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Consider two links, $(A,B)$ and $(B,C)$, with propagation delays of $d1$ and $d2$, respectively. Assume that $\text{host A}$ sends $\text{M packets}$ to $\text{host C}$ using a sliding window flow control protocol with a window of size $W$.
What is the minimum time it take to send all packets from $A$ to $C$ when the flow control protocol is implemented end-to-end between $A$ and $C$ ?
Note: Transmission delays are negligible
  1. $2(d1+d2) \left \lceil \left ( \dfrac{M}{W} \right ) \right  \rceil$
  2. $(d1+d2) MW$
  3. $2(d1+d2)\left \lceil \left ( \dfrac{W}{M} \right ) \right  \rceil$
  4. $2(d1+d2)W$
asked in Computer Networks by Boss (17.2k points) | 97 views
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getting C. please explain how A is correct.
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Please expllain .
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Meanwhile, please, share your solution.
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A is the answer because

in question it is given that window size is w means window can have w packets at a time

so number of times full window needs to be send is  M / W

so answer should be A

1 Answer

+4 votes
Best answer

Size of window is W, so we can send maximum upto W packets.

We have M number of packets so number of times we need to send the window is $\frac{M}{W}$

Since transmission time is negligible, first window will reach from A to C in $d1+d2$ time, An ack will come from C to A in $d1+d2$ time.

We need to do this $\frac{M}{W}$ number of times. 

Hence total time taken is $2(d1+d2)(\frac{M}{W})$ 

PS: If W is not multiple of M that's why ceil is taken

answered by Boss (15k points)
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