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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

  1. $1- \dfrac{  N}{i}$
     
  2. $\dfrac{i}{(N + i)}$
     
  3. $1$
     
  4. $1 - e^{\left(\frac{i}{N}\right)}$
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Best answer
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109 votes

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)

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44 votes

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

 

2 votes
2 votes

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:

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