Made Easy test series

Question

Consider a system using Go-Back-N protocol over fibre link with bandwidth 1Gbps and packet size 512B. Propagation speed of fibre is 2*10⁸m/sec with length 1Km. Assume error free and duplex communication with ACK packet size 64B. Then the value of N that yields maximum utilization of the sender is______?

Comments

For maximum utilization, I am getting the value of N as 3.56. So we should take the floor value, ie. N should be 3. But in the solution they have taken the ceil value, ie, 4..Which one is correct?

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It's saying maximum then take the floor, According to the rule.

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According to me for maximum utilization

(L/B)*N>=RTT going by this formula where L=length of frame B is bandwidth N is the maximum sender size and RTT is round trip time  I am getting N=3.54 now if we send 3 frames this formula will come wrong so for maxmium utilization sender window size will be 4

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What r u trying to say ???

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I am trying to tell the value of N for which maximum utilization of link will be there....and the formula given above is for maximum utilization

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@Somoshree,

In case of GoBack-N, maximum utilization is reached when window size is at least equal to (1+2a) i.e. N≥(1+2a) [This formula doesn't consider the ack transmission time].

What is channel utilization?

The fraction of time the sender stays busy in transmitting the frames until it gets back an Ack. Because after getting back the ack then the sender will be again busy responding accordingly(i.e. if ack is negative then resend the same frame or else transmit another frame).

In case of Stop and Wait, sender is busy for tx time out of total (tx+2tp) and after transmitting one frame keeps idle until it gets an ack.

In case of Go Back N, sender is busy transmitting upto N frames and then waits until it gets back an Ack after (tx+2tp).

Maximum utilization is achieved when the sender doesn't become idle for any moment. When will that happen?

1) When the sender has just stopped transmitting all it's N frames and the Ack comes so sender sends the next frame immediately.

2) When sender hasn't finished transmitting all the N frames and before that Ack comes. So sender has to finish it's current window and then slide to next frame.

In both the cases sender is busy giving the max. channel utilization.

If I could clear you up to this, then can we say that it's always better if N is more? Because in that case sender will always keep sending frames and never stop.

If you take N=3 then there might be atleast some idleness as you got N=3.56. So better you take N=4 to guarantee max utilization.

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kumar.dilip

https://gateoverflow.in/174829/sliding-window

they take floor

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MiNiPanda sir if we calculate the optimal window size for full utilization by this formulae((L/B)*N)>RTT) we would also get 3.54 and sender has optimal window size 3 the link would not be full utilized if the sender has optimal window size of 4 then link utilization will be full

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@Navneet Kalra

Yes I supported your previous comment with explanation that I knew..

And don't call me Sir _/\_

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MiNiPanda  Navneet Kalra  read the comments section of the selected answer in this link https://gateoverflow.in/39577/gate2016-2-55

i got totally confused with this :( can u provide some resource where it is mentioned that we need to take ceil value in such cases?because till date in all the questions that i have solved regarding max utilization, all of them have taken the floor value

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Somoshree Datta 5 mam please go by the concept...for maximum link utilization sender has to use all the time for sending the frames(it cannot set idle) so according to my formula answer is 3.54 means if you take 3 then some amount of  time the sender will sit idle and thus the link will not be effectively utilized....now if we make the value to 4 then sender after sending the 4th frame will get the acknowledgement for 1 and will again load a frame into its buffer and send....it does not have to sit idle waiting for the acknowledgement of the first frame if no corruption and loss occours so....for maximum link utilization the sender buffer size will be 4 here 

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When we want to choose N large enough to allow continuous transmission while waiting for an ACK for the first packet of the window, then N >(DTP + 2DP + DTA) / DTP where [DP = prop delay    DTA = ACK trans. Time    DTP = packet trans. time].

so using this N>3.566

so N = 4

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I am getting answer as 3

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Navneet Kalra 

Yes u were right..for any number of packets larger than or equal to 1+2a, the sender will always be busy sending the frames while the acknowledgements keep on coming..efficiency in that case will always be 100%..

Ref: http://web.cse.ohio-state.edu/~athreya.14/cse3461-5461/Cse3461.D.DataLink.09-11-2012-cont.pdf

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Utkarsh Joshi please check your RTT time..there will be some problem out there

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Some cal mistake got 4 now.(3.56)

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3 should be the answer

not practical to send 4 since 3.56 packets can be accomodated hence 3

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Here it says duplex connection, meaning full duplex, so shouldn't we use N=1+a here instead of N= 1+2a?

Answer 1

When we want maximum utilization the numerator >= denominator (in mathematical terms)