finalsol_2005

finalsol_2005 - MS&E 221 Final Examination Ramesh...

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Unformatted text preview: MS&E 221 - ' Final Examination Ramesh Johari - March 14, 2005 Instructions . 1. Take altemate seating if possible. 2. Answer all questions in the spaces provided on these sheets. If needed, additional paper will be avail— able at the front of the room. Answers given on any other paper will not be counted. 3. You may use a calculator, but no notes or books. 4. The examination begins at'82'30 am, and ends at 11:30 am. Honor Code In taking this examination, I acknowledge and accept the Stanford University Honor Code. NAME (signed) ' gal/MTIONS . ' - NAME (printed) Useful Formulas 1. If T is an exponentially distributed random variable with mean 1/A, then the density of T is given by fT, where: . fT(t) = Ae’”, t > 0. 2. If S is a random variable with a gamma distribution of parameters n and A, where n is a positive integer, then the density of S is givenrby f3, where: ' ~As n—l h®=§%é%%+yszfl 3. For an M /M / 1 queue with arrival rate A and service rate ,u with A < p, the equilibrium distribution is: WQ=fi=U—MH,j=QLZm, where p = A/u. 4. For an M /M / oo queue with arrival rate A and service rate p, the equilibrium distribution is: . 6"" . I 1:0’1,2,"', where p = /\/,u. 5. For anM /M / K / K queue with arrival rate A and service rate ,u,’ the equilibrium distribution is: Q WM . P(Q=J)=Wa 9:0,1,2a---,Ka where p '= A/ PART A —' Short answer questions. (60 points; 12 points per question) 1. Either justify the following statement if true, or provide a counterexample if false: If a state 2' is recur- rent for a continuous time Markov chain Xt, then it is also recurrent for the jump chain corresponding t0 Xt. . (I ream/Mi" fly X; a; it an, ave 1w 4w: J 74,4m P(,X{.=l (yo‘l)=1- 2. Suppose that Xn- is a discrete time Markov chain on'the finite state space X = {1, . . . , K Assume that all states are aperiodic. Suppose also that this Markov chain has exactly two closed communicat— ing classes, A1 and A2 (although there may be other communicating classes that are not closed). For any two states 2' and j does the limit: ' - limIP(Xn='j|X0=z') = Lg. n—voo J always exist? Justify your answer. we mm remitted {to M .‘s ivveducilole M KM sled-c Space => ’ V‘V‘rqme ivwm‘awr distribution was, if?“ Haiti) M MD- = Wm Ak l X0=U k;=l-,Z glmce fifile alt/VA mo’r closed 3' rfmwflimi’r mm + mm = I lww" git/Ice i' #4? cbm‘m I .‘5 'apm‘l’df‘li, "fl )6 AU “"6” lag = NO 1mg); he )‘e A. mm L;)~= hzfihlwi‘); “MA . Cfi A7 UA’Z, 2 56 Hm Wit ADA/mgr exx‘g’g, .3. Consider an infinite server queue, where customers arrive according to a Poisson process of rate A. (Thus every arriving customer gets assigned his own server.) The service times of customers are independent, with expOnential distribution of mean 1/“. Assume that the queue starts in equilibrium at time —00; i.e., the number of customers in the queue is a two-sided stationary process. Suppose we note the times at which customers depart from the queue. What is the distribution of the length of time between successive departures? Justify your answer. The “MW/'00 (7W {5 rave-agitate. (I) Twat; ' 42mm rm: 6% mm at we a. (a) Wepar’rwes M rammed Pam: are 170 GEM a! rake} 3 ~ ( 3) {’MQS I)” "imam! [W'CC’B are Patron 0? raft-g R % Wmtcmy) . 2?? Lat/9H1 a? «Hm Latvia/w Successive Wat-Writ ~ cxfa (9‘). 4. Either justify the following statement if true, or provide a counterexample if false: A continuous time Markov chain is reversible if and only if the corresponding jump chain is reversible. (You should assume that the Continuous time Markov chain is irreducible, and that the state space is finite.) 4M1?- jnredotC-rtu'c ~“' thilc Sink 5/1066 5") [/L‘M4“?l/L[ ,4, , CTMC (“WWW-t Afi+n17w+7m T l W 9x1 : habit-Mg Faramgjer fi/ (Shiv ( m . . “ - - Z) 1r < * IWMMTL «w. .m- m J'qu Chm («M W t 'L k °° since rH/xe sink jffice «‘5 finale). AM; 10,), My (j/j; Tn. , 15.62,; [CT-MC (Mailed tawch <9 A; c. Tr). a). _ €=> . fig —— fiipjb [pm/1c 0Wle balawte] 4 S9 i5 waibk @ Jam? 00min {3‘ rmwnfle‘ 5. You want to estimate the joint distributionof two random variables X and Y. Each of the random variables takes values on the finite state space {1, . . . , 10}. You are given the following conditional distributions, for all values of a: and y: IP<X = W = y) = my); WY = le = m) = you). You should assume that f(:v|y) > 0 and g(y|ac) _> 0 for all values of a: and y. We denote the joint distribution of X and Y by IP(X = x, Y = y) = h(m, y). Using only f and 9, construct a Markov chain with invariant distribution given by h. Justify your answer. L. . (Aw7ue [P (yum % xer, yum : 3M: ll 5m My) >0, jrxlgxw v 76.3, m cm is {wet/{Wilde 0w £me 34qu SFMc 53> (AWN imvwiawl 413+, Check. (bladed Enigma, (9W9 Md *0 check (“Wang/y),- (OWN —? (793W '(wlm x'a‘x, y“): my WW I M) :gwwilrx'w _ 177le g : _-, MUG”) T](X/j {XII Similarly 093),“? [1,)”. 50, char“ )5 M% (/{l/l‘r71/Le (In Val/I‘MVH’ (X13) - 5 PART B. (120 points) Problem 1. (35 points) A gambler needs to win $10 quickly, and he plays the following game. At each stage, the gambler chooses a wager, and a fair coin is tossed (i.e., the coin has probability 1 / 2 of coming up heads, and probability 1/2 of coming up tails). If the coin Comes up heads, the gambler wins an additional amount equal to his Wager. If the coin comes up tails, the gambler loses his wager. The gambler chooses the following (aggressive) policy. If the gambler’s CUrrent wealth is less than or equal to $5, the gambler wagers everything at the next stage. If the gambler’s current wealth. is greater than $5, he wagers just enough on the next stage so that if the coin flip comes up heads, his resultingtotal wealth would be exactly $10. - (a) (5 points) Describe the gambler’s current wealth using a Markov chain model, and identify the com— municating classes. ,wmmmmM ' m ' E 0 i - O t 2 * 4 9’ '6 ‘ D F“ WNW-{ad we Fm,“ lav/.117 '/; _. P00 = (710.10 :— I - For (' § 6', PM = PL‘IZJ '1 FM 1. >6) Pg“): _ ' gIR‘F’IO ’1 '/Z . CUWWMV‘iCWg cum; = M, {10%, m, m, 261, :71, W, {2, 4,6,9 E, (b) (10 points) Assuming that the gambler starts with $2, calculate the probability that he eventually -reaches $10. I q I in? = Wm l0 7t, =15). , L0 : O ‘ - bl” : l a l hi ” 144' ~—9 h a 414 [/1 = i? 8 ‘ 2 r 144 {its . " M. +3 “'9 W a. g, = “ z + , Z 2 -—9 bub = 3' in, +3 T 63 H w— 3‘ <>\ .1. (c) (10 points) Assuming that the gambler starts with $8, calculate the probability that he eventually loses all his wealth. le Cz’flll/t/l is gamma so (ll/Le OMKWM (Almalivelyl [_ “Lb : 1,4142 '/ If; (d) (10 points) If the gambler starts with $2, what is the expected number of stages until the game ends (i.e., until either the gambler loses all his wealth, or he earns $10)? M’ ear/l1 sage, W45 W/Fwé. "/2, 0/ l {5 jail/MW r?) mum f’fl Slt'djpf 4v Problem 2. (45 points) You have an e-mail program that has an inbox that can hold at most three e—mail messages. Email messages arrive/to your inbox according to a Poisson process of rate A. For any given message, there is a probability p that the message is spam; in this case your spam filter always catches the message and sends it to your trash folder, so that it never enters your inbox. When your inbox is full, any messages that arrive are blocked. (a) (5 points) For the first four parts of this question, you should assume that whenever you check e-mail, you instantaneously empty the inbox (i.e., you take zero time to process and delete all your e-mail). After checking your e—mail, you wait an exponentially distributed amount of time with mean 1/?” before checking your e-mail again. Give a Markov chain description of the number of messages in your inbox. Staff” lb passages (in )1"be . I (NM "5pm «WWW W? I 170165in [AU—I7») (b) (10 points) What is the long-run fraction of time that your inbox is full? ' 7!; 2617) = [11“. +1; +19% : (lerrolr f9 : V“ my = T]; 2(ivr) ; o . , ). '3 i t, (+1“? £971}: gar) 3 Tl,l (f+ 7‘0"?» 5 Th) “((1’) (Yvi )UT» 'Mi-F) ‘ 5) Tr, :- V ‘ '1‘ t (r + )(H’fiz Eff/:44 fig“, 1T - ' 0x4 TL [rt fiU-fn : Tit/MP?) L “A 3' Lita—.— 3 IF; 3 3 (tr—t 2(1’FD 8 .. (c) (5 points) Assume that the Markov chain is in equilibrium. What is the probability that an arriving message does not enter your inbox? ' W 7 l Nflmge is ‘ Mexajc (‘5 -v~0+‘ I spam ’ 54?an lam/i (vaax is: div” (Ly Pkg—k) 5 r (d) (10 points) Now suppose that whenever you check your e-mail, if your inbox is not empty, you require an exponentially diStributed amount of time with mean 1/3 to process and delete all yOur e-mail (regardless of the number of messages in your inbox). For simplicity, assume that you take your computer off-line while you are checking e—mail, so that no messages arrive while you are checking e-mail. After having finished processing your e—mail, you wait an exponentially distributed amount of time with mean 1/?” before checking your e-mail again. What is the long-run fraction of time that you spend checking e-mail? C z: 0%!ch WW“ - 0 LZJ‘ #41 “Ways I‘vv‘ ' I Mbux_ “Fag = (1T,+1rz+W3)Y : (I'm-TQM? m Mth = TQS V l =i ‘ 3‘ 4g. %‘ (e) (10 points) Now suppose that whenever you check your e-mail, each message in your inbox requires ' an exponentially distributed amount of processing time with mean 1 /,u. As before, we assume that no messages arrive while you are checking e-mail. We also continue to» assume that after you are finished _ emptying your inbox, you wait an exponentially distributed amount of time with mean 1/1" before I checking your e-mail again. Write down a system of equations we can use to determine the long—run fraction of time that you spend checking e—mail. Express the long-run fraction of time that you spend checking e-mail in terms of the solution to these equations; you do not need to solve the equations. 35/2393 gt: g "4556323 mt (mung Wu. NW) I V, "g! i Nitajfl, CMCll/lldj mail. ‘9" r I M— l: 11 » _ . ./-’~>o as 1T0 {7 ml" . : (l’ ) ' m’l’) M V l‘ _' fill (“17) l”) T“ 7‘“ l’ y fi>o Trzfl +V) : W’A l’r ' " 1L2 _. , war) / . 1r” v 3 «fl ’Mt/f) 00 f‘ Tn; ’4 = Tl’m‘V + Was It ' Trust» “rum-r may {MP = $3an Haw/’70" at 4W WM? Mm =- m + 193 + 1% . (f) (5 points) In terms of the solution to the quations in part (f), what is the long-run rate at which you process messages (i.e., the long-run average number of messages processed per unit time)? Faxes; msgafiwfi- (are r wmg Check/(by Mail, WHO 0 ($67“? 10 Problem 3. (40 points) Consider a customer-service center to which calls arrive according to a Poisson process of rate A. Arriving customers have two choices: they can either immediately request to speak with a customer-service agent, or they can request to use an automated customer—service system. The customer- service center has only one human customer-service agent, so' customers who request to speak with this agent must queue fer service. The agent can serve customers at a rate of ,u per hour; service times are independent and exponentially distributed. If customers choose to use the automated system, they immediately begin service; calls which use the automated system last an exponentially distributed amount of time with mean 1/01. The customer-service center can hold infinitely many calls, including both those that are queuing for , the agent, and those Which are using the automated system. V Assume that arriving customers choose to wait for the agent with probability p, and otherwise with probability 1 — p they choose to use the automated system. (a) (8 points) Give a Markov chain model we can use to analyze this system. Determine the necessary and sufficient condition(s) that must be satisfied by A, ,u, (1, and/or p under which this Markov chain is positive recurrent. 5 (#1 0% WSW m/fii'fil/Lfl. (law/1%, # 04th Vian git/1%., system) 3 (703) e' {0,1,2vwi‘xMILL-"‘3 Q{((?€39), bur-W) = AI, , y: ,i.z,. . .‘;3=o,',1~—,-~- Qih’rhghfinjl) = ’4, x=0,\,Z,. .'./'v=.0,\,‘Z/... a; z’ 7T0!!! )l 3: OI’IZ'I' “ Q((’V,5H), I Opel)“, 7=0,|/2,.-1-,lj:OI))ZI._. CQUVGy), my») > o Nth/me. » (a, (we + (7.3»). Tim. age/mi": qw i: IM/M/yf' m awmmd 5414M r: M/M /ao. fle’. 4-7WS we F‘ k M (K. SW 19,8 . q M/M /( ("2 isiaLie MAW WM < W'e (“it £9 Need ( ’XE < '5 Ly (Fm/Hm recrAwe/Me. 11 (b) (7 points) If the assumption in part (a) is satisfied, what is the long-run average number of customers in the customer—service center? : Mum il’ llfl M /M /l W/ ' MMV'M MIL: A F1 Wile rat}! I I; V I + ‘ MM Hull: M M/M/no 4/ nan/9’01? ml: (A07) WM rah. o< U Jfiq+ flip). r”? r (c) (5 points) Calculate the value of p such that the mean waiting time of a customer arriving in equilib- ‘ rium will be the same regardless of whether they choose to wait for the agent, or they choose to use the automated system. (For this part of the question, assume that A < ,u and a < ,u.) ~ MW" Warlfij limit M M/M/l alga/n" I . rilr Ill/lwa Miami/1.3 4am uh M/M/m outan f. 51% 12 (d) (10 points) If a customer is using the automated system, they may become impatient. Suppose that each customer who chooses the automated system is willing to wait at most an exponentially dis~ tributed amount of time with mean 1/;6; if after this time they have not completed service in the autOmated system, they quit and join the queue for the human agent instead. What is the long-run fraction of time that the human agent is busy serving customers? (For this part of the question, as- sume that A < ,u.) rot/f 40 wamm age/wt 714.4% a 9x1; + arrpfizfl), MM" gem/fa? «hwb. WWW agwi' 1, i“ :z) i5 man v’lYM-HW lawman mgw’l‘ ('5 WWI/lg a At l— )(l’E)/F//K+F))I F (e) (10 points) Suppose that currently, one customer is being served by the agent, and two customers are using the automated system. Find the probability that the entire system empties before another call arrives to the center. F13! Sjslewt (iv «M59411, W 4km MMfiL' MON/C lamb/c m-mav'al " lm’r +IM Walla/V Wm}? ’wmliW-f . Famed/13 : 2:” )(dfa) o( iii,“ :1 4. ’A+_2¢_+') [xvi-Mt) 0”") jg « « _L f/{er-l-‘g 'pk-ot+) fit) 13 ...
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This note was uploaded on 02/03/2011 for the course MS&E 221 taught by Professor Ramesh during the Winter '11 term at Stanford.

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finalsol_2005 - MS&E 221 Final Examination Ramesh...

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