 3 ] = λ(1 − (1 − p 1 )(1 − p 2 )) ,
so that
E[X 1 + X 2 + X 3 ] = λ(1 − (1 − p 1 )(1 − p 2 )) = λ(1 −
E[X 2 ]E[X 1 ]
(E[X 2 ] + E[X 3 ])(E[X 1 ] + E[X 3 ])
Hence we can estimate λ by
λˆ = (X^1 +^ X^2 +^ X^3 ) (1 − (^) (X 2 +XX 32 )(XX^11 +X 3 ) )
(d) Give an estimate of E[X 4 ] the expected number of errors not found by either proof- reader.
——————————————————————— Note that E[X 4 ] = λ − (E[X 1 ] + E[X 2 ] + E[X 3 ]) ,
so we can estimate E[X 4 ] by
λˆ − (X 1 + X 2 + X 3 ) = (X 1 + X 2 + X 3 )( Z (1 − Z)
where
Z ≡
X 2 X 1
(X 2 + X 3 )(X 1 + X 3 )
(e) Suppose X 1 = 60, X 2 = 30 and X 3 = 40. What is the estimated distribution of X 4?
——————————————————————— As stated above, we know that X 4 has a Poisson distribution. We estimate its mean E[X 4 ] by
130(
(e) Now we contemplate hiring the third proofreader who independently finds errors with probability q = .9. How much do we reduce the expected number of uncovered errors by using this third proof reader.
———————————————————————
The number of undiscovered errors before using this new proofreader is X 4. The number of undiscovered errors after using this new proofreader is Poisson distributed with mean E[X 4 ](1− q). We anticipate that the a proportion (1 − q) of the remaining errors will remain. We will expect to delete E[X 4 ]q errors. That is, we will discover a proportion q of the remaining errors if we use the third proof reader. We expect to reduce the number of undiscovered errors from 45 to 4.5.
———————————————————————
