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Unformatted text preview: Homework 9 – Student: Son Tran – ID: 2007-23576 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 steady-state 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: 2007-23576 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- - – - + – - – - - –...
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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