GATE CSE 2015 | Set 2 | Question: 8

Question

A link has transmission speed of $10^6$ bits/sec. It uses data packets of size $1000$ $\text{bytes}$ each. Assume that the acknowledgment has negligible transmission delay and that its propagation delay is the same as the data propagation delay. Also, assume that the processing delays at nodes are negligible. The efficiency of the stop-and-wait protocol in this setup is exactly $25\%$. The value of the one way propagation delay (in milliseconds) is_____.

Comments

What is the significance and meaning of one way propogation delay?

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GO to the Root of Concept: Video Explanation with timestamp 15 Questions on Stop Wait Protocol| GATE PYQs| MIT, Berkeley |Efficiency & Throughput| With NOTES

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$BW=10^6\;bits/sec$
$L=1000\;bytes=1000\times8\;bits$

$\therefore T_t=\frac{L}{BW}=\frac{1000\times 8}{10^6}\times 10^3=8\;msec$
$\eta=25\%=\frac{1}{4}→ \frac{1}{1+2a}=\frac{1}{4}→ 4=1+2a→ a=\frac{3}{2}$

$a=\frac{T_p}{T_t}=\frac{3}{2}→ T_p=\frac{3}{2}\times8=\color{DarkGreen}12\;msec$

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Tt = 8 msec
0.25  =  1 /  [1 + 2*(Tp / Tt)]  ---> effeciency
Tp = 12 mec

Answer 1

In stop and wait, a frame is sent and next frame will be sent only after ACK is received.

$\text{Efficiency} =\dfrac{\text{Amount of data sent}} {\text{Amount of data that could be sent}}$

$=\dfrac{\text{Amount of data sent}}{ RTT \times 10^6}$
 

$= \dfrac{\text{Amount of data sent}}{ \left( \text { Prop. delay for data}+\text{Prop.delay for ACK}+ \text{Transmission time for data} +\text{Transmission time for ACK} \right) \times 10^6} $

$= \dfrac {1000 \times 8}{ \left( p + p + 1000 \times \dfrac{8}{10^6} + 0 \right) \times 10^6}$

$= \dfrac{8}{2p+8\;\text{ms}} \text{ (where p is the prop. delay in milli seconds)}$

$= \dfrac{4}{p+4} = 0.25 \text{ (given in question)}$

So, $p + 4 = 16, p = 12 \;\text{ms}$.

Comments

@Arjun sir why you have multiplied RTT with 10^6 in the first formula.

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Cz it’s a round trip !!
Sender to receivr( data transmission ) and from receiver to sender( acknowledgement transmission)

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For do this question in less time..

efficiency(n) = 1/(1+2a)

=>  1/4 = 1/(1+2a)

=>  1+2a = 4 => 2a = 3

=>  a=3/2

=>  Tp/Tt =3/2

=>  Tp = (3/2)*Tt

Answer 2

Transmission time Tt=1000*8/10^6=8/10^3 sec=8 ms

efficiency =25%=1/4=Tt/(Tt+2*Tp)

8/(8+2*Tp)=1/4

Tp=12

so propagation Delay is 12 ms

Comments

सबसे बढ़िया explaination है

Answer 3

as per the question Bandwidth$= 10^{6}bps$

data size$=1000 Bytes$

by these two we can calculate transmission time by using formula $\frac{Data size}{BW}$

as data size in bytes convert to bits hence $T_{t}= 8ms$

Efficiency$=25\% i.e$ $0.25$

efficiency$= \frac{T_{t}}{T_{t}+2T_{p}}$         (I am not taking td ack time as it is negligible)

$0.25= \frac{T_{t}}{T_{t}+2T_{p}}$

$0.25(T_{t}+2T_{p}) = T_{t}$

$0.25T_{t} + 0.5T_{p} = T_{t}$

Hence $0.5T_{p} = 0.75T_{t}$

$T_{p} = 0.75T_{p}/0.5$

$T_{p} = 1.5 T_{t}$

$1.5\times8 = 12ms$

$T_{p} = 12ms$

Answer 4

Given Bandwidth (B) = 10^6 bits/sec , L = 1000 * 8 bits ,  η = 25% = 1/4

Tt = L/ B = 1000* 8 / 10^6 = 8ms  (Transmisson delay of acknowledgement is negligible not  Packet )

 η=1/1+2*a

1/4=1/1+2*a => a = 3/2 => Tp = 3/2*Tt => Tp = 3/2 * 8 = 12 ms

Answer 5

☺

Comments

for any doubt comment below

Answer 6

Answer: 12 ms

 

$$\text{Transmission time, } T_t = {Packet \space size \over Bandwidth} = 1000 bytes / 10^6 / 8 Bps = 8ms$$

 

$$\text{Efficiency of Stop and Wait Protocol, } \eta = {1 / 1+2a} \text{ where a = } T_p/T_t$$

 

$$\text{Given efficiency, } \eta = {0.25}$$

 

$${1 \over 1+2a} = {0.25}$$

 

$$1=0.25+0.5a$$

 

$$0.5a = 0.75$$

 

$$a = 1.5$$

 

$$T_p/T_t=1.5$$

 

$$T_p=1.5 \times T_t = 1.5 \times 8 = 12 ms$$