36 votes 36 votes 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$$\text{%}$. The value of the one way propagation delay (in milliseconds) is_____. Computer Networks gatecse-2015-set2 computer-networks mac-protocol stop-and-wait normal numerical-answers + – go_editor asked Feb 12, 2015 • edited Jun 16, 2018 by Pooja Khatri go_editor 11.9k views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply behl1anmol commented Jan 8, 2022 reply Follow Share What is the significance and meaning of one way propogation delay? 0 votes 0 votes GO Classes Support commented Dec 15, 2023 reply Follow Share GO to the Root of Concept: Video Explanation with timestamp 15 Questions on Stop Wait Protocol| GATE PYQs| MIT, Berkeley |Efficiency & Throughput| With NOTES 0 votes 0 votes Akash 15 commented Jan 23 reply Follow Share $BW=10^6\;bits/sec$ $L=1000\;bytes=1000\times8\;bits$ $\therefore T_t=\frac{L}{BW}=\frac{1000\times 8}{10^6}\times 10^3=8\;msec$ $\eta=25\%=\frac{1}{4}→ \frac{1}{1+2a}=\frac{1}{4}→ 4=1+2a→ a=\frac{3}{2}$ $a=\frac{T_p}{T_t}=\frac{3}{2}→ T_p=\frac{3}{2}\times8=\color{DarkGreen}12\;msec$ 1 votes 1 votes Please log in or register to add a comment.
Best answer 45 votes 45 votes 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, $p + 4 = 16, p = 12 \;\text{ms}$. Arjun answered Feb 12, 2015 • edited Jun 19, 2021 by Lakshman Bhaiya Arjun comment Share Follow See all 9 Comments See all 9 9 Comments reply Show 6 previous comments dharmesh7 commented Dec 17, 2018 reply Follow Share @Arjun sir why you have multiplied RTT with 10^6 in the first formula. 0 votes 0 votes Jeel Dasvani commented Jul 18, 2021 reply Follow Share Cz it’s a round trip !! Sender to receivr( data transmission ) and from receiver to sender( acknowledgement transmission) 0 votes 0 votes satyaAchar commented Mar 14 reply Follow Share 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 0 votes 0 votes Please log in or register to add a comment.
19 votes 19 votes 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 Pranabesh Ghosh 1 answered Jul 27, 2016 Pranabesh Ghosh 1 comment Share Follow See 1 comment See all 1 1 comment reply mahendrapatel commented Jan 4, 2023 reply Follow Share सबसे बढ़िया explaination है 0 votes 0 votes Please log in or register to add a comment.
5 votes 5 votes 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) $0.25= \frac{T_{t}}{T_{t}+2T_{p}}$ $0.25(T_{t}+2T_{p}) = T_{t}$ $0.25T_{t} + 0.5T_{t} = T_{t}$ Hence $0.5T_{p} = 0.75T_{t}$ $T_{p} = 0.75T_{p}/0.5$ $T_{p} = 1.5 T_{t}$ $1.5\times8 = 12ms$ $T_{p} = 12ms$ Deepak Raj 1 answered Oct 9, 2018 • edited Jan 23, 2019 by Lakshman Bhaiya Deepak Raj 1 comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes 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 Utkarsh Pathak answered Apr 19, 2019 Utkarsh Pathak comment Share Follow See all 0 reply Please log in or register to add a comment.