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Consider two hosts $X$ and $Y$, connected by a single direct link of rate $10^6\; \text{bits/sec}$. The distance between the two hosts is $10,000\;\text{km}$ and the propagation speed along the link is $2 \times 10^8 \;\text{m/sec}$. Host $X$ sends a file of $50,000\;\text{bytes}$ as one large message to host $Y$ continuously. Let the transmission and propagation delays be $p\; \text{milliseconds}$ and $q \; \text{milliseconds}$ respectively. Then the value of $p$ and $q$ are

  1. $p=50$ and $q=100$
  2. $p=50$ and $q=400$
  3. $p=100$ and $q=50$
  4. $p=400$ and $q=50$
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5 Answers

Best answer
31 votes
31 votes

$T_t = \dfrac{\text{Length}}{\text{Bandwidth}}$

$\quad =\dfrac{50000\times 8}{10^6}$

$ \quad = \dfrac{40}{100} = 0.4s = 400\; \text{ms}$

$T_p = \dfrac{\text{Distance}}{\text{Velocity}}$

$\quad =\dfrac{10000\times 10^3}{2\times 10^8}$

$ \quad = \dfrac{1}{20} = 0.05s = 50\; \text{ms}$

 

Hence, answer (D) $p=400,q=50.$

edited by
8 votes
8 votes
Given Bandwith $= 10^6 bps$

Length$= 50000 bytes= 8\times50000 bits$

Distance $= 10000km = 10^7 m$

Speed $= 2\times10^8m/s$

$p=\frac{L}{BW} =0.4s = 400ms$

$q=\frac{D}{S}=0.05s = 50ms$

So,the correct answer is $(D)$
edited by
6 votes
6 votes
$Transmission Delay(T_{t})=\frac{File size}{Bandwidth}=\frac{50000 bytes}{10^{6}bits/sec}=\frac{50000*8 bits}{10^{6}bits/sec}= 0.4 sec = 400 msec$

$Propagation Delay(T_{p})=\frac{Distance between to hosts}{Propagation Speed}=\frac{10000 km}{2*10^{8}m/sec}=\frac{10000*1000 m}{2*10^{8}m/sec} = 0.05 sec = 50 msec$

So, Correct answer is (D)
edited by
1 votes
1 votes
The question asks for two simple questions

p - transmission delay

q- Propogation delay

 

p = L/B

$\frac{10000*8}{10^{6}}$

= 400ms

 

dont even solve for Propagation delay as no other option has p=400

just tick it

That should be the approach
Answer:

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