This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Homework 9 – Student: Son Tran – ID: 200723576 Data Networks – Spring 2008 4.1 a) Let p n ( k ) be the probability to have n backlogged nodes during time slot k . The transition probability P i , j is the probability to have j backlogged nodes in next time slot given that we have i backlogged nodes in the current time slot. Now, computing p n ( k + 1) by decomposing into conditioning on all possible previous states give us + = ( ) , = = + ( ) , ( ) pnk 1 ipi k Pn i i 0n 1pi k Pn i 1 since P n ,i = 0 if i > n + 1 because it is impossible to successfully transmit more than one packet during a time slot. Assuming that the steadystate probabilities exists, = →∞ ( ) pn limk pn k , taking the limit as →∞ k of (1) gives us = = + , pn i 0n 1piPn i . Furthermore we have to have total probability 1, = = i 0mpi 1 b) = = + , + + + , pn i 0n 1piPi n pn 1Pn 1 n + = pn 1 pn = + , + , i 0n 1piPi nPn 1 n c) Using the result from b) = p1 p0 , , = ( p0P0 0P1 0 p0 1 , ) , P0 0 P1 0 = ( p2 p1 1 , ) P1 1 , , = ( p0P0 1P2 1 p0 1 , ) , ( P0 0 P1 0 1 , ) P1 1 , , = [( p0P0 1P2 1 p0 1 , )( P0 0 1 , ) P1 1 , , ] , , P0 1P1 0 P1 0P2 1 d) Subtitution the result from b) into equation p 0 + p 1 + p 2 = 1 +( p01 1 , ) , +( P0 0 P1 0 1 , )( P0 0 1 , ) P1 1 , , , , = P0 1P1 0P1 0P2 1 1 using common denominator and solving for p gives us = , , , , + , p0 P1 0P2 1P1 0P2 1 P2 11 , +( P0 0 1 , )( P0 0 1 , ) P1 1 , , = , , , ( , P0 1P1 0 P1 0P2 1P1 0 P2 1 , )+ P0 1 1 , ( , + P0 0 P2 1 1 , )= , , , , + P1 1 P1 0P2 1P1 0P2 1 1 , ( + , P0 0 1 P2 1 , ) P1 1 Since P 0,1 = 0 because we can’t go from no backlogged nodes to just 1 backlogged node. 4.4 a) Backlogged nodes do not accept any new packets. Thus the number of nodes which can accept packets are the currently unbacklogged ones (i.e. m − n k ). Each of these gets a new arrival with rate q a . Then, the expected number of new arrivals is given by E [ N a ] = E [( m − n k ) q a ] = ( m − n ) q a Homework 9 – Student: Son Tran – ID: 200723576 Data Networks – Spring 2008 b) In steady state, the expected number of new arrivals should equal the expected number of departures. Thus E [ N d ] = E [ N a ] = ( m − n ) q a c) At the beginning of a slot, the number of packets in the system is given by the number of backlogged nodes plus the number of new arrivals. Thus E [ N sys ] = E [ n k + N a ] = n + ( m − n ) q a d) e) 4.6. a) The probability of a successful transmission (when there are n backlogged nodes) given in Equation (4.5) on p.280 is equal to P succ = (m − n)q a (1 − q a ) m−n−1 (1 − q r ) n + (1 − q a ) m−n nq r (1 − q r ) n−1 This is just the probability that there is either exactly one new transmission and no retransmissions from a backlogged node or exactly one retransmission from a backlogged node and no new transmission. For given m , n and q a , differentiating P succ with respect to q r gives ∂∂ =  – qrPsucc m nqa1 qa  –  + –  –   –...
View
Full
Document
This note was uploaded on 06/15/2008 for the course COMPUTER S 853 taught by Professor Choi during the Spring '08 term at Seoul National.
 Spring '08
 Choi

Click to edit the document details