OF UNIVERSITY SOUTHERN CALIFORNIA, SPRING 2008
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EE549: Problem Set #2 Solutions
I. D EPARTURES a) (See Fig. 1). The total departures before time t is dened as D(t). These departures represent packets that D(t) were fully served (in FIFO order), and the service times thus expended a total time of i=1 Si throughout the interval [0, t]. It follows that this time duration is less than or equal to t.
S1 0
S2
S3 S4 S5
SD(t) SD(t)+1 t time
Busy Periods
Fig. 1. A timeline illustrating the inequality of part 1(a).
b) If D(t) does not increase to innity as t , then the departure rate is 0, which is certainly less than or equal D(t) to min[1/E {S} , ]. Here we consider the case when D(t) as t . From the inequality i=1 Si t (which holds for all t 0), we have: D(t) 1 t Si D(t) D(t)
i=1
Taking a limit as t and using the law of large numbers yields:
E {S} lim t t D(t)
and so the departure rate is less than or equal to 1/E {S}. However, because D(t) N (t) for all t (as the departures cannot be more than the arrivals), we know that:
t
lim D(t)/t lim N (t)/t =
t
because N (t) has rate . Therefore, the departure rate is less than or equal to and it is also less than or equal to 1/E {S}. Thus, the departure rate is less than or equal to min[1/E {S} , ]. c) If > 1/E {S}, we know that the departure rate satises limt D(t)/t min[1/E {S} , ] = 1/E {S} with prob. 1. We have that L(t) = N (t) D(t). Thus:
L(t) t t lim N (t) D(t) t t t = lim D(t)/t = lim
t
1/E {S} with prob. 1
Because the limit of L(t)/t is greater than or equal to the positive value 1/E {S} with probability 1, it follows that L(t) with probability 1.1
1 In fact, for any sample path L(t) that satises L(t)/t c > 0, for any t t . Thus, L(t) (c )t for all t t .
> 0 there exists a time t such that L(t)/t c
for all
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II. F IXED L ENGTH PACKETS AND THE M ULTIPLEXING I NEQUALITY a) Proof for Lsingle (t): Fix a time t. Suppose that Usingle (t) = 0. Then the single-server system is empty and so we also have Lsingle (t) = 0 (so that Lsingle (t) = Usingle (t) holds for this case). Now suppose that B Usingle (t) > 0. Then Lsingle (t) > 0. Also, because of FIFO service, there are exactly [Lsingle (t) 1] packets in the buffer (none of which have started their transmission), and there is exactly 1 packet in the server. Thus: Usingle (t) = (Lsingle (t) 1)B + R(t), where R(t) is the residual amount of bits to be processed in the server, and satises 0 < R(t) B . Thus:
Lsingle (t) = Usingle (t) R(t) Usingle (t) R(t) +1= + (1 ) B B B
Note that 1 > 1 R(t) 0. Because Lsingle (t) is an integer and it is greater than or equal to Usingle (t)/B and B less than Usingle (t)/B + 1, it follows that it is the smallest integer greater than or equal to Usingle (t)/B . L(t) Proof for Lmulti (t): Fix a time t. Note that Umulti (t) = i=1 Ri (t), where Ri (t) is dened as the residual amount of unprocessed bits of packet i in the system at time t. We thus have:
Lmulti (t)
Umulti (t) =
i=1
Ri (t) Lmulti (t)B
where the nal inequality follows because Ri (t) B for all i. Thus, Lmulti (t) is an integer and Lmulti (t) Umulti (t)/B . Therefore, Lmulti (t) Umulti (t)/B . b) We know by the multiplexing inequality that Usingle (t) Umulti (t) for all t. Therefore Usingle (t)/B Umulti (t)/B for all t. From part (a) we also have that Lsingle (t) = Usingle (t)/B and Lmulti (t) Umulti (t)/B for all t. Thus, Lsingle (t) Lmulti (t) for all t. c)
X1(t) (Poisson)
X1(t) (Poisson) X2(t) (Periodic)
L(t)
= 6 kb/sec
L1(t) L2(t)
2=4 kb/sec 2=2 kb/sec
X2(t) (Periodic)
Fig. 2.
The packet version of the multiplexing inequality for problem 2c.
From part (b), we know that L(t) L1 (t) + L2 (t), where L1 (t) and L2 (t) are the total number of packets in the multi-queue system shown in Fig. 2, with constant transmission rates 1 and 2 such that 1 + 2 = 6. Therefore:
L L1 + L2
(1)
Let 1 = 4 and 2 = 2. The queue L1 (t) is thus M/D/1 with loading = 1 B/1 = 3/4. Thus, by the M/D/1 formula, we have: (3/4)2 + 3/4 = 15/8 = 1.875 L1 = 2(1 3/4) The queue L2 (t) has a periodic packet of size 2 kb arriving every second, and has a service time of 1 second, and so there is never more than one packet in queue 2 (i.e., L2 (t) = 1 for all t, and so L2 = 1). Combining these expressions for L1 and L2 in the inequality (1) yields:
L 23/8 = 2.875 packets
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III. B ROADCASTING WITH THE L AW OF L ARGE N UMBERS a) Each state of the Markov chain represents a number of users that have the current packet (so that no users have the packet in state 0, and all users have the packet in state 3). The rst equation states that T 0 is equal to the sum of 1 (as we must transmit the packet at least once), plus the expected remaining time (conditioning on all possible states that we can be in after the rst transmission). No users receive the rst transmission with probability (1 p)3 , and hence with this probability we are still in state 0 after the rst transmission, so the expected remaining time is still T 0 . One user successfully receives the packet after the rst transmisison with probability 3p(1 p)2 , and hence with this probability the expected remaining time is T 1 , etc. b) Solving the equations gives: S 2 = 1/p = 1.11111, S 1 = 1.21212, S 0 = 1.3040. c) Let Zi be the service time of packet i, dened as the time required for all 3 users to receive packet i, starting from when it is rst transmitted by the satellite. Clearly {Zi } are i.i.d. and have mean E {Zi } = S 0 = 1.3040. i=1 Thus, D(t) is a renewal process with i.i.d. inter-renewal times Zi , and so the time average departure rate is equal to 1/S 0 = 0.7669 packets/sec. IV. M ULTIPLEXING I NEQUALITY, S LOW T RUCKS ,
AND
D ELAY
a) If the multi-server system never empties, then we are done, as we trivially have that the single-server system empties before the multi-server system. Else, let t be the time when the multi-server system empties. We know by the multiplexing inequality that Usingle (t) Umulti (t) for all t, and hence:
Usingle (t ) Umulti (t ) = 0
It follows that Usingle (t ) = 0, and so the single-server system is also empty at time t . Thus, it must have emptied either on or before the time t , the multi-server system emptied. b) Let B1 = B kb. Let B2 = B3 = . . . = B100 = 1 kb. In the single-server system, we have:
single single single single = B/2 + 2/2, . . . , W100 = B/2 + 99/2 = B/2 + 1/2, W3 = B/2, W2 W1
That is, Wisingle = B/2 + (i 1)/2 for i {2, 3, . . . , 100}. Therefore:
W single = 1 100
100
[B/2 + (i 1)/2]
i=1
= B/2 +
1 200
99
i
i=0
1 = B/2 + (99)(100)/2 200 99 = B/2 + 4 In the multi-server system: Suppose that B/1 99/1, so that the rst packet takes longer to serve in the multi-server system than all 99 other packets. Thus, we have:
single single single single W1 = B, W2 = 1, W3 = 2, . . . , W100 = 99
and so:
W multi = = = = B 1 + 100 100 B 1 + 100 100
100
Wimulti
i=2 100
(i 1)
i=2
B 1 + (99)(100)/2 100 100 B 99 + 100 2
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Therefore:
W multi B/100 + 99/2 1/50 + 99/B = = B/2 + 99/4 1 + 99/(2B) W single
Let B = 9900. Then 99/B = 1/100, and we have:
W multi 1/50 + 1/100 = 6/201 < 1/30 = 1 + 1/200 W single
V. DYNAMIC ROUTING TO PARALLEL S ERVERS a) The JSQ strategy routes to the queue i {1, . . . , K} such that:
i = arg min
i{1,...,K}
Ui (t )/i
(JSQ)
The Greedy strategy routes to the queue i {1, . . . , K} such that:
i = arg min
i{1,...,K}
(Ui (t ) + B)/i
(Greedy)
If i = (equal rates) for all i {1, . . . , K}, then the Greedy strategy chooses the queue i with the smallest value of Ui (t )/ + B/, which is the same as choosing the queue with the smallest value of Ui (t )/, which is the same as the JSQ strategy. b) Due to a typo, there was no part (b) for this problem. c) Suppose that we use the Greedy policy for all time, and that Ui (t) = 0. Consider any queue j . We want to show that Uj (t) Bmax j /i . If Uj (t) = 0, then we are done. Else, let B be the size of the packet in queue j that is last in line (so that it is either last in the buffer, or there are no packets in the buffer and it is being processed in the server. The time that this packet will exit queue j is exactly t1 = t + Uj (t)/j . Because this packet was routed according to the Greedy strategy, this time t1 must be less than or equal to the time it would have emptied queue i if it was routed there, which is less than or equal to t2 = t + B/i (because it would have started its service in queue i at least by time t). Therefore, t1 t2 , and so:
t + Uj (t)/j t + B/i t + Bmax /i
Rearranging terms in the above inequality yields:
Uj (t) Bmax j i
d) Let Ugreedy (t) be the total unnished work under the greedy strategy, and let Uother (t) be the total unnished work under any other routing policy. Fix a particular time t. Case 1: Suppose time t is not in a fully loaded interval of the Greedy system. In this case, the Greedy system has at least one idle server (so that Ui (t) = 0 for at least one queue i), and so Uj (t) Bmax j /i for all j = i. It follows that:
Ugreedy (t) =
j=i
Uj (t)
j=i
j Bmax i
Bmax
j=i
j Bmax 1
K j=1
j Bmax 1
K j=1
j Uother + (t) 1
Thus, the result holds for Case 1. Case 2: Suppose time t is in a fully loaded interval of the Greedy system.
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In this case, let tf be the start of the current fully loaded interval, and note that tf t and that all servers of the Greedy system transmit at the full offered transmission rate during the interval from tf to t. Thus:
Ugreedy (t) = Ugreedy (t ) + X[tf , t] Ygreedy [tf , t] f = Ugreedy (t ) + X[tf , t] f
K t
(2) (3) (4)
( )d
tf t
Bmax
j=1 K
j + X[tf , t] 1
( )d
tf
Bmax
j=1 K
j + X[tf , t] Yother [tf , t] 1 j + Uother (t ) + X[tf , t] Yother [tf , t] f 1 j + Uother (t) 1
(5)
Bmax
j=1 K
(6)
= Bmax
j=1
(7)
where (2) holds by the I-O equation, (3) holds because the Greedy strategy is fully loaded during [tf , t], (4) holds because the time just before tf is not a fully loaded time, so that the total unnished work is upper bounded by j Bmax K 1 as shown in Case 1, and (5) holds because the other policy cannot have more bit departures over j=1 the interval [tf , t] than the sum of the integrals of the offered transmission rates over that interval. Finally, (6) and (7) follow by the I-O equation. Thus, the result also holds for Case 2, and we are done. e) The JSQ strategy ensures that it keeps unnished work is no more than (K 1)Bmax beyond that of any other policy. The Greedy strategy ensures that unnished work is no more than Bmax K (i /1 ) beyond any i=1 other strategy. This bound can be arbitrarily large (depending on the ratio of the largest rate to the smallest rate). Thus, the JSQ bound is better. However, this does not mean that one policy is better than the other. Indeed, the greedy policy can perform better in terms of delay in many cases (such as when only one packet arrives), even if it has a worse bound. The Greedy and JSQ policies are exactly the same if i = for all i. VI. R ATE S TABILITY FOR M ULTI -S ERVER , S HARED B UFFER S YSTEMS a) Note by the tracking inequality we have:
Usingle (t) Umulti (t) Usingle (t) + (K 1)Bmax for all t
where Usingle (t) represents a single-sever queue with the same input process but with a transmission rate process (t) = 1 (t) + . . . + K (t). Therefore:
t
lim
Because (K 1)Bmax
Usingle (t) Usingle (t) Umulti (t) (K 1)Bmax lim lim + lim t t t t t t t is a constant, we have (K 1)Bmax /t 0 as t , and so:
t
Usingle (t) Usingle (t) Umulti (t) lim lim t t t t t Now note that the time average rate of (t) is given by 1 +. . . K . Thus, by the rate stability theorem, Usingle (t) is rate stable if and only if r 1 + . . . K . Thus, if r K i , we have limt Usingle (t)/t = 0 with prob. i=1 1, and so with prob. 1 we have: 0 lim Umulti (t)/t 0 lim
t
which means that limt Umulti (t)/t = 0 with prob. 1, and so the multi-server system is rate stable. b) Suppose that r > K i . Note that: i=1
t K
Umulti (t) X(t)
0 i=1
i ( )d
(8)
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That is, Umulti (t) is at least as large as the total bit arrivals minus the maximum possible bit departures. Dividing (8) by t and taking limits yields (with prob. 1):
Umulti (t) lim t t X(t) lim t t
K K i=1
1 t t lim
t
i ( )d
0
= r
i=1
i > 0
Therefore, Umulti (t)/t does not tend to 0 as t , and so Umulti (t) is not rate stable. VII. R ATE S TABILITY FOR A S ERVER S CHEDULING P ROBLEM Suppose we can design a server allocation rule that yields well dened time average server rates i for all queues i. Note by the rate stability theorem that Ui (t) is then rate stable if and only if ri i . It follows that we just need to design the policy to achieve time average server rates i such that i ri for all i. The following lemma about probabilistic server allocation is useful: Lemma 1: Suppose that every slot, a server is placed to queue i i.i.d. with probability p. Then i = i p. Proof: Let tk = kT be the time at which the k th timeslot ends (for k {1, 2, . . .}). Then:
1 tk
tk
i ( )d
0
=
1 kT 1 kT
k
mT
i ( )d
m=1 (m1)T k
=
i T 1m
m=1
= i
1 k
k
1m
m=1
where 1m is an indicator function that is 1 if queue i is selected on slot m, and zero else. Note that {1m } are m=1 i.i.d. with E {1m } = p. Then:
1 k tk lim
tk
i ( )d
0
= i
1 k k lim
k
1m
m=1
= i p with prob. 1
and thus i (t) has a well dened time average rate that is given by i p. a) Let r1 = .9 kb/sec, r2 = .4 kb/sec, and r3 = .5 kb/sec. Suppose that 1 = 2 = 3 = 1 kb/sec. Independetly every slot, ip a coin and make the following decisions: Serve queues 1 and 2 with probability 0.5. Serve queues 1 and 3 with probability 0.5. Let pi = P r[Serve queue i]. Then clearly p1 = 1, p2 = p3 = 0.5. Therefore: 1 = p1 = 1 kb/sec, 2 = p2 = 0.5 kb/sec, 3 = p3 = 0.5 kb/sec. Then ri i for i {1, 2, 3}, and so all queues are rate stable. b) Let r1 = .9 kb/sec, r2 = .45 kb/sec, and r3 = .6 kb/sec. Supppose that we have the same system as part (a), so that 1 = 2 = 3 = 1 kb/sec. Independently every slot, ip a coin and make the following decisions: Serve queues 2 and 3 with prob. 0.1. Serve queues 1 and 2 with prob. 0.4. Serve queues 1 and 3 with prob. 0.5. Then:
p1 = 0.9 p2 = 0.5 p3 = 0.6 = = = 1 = 0.9 r1 2 = 0.5 r2 3 = 0.6 r3
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Thus, by the rate stability theorem, all queues are rate stable. c) For all time t 0 and for any i {1, 2, 3}, we have:
t
Ui (t) Xi (t)
0
i ( )d
Xi (t) t
where the nal inequality holds because i ( ) 1 for all . Dividing by t yields:
Xi (t) Ui (t) 1 t t Taking a limit yields limt Ui (t)/t ...