Question 4) Sliding Window Protocol

Question

Consider source and destination in a 10Mbps network with senders window 5 packets .Transmission Delay for a packet is 10 micro second  Propogation Delay form source to destination is 250 micro sec .How long sender is ideal once a window os transmitted before being starting second window . assume ACK transmissioon time is negligible


My Approach

Time taken for 1 packet  = 10 us

Time taken for 5 packets = 50us

The sender cannot receive acknoledgement before round trip time which is 500ms
After sending 5 packets , the minimum time the sender will have to wait before starting transmission of the next packet = 500 – 50 = 450us

answer given as 400us  . Is my approach correct ?

Comments

@akhilnadhpc.,, you r right.. min. time will be 450 us.

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What is the final answer? 450 or 460?

I too get 450. @srestha why 510?

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why next window will be sent after receiving the ack of first packet of the window ?

Answer 1

Round-trip delay time (RTD) or round-trip time (RTT) is the length of time it takes for a signal to be sent plus the length of time it takes for an acknowledgment of that signal to be received.

RTT doesn't mean 2*Propagation Time.
RTT = 510us, 1 Packet TT = 10us
5 Packet TT = 50 us.

Idle Time = 510us - 50us = 460us

Comments

that means 1st packet total takes 510 micro sec. After 1st packet fully received , it could send next packet. rt?

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Ack of first packet get atleat 510 micro sec..

If we consider only ack  delay from receiver to sender...

First 50us sender is busy after it will wait till ack  not received...

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Hi Digvijay Landry,

Can u just tell me how u got RTT=510.. for 10 micro sec what you have considered here.. if it is TT then it is already included in 50 na( I.e 5*10).. pls explain

Answer 2

1st packet would travel 40us by the time source has put all 5 packets on the channel.

Which means after 1 packet is put on channel, sender is busy for another 40us.

Now from this point, 1st packet will take another 210us (250-40) to reach receiver.

And add another 250 us (for ack) to reach the sender.

SO after 210+250 =460us of inactivity, the sender can start transmitting again.

I think the answer should be 460 us. 

Comments

Can anyone explain why can't we consider transmission of packets in a pipelined fashion as we do in a packet switched network?