{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# ch4_add_1 - = 1 9(c What is the pmf of X p k = P X = k = Î...

This preview shows pages 1–2. Sign up to view the full content.

Birth and Death Processes Example: Communication System A communication system has two processers that decode messages and a buffer that holds at most two messages. A message that arrives when the buffer is full is lost. Each processer needs an average of 2 minutes to decode a message. Messages arrive at a rate of 1 per minute. Assume that inter-arrival times and processing times are independent exponential random variables. (a) Draw a state diagram with possible states and corresponding birth/death rates. Solution: 1 1 0 1 2 3 4 1 1 .5 1 1 1 Let T 1 = time until processer 1 finishes and T 2 = time until processer 2 finishes. E [ T i ] = 2 i = 1 , 2 implies that rate parameter for both distributions are 1/2 i.e., T 1 , T 2 iid Exp (1 / 2) For k = 1, implying that μ 1 = 1 2 = . 5 For k 2, D = min { T 1 , T 2 } giving D Exp ( 1 2 + 1 2 = 1); thus μ k = 1 , k = 2 , 3 , 4 (b) What is the (large t ) probability that the system is empty? S = 1 + 1 . 5 + 1 · 1 1 · . 5 + 1 · 1 · 1 1 · 1 · . 5 + 1 · 1 · 1 · 1 1 · 1 · 1 · . 5 = 9 implying that p 0 = 1 S = 1 9 (c)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: = 1 9 (c) What is the pmf of X ? p k = P [ X = k ] = Î» Î» 1 ...Î» k-1 Î¼ 1 Î¼ 2 ...Î¼ k p k 1 2 3 4 p k = P ( X = k ) 1 9 2 9 2 9 2 9 2 9 1 (d) What is the (large t ) expected # of jobs in the system? E [ X ] = 0 Â· p + 1 Â· p 1 + 2 Â· p 2 + 3 Â· p 3 + 4 Â· p 4 = 0 Â· 1 9 + 2 9 Â· (1 + 2 + 3 + 4) = 20 / 9 (d) What is the probability that an incoming in message is turned away? An incoming message is lost when system is full i.e., X = 4 P (message lost) = P (system full) = P ( X = 4) = p 4 = 2 / 9 (e) What is the expected number of messages in the buffer? Let Y = # of messages in the buï¬€er; note that Y is a function of X X 1 2 3 4 Y 1 2 p k 1 9 2 9 2 9 2 9 2 9 E [ Y ] = 0 Â· p + 1 Â· p 1 + 2 Â· p 2 + 3 Â· p 3 + 4 Â· p 4 = 0 Â· 1 9 + 0 Â· 2 9 + 0 Â· 2 9 + 1 Â· 2 9 + 2 Â· 2 9 = 5 / 9 2...
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

ch4_add_1 - = 1 9(c What is the pmf of X p k = P X = k = Î...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online