EE555-USC-Spring08-Silvester-HW4 Solutions

EE555-USC-Spring08-Silvester-HW4 Solutions - EE555...

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Unformatted text preview: EE555 Broadband Network Architectures (Spring 2009) "Last Homework" Solutions Section A: Access Networks Question A.1. Power: The following table shows the received power at each house based on an output power of 1 from the CO. We see that there is more loss to the last house in Plan B. Consequently more power would have to be inserted into the system in Plan B; also the variability in received powers is larger in plan B. PLAN A House CO out Splitter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Cost: Plan A requires the splitter and home runs to all the houses. Total length of fiber is (100+200+...+3200) = 52,800 metres. Plan B requires 32 taps and 3,200 metres of fiber. Raw fiber costs are ~$0.25 per metre but installation costs would likely dominate. Plan B is likely to have a higher installation and maintenance cost due to the taps even though they do not fail often as they are passive optical devices. Reliability: Plan A is more reliable since each house has its own connection whereas in B any failure or break will affect a large number of houses. Loss factor Power (dB) level 1 0.0158 0.0157 0.0156 0.0155 0.0154 0.0153 0.0152 0.0151 0.0150 0.0149 0.0148 0.0147 0.0146 0.0145 0.0144 0.0143 0.0142 PLAN B Power Level 1.0000 0.0400 0.0380 0.0361 0.0343 0.0326 0.0310 0.0294 0.0279 0.0265 0.0252 0.0239 0.0228 0.0216 0.0205 0.0195 0.0185 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 18.5100 18.5400 18.5700 18.6000 18.6300 18.6600 18.6900 18.7200 18.7500 18.7800 18.8100 18.8400 18.8700 18.9000 18.9300 18.9600 0.0141 0.0140 0.0139 0.0138 0.0137 0.0136 0.0135 0.0134 0.0133 0.0132 0.0132 0.0131 0.0130 0.0129 0.0128 0.0127 0.0176 0.0167 0.0159 0.0151 0.0143 0.0136 0.0129 0.0123 0.0117 0.0111 0.0105 0.0100 0.0095 0.0090 0.0086 0.0082 House PLAN A Loss factor Power (dB) level PLAN B Power Level 18.0000 18.0300 18.0600 18.0900 18.1200 18.1500 18.1800 18.2100 18.2400 18.2700 18.3000 18.3300 18.3600 18.3900 18.4200 18.4500 18.4800 Convenience: Plan B The taps would likely be installed at time of deployment and would be located on the poles (or in the vault in front of each house). Connecting up a new house would require a connection from the tap to the house. Plan A when a new house is activated a new fiber has to be run from the BDU to the house (when the fiber is aerial) which is more timeconsuming; or if the fibers are stubbed to each house (aerial or subterranean) a connection has to be made from the stub to the house. (FYI Verizon uses Plan A.) Question A.2 a) The arrival process (frame to frame) is Binomial. qi (t + 1) = max{qi (t ) - 1, 0} + ai (t + 1) = qi (t ) - qi (t ) + ai (t ) N ai (t ) = p i (1 - p ) N -i i You can solve this using the same approach as we used for M/G/1 since the equations are similar. (1 - q0 ) + a 2 - 2a (1 - q0 ) a + a (1 - p ) + a 2 - 2a 2 a (2 - p ) - a 2 N= = = 2(1 - a ) 2(1 - a ) 2(1 - a ) Since q0 = 1 - a ; a = Np; and a 2 = Np (1 - p ) + ( Np )2 This is the mean number is system. So the delay is T = N / a measured at the frame time scale. To adjust to slot time scale we first compute the mean waiting time (in frames) by W = T - 1 , and compute the system time in slots as: S = WN + 1 + N ( N - 1) where the first term converts the waiting to slots, 2 the second term is packet transmission time (1 slot) and the third term is the average time from when an arrival occurs (sometime during a frame) until it is recognized in the frame level system process. b) This is trickier than I had intended (I was thinking about DFT see simulation comment below). It seems that the state needs to contain the buffer occupancy and the S/B state for each node so we need a full vector state description in order to be able to write the system evolution equations. X (t ) = [q (t ), s (t )] q (t ) = [q1 (t ), q2 (t ),... , qN (t )] where qi (t ) = queue size at i at time t s (t ) = [ s1 (t ), s2 (t ),... sN (t )] where si (t ) {S , B} is state of i at time t From which you can write a set of system state equations. BONUS: These are simulation results for the fixed TDM solution, the Slotted Aloha (IFT Immediate first transmission) which is the protocol described in the homework, and Slotted Aloha (DFT delay first transmission) where a node having a packet to send transmits with probability even on the first attempt. DFT is inferior to IFT in terms of delay but has the same limiting throughput and is easier to analyse, so it is often studied for stability analysis. In the simulations I used =1/10 which is necessary for stability of SA; and varied the arrival probability, p, to vary the throughput (successful transmissions). Also shown are the analytical results for TDM. Tput Delay SADFT SAIFT 0.01 10.966 1.200 0.05 11.160 1.618 0.10 12.402 2.294 0.15 14.060 3.268 0.20 16.942 4.634 0.25 21.780 6.392 0.30 29.776 11.006 0.35 50.018 21.462 0.40 0.50 0.60 0.70 0.80 0.90 TDM 5.366 5.580 6.004 6.266 6.690 7.006 7.616 7.902 8.582 10.166 12.302 15.980 24.812 42.904 Analytical S (TDM) 5.545 5.737 6.000 6.294 6.625 7.000 7.429 7.923 8.500 10.000 12.250 16.000 23.500 46.000 60 Delay SADFT 50 Delay SAIFT 40 Delay Delay TDM 30 20 10 0 0.00 0.20 0.40 Load 0.60 0.80 1.00 Section B: Switches and Routers Question B.1 The number of bytes entering the memory per second is nC where C is the line speed in bytes per second. Similarly the memory must deliver nC bytes to the outputs every second. Each memory read/write is for w bytes and takes t*109 seconds; so the maximum memory rate is w*109/t bytes per second. 50 10 150 MBytes per second 1.2 Gbps 2 320 [The packet size does not really figure in this solution since it is a multiple of the word size, it would be a factor if b/w was not an integer] 2 10 10 Question B.2 i) f1 f2 P = 12 f1 0 f2 / 2 f1 f 2 f1 f 2 f1 / 2 f2 2 f1 f 2 f1 f 2 f1 / 2 0 f 22 f 22 f2 ii) Solve = P either algebraically or numerically. iii) 1 = 11 + 12 + 21 and 2 = 12 + 21 + 22 The algebra is tedious but you can verify the fractions match can also look at a numerical solution to verify. iv) See below. Note that at f1 = 0.5 the total through is 2*0.75 as we saw in class and that for any case the total throughput is lower. tput 1.60 1.40 1.20 1.00 0.80 0.60 0.40 0.20 0.00 0.0 0.2 0.4 0.6 0.8 1.0 Question B.3 a) A switch is strictly nonblocking if any idle input can be connected to any idle output without any rearrangement of existing connections, and rearrangeably nonblocking if this can be accomplished with rerouting of some of the existing connections. b) No. Suppose the packet on input 00 has a packet for output 00 and this was accomplished by setting all 3 switches in the top row to be straight through; and the packet on input 11 has a packet destined for output 10, which was accomplished by setting the first two switches in the bottom row to straight and the last to crossed; it is now not possible to route a packet destined to output 11 from input 01 without rearranging the 1110 connection. It is rearrangeably nonblocking can base this on the fact that this is a Benes design. Question B.4 A banyan switch can route any input to any output (but not necessarily all at the same time). So with two in series, there will be n paths between any source to any destination (pick any intermediate destination and there is a route from the input to it and also from it to the desired output; and you can use any of the intermediate destinations. Proof of rearrangeably nonblocking property (RNB). This is easy to see for a small Banyan (say 4x4) and with some rearranging (make the second a mirror image of the first), 2 tandem 8x8 switches can be drawn in a Benes design with the core consisting of 4x4 blocks. Similar for 16x16 etc. And Benes designs are RNB. Question B.5 a) HOL blocking arises when packets arriving at different input ports are destined for the same output port. If a HOL packet of a certain buffer at the input cannot be switched to an output because of contention, the rest of the packets in that buffer are blocked by the HOL packet, even if there is no contention at the destined output ports for those packets. It has severe performance degrading effects. b) By using virtual output queues; by speeding up the switch fabric. c) Internal blocking arises when certain paths through the switch cannot be supported. d) To provide alternate paths (either by using parallel fabric or having more stages). (note you also needed to give examples for a and c) Question B.6 You need to draw an 8x8 optical switch with 4 external input and 4 recirculating inputs going through delay lines. You may have one delay line of length 4, one of delay 2, and 2 of delay 1 (other combinations are possible). When contention for an output occurs one of the packets is routed to a delay line. Which delay line to use is selected by trying to avoid future collisions with other packets that are recirculating. We gave an example of this in class for a smaller switch. Section C C: Traffic Ma anagement Question C.1 Markov Chain for the num mber of toke available at the beginning of a s ens slot (assumes that s a) M pa ackets arrivin ng during a slo ot consume a available toke ens before the e new token i is added): g that the system starts wi ith a nonemp pty token buc cket). Selfloo ops are not sh hown. (assuming k -1 p p2 k +1 = po k = 0 p2 0 1 k 5 k =1 5 k =1 k -1 p0 p2 k = 1 2 3 4 p0 p0 p0 p0 1+ + + + p2 p2 p2 p2 b) # . Question C.2 me opens" the pa acket is imme ediately transm mitted (i.e. go to oes Bucket rate is r. Assum that once the bucket "o t nd er buffer. the output line card) an is no longe counted as being in the b a) Let q ( n) = Number of packets i the buffer at the nth d r in r departure (in a busy perio od) Then q ( n + 1) = max{ {min{q ( n) + a , B} - 1, 0} where B is the bucket depth and a is a random v variable descr ribing the num mber of arrivals is an inter rval nding to the t time between n packet depa artures as det termined by t the bucket lea ak rate r. correspon There are e three states in this Marko ov Chain {0,1, ,2} and the transition matr rix is: a0 + a1 a2 1 - a0 - a1 - a2 P = a0 a1 1 - a0 - a1 0 1 - a0 a0 and ai = by: ( t ) i - t e i! where t = 1/ r and r is th bucket rat . The expe w he te ected number of losses is given L = 0 ai (i - 3) + 1 ai (i - 2) + 2 ai (i - 1) i=4 i =3 i=2 And the lo oss rate is B = L / a = L / t . Finding t mean delay is tricky s the since arrivals do not see a queue dist s tribution corresponding t the to states giv above. Also, the "res ven A sidual life" of the packet in service (a actually the r residual life o the of token gen neration) is hard to compu ute since the ere are losses. I think you need to mod del this differ rently to solve th his part. nsider the fee ed back system m as shown b below. b) We con satisfy (base ed on B which h is a function n of '). We need to find ' to s Question n C.3 a) FC CFS Packet id d (group.packet#) P1.1 P2.1 P1.2 P1.3 P2.2 P1.4 P2.3 P2.4 b) Non preemptive priority Packet id d (group.packet#) P1.1 P2.1 P1.2 P1.3 P2.2 P1.4 P2.3 P2.4 Arrival time 1 2 3 4 6 8 9 10 Av verages Departure time 4 7 19 22 10 25 13 16 W1 0 13 15 14 10.5 W2 2 1 1 3 1.75 W W 0 2 13 1 15 1 1 14 1 1 3 6.1 125 Arrival time 1 2 3 4 6 8 9 10 Av verages Departure e time 4 7 10 13 16 19 22 25 W1 0 4 6 8 4.5 W2 2 7 10 12 7.75 W 0 2 4 6 7 8 10 12 6. .125 c) Weighted Fair Queueing WFQ t 0.0 1.0 2.0 3.0 4.0 4.0 6.0 7.0 8.0 9.0 10.0 10.0 13.0 16.0 19.0 22.0 25.0 U(t) 0.0 3.0 5.0 7.0 6.0 9.0 10.0 9.0 11.0 13.0 12.0 15.0 12.0 9.0 6.0 3.0 0 dV/dt V(t) A/D 3.0 0.0 A 1.0 3.0 A 1.0 4.0 A 1.0 5.0 D 1.0 5.0 A 1.0 7.0 A 1.0 8.0 D 1.0 9.0 A 1.0 10.0 A 1.0 11.0 D 1.0 11.0 A 1.0 14.0 D 3.0 17.0 D 3.0 26.0 D 3.0 35.0 D 0.0 D Gp Pk ID 1 2 1 1 1 2 2 1 2 2 2 1 2 2 1 1 ID 1 1 2 1 3 2 1 4 3 2 4 3 3 4 3 4 Sz 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 S F 0.33 0.67 1 0 1 1 2 1 2 2 2 3 3 3 3 2 2 2 1 0 2 0 0 1 1 1 1 2 1 1 2 1 2 2 1 0 0 0 W BAR W1 W2 W 1.1 arrives, enters service 2.1 arrives 1.2 arrives 0 1.1 departs, 2.1 enters service 1.3 arrives 2.2 arrives 2 2.1 departs; 2.2 enters service 1.4 arrives 2.3 arrives 1 2.2 departs; 1.2 enters service 2.4 arrives 7 1.2 departs; 2.3 enters service 4 2.3 departs; 2.4 enters service 6 2.4 departs; 1.3 enters service 15 1.3 departs; 1.4 enters service 14 1.4 departs; end of busy period 6.125 0.0 9.0 3.0 7.5 9.0 18.0 18.0 27.0 7.5 12.0 27.0 36.0 12.0 16.5 16.5 21.0 0 2 1 7 4 6 15 14 9 3.25 (*) You could also select 2.3 here since there is a tie in S(2,3) an S(1,3) Note that the overall average wait (W) is the same for all. Question C.4 a) This is an M/M/1/K queue. For /=1, we find that all states are equally likely, so: i = 1/ ( K + 1) Blocking probability: B = K = 1 K +1 0i K Throughput: = (1 - B) = K K +1 1 K Alternately, = (1- 0 ) = 1 - = K +1 K +1 since = b) k +1 = So, K -k k K k K! k = 0 ( K - k )! K 0k K and find 0 by summing over all states. The throughput is most easily computed as = (1 - 0 ) . For the case of /=1 k = K! 1 0 ( K - k )! K k 1 k K -1 i -1 K K! 1 = 1 - i =0 ( K - i )! K Question C.5 The Markov Chain is as follows: Denotes a state with i packets in the buffer before congestion has occurred Denotes a state with i packets in the buffer after congestion has occurred You were not asked to solve the model. ...
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