GATE IT 2006 | Question: 64

Question

Suppose that it takes $1$ unit of time to transmit a packet (of fixed size) on a communication link. The link layer uses a window flow control protocol with a window size of $N$ packets. Each packet causes an ack or a nak to be generated by the receiver, and ack/nak transmission times are negligible. Further, the round trip time on the link is equal to $N$ units. Consider time $i > N$. If only acks have been received till time i(no naks), then the goodput evaluated at the transmitter at time i(in packets per unit time) is

• A. $1- \dfrac{  N}{i}$
   
• B. $\dfrac{i}{(N + i)}$
   
• C. $1$
   
• D. $1 - e^{\left(\frac{i}{N}\right)}$

Comments

The question seems to be somewhat incorrect because RTT is given as N units and window size=N, as 1 unit of time is required for transmitting 1 packet so total transmission time of window will become N units there is no scope left for any other time consumption so how the RTT can be N units.

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Why 1st ACK @ N+1?

Since RTT is given as N units so it should be relieved within N units. Not only 1st RTT of whole window is N so all ACK's should arrive in N units. But that doesn't seems to be possible.

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GO to the Root of Concept: Video Explanation with timestamp: Sliding Window PYQs| GATE 2009 Q57, 58| 2006 Q46, 64| 2003, 2014, 2015, 2004, 2016, 2007| With NOTES

Answer 1

In computer networks, goodput is the application level throughput, i.e. the number of useful information bits delivered by the network to a certain destination per unit of time. (From wikipedia).

So, successful delivery of packet can be assured if $\text{ACK}$ has been received for it.

So till time $'i'$ we would have transmitted $'i'$ packets but only $(i-N)$ can be acknowledged as minimum time for a packet to get Acknowledged is $N$ (since $\text{RTT}$ is $N$ which is equal to the window size, there is no waiting time
for the sender).

So, successfully delivered packets $=(i-N)$

Time for transmission $=i$

Goodput $=\dfrac{\text{Successfully delivered data}}{\text{Time}}$

$=\dfrac{(i-N)}{i}$

$= 1- \dfrac{ N}{i}$

Therefore (A)

Comments

thanks man for a better explanation

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perfect solution

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why there is no waiting time?  I think the waiting time is 1 unit.

Because here when RTT is N and the window size is N+1 then only waiting time should be zero.

Correct me anyone if i am wrong.

Answer 2

I tried to make it easier by taking an example that how it is actually working. 

 

Comments

Good explanation.

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just in case if anyone thinking about the accumulative

acknowledgements.

 

"Each packet causes an ack or a nak to be generated by the receiver"

 

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The first acknowledgement will arrive at the time when 5th frame is being sent not at 6th time, beacuse no. Of frame sent = round trip time..

Answer 3

1 unit of time —  packet.

N units of time — N packets.

i units of time — i packets.

 

If only acks have been received till time i (no naks)

It means till time i, i packets are ACKed?

No.

Whatever i is, ACKed packets would always be (i - N) because it takes N units of time to receive an ACK for a packet (because N is the RTT). So from total packets (ie i packets) ACKed packets are (i - N).

 

So, Goodput = $\frac{i-N}{i}$

=> $\frac{i}{i} - \frac{N}{i}$

=> $1 - \frac{N}{i}$

 

Option A

Answer 4

 

The image explains

everything clearly