Math 597/697: Solution 5

Transcription

1 Math 597/697: Solution 5 The transition between the the ifferent states is governe by the transition matrix P = 4 0 5, () an v 0 = /4, v = 5, v 2 = /3, v 3 = /5 Hence the generator is given by (2) 2 The state space consists of all pairs (m, n) of nonnegative integers The time until a given pair of male an female prouces an offspring is an exponential RV with parameter λ If the population at time consists of m males an n females there are nm possible pairs an hence we obtain v (m,n) = λmn, (3) q (m,n)(m+,n) = pλmn, q (m,n)(m,n+) = ( p)λmn (4) P (m,n)(m+,n) = p, P (m,n)(m,n+) = ( p) (5) 3 (a) The number of computers X t in operating conition at time t is either 0,, 2 an the number of computers operating is either 0 if X t = 0 or if X t = or 2 The rate of failure of the operating computer is an the rate of repair is λ There is only one repair facility so that at most one computer can be repaire at a time We fin an v 0 = λ, v = λ +, v 2 = (6) P 0 =, P 0 = λ +, P 2 = λ λ +, P 2 = (7)

2 Hence the generator is λ λ 0 λ λ 0 The backwar equations are, for j = 0,, 2, (8) t P 0j = λ(p 0j P j ), (9) t P j = (P 0j P j ) λ(p j P 2j ), (0) t P 2j = (P j P 2j ), () an the forwar equations are t P j0 = λp j0 + P j, (2) t P j = λp j0 (λ + )P j + P j2, (3) t P j2 = λp j P j2 (4) The stationary istribution is π j = (λ/) j + (λ/) + (λ/) 2 (5) It is also a limiting istribution since the state space is finite an the chain is irreucible In the long run, the proportion of time when the system is operating (ie in state or 2) is (λ/) + (λ/) 2 + (λ/) + (λ/) 2 (6) (b) The ifference with (a) is that the two computers can be in operation at the same time an so the rate of failures when X t = 2 is twice the rate of (a) v 0 = λ, v = λ +, v 2 = 2 (7) Then λ λ 0 λ λ 0 2 2, (8) 2

3 an the stationary istribution is in that case π j = j! (λ/)j + (λ/) + 2 (λ/)2 (9) an so, in the long run, the proportion of time when the system is operating (ie in state or 2) is (λ/) + 2 (λ/)2 + (λ/) + 2 (λ/)2 (20) (c) Let us consier first the case (a) We compute the expecte time to reach 0 starting from 2 We write T = T 2 + T 0 where T 2 is the first time the chain reach state stating from state 2 an T 0 is the first time the chain reach state 0 starting from Clearly T 2 is an exponential RV with mean To compute E[T 0] we conition on the event A where the first transition is to 0 (probability /(λ + ) or the event B where the first transition is to 2 (probability λ/(λ + ) Since the first transition occurs in average after a time /(λ + ) we fin E[T 0 A] = E[T 0 B] = λ +, (2) λ + + E[T 2] + E[T 0 ] (22) an so, E[T 0 ] = = λ + + λ + ( + λ ) + λ λ + (E[T 2] + E[T 0 ]) (23) λ λ + E[T 0]) (24) Solving gives E[T 0 ] = + λ an E[T ] = 2+ λ In case (b) one fins λ 2 instea 4 Each component is an on/off system with generator ( λi λ i i i ) (25) 3

4 For each component the stationary an limiting istribution is for i = A, B, π (i) i 0 =, π (i) λ i = (26) λ i + i λ i + i By inepenence the stationary istribution of the two componentsystem is π (A) i π (B) j (a) If the components are working in parallel the is operating if at least of the components is in state 0, hence the esire probability is π (A) π (B) = λ A λ B (λ A + A )(λ B + B ) (27) (b) If the components are working in series, both must work for the system to be operating, hence the esire probability is π (A) 0 π (B) 0 = A B (λ A + A )(λ B + B ) (28) 5 Consier a continuous time branching process efine as follows A organism lifetime is exponential with parameter λ an upon eath, it leaves k offspring with probability p k, k 0 The organisms act inepenently of each other We assume that p = 0 so that, at eath, there are no transition from a state into itself Let X t be the population at time t A transition occurs when an organism ies an if there are n organisms the average time until a eath is /nλ an so v n = nλ When an organism ies it is replace by 0, 2, 3, 4, organisms an therefore we have P nn = p 0 P nn+ = p 2, P nn+k = p k+ (29) Therefore the generator is given by λp 0 λ λp 2 λp 3 0 2λp 0 2λ 2λp 2 (30) The backwar equations are t P 0j = 0, (3) 4

5 t P j = λ (p 0 P 0j + p 2 P 2j + p 3 P 3j + P j ), (32) (33) t P nj = nλ (p 0 P n j + p 2 P n+j + p 3 P n+2j + P nj ) (34) an the forwar equations are t P j0 = λp 0 P j, (35) t P j = λp j + 2λp 0 P j2, (36) t P j2 = λp 2 P j + 2λP j2 + 3λp 0 P j3, (37) t P j3 = λp 3 P j + 2λp 2 P j2 3λP j3 + 4λp 0 λp j3, (38) (39) In case of binary splitting we have if the probability of eath without offspring is p λp λ λ( p) 0 0 2λp 2λ 2λ( p) (40) As ones sees from the matrix A the Markov chain is not irreucible, 0 is an absorbing state an all other states are transient The istribution π = (, 0, 0, 0, ) is a stationary state, which escribe the situation where noboy lives Note that the transition matrix P ij is given by a ranom walk, so one might expect that if p < /2 every initial istribution will converge to π while if p /2 there will be a positive probability to escape to infinity 6 Consier the M/M/ queue with arrival rate λ an service rate (a) The M/M/ queue is a Birth an Death process with λ n = λ 5

6 an n = n Hence the generator is λ λ λ λ λ 2 λ λ 3 λ (4) (b) Using the general solution for the stationary an limiting istribution for a Birth an Death process (the M/M/ is irreucible) one fins λ n λ 0 λ n + = + n= n n! n= n = e(λ/) (42) an so ) n e (λ/) (43) π n = n! ( λ The M/M/ is positive recurrent an the stationary istribution is Poisson with parameter λ/ (c) Since we have P {X t+h = n + X t = n} = λh + o(h), (44) P {X t+h = n X t = n} = nh + o(h), (45) P {X t+h = n X t = n} = ( λh nh) + o(h),(46) E [X t+h X t = n] (47) = (n + )λh + (n )nh + n( λh nh) + o(h),(48) = λh nh + n + o(h) (49) an Therefore an E [X t+h ] = λh he [X t ] + E [X t ] + o(h) (50) t E [X t] = λ E [X t ] (5) E [X t ] = E [X 0 ] e t + λ ( e t) (52) 6

7 7 Consier a queuing system with one single server, arrival rate λ an serving rate (a) When N customers are in the system, the arriving customers give up an o no enter the system So the state space is {0,,, N}, the generator is λ λ 0 0 λ λ 0 0 (53) 0 0 λ λ 0 0 The limiting istribution can be foun using the general formula for the birth an eath process with λ n = 0 for n N One fins for 0 j N π j = (λ/) j + N k= (λ/) k = (λ/) j ( (λ/)) λ (λ/) N+ N+ λ = (54) (b) When n customers are in the system, an arriving customer will join the system with probability /(n + ) We have n = for n an λ n = λ/(n + ) for 0n 0 Then λ n λ 0 λ n + = + n= n n! n= n = e(λ/) (55) an the stationary istribution is Poisson with parameter λ/ 8 Consier a Yule process (pure birth with linear growth) an let T i be the time it takes for a population of size i to reach i + (a) Since it is a pure birth process T i is exponential with rate iλ (b) Let us suppose we have n components with ii exponential lifetime X i The RV max(x,, X n ) is the time it takes for the n components to fail Starting with n components, the time until the first failure is the minimum of n exponential RV with parameter λ which is an exponential RV T n with parameter nλ By the memoryless property of the exponential, the time between the first failure an the secon failure is, again, the minimum of 7

8 n exponential RV with parameter λ which is is an exponential RV T n with parameter (n )λ Hence max(x,, X n ) = T n + T n + + T (56) (c) We have, by (b), an inepenence P {T + T 2 + T n < t} = P {max(x,, X n ) < t} = P {X < t} P {X n < t} = ( e λt ) n (57) () Since X t is a pure birth process the Markov chain X t only makes transition from j to j + So T + +T n is the time necessary to reach n starting from an P {T + + T n < t} is the probability that X t n P n (t) = P {X t = n X 0 = } = P {X t n X 0 = } P {X t n + X 0 = } = P {T + + T n < t} P {T + + T n < t} = ( e λt ) n ( e λt ) n = e λt ( e λt ) n (58) Hence P {X t = n X 0 = } has a geometric istribution with parameter e λt (e) By () X t conitione on {X 0 = } is a geometric ranom variable By efinition of a Birth process, the iniviuals reprouce inepenently of each other, an therefore X t conitione on {X 0 = m} is the sum of m geometric ranom variables The sum of m geometric RV with parameters p is calle a negative binomial RV with parameters (m, p) A geometric RV is the number of inepenent trials necessary to obtain one success, so that a negative binomial RV is the number of trials necessary to obtain exactly m successes The probability istribution of the negative binomial RV is, for k m, (at least m trials are necessary) p(k) = ( ) k p m ( p) k m (59) m 8

9 The combinatorial factor is obtaine by noting that the last trial must be a success an by counting the number of ways of having the remaining m successes Therefore we obtain, for n m ( ) n P mn (t) = (e λt ) m ( e λt ) n m (60) m 9