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Urban Operations Research Prepared by Michael D. Metzger

Fall 2004 Problem Set 1 Solutions 9/27/04

1. LO Problem 2.3 (Ingolfsson, 1993; Kang, 2001)

Anyone who arrives at the transfer station observes two independent Poisson processes, A and B. λAEach arrival of the combined Poisson process comes from process A (B) with probability λA+λB

( λB ), independently of all other arrivals. Thus the combined Poisson process has an embedded λA+λB

Bernoulli process. To solve this problem, you need a good grasp of the fundamental properties of

Poisson and Bernoulli processes. If you feel uncomfortable with the answers below, now is a good

time to review Poisson and Bernoulli processes, for example, by reading Chapter 4 of “Fundamentals

of Applied Probability Theory” by Alvin Drake.

(a) (i) The times between successive A train arrivals are independent exponential random vari­

ables with parameter λA = 3/hour. By the memoryless property, the fact that Bart

arrives at a random time has no relevance. The time he has to wait until the next A

train is still an exponential random variable with parameter λA. So if we call his waiting

time X, then the PDF of X is given by

−3xfX(x) = λAe −λAx = 3e , x ≥ 0.

This is in fact a random incidence question. So we can also solve this question by using

formula (2.65) in the textbook. Let Y be the interarrival time of A trains, then

1 − FY (x)fX(x) = .

E[Y ]

x1E[Y ] = λ1 A

= 3 . FY (x) = P{Y ≤ x} = λAe−λAydy = 1 − e−λAx = 1 − e−3x . Hence 0

−3x)1 − (1 − efX(x) = 1 = 3e −3x , x ≥ 0.

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(ii) An easy way to answer this question is to first translate it into a Bernoulli process that

one is familiar with, for example, a sequence of coin tosses. An A train then becomes a

“head” and a B train becomes a “tail”. The problem now reads: What is the probability

that one obtains at least 3 tails before a head is obtained? This probability is the same

as the probability that the next three tosses result in tails (i.e. the next three trains

are B trains). The outcomes of the fourth and subsequent tosses are irrelevant for the

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purposes of answering this question. Thus,

P{at least 3 B trains arrive} = P{next 3 trains are B trains} � �3 � �3 � �3λB 6 2

= = = . λA + λB 3 + 6 3

Another way to answer this question is to consider the complementary event. Let NB

be a random variable denoting the number of B trains that arrive while Bart is waiting.

Then

P{NB ≥ 3} = 1 − P{NB = 0} − P{NB = 1} − P{NB = 2} � �2 � �3λA λB λA λB λA 2

= 1 − − · − = . λA + λB λA + λB λA + λB λA + λB λA + λB 3

(iii) In order for exactly 3 B trains to arrive while Bart is waiting, the next 3 trains should

be B trains and the fourth one should be an A train.

P{exactly 3 B trains} = P{next 3 trains are B trains, the 4th train is an A train} � �3 � �3 � �

λB λA 2 1 = = .

λA + λB λA + λB 3 3

(b) The combined Poisson process of train arrivals has an arrival rate of λ = λA + λB = 9/hour.

The probability of having exactly 9 arrivals during any hour (t = 1) in this process is obtained

by using the Poisson PMF formula.

−λt 99 −9(λt)9e ePK(9) = = � 0.1318.

9! 9!

(c) To answer this question, one needs to consider another Bernoulli process associated with the

combined Poisson process. In this Bernoulli process, each hour is an independent trial and

an hour is a success if exactly 9 trains arrive during that hour. From (b), a success occurs

with probability PK(9) � 0.1318. The question asks what the expected number of trials until

the first success is. The answer is the expected value of a geometric random variable with

parameter PK(9), i.e.

1 1 E[number of hours until exactly 9 trains per hour] = � = 7.59.

PK(9) 0.1318

(d) (i) An A train will be delayed if the time, denoted by Z, from the arrival of the A train to

the moment of the arrival of the next B train is less than 30 seconds. By the memoryless

property, this time has an exponential PDF with parameter λB. Thus the probability

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that an A train is delayed is obtained by

� (30 secs)(1 hr/3600 secs) PD = P{Z ≤ 30 seconds} = λBe −λBt dt

0 � 1/120

−1/20 = 6e −6t dt = 1 − e . 0

Since probabilities equal long-run frequencies, this is also approximately the fraction of

A trains that are delayed over some reasonably long period, say a month.

(ii) A B train passenger that benefits from the delay policy does not have to wait at all for

an A train, so her expected waiting time under the policy is zero. Without the policy, 1her mean waiting time would be λ

1 A

= = 20 minutes. Therefore such a passenger’s 3

mean waiting time reduction is 20 minutes.

Next, consider a passenger in an A train. With probability PD computed in (i), the

passenger’s travel time will increase by the amount of time that the A train waits for a

B train. Note that the mean increase in travel time for a passenger in an A train that

is held for a B train is equal to E[Z | Z ≤ 1/120]. To compute this quantity, we invoke

the total expectation theorem:

E[Z] = E[Z | Z ≤ 1/120]P{Z ≤ 1/120} + E[Z | Z > 1/120]P{Z > 1/120}.

1 1Since E[Z] = λ1 B

and E[Z|Z > 1/120] = 120 + , we have λB

1/λB = E[Z | Z ≤ 1/120]PD + (1/120 + 1/λB)(1 − PD).

Rearranging terms,

1/λB − (1/120 + 1/λB)(1 − PD)E[Z | Z ≤ 1/120] =

PD

� 0.00413 hours = 14.9 seconds.

Therefore the mean increase in travel time for an A train passenger is

E[Z | Z ≤ 1/120] × PD = 14.9 seconds × (1 − e −1/20) = 0.73 seconds.

One might attempt to obtain the mean increase in travel time (i.e. the expected waiting

time, denoted by E[W ]) for an A train passenger as follows:

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E[W ] = E[W | Z ≤ 1/120]PD + E[W | Z > 1/120](1 − PD)

? 1 1 = · × (1 − e −1/20) + 0 .

2 120

= ·This is not correct because E[W | Z ≤ 1/120] � 1 1 120 . The restriction, Z ≤ 1/120,2

cuts off the right tail of the exponential density curve, fW (w), but what is left is NOT

uniform.

Let us assume that it never happens that two or more A trains are delayed waiting for

the same B train, i.e. we ignore the possibility that two A trains may arrive within

30 seconds of each other. Under this assumption, one A train is held for every B train

receiving the benefits. Therefore the policy will lead to a net global travel time reduction

if

E[total time reduction] > E[total time increase]

1 ⇒ E[NBA] × hours > E[NA] × 0.00413 hours

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⇒ E[NBA] > 0.012 × E[NA],

where NBA is the number of people on a B train who wish to transfer to an A train and

NA is the number of people on an A train being held. In words, the policy is favored if

the average number of people on a B train who wish to transfer is at least 1.2% of the

average number of people on an A train.

This is a condition that one would expect to hold true for most subway transfer stations.

You might want to think about the effect of ignoring the possibility of two A trains

arriving within 30 seconds of each other. How would one assess the reasonableness of

this simplification? How would the analysis change if one did not make this simplifying

assumption?

3.1 (Pinker, 1995)

(a) Let Ji, i = 1, 2 be the event that Jones patrols beat i, let Ei, i = 1, 2 be the event that Elmer

burglarizes beat i, and let S be the event that Elmer and Jones work on the same beat. Then

Pr(S) = Pr(S|J1) Pr(J1) + Pr(S|J2) Pr(J2)

= Pr(E1|J1) Pr(J1) + Pr(E2|J2) Pr(J2)

= (1/3)(1/2) + (1/3)(1/2) = 1/3

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� �

� �

� �

(b) Let XE and XJ be Elmer and Jones’s positions during the burglary. We are told that these

two random variables are independent and both have a uniform distribution over [0, l]. Let

W = |XE −XJ |. We are asked to compute Pr{W > l/2|W > l/4}. The two events of interest

are shown in the sample space for (XE , XJ) below. Since (XE , XJ) is uniformly distributed

over [0, l]2, the desired probability is the ratio of the areas of the events {W > l/2}∩{W > l/4}

and {W > l/4}. From the picture of the sample space,

Pr({W > l/2} ∩ {W > l/4}) l2/4 4 Pr{W > l/2|W > l/4} = = =

Pr{W > l/4} 9l2/16 9

Figure 1: Sample Space

(c) In (b) we computed the areas corresponding to Pr{W > l/2} and Pr{W > l/4}. By general­

izing the argument, we can compute

Pr{W > w} = (1/l2)(2)(1/2)(l − w)2 = (1 − w/l)2 for 0 ≤ w ≤ l

The cdf is FW (w) = Pr{W ≤ w} = 1 − Pr{W > w} = 1 − (1 − w/l)2 and the pdf is

d d fW (w) = FW (w) = (1 − (1 − w/l)2) = (2/l)(1 − w/l) for 0 ≤ w ≤ l

dw dw

(d) The conditional expectation rule applies to probabilities of events also. This is easily seen by

letting IA be an indicator random variable which takes on the value one if event A occurs

and zero otherwise. Then E[IA] is seen to equal PA. As a result,

� l

PA = Pr(A|w)fW (w)dw0

� d � l

= Pr(A|w)fW (w)dw + Pr(A|w)fW (w)dw 0 d

� d � l

= (1 − w/d)(2/l)(1 − w/l)dw + (0)(2/l)(l − w/l)dw 0 d

� d

= (2/l) (1 − (1/l + 1/d)w + (1/dl)w 2)dw 0

= (2/l) d − (1/l + 1/d)d2/2 + (1/dl)d3/3

= (2/l) d − d2/(2l) − d/2 + d2/(3l)

= d/l − (1/3)d2/l2

= (d/l)(1 − (1/3)(d/l))

When d = 0, then Jones will apprehend Elmer only if they are at exactly the same location,

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� �

which is an event of probability zero. Accordingly, PA = 0 when zero is substituted for d.

When d = l is substituted, PA = 2/3 is obtained. Note that this is equal to Pr(A|E[W ]) =

Pr(A|l/3) = 1 − (l/3)/l = 2/3. This happens because when d = l, then Pr(A|w) is a linear

function of w, say aw + b. Then

� l

PA = Pr(A|w)fW (w)dw 0

� l

= (aw + b)fW (w)dw 0 � l � l

= a wfW (w)dw + b fW (w)dw 0 0

= aE[W ] + b

= Pr(A|E[W ])

When d < l, then Pr(A|w) is piece-wise linear, i.e., non-linear over [0, l].

(e) Let S be the event that Elmer and Jones are on the same beat and A be the event that Jones

apprehends Elmer. Then

d d Pr(A) = Pr(A|S) Pr(S) + Pr(A| not S)(1 − Pr(S)) = PA(1/3) + 0 = 1 − ≡ p

3l 3l

Furthermore, event A is independent of whether Elmer has been apprehended on previous

nights, i.e., the process of Jones apprehending (success) or not apprehending (failure) Elmer

on successive nights is a Bernoulli process. In Bernoulli process terminology, we are asked

to compute the probability that the first success occurs on the third trial. This equals the

probability of two failures followed by a success, i.e.,

(1 − p)2 p = (1 − PA/3)2(PA/3)

(f) Let Ai and Si be the events that Elmer is apprehended on the i-th night and that Elmer and

Jones are on the same beat on the i-th night. Let IAi and ISi

be indicator variables for those

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�

� �

events. Then we want

Pr

�

10 �

1

ISi = 3|

10 �

1

IAi = 0

�

= Pr{

�10 1 IAi

= 0| �10

1 ISi = 3}Pr{

�10 1 ISi

= 3}

Pr{ �10

1 IAi = 0}

= (1 − PA)3

�

10 3

�

(1 − Pr(S))7 Pr(S)3

(1 − PA/3)10

(1 − PA)3(120)(2/3)7(1/3)3

= (1 − PA/3)10

= 15, 360 (1 − PA)3

(3 − PA)10

(g) As before, Pr(A) = Pr(A|S)(1/3), and

� l

Pr(A|S) = Pr(A|w)fW (w)dw 0

Given that Elmer is to Jones’s left, W will be uniformly distributed over [0, l/2]. The same is

true if Elmer is to the right of Jones. Therefore, unconditionally, W is uniformly distributed

over [0, l/2]. First, suppose that d < l/2. Then

� d

Pr(A|S) = (1 − w/d)(2/l)dw = d/l 0

Second, suppose that d ≥ l/2:

� l/2

Pr(A|S) = (1 − w/d)(2/l)dw = 1 − (1/4)(l/d) 0

Therefore, d if d < l/2

Pr(A) = �

3l �

1 1 − l if d ≥ l/23 4d

As in part (d), Pr(A) = 0 if d = 0. If d = l, Pr(A) = (1/3)(1− 1/4) = 1/4, which agrees with

1 1 1 l/4 1 Pr(A|E[W ]) = Pr(A|l/4) = 1 − =

3 3 3 l 4

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Problem 3 (Metzger 2004)

PART A:

Let V=Number of left to right pedestrians crossing on a green light.

We can divide the quantity into two parts: W and Z

Let V=W+X+Z

Where:

W=The probability the first pedestrian to arrive after a green light is a left to right

X= The probability the second pedestrian to arrive after a green light is a left to right

Z=The number of left to right pedestrians to arrive during the Q min between the second

arrival(X) and the green light.

Thus we can take the expectation of each side:

E[V]=E[W+X+Z]

We can split the above up using the additive property of expectations as follows:

E[V]=E[W]+E[X]+E[Z]

As shown in class for case 3, we can write W as follows:

W =

λL

λ .. . .. 0 R p w

λ + λR L

What the above is saying is the probability the first arriving pedestrian is a left to right is the

proportion of arrivals that are left to right. Thus the expected value for W:

λ λ λ W E ] = L 1 * + R 0 * = L[

λ + λ λ + λ λ + λR L R L R L

Since arrivals are independent, based on the properties of a Poisson process(see chapter 2 for

more information) we can write an expression for X in the same manor

X =

λL

λ .. . .. 0 R p w

λ + λR L

¤ ŒŒ‹ ŒŒ›

ŒŒÕ

¤ ŒŒ‹

À p w .. . .. 1

λ + λR L

ŒŒÃ

Àp w .. . .. 1

λ + λR L

ŒŒÃ

8

ŒŒ›

ŒŒÕ

We can similarly write the expectation out in the same fashion.

λ λ λ X E ] = L 1 * + R 0 * = L[

λ + λ λ + λ λ + λR L R L R L

We know from class that Z is has a Poisson PDF, with time parameter Q and can be written as

follows.

(λ Q) z

−λ QZ P = z) = e(

z!

Thus the expected value of Z is:

Z E ] = λ Q[

Thus we can now sum these three quantities to find E[V]

λ λ 2λ [ LV E ] = L + + λ Q = L + λ Q

λ + λ λ + λ λ + λR L R L R L

Part B:

In this part we need to find the prob V=0. Based on our above equation V=W+X+Z, the

probability V=0 is the probability that each of V’s components are zero. Thus we can write this

mathematically as:

( ( ( (V P = )0 = W P = * )0 X P = * )0 Z P = )0

The P(W=0) is simply the probability that the first pedestrian to arrive after a green light is a right

to left pedestrian. Thus:

λ W P = )0 = R(

λ + λR L

Since arrivals are independent the probability that X=0 is equal to the probability that W=0. Thus:

λ X P = )0 = R(

λ + λR L

Lastly the probability that ZX=0, is the probability that we have no arrives in the Q time units

after ‘X’ arrives. Thus we can plug z=0 into the above formula for the distribution of Z to find

this quantity.

Z P = )0 = e −λQ(

Thus we can now find the probability that V=0

9

C

2

≈ λR ’ −λQ÷÷V P = )0 =( eΔΔ

«λ + λR L ◊

Part C:

WE next need to find the probability density for the time between green lights. Clearly this time

we can again call A is composed of three intervals. The time until the first pedestrian to

arrive(call this time B), plus the time between the first and the second pedestrian to arrive(call

this time C), plus Q time unites after the second pedestrian arrives.

Thus:

A=B+C+Q.

We know that the arrivals are a merged Poisson process, and thus from basic prob we know that

the inter-arrival times of a Poisson process are exponentially distributed. Thus we can write the

distributions of B and C as follows:

− tf B ) ( = λe λ t

tf ) ( = λe −λt

Thus in order to calculate the time distribution of V we need to sum two exponentially random

variables plus a constant. From probability we know the sum of two exponential random

variables is an earlang random variable of order two. Two see this once can convolve the two

distributions. Lastly we need to account for the Q time units after the second pedestrian arrives

thus we add a shift factor of Q by replacing every t with a t-Q in the earlang distribution. Thus

we get:

2 tf ) ( = λ (t − Q)e −λ (t−Q)

A

Part D: We need to calculate the expected amount of time a random pedestrian waits for a green light.

There are three types of pedestrians.

Type 1: Those who are the first to arrive after a green light

Type 2: Those who are the second to arrive after a green light

Type 3: Those who arrive during the Q time units between the second arrival and the green light.

Based on part A we can calculate the probability a random pedestrian is of each type.

During each interval we have one type one, one type to, and we expect on average λQ Type 3’s.

Thus using the total probability theorem:

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1 P(Type )1 =

2 + λ Q

1P(Type )2 =

2 + λ Q

λ QP(Type )3 =

2 + λ Q

Thus now we need to calculate the expected wait time for a random arriver in each group. Let us

start with type 3. Pedestrians who are in type 3 are uniformly distributed over the last Q time

Q units between each green light. Thus on average a pedestrian in this class waits, . A pedestrian

2 in type two waits always exactly Q time units till the light turns green. A pedestrian in type one

waits the time between the 1st

and second arrival plus Q time units thus on average the time

1 between arrivals in a passion process is , plus the Q time units after the second pedestrian

λ 1

arrives until the light turns green, thus his average waiting time is + Q . λ

Thus the expected waiting time of a random pedestrian is simply the weighted average of these.

E[RandomPedestrian ] =≈Δ «

’+≈Δ «

’ λ Q1 1 1 Q( Q)+ Q÷◊

÷◊

+ λ + λ Q 2 + λ Q + λ Q2 2 2

Part E:

In order to calculate the expected wait time for a random observer we need to use the following

formula derived in the book (eq. 2,66) and in class.

Let Z be the expected wait time of a random observer:

[ xZ E ] =σ x

2

+µ

2µ 2 x

Let E[A] be the time till the first pedestrian after a green.

Let E[B} be the time between the first and second pedestrians after a green.

Since expectations are additive we know:

1 1 2 [ [µ = A E ] + B E ] + Q = + + Q = + Q

λ λ λ

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x

We can similarly find σ x

2.

1Let var( A) = (Poisson Property)

λ2

1Thus var( B) =

λ2

Since the time between the second arrival and the next green is a constant it has a variance of

zero.

Thus since we can add variances:

1var( A) =

λ2

1 1 2σ 2 = var( A) + var( B) = + = x

λ2 λ2λ2

Thus if we plug these two equations into the equation for E[Z] we obtain the desired quantity.

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