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Unformatted text preview: IEOR4106 Intro to OR  Stoch Models DavidD. Yao. Fall 2011 Midterm Examination (Part I; 75 minutes) a. Please return this sheet together with yoyr solutions. All problems are equally weighted. 1. Suppose X relates to Y as follows: given Y = y, X follows a Poisson distribution with
mean y. Suppose ,u :2 E(Y) and a2 := Var(Y) are knoWn. Derive E(X), Var(X) and
Cov(X, Y), i.e., express them in terms of the known quantities. 2. A coin shows head with probability p on each flip. (i) Flip the coin n times; let X be the number of heads. Next7 ﬂip the coin X times, let
Y be the number of heads. Let W := X + Y be the total number of heads (over the two
rounds of ﬂipping). Derive (ii) Flip the coin until the ﬁrst time a head shows; let X be the corresponding number of
ﬂips. Next, as in (i), ﬂip the coin X times and let Y be the number of heads. Let W := 1+Y
be the total number of heads; Derive Is it equal to or larger/ smaller than E(W) in
(i), assuming 71 = 1? (iii) Derive Var(WS) in (i). 3. A system of two parallel servers serve a. stream of jobs that arrive at the system following
a Poisson process of rate A. Each job requires an independent exponential service time with
mean 1 / u from either server. Any arriving job that ﬁnds both servers occupied will be
blocked and lost. For 72 = 1, 2, let T2 denote the time until the system empties starting from
1' jobs in the system. Derive T... David D. Yao, Fall 2011 IEOR4106 Intro to OR — Stoch Models Midterm Examination (Part II; 75 minutes) Q Please return this sheet together with your solutions. All problems are equally weighted. 4. X and Y are two independent exponential random variables with mean 1//\ and 1/;1,,
respectively. (i) Derive P(X Z :rlX < Y), for a: 2 0. What distribution does the conditional
random variable W := (X IX < Y) follow? Give your answer an intuitive interpretation. (ii) Are the following two expressions correct? If yes, explain; if no, provide a correction. E[X  1(X < Y)] = E(X)  P(X < Y), E[XIX < Y) = E(X)  P(X < Y).
L/
X
Note: 1( denotes the indicator function. 5. (i) Passengers arrive at a. train station following a Poisson process with rate A. The train
leaves at a given time t > 0, and all waiting passengers get on board. Let W denote the
total waiting time of all passengers on board. Derive E(W) and Var(W). (ii) Suppose the train leaves at a time T that is uniformly distributed over [0,0], where
a > 0 is given. Derive E(W).
(iii) Derive Var(W) in (ii). 6. Let 5,, be the sum of n independent rolls of a fair die. For a given integer k > 1 (e.g.,
k = 17), let 1),, := P(Sn is divisible by k). Derive the limit 1) := limn_.oo p" by going through
the following steps. (i) To be speciﬁc, consider k = 4. Let X" be the remainder of 3,, divided by k = 4. (Hence,
the possible values for Xn are 0, 1,2, 3; with X1, = 0 corresponding to 8,, being divisible by
k —_: 4.) Clearly, {Xn} is a. Mark0v chain. Write down its onestep transition matrix P, and
derive its stationary distribution 7r. (ii) Repeat (i) for k = 7.
(iii) From the above analysis, what can you conclude about p; specifically, what. is p for
k = 17? ' I: L 1. Solution. We know E(X'IY) z Var(XlY) = Y. Hence, E(X) .: E[E(XY)] =3E(Y), and
Var(X) = Var[E(XY)] + E[Var(XY)] = Var(Y) + E(Y).
Similarly,
E(XY) = E[E(XYY)] = E{YE(XlY)] = E(Y2),
whereas E(X) = E(Y). Hence, Cov(XY) = E(XY) — E(X)E(.Y) = Var(Y). 2. Solution. We have E(YIX) 2 X1) and E(X) = np; hence E(W) = E(X) + E(Y) =
up + np2. h (ii) Here, E(X) = 117; hence, E(Y) = 1, and E(W) 2 1+%, which is larger than E(W) = 1+p
in with up = 1. Only when p = 1, the two are equal. (iii) We have ElVar(WIX)l = E[0+Xp(1—p)l = mom—p), Var[E(WX)l = VarlX(1+p)l = (1+p)2np(1p),
Hence, Var(W) = np(1 — p)[p + (1 + p)2]. 3. Solution. Conditioning on the next event being an arrival or a service completion, we have
E(T)——1 +E(T) A
'1 /\+p 2A+u Clearly, due to blocking, T2 must ﬁrst reach a service completion, i.e., E(Tg) :2 + E(Tl).
Hence, 1 ,\ 1
E T = —— + —— — + E T ,
leaders P9,.E,(,T1 ,Euﬁ,,.+.ﬁzi_and E(T2) = % + 4. Solution. Since P[X > atX < Y] = P[X > :c,X < Y]/P[X < Y], and P[X > $.X < Y] 2/ ((3—AI — e_>””),ue_“ydy = A
I /\ +p +00
e—(A+u):t. 7 dividing the above by P[X < Y] = ,1", we have P[X > xlX < Y] = e(HWI. That is, W :2 (XIX < Y) follows an exponential distribution with mean 1/(A +11). Intuitively, this
result says that given X < Y, X follows the same distribution as min(X, Y). Hence,
1 1 A (i.e.. the second expression is correct); and the ﬁrst one is wrong, the correction is:
 /\ EX1X<Y :EXX<YPX<Y:—.
i < )1 l l ) < > (My)? 5. Solution. Conditioning on N(t) = n, the 71 passengers arrive i.i.d. uniformly over
[0,t]. Hence, E(W) = E[N(t)§] 2 L51, and ~
' 2 t /\t3 M3 _ M3 Var(l/V) = E[Var(WN(t))] +Var[E(l/V1Y(t)~).]':' = 1‘3 ; (ii) Let T denote the time the train leaves. Condition upon T = t, the results are already
derived in Hence, E(l/V) = E[E(WT)] = E[)\T2/2] = Aa2/6. Furthermore,
,\ 3 A2 4 Var(W) = E[Var(WT)] + Var[E(WT)] = E[§T3] + Var[—;:T2] = Ea, + Ea , where we note E(T3) = f8 t3dt 2 % and Var(T2) = E(T“) _ E2(T2) = 6. Solution. {Xn} is a doubly stochastic M.C., regardless of what is k. For instance, when
k = 4, consider pij for j = 2, the four transition probabilities associated with 2' = 0, 1, 2, 3
are: %, g, %, %, respectively. Hence, the stationary distribution of the MC. is uniform
7r, : l/k: for all i = 0, 1, . . . , k— 1. Consequently, p = we = l/k, and p = 1/17 when k: = 17. u ,' , at "am 4 N ._, LAM“... r‘z I: it». w: ,. ...
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 Summer '12
 Stochastic
 Poisson Distribution, Exponential distribution, var, Poisson process, David D. Yao, Stoch Models

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