Data Link Layer

Question

Compute the fraction of the bandwidth that is wasted on overhead (headers and retransmissions) for protocol 6 on a heavily loaded 50 kbps satellite channel with data frames consisting of 40 header and 3960 data bits. Assume that the signal propagation time from the earth to the satellite is 270 msec. ACK frames never occur. NAK frames are 40 bits. The error rate for data frames is 1% and the error rate for NAK frames is negligible. The sequence numbers are 8 bits.

Comments

i think 1% tell if 100 frame send then 1 frame lost ..

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Ques. is incomplete no. of seq. Bits are 8 bits

Answer 1

Let's take total 100 packets has to send 1% error means 1packet lost...

total overhead = 40*100+ 4040(retransmission of packet)

Total overhead / total data sent=

(8040/100*4000+4040)*100%

=1.989% overhead

 

So 100% equals 50kbps then 1.989%

Gives almost 1kbps

Comments

why r u not considering error rate for data frames??u r just considering error for NAK,,is retranmission for NAK only is overhead not for data frames??

Answer 2

Out of 100 frames send 1 will be lost.

Total data bits = 3960*100 bits
total size of Hader= 40 *100 bits
Nak= 40 bits

when the sender receives NAK,

It will send a data frame again,( the frame size is 4000 bits)

Total wastage= (4000) lost data frame + (4000) retransmit lost data frame + (40) Nak bits = 8040 bits

Comments

The retransmitted frame must not be termed as wastage so it should be 4040 bits.

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Why are you not considering sequence number?

Answer 3

unused bandwidth=(1-n)*B

efficiency(n)=useful bits/total bits

2^8 frames are sent among which 2.56(1%) are corrupted

total bits=256*4000  +  2.56*4040 =1024102.4

useful bits=3960*256=1013760

n=0.98990101

unused bandwidth=0.5 (app)