0:6 0:3 0:1 0:4 0:2 0:4

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1 UTKAST EKSAMEN I: MOTA STOKASTISKE PROSESSER ENGLISH VERSION VARIGHET: 4 TIMER DATO: 22. november 26 TILLATTE HJELPEMIDLER: Kalkulator; Tabeller og formler i statistikk (Tapir forlag): Rottman: Matematisk formelsamling. OPPGAVESETTET ESTAR AV 3 OPPGAVER PA 3 SIDER Problem The price X t of natural gas from a eld in the North Sea is roughly described by three levels: : low, : medium and 2: high. The development of X t in time from one week to another is assumed described by a Markov chain with transition matrix P = :6 :3 : :4 :2 :4 A ; P 2 = : :2 :7 :49 :26 :25 :36 :24 :4 A : :2 :2 :58 a) Find P (X 2 = jx = ), P (X 3 = 2jX = ). Also, nd the probability for X 9 = X = X = given that X 8 =. b) Set up the equilibrium equations for the chain and nd the corresponding equilibrium probabilities. What concrete interpretation can you give of the equilibrium probabilities? c) Given X 8 =, use the absorption technique to nd the probability that the price gets to state 2 during 3 weeks. Note that you do not need all of the elements in the last matrix multiplication when using the absorption technique. d) If the memory of the chain extends 2 weeks back in time, dene new states and set up the transition matrix for this chain by marking with a cross the possible transitions. If the memory goes three weeks back in time, what is the dimension of the transition matrix in this case? For the state 2, which means high gas price in the current week, medium gas price in the last week and low gas price two weeks ago, what are the possible states next week?

2 Problem 2 A wind power system consists of a number of wind mills. Some of the wind mills suer from a minor problem that occurs according to a Poisson process on the average once per month for each of them.the time of repair is so small that it can be neglected here. a) We look at one of these wind mills. Let the unit of time be one month, and let T i be the time between the time points when the problem occurs for the (i )-th and ith time. Further, let S n = T + T T n be the time point when the problem occurs for the n-th time, and let N(t) be the number of times it occurs during a time period of length t. Find P (T 2jT ), E(S 5 ), P (N(2) = 2jN(:5) = ) and E(S 5 jn() = 2). b) The expense for each time that this problem occurs is a stochastic variable X that is normally distributed with an expectation of kr. and standard deviation 2 kr. The total expense during one year then becomes Y = N(2) X i= where X i is the expense for the i-th repair, and N(2) is the number of times the problem occurs during this year. Find the expectation and variance of Y. c) The rate at which this problem occurs in a more realistic model depends on the time of the year. In the rst part of a year the occurrence rate decreases from 3 2 to 2 according to the formula X i ; (t) = 3 2 t; t 6; 6 and in the last part of the year it increases from 2 to 3 2 according to (t) = (t 6); 6 t 2: La N and N 2 be the number of times the problem occurs in the rst and second part, respectively. Which distribution does N and N 2 have? Find the expectation and the mean for these two variables.

3 Problem 3 A medium sized company has a computer park consisting of 3 main computers that serve the employees that each have their own P. There are two computer engineers that look after this equipment. They take care of current problems for the P-s of the employees and the main computers. (The latter ones have a higher priority in case of trouble). The problems that the employees have, are registered by the engineers that handle them as soon as they have time. a) First assume that for the problems that the employees have with their P-s, there is an exponential xing time with expectation 2 minutes. If you as employed in this rm has a problem with your P, and there are 4 persons ahead of you in the line waiting for help from the engineers in addition to the 2 employees currently being helped, what is the expected time until you have your problem solved? In the long run it has been seen that one of the engineers on the average only needs 5 minutes to solve a problem (whereas the other one still needs 2 minutes). What is now the expected time until you have your problem solved? If 2 employees both are being helped by the engineers, what is the probability that the person being helped by the slow engineer, nevertheless is the one being nished rst. (In all of this part of the problem it is assumed that the two engineers are only busy with the employees' P-s). b) The three main computers breaks down and are being repaired according to exponential distributions. Assume that all of them have the same breakdown rate and repair rate (there is no dierence in the eciency of the engineers here, and they work separately and only at one machine at a time). Let the states,, 2 and 3 denote that respectively,, 2 and 3 main computers work. It is informed that computer, 2 and 3 break down independently of each other and on the average 5 times per half a year for each of them, whereas expected repair time for each breakdown is day. You can take one month as your time unit and you may assume that there are 3 days a month. Explain why you have = 3 and = 5=6. Set up the equilibrium (balance) equations for this system and solve them given the information on breakdown and repair rate c) What is the probability that all of the main computers are down? What is the expected number of main computers that are up, and what is the expected number of engineers working at the main computers at anyone time? What is the proportion of time for which both of the engineers are busy with the main computers? (If you are not able to solve part b) of this problem, assume that the equilibrium probabilities P ; P ; P 2 and P 3 are known and answer the questions in this part by means of P ; P ; P 2 and P 3 ).

4 Dag Tjstheim

5 (Forbehold om feil). Utvidet fasit Oppgave a) P (X 2 = jx = ) = :3; P (X 3 = 2jX = ) = :25: P (X 9 = X = X = jx 8 = ) = (:6) 3 = :26 b) Likevektslikningene blir: :6 + :4 + : 2 = som har lsningen :3 + :2 + :2 2 = = = 6 47 ; = 47 ; 2 = 2 47 : Likevektssannsynlighetene gir den andelen av tiden som Markov kjeden er i de ulike tilstandene. Prosessen benner seg for eksempel i tilstand i 6/47 deler av tiden i det lange lp. c) Teknikken med absorberende tilstand gir: Q = :6 :3 : :4 :2 :4 A ; Q 2 = :48 :24 :28 :32 :6 :52 A ; Q 3 = og sannsynligheten for at prisen kommer i tilstand 2 i lpet av tre uker blir.324. d) Overgangsmatrisen nar hukommelsen gar to trinn tilbake blir: P = x x x x x x x x x x x x x x x x x x x x x x x x x x x Nar hukommelsen gar tre trinn tilbake blir dimensjonen pa overgangsmatrisen 27. Mulige overgangstilstander for tilstanden 2 er 2, 2 og 22. A x x :424 x x x A

6 Oppgave 2 2a) P (T 2jT ) = P (T ) = e = :37, E(S 5 ) = 5, P (N(2) = 2jN(:5) = ) = P (N(:5) = ) = :33, E(S 5 jn() = 2) = + E(S 3 ) = 3. 2b) E(Y ) = E(N(2))E(X) = 2 = 2. Videre, Var(Y ) = VarfE(Y jn(2)g+efvar(y jn(2)g = (E(Y )) 2 VarfN(2)g +Var(Y )EfN(2)g = 2 () (2) 2 = c) La N (t) og N 2 (t) vre Poisson-prosessene denert for t 6, og 6 t 2, henholdsvis. De tilsvarende Poisson-variable N = N (6) og N 2 = N 2 (2) har parametrene Z 6 = (3 Z t)dt = 6; og 2 = ( (t 6))dt = 6: 6 Det flger at E(N ) = Var(N ) = E(N 2 ) = Var(N 2 ) = 6. Oppgave 3 3a) Forventet ventetid er = 7 minutter. Dersom den ene P-ingeniren bruker 5 minutter,er systemet bestaende av de to et eksponsielt system med = og forventet tid mellom utgang av kunder er 6 7 < minutter. Forventet total tid blir na = 6min La T og T 2 vre betjeningstid for hurtig og langsom betjener, henholdsvis. Vi har 2 P (T 2 < T ) = = b) Likevektslikningene for de re tilstandene blir: = 3 7 : lses disse, far en : 2P = P ; : (2 + )P = 2P 2 + 2P ; 2 : ( + 2)P 2 = 3P 3 + 2P ; 3 : 3P 3 = P 2 : P = ; P = ; P 2 = ; P 3 = :

7 3c) P (Alle ute) = =33769 = : E(Antall i drift) = = = 2:92: E(Antall dataingenirer i virksomhet) = = = :8: Endelig gir P + P = 73=33769 = :26 som den andelen av tiden begge ingenirene er sysselsatt med dette.