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Due: April 8, 2004 Spring 2004

ENEE 426: Communication Networks

Dr. Narayan

TA: Quang Trinh

Problem Set 2 Solution

1. (3.57)

An early code used in radio transmission involved using codewords that consist of binary bits and contain

the same number of 1s. Thus, the 2-out-of-5 code only transmits blocks of 5 bits in which 2 bits are 1 andthe others 0.

Solutions follow questions:

a. List the valid codewords.

11000

1010010010

10001

0110001010

01001

00110

00101

00011

b. Suppose that the code is used to transmit blocks of binary bits. How many bits can be transmitted

per codeword?

There are 10 possible codewords. Three bits per codeword can be transmitted if eight codewords areused.

c. What pattern does the receiver check to detect errors?

Each received codeword should have exactly two bits that are ones and three bits that are zeros to be a

valid codeword.

d. What is the minimum number of bit errors that cause a detection failure?

A valid codeword can be changed into another valid codeword by changing a 1 to a 0 and a 0 to a 1.

Therefore, two bit errors can cause a detection failure.

2. (3.63)

Letg1(x) = x + 1 and letg

2(x) = x

3

+ x2

+ 1. Consider the information bits (1,1,0,1,1,0).

a. Find the codeword corresponding to these information bits ifg1(x) is used as the generating

polynomial.

Codeword = 1101100

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b. Find the codeword corresponding to these information bits ifg2(x) is used as the generating

polynomial.

Codeword = 110110111

c. Cang2(x) detect single errors? double errors? triple errors? If not, give an example of an error

pattern that cannot be detected.

Single errors can be detected since g2(x) has more than one term. Double errors cannot be detected

even though g2(x) is primitive because the codeword length exceeds 2

n-k

-1=7. An example of such

undetectable error is 1000000010. Triple errors cannot be detected since g2(x) has only three terms.

d. Find the codeword corresponding to these information bits ifg(x) = g1(x) g

2(x) is used as the

generating polynomial. Comment on the error-detecting capabilities ofg(x).

Codeword = 1101100011

The new code can detect all single and all odd errors. It cannot detect double errors. It can also detect

all bursts of length n k = 4 or less. All bursts of length 5 are detected except for the burst that equals

g(x). The fraction 1/2

n-k

= 1/16 of all bursts of length greater than 5 are detect

3. (5.18)

A 64-kilobyte message is to be transmitted from the source to the destination. The network limits packets to

a maximum size of two kilobytes, and each packet has a 32-byte header. The transmission lines in the

network have a bit error rate of 106, and Stop-and-Wait ARQ is used in each transmission line. How long

does it take on the average to get the message from the source to the destination? Assume that the signal

propagates at a speed of 2 x 105 km/second.

Solution:

Message Size 65536 bytesMax Packet Size 2048 bytes

Packet Header 32 bytes

Available for info 2016 bytes# of packets needed 32.51 packets

Total 33 packets

bit error rate 1E-06bits/packet 16384

Probability of error in packet 0.016251 1 (1 bit_error_rate) ^ (bits/packet)

Propagation speed 2E+05 Km/s

Distance 1000 Km Bandwidth 1.5 Mb/s

We assume that the ACK error, the ACK time, and processing time are negligible.

Tprop = distance / propagation speed = 0.0050 s

Tf = packet size / bandwidth = 0.0109 s

T0 = Tprop + Tf= 0.0159 s

Pf = probability of error in packet = 0.016251 E[Ttotal] = T0/ (1 - Pf) = 0.0162

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There is pipelining effect that occurs as follows: After the first packet arrives at switch 1, two

transmissions take place in parallel. The first packet undergoes stop-and-wait on the second link whilethe second packet undergoes stop-and-wait in the first link. The packet arriving at the switch cannot

begin transmission on the next link until the previous packet has been delivered, so there is an

interaction between the transmission times of the two packets. We will neglect this effect. The time to

send every packet over two links is then the initial packet transmission time + 33 additional packet

timess, and so the average time is E[Ttotal] * 34 = 0.522 seconds.

4. (5.20)

The Trivial File Transfer Protocol (RFC 1350) is an application layer protocol that uses the Stop-and-Wait

protocol. To transfer a file from a server to a client, the server breaks the file into blocks of 512 bytes and

sends these blocks to the client using Stop-and-Wait ARQ. Find the efficiency in transmitting a 1 MB fileover a 10 Mbps Ethernet LAN that has a diameter of 300 meters. Assume the transmissions are error free

and that each packet has 60 bytes of header attached.

Solution:

The propagation delay in an Ethernet LAN is negligible compared to the total transmission time of a packetfrom start to finish. Ignoring processing time and using the terminology in the chapter, we have:

00

6

4

4

66

3.888828.010101064.4

5128

1064.41010

64

1010

)60512(8

==

=

==

=

+

+=+=

R

t

nn

R

ttt

o

of

effo

o

ackfo

One more source of overhead occurs because the last packet is not full. However, this additional overhead

accounts for a very small fraction of the total overhead and does not affect the above result.

5. (5.33)

A telephone modem is used to connect a personal computer to a host computer. The speed of the modem is

56 kbps and the one-way propagation delay is 100 ms.

Solutions follow questions:

a. Find the efficiency for Stop-and-Wait ARQ if the frame size is 256 bytes; 512 bytes. Assume a bit

error rate of 104.

First we have the following:

Pf= 1 (1 104)nf

nf= 256 8 = 2048 or nf= 512 8 = 4096

tprop = 100 ms

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no = 0

na = 64 bits

tproc = 0

Using the results in Equation 5.4,

Rn

tt

n

n

n

n

P

f

procprop

f

a

f

o

f )(21

1

)1(+

++

=

= 0.125 (nf= 2048)

= 0.177 (nf= 4096)

b. Find the efficiency of Go-Back-N if three-bit sequence numbering is used with frame sizes of 256

bytes; 512 bytes. Assume a bit error rate of 10

4

.

Given that WS= 23 -1= 7, we can calculate that the window size is:

R

Wn sf = 256 ms

Since this is greater than the round trip propagation delay, we can calculate the efficiency by using the

results in Equation 5.8.

fs

f

o

fPW

n

n

P)1(1

1)1(

+

=

= 0.385 (nf= 2048)

= 0.220 (nf= 4096)

6.

Each arriving batch of N packets contains a total of NL bits, requiring a transmission

time of exactly NL/R secs. Hence, all these N packets would have been just completelytransmitted at the arrival instant of the next batch of N packets. (Thus, there are no

more that N packets in the buffer at any time.) It is now clear that all the last (i.e., Nth)

packets in each batch will incur the (same) longest delay, caused by the transmission of

the preceding (N 1) packets. This delay = [(N1)L]/R secs.

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7.All the nodes in the network can communicate with each other provided that no more

than 1 link fails. Hence, the desired probability is:

P[{no link failure} U {1 link failure}]

= P[no link failure] + P[1 link failure]

= (1 p)6+

1

6p(1 p)5 = (1 p)6+ 6p(1 p)5 = (1 p)5(1 + 5p).

8.

[i]If only one link fails, clearly network connectivity is still maintained. If two links fail, network

connectivity may be lost (e.g., if links ad and ab fail) or still retained (e.g., if ab and bd fail). On the other

hand, if three links fail, then regardless of the failure locations network connectivity is lost. Note thatthree link failures must consist of the failure of three outer links or two outer links and the diagonal link; in

either case, network connectivity is lost. Thus, the required minimum number of link failures = 3.

[ii]There are 5 links in the network. Network connectivity may be lost if 2 links fail, butis definitely lost if 3, 4 or 5 links fail. Using the abbreviated notation: NC lost = network

connectivity lost.

P[NC lost] = P[ 2 fail i.e. ab, ad OR bc, cd ] + P[ 3 fail ] + P[ 4 fail ] + P[ 5 fail ]

= 2p2(1-p)3 +

3

5p3(1-p)2 +

4

5p4(1-p) +

5

5p5

= 2p2(1-p)3 + 10p3(1-p)2 + 5p4(1-p) + p5

= 4p5 - 9p4 + 4p3 + 2p2

Note that of the

2

5= 10 distinct ways of 2 link failures, network connectivity is lost only in the case when

ab, ad fail or bc, cd fail.

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