ee132b final review sol

EE132B-Final Sample - Sol. UCLA 2014 Fall Prof. Izhak Rubin

Problem 1(a) Use Dijkstras algorithm, showing all steps, to derive the shortest path tree

rooted at node 1 for the network graph with indicated link weights as shownbelow.

5 4

73

P

6

3

12

2

1

3 6

3

4

5

Figure 1: The network graph with indicated link weights

(b) Assume each link to (independently) fail, over a specified period of time, withprobability 0.15. Calculate the probability that the shortest path from node 1to node 6 stays intact over this period of time.

Ans:

(a) Step 1:

P = {1}, T = {2, 3, 4, 5, 6} d(1, 2) = 5 d(1, 3) = 3 d(1, 4) = d(1, 5) = d(1, 6) =

Step 2: i = 3 P = {1, 3}, T = {2, 4, 5, 6}

Step 3: d(1, 2) = min{5, 3 +} = 5 d(1, 4) = min{, 3 + 7} = 10 d(1, 5) = min{, 3 +} =

1


d(1, 6) = min{, 3 + 12} = 15 Step 2:

i = 2 P = {1, 2, 3}, T = {4, 5, 6}

Step 3: d(1, 4) = min{10, 5 + 4} = 9 d(1, 5) = min{, 5 + 6} = 11 d(1, 6) = min{15, 5 +} = 15

Step 2: i = 4 P = {1, 2, 3, 4}, T = {5, 6}

Step 3: d(1, 5) = min{11, 9 + 3} = 11 d(1, 6) = min{15, 9 +} = 15

Step 2: i = 5 P = {1, 2, 3, 4, 5}, T = {6}

Step 3: d(1, 6) = min{15, 11 + 3} = 14

Step 2: i = 6 P = {1, 2, 3, 4, 5, 6}, T =

Nodes Preceding node along the shortest path, Shortest distance2 (1,5)3 (1,3)4 (2,9)5 (2,11)6 (5,14)

2


(b)The shortest path from node 1 to node 6: 1 2 5 6. The probability that

the the path is intact over the period is given by

p = (1 0.15)3 = 0.6141. (1)

Problem 2Consider a communications link that employs two (time simultaneously used)

communications channels, Channel 1 and Channel 2, operating at data rate 10 Mbpsand 20 Mbps, respectively. The bit error rates measured across Channel 1 and Chan-nel 2 are equal to 5 105 and 104, respectively. Each message frame contains4000 information bits and 400 overhead bits. A Select-and-Repeat ARQ error-controlscheme is employed.(a) Calculate the maximum throughput rates realized across each channel and

across the link.

(b) For a randomly selected message (that is equally likely to use either channel),calculate the average number of transmissions incurred until it is correctly re-ceived.

(c) Assume that resources are provided to enable the reduction of the bit error rateof one of the channels by 50%. Determine which channel should be improvedand compute the percentage improvement in link maximum throughput levelthat is then achieved.

Ans:

(a)Under selective repeated ARQ scheme, the throughput rate is given by

= [K n(h)] (1 PE)RK

. (2)

For channel 1, PE1 = 1 (1 5 105)4400 = 0.1975. The throughput rate of channel1 is given by

1 =4000 (1 PE1) 10 106

4400 7.3 Mbps. (3)

For channel 2, PE2 = 1 (1 104)4400 = 0.3560. The throughput rate of channel2 is given as

2 =4000 (1 PE2) 20 106

4400 11.71 Mbps. (4)Thus, the throughput rate across the link is given by

link = 1 + 2 = 19.01 Mbps. (5)

3


(b)

E[NT ] =12E[NT1 ] +

12E[NT2 ] =

12

11 PE1

+ 121

1 PE2= 1.4. (6)

(c)

By improving channel 1, we have P E1 = 1 (1 51052

)4400= 0.1042 and the

improved throughput rate is given by

1 =4000 (1 P E1) 10 106

4400 8.144 Mbps. (7)

By improving channel 2, we have P E2 = 0.1975 and the improved throughput rate isgiven by

2 14.591 Mbps. (better!) (8)

Thus, the improved throughput rate across the link is given by

link = 1 + 2 21.89 Mbps, (9)

which is improved by 15.16%.

4


Problem 3(a) Consider the following operation of a transport layer processor at a user ter-

minal as it transmits messages across a TCP transport connection. Assumemessages directed by the application entity for transmission across this trans-port connection to arrive in accordance with a Poisson point process at a rateof 420 Kbps. A message transmission time across the transport connection isassumed to take random time and to follow an exponential distribution. On av-erage, a message contains 6000 bits. Compute the minimum data rate requiredfor transmission across the connection to ensure the message total average delay(in waiting and in transmission) to be no higher than 25 msec.

(b) For the system designed in (a), calculate the probability that a message willhave to wait in the FIFO queue for longer than 50 msec.

Ans:

(a)

This is M/M/1 queueing system. E[L] = 6000 bits, = 420 kbps6000 bits/msg = 70[msg/sec] and = R6000 [msg/sec]. As a result,

E[D] = 1(1 ) =

1 =

1R

6000 70 25 103. (10)

Thus, R 660 bps.

(b)

P (W > 50 103) = 1 P (W 50 ms)= 1

(1 e(1)50ms

)= 0.0861.

(11)

5


Problem 4Consider an access system that connects user sessions to the Internet. The system

uses two outgoing communications lines to connect to the Internet; each line operatesat a data rate of 2 Mbps. Session connection requests arrive at the access system inaccordance with a Poisson process. The system provides for no queuing of requestsfor session access. If both outgoing lines are busy, a session access request is blocked.Once a session requested is granted, the session will occupy the assigned outgoingline for a random period of time that is assumed to be exponentially distributed.On the average, a user will transmit across the line an average of 6000 bits beforeit terminates the session, freeing the line for use by a future arriving session. Thesystem designer aims to ensure that the blocking probability of session request is nothigher than 0.1 (so that no more than 10% of session requests are blocked). Calculatethe maximum session arrival rate that the designer should plan to support to assureclients with his declared session blocking probability objective.Ans:This is an M/M/2/2 QS. = R6000 = 333.33 [msg/sec], f =

and =

m. As a

result, we have

PB = P (2) =a22i=0 ai

= a2a0 + a1 + a2

=f2

21 + f + f22

0.1. (12)

Then, 0.45f 2 0.1f 0.1 0 f 0.5954 0.5954 = 198.46 [msg/sec].

6

ee132b final review sol

Documents