MadeEasy Test Series: Computer Networks - Flow Control Methods

Question

Please help me to solve problem 8.

Answer 1

$\begin{align*} 10^6 bits &\leftrightarrow 1sec\\ 100 bits &\leftrightarrow \frac{100}{10^6} = 0.1ms \end{align*}$

0.1ms is the transmission time

$$\begin{align*} \eta &= \frac{TT}{TT+2PT}\\ &=\frac{0.1ms}{0.1ms+2\times 1sec}\\ &=\frac{0.1ms}{0.1ms+2\times 1000ms}\\ &= 4.99975\times 10^{-5} \end{align*}$$

answer for Q.9 = $\eta$ and if that's the efficiency then the actual bandwidth that was utilized is $\eta \times C$

Hence, answer = option B

Comments

RTT is defined as in this question every where- time for ACK to come back. 

http://www.eecs.yorku.ca/course_archive/2010-11/F/3213/CSE3213_11_FlowErrorControl_F2010.pdf

Now, in utilization formula, most good sources avoid the usage of "RTT" as it becomes ambiguous. 

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yes, that is the definition for that. In our previous conversations I wrote this def. only.

Tried searching on google for such question from top universities and this is the best I came up with, till now : http://goo.gl/b3ib89

Ctrl+F search for RTT

& if we add RTT to transmission time we get, Transfer time (from a ppt.)

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yes, as per that definition, here we have a packet of size "100 bits" and so ACK must start only after receiving 100 bits (transmission time). And I'm not saying to add TT to RTT or not and this should be done as per the question- the point is ACK will be sent only after entire frame is received. Sometimes RTT might be given as for 1 bit frame. There might be packet processing delay given in question. So, remembering formula is a hard task and just the above definition of RTT is enough. When I do networking questions I'll see for similar ones from standard sources- hard to find with google.

Answer 2

In Stop and Wait we can transmit only one packet in 1 RTT.

As the name suggests, after sending 1 packet, Stop & Wait protocol stops & waits until the ACK is received or a time out occurs.

So we can transmit 1 packet or 100 bits in 1 RTT.

That is transmission rate is 100 bits per RTT, or 100 bits per 2 second. 

So Transmission rate = 50 bps.

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Now link utilization = Current transmission rate / rate at which we could have transmit the data using the full capacity of the link (or channel)

so link utilization = 50 bps / 100,000 bps = 0.0005

We can see here Stop & Wait can be astoundingly inefficient, hence we need sliding window protocols.

Choosing appropriate windows size in sliding window protocol leads to quite higher link utilizations.