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If the bandwidth of a medium is 100 Mbps and round-trip time is 50 microseconds then calculate the sequence bits in Go Back N ARQ flow control policy is applied and frame size 100 bits has to be transfer.

A )     5

B)      6

C)      7

D)      None of these
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what is d answer? 5 or 7
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given answer is 6
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Anwer will be 6. We have 50 microsec and we have bandwidth 100 mbps. So,we can send 5000 bits in that round trip time. So,each frame cntains 100 bits so,5000/100=50 frames. So, for 50 frames we must have log(50)=6 sequence bits.
+1
how r u processing this question.

ws+wr<=ASN(AVAILABLE SEQUENCE NUMBER)

we have to find ws(sender window size)

capacity of channel/frame bits will give the same

ceil of log(ws+1) will give the answer
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We can send 50 frames. So,window size =50. So,sequence number =window size+1.So 51. Taking ceil(log(51)) will give 6 as the answer.
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thanks rahul

i am doing a mistake  now problem has been solved

i am taking RTT as propagation time which is  wrong
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L>=2*tp*B

=>RTT*B

=>50*100 bits=> 5000 bits

given frame size = 100 bits

no of seq required => 5000/100==> 50 +1=> 51

no of bits => log(51)==> 5.6==> 6 bits are required

Correct me if I'm wrong
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yaa correct bro :)

## 1 Answer

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Best answer

Round trip time = 2* Tp (propagation delay) = 50 * 10-6 sec

Transmission time ( Tt)= frame size(L) /Bandwidth (B) = 100 / (100 * 106 ) sec = 10-6 sec

Total window size = 1+ 2* (Tp / Tt )

=  1+ (50 * 10-6 / 10-6)

= 51

No of bits for sequence number = ceil of (log2 51) = 6

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Here Will the bit sequence will be Ceil( log2(1+2a))

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1
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7