0:4 0:4 0:2 0:2 0:6 0:2 P = Moreover, it is informed that 0:28 0:46 0:26 0:24 0:50 0:26 0:24 0:41 0:35

Transkript

1 EKSAMEN I: MOTA STOKASTISKE PROSESSER ENGLISH VERSION VARIGHET: 4 TIMER DATO:. DESEMBER 4 TILLATTE HJELPEMIDLER: Kalkulator; Tabeller og formler i statistikk (Tapir forlag): Rottman: Matematisk formelsamling. OPPGAVESETTET BESTAR AV 4 OPPGAVER PA 3 SIDER Problem Daily wear and tear on an oil platform is characterized as a Markov chain fx n g with the states : Low; : Medium and : High wear and tear. The transition matrix P is given by P = Moreover, it is informed that P = :4 :4 : : :6 : C A : : :3 :5 :8 :46 :6 :4 :5 :6 C A : :4 :4 :35 a) Write up P (X = jx = ), P (X = jx = ) and P (X = jx = ) b) Set up the equations for the equilibrium probabilities, og and nd these. In the equilibrium situation also nd P (X n = jx n+ = ). Hint: Use Bayes formula. Problem A consulting rm does service on home PC-s. Error messages are coming in to the rm according to a Poisson process with on the average 5 error messages per hour. Let N(t) be the number of error messages during a time period t. Let T i be the time between the (i-)-th og i-th error message and S n = T + T n be the time to the n-th error message. a) Find the following quantities: P (N() = 5), P (N() N() = 5jN() = 5), E(T ) and E(S jn(3) = 3).

2 b) The cost of reparations is included in the customer contract that the pc-users have. The cost X for the consulting rm for a service assignment is normally distributed with expectation = kr og standard deviation = 5 kr. The number of service assignments N during one week is Poisson distributed with an average of per week. Find the expected cost during one week. Also nd the corresponding standard deviation when it is informed that Var( NX i= X i ) = E[Var( NX i= X i jn)] + Var[E( NX i= X i jn)] c) The consulting rm is going to recruit new personel. Job candidates rst go trough a pc-technical interview and then immediately afterwards, if the next interviewer is not busy, a personality interview.. The time spent on the technical interview is exponentially distributed with mean 3 minutes. The same is the case with the personality interview. When Bella arrives for the technical interview, Anita is at the personality interview. What is the probability that Anita is nished with the personality interview before Bella is nished with the technical interview. If this event occurs, what is the expected time until Bella has nished both of the interviews? If Bella is nished with her technical interview before Anita is nished with her personality interview, what is then the expected time Bella spends before she is nished with everything? Use this to nd the expected time until Bella is nished with both interviews. Is the exponential assumption realistic here? d) Carl, who works in the rm, has a bad habit of oversleeping. The rm opens at 9 o'clock. If he oversleeps, the others have found that his arrival time is uniformly distributed between 9 and.3. By means of the information in a) nd the expected number of error messages that come in in his absence. Hint: First write up the expected number of messages that come in in the time period (; t), where t is between and.5. Problem 3 The consulting rm in problem also receives a large number of other requests and the home-pc customers must stand in a telephone queue with other customers. a) Assume that incoming telephone calls occur according to a Poisson process with an average arrival rate of one per minute. There are two consultants that take care of the customers. The lenght of a telephone conversation for both of them is exponentially distributed with an expectation of one and a half minute. What is the probability that

3 a conversation lasts more than minutes? Given that a conversation has lasted in minutes, what is the probability that it will last for still more minutes? Given that the conversation starts simultaneously for customers, what is the probability that the shortest conversation lasts for more than minutes? b) Assume that both of the consultants are busy when you call, and that you in addition have 5 persons ahead of you in the line. What is then the expected time until you are nished with your conversation? What is the probability that one of the two customers that talks to a consultant when you join the line, is still in the telephone when you are nished with your conversation? What would the expected time had been if there had been 3 consultants who were all busy, and that you addition had 4 ahead of you in the line c) Now assume again that there are only consultants. Further, assume that customers that call when there are customers getting service and in a waiting position, leave the line. Set up the balance equations for this system. Find the expected number of customers in the system, and nd the proportion of time for which both of the consultants are busy. Problem 4 An institute at a college has copy machines, machine A and machine B. Machine A is new, and because of beginner trouble it breaks down relatively often, actually times per month, such that per hour the breakdown rate is set to =. The down time is exponentially distributed with mean = 4 hours. Machine B breaks down once every two month in average, that is, the breakdown rate is set to = per hour, whereas down time 4 is = 6 hours. Just one repair person is available. This system denes a Markov chain with states: A : Both machines are down, A under repair. B : Both machines are down, B under repair. A : A is up, B is down. B : B is up, A is down. : Both are up. Find the intensity rates for going out from each of these 5 states and set up the transition matrix between the states. From this matrix nd the probability that A should be going down while B is down, and the probability that B should be going down while A is down.

4 Utvidet fasit MOTexh4 (forbehold om feil). Oppgave a) P (X = jx = ) = :3; P (X = jx = ) = :5; P (X = jx = ) = :5: b) Likevektslikninger: :4 + : + : = :4 + :6 + :3 = : + : + :5 = + + = Lsning blir: = 4 ; = 3 8 = 7 : P (X n = jx n+ = ) = P (X n = )P (X n+ = jx n = ) P (X n+ = ) = 3 : = :5: Oppgave a P (N() = 5) = 55 5! e 5 = :75; P (N() N() = 5jN() = 5) = P (N() = 5) = :75; b E(T ) = 5 ; E(S jn(3) = 3) = = 5 : K: Kostnad. Antar uavhengighet mellom N og X i, K = P N i= X i E(K) = E(N)E(X i ) = = 4: Var(K) = E(NVar(X i ))+Var(NE(X i )) = E(N)Var(X i )+(E(X i )) Var(N) = (5) +() = 45; dvs SD(X) = p 45 = 9: c) T : Tid pa fagintervju, T : Tid pa personlighetsintervju. P (T < T ) = = : T < T medfrer E(T Bella ) = = 6, mens T > T medfrer (p.g.a hukommelsesegenskapen) E(T Bella ) = = 9. Hvis begge muligheter tas hensyn til, E(T Bella ) = (6 + 9) = 75:

5 d) N(t): Antall samtaler i [; t)., E(N(t)) = 5t N: Antall feilmeldinger i Carls fravr. T : Carls ankomstid. E(N) = Z 3= E(N jt = t)f T (t)dt = Z 3= 5t h 3 dt = i 3= 3 5t = 5 4 : Oppgave 3 3a) T : Lengde pa samtale. Betjeningsrate er = E(T ) = 3 = 3 P (T ) = e 3 = e 4 3 = :6; P (T 4jT ) = P (T ) = e 4 3 ; P (T min ) = e 3 = e 8 3 : 3b) T : Tid til du er ferdig. E(T ) = = 6 (Forventet tid til 4 systemet er tmt gitt at det ikke er ankommet nye kunder er ). 4 P (Fremdeles en av to opprinnelige kunder i k) = (P (Fremdeles en bestemt av de to opprinnelige kundene, f.eks Ola fremdeles i k) = 7): 3 betjenere: E(T ) = = 7 (Forventet tid til system er tmt 6 gitt at det ikke kommer nye kunder er ) c) Tilstand n: n kunder i systemet. Balanselikningene gir P = 3 P ; P = 3 4 P = 9 8 P ; P 3 = 3 4 P = 7 3 P : P + P + P + P 3 = gir P = 3 43 ; P = ; P = ; P 3 = 7 43 : Forventet antall kunder i systemet: = Andel av tid begge opptatt: P + P 3 = 36+7 = 63 : Oppgave 4 Intensitetsrater ut av tilstander: A :, B :, A : +, B : +, : + Overgangsmatrise blir da: 6

6 P = P (A gar ned nar B er nede) = + = P (B gar ned nar A er nede) = + = C A = 3 53 : = : Dag Tjstheim