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Suppose two hosts are connected through two intermediate switches
Suppose each link (one way) propagation delay is 20 ms and each
link data transfer rate is 1 Mbps.
If packet size is 1000 bytes then the amount of time required to send
one file of size 5000 Bytes from sender to receiver is______ ms.
edited | 87 views
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196ms??
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YES 196 should be correct!
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how?
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Transmission time (Tt) =1000Bytes/1Mbps.    = 8 ms

Tp = 20 ms
File size=5000 Bytes
Number of packets in file,N = 5000Bytes/1000Bytes
N=5
Total time required to send file from sender to receiver
= (N* Tt+Tp) + (Tt + Tp) +(Tt + Tp)
= (5* 8 + 20) + (8 + 20) + (8 + 20) ms
= 116 ms
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First packet will take 3*Tp and 3*Tt which is (3*8)+(3*20) = 24+60 = 84ms

Rest 4 packets will follow the first one and we will apply pipelining:

4*(Tt + Tp)

= 4*(8+20)

= 4*28

=112ms

So total time will be 84+112 = 196ms.

Realize that two hosts connected via 2 switches creates 3 links in between.
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@Nidhi Budhraja

116 ms is the correct answer.

think it this way.. all the packets are transferred in pipeline. so find the transmission time of all the packets at the source which is 5*1000B/1Mbps=40ms.

now while the 5th packets is getting tranfered, remaining packets would be on theiir way, so we need to find the propagation delay and delays at the 2 switches only for the last packet.. which is 20*3ms=60ms(propagation delay through link)+ 2*1000*8/1Mbps=16ms(transmission delay at 2 switches)

summing up all= 40ms+60ms+16ms=116ms

compare with the answer you posted.. both are same.

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Just for the sake of matching with the answers provided, we should not change the standard concepts.

@Utkarsh Joshi which approach do you think is correct. Am I doing a mistake somewhere?

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@Gupta731

i'm not changing the concept. this is concept for the pipelining. there is a previous year question based on this question, i don't remember. verify it and you will get to know.

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okay.
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Even I am getting 116.

@Utkarsh Joshi

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