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3 years ago

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Suppose you are designing a sliding window protocol for a 1 Mbps point-to-point link to the moon, which has a one-way latency of

1.25 seconds. Assume that each frame carries 1 KB of data. (a) How many frames need to be in transit to keep this pipe full? (b) What is the maximum reasonable size for SWS and RWS needed for such a system? (c) For the SWS and RWS sizes found in part b, what is the number of bits needed for the sequence number?

1 answer:

mylen [45]3 years ago

7 0

Answer:

approx 306 frames need to be in transit to keep this pipe full

maximum reasonable size is 306 frames

sequence number = 10 bit

Explanation:

given data

transmission rate = 1 Mbps

size of frame Tt = 1 KB

one way latency Tp = 1.25 s

to find out

How many frames need and What is the maximum reasonable size for SWS and RWS and what is the number of bits needed for the sequence number

solution

we know useful time is = $\frac{1 KB}{1 Mbps} = \frac{1024*8}{1000*1000}$

so total time is

total time = Tt + 2Tp

total time = $\frac{1024*8}{1000*1000}$ + 2 × 1.25

so

possible frames are

$\frac{n*Tt}{Tt+2Tp}$ ≤ 1

n ≤ 1 + $\frac{2.5*1000000}{1024*8}$

n ≤ 306.17

so approx 306 frames need to be in transit to keep this pipe full

and

we know here

size of sending window is equal to size of receiving window

so maximum reasonable size is 306 frames

and

bit need for sequence number is

sequence number = $log^{306+306}_2$

sequence number = 10 bit

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