Answers to the problems from problem solving classes

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1 Answers to the problems from problem solving classes Class, multiaccess communication 3. Solution : Let λ Q 5 customers per minute be the rate at which customers arrive to the queue for ordering, T Q 5 minutes be the average time each customer spends waiting in the queue. By Little s theorem we get N Q λ Q T Q 25 customers in the queue on average. Let λ E 0.5λ Q 2.5 customers per minute be the rate at which customers arrive from the ordering queue to sit down and eat in the restaurant, T E 20 minutes be the average time to eat the meal. Little s theorem gives us N E λ E T E customers on average eating in the restaurant. So, in total there are N N Q + N E customers in the restaurant on average. Solution 2: Let λ 5 customers per minute be the rate at which customers arrive to the restaurant. P(eat in) P(take out) /2, Tqueue 5 minutes, T eat 20 minutes, average time per customer Tavg TqueueP(take out)+ (Tqueue + T eat )P(eat in) 5 /2 + (5 + 20) /2 30 /2 5 minutes. By Little s theorem we get average number of customers in the restaurant as N λtavg (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 + ) by decomposing into conditioning on all possible previous states give us p n (k + ) i n+ p i (k)p i,n p i (k)p i,n () since P i,n 0 if i > n + because it is impossible to successfully transmit more than one packet during a time slot. Assuming that the steady-state probabilities exists, p n lim k p n (k), taking the limit as k of () gives us p n n+ i0 p ip i,n. Furthermore we have to have total probability, so m i0 p i. (b). Separating last term of sum, p n n i0 p ip i,n +p n+ P n+,n and solving for p n+ yields p n+ n p n p i P i,n P n+,n i0 i0

2 (c). Using the results from (b) p p 2 (p 0 p 0 P 0,0 ) p 0 ( P 0,0 ) P,0 P,0 (p p P, p 0 P 0, ) P 2, (p ( P, ) p 0 P 0, ) P 2, p0 ( P 0,0 )( P, ) p 0 P 0, P 2, P,0 p 0 (( P 0,0 )( P, ) P,0 P 0, ) P 2, P,0 (d). From above we have the equations p p 0 ( P 0,0 ) P,0 p 0 p 2 (( P 0,0 )( P, ) P,0 P 0, ) P 2, P,0 inserting them into p 0 + p + p 2 yields ( p 0 + P 0,0 + ( P ) 0,0)( P, ) P,0 P 0, P,0 P 2, P,0 using common denominator and solving for p 0 gives us p 0 P 2, P,0 P 2, P,0 + P 2, ( P 0,0 ) + ( P 0,0 )( P, ) P,0 P 0, P 2, P,0 P,0 (P 2, P 0, ) + ( P 0,0 )( + P 2, P, ) P 2, P,0 P,0 P 2, + ( P 0,0 )( + P 2, P, ) since P 0, 0 because we can t go from no backlogged nodes to just backlogged node. We can go further than the problem stated in the book by expressing the transition probabilities in terms of retransmission probability q r and packet arrival probability q a, we have P 0,0 ( 2 ) ( q a )q a + ( 2 0 ) ( q a ) 2 2( q a )q a + ( q a ) 2 ( q a )(2q a + q a ) ( q a )( + q a ) qa 2 P,0 ( q a ) q r ( q a )q r 0 2

3 P, q a ( q r ) + ( q a )( q r ) 0 0 q a ( q r ) + ( q a )( q r ) q r 0 2 P 2, ( q r )q r 2( q r )q r 0 and inserting those into our expression for p 0 we obtain p 0 p p 2 2( q a )( q r )q r 2( q a )( q r )q r + q 2 a (3 2q r) 2( q r )q 2 a 2( q a )( q r )q r + q 2 a(3 2q r ) q 2 a 2( q a )( q r )q r + q 2 a(3 2q r ) 4.3 (a). The arrival rate, as a function of the attempt rate, is a straight line from (/e, /e) to (, 0). (b). The probability for a successful transmittion is Ps Q a (,n)q r (0,n)+ Q a (0,n)Q r (,n) and when n m we have Ps Q a (,m)q r (0,m) + Q a (0,m)Q r (,m) m 0 + ( q r ) m q r m( m )m m ( m )m and in case q r and q a are small we have Ps e (m) and (m) (m m)q a + mq r mq r, and the first expression we derived indeed e as m. (c). Any straight line from (,0) to (mq a,mq a ) will pass the departure curve only once. (d). We have (n) (m n)q a + nq r mq a + n(q r q a ) e + n( m me ) e + n m ( e ) so e n m ( e ) solving for n we get n m( e ), which is the number of backlogged nodes that corresponds e to any given attempt rate. Now we can compute the arrival rate (m n)q a nq r by inserting that expression for n and obtain ( e ) m m e e e e e which is our arrival rate expressed in terms of the attempt rate. In the equilibrium point this has to equal the departure rate e, so the equation we want to solve numerically 3

4 is e e which is (e )e or (e )e which we can iterate a few times from starting value to obtain the approximate solution (e). From (d) we have n m e e 4.5 e e so almost 9% of the arriving packets are not accepted into the system but discarded due to the node they are arriving to being backlogged. (a). iven that a specific packet has not been successfully transmitted before time slot i it will be transmitted successfully with probability pq r. Let T q be the the time a packet is queued up in the system, we then have P(T q 0) p P(T q ) P(T q T q > 0)P(T q > 0) pq r ( p) P(T q 2) P(T q 2 T q > )P(T q > ) pq r ( p)( pq r ) P(T q 3) P(T q 3 T q > 2)P(T q > 2) pq r ( p)( pq r ) 2. P(T q k) P(T q k T q > k )P(T q > k ) pq r ( p)( pq r ) k so we can compute the expected queueing time as E[T q ] kpq r ( pq r ) k ( p) p( p)q r k( pq r ) k k now consider k kxk d dx x p( p)q r (pq r) 2 k d dx xk d dx k k xk d dx x x ( x) 2. Inserting x pq r now yields E[T q ] ( p) pq r. After the queueing time it takes another time slot to transmit the packet so in total we have the expected delay T + T q + p pq r. (b). For a given backlogged packet, when we have n backlogged nodes, the probability for successful transmission is p ( q a ) m n ( q r ) n, since the m n not backlogged packet can t have any arrivals and the remaining n backlogged nodes should not retransmit. Thus p q r ( q a ) m n ( q r ) n e (qa(m n)+qrn) e (n). Similarly if we are given a newly arrived packet, the probability for successful transmission is p ( q a ) m n ( q r ) n q a ( q a ) m n ( q r ) n e (qa(m n)+qrn) e (n). 4

5 (c). Substituting p e into E[T q ] from (b) gives us E[T q ] e q re. In equilibrium we have e (m n )q a, i.e. e (m n )q a. Using that we now obtain E[T q ] (m n )q a q r (m n )q a (m n )q a (m n )q a q r (m n )q a + n q r (m n )q a (m n )q a q r n (m n )q a Thus, since T + T q we have the claimed result. (d). Attempt rate (n) (m n)q a +nq r mq a +n(q r q a ) n m, at equilibrium we have e (m n)q a, so we get ( n m )e ( n m ) 0.3( n m ) solving numerically for n m we get the stable equilibrium state n as n m 0.243, which is n n m/8, and we get T + 0.3( n m ) m + (8 ) m 2. m 2. 5