MadeEasy Full Length Test 2019: Computer Networks - Sliding Window

Question

• A. A 1 Mbps satellite link connects two ground stations. The altitude of the satellite is 6000 km and speed of the signal is 3 × 10^8   m/s. What should be the packet size for a channel utilization of 50% for a satellite link using go-back-63 sliding window protocol? Assume that the acknowledgment packets are negligible in size and that there are no errors during communication.

• A. 81031 Bytes
• B. 83101 Bytes
• C. 801301 Bytes
• D. 81301 Bytes

 

Comments

They have provided the wrong solution. Answer must be 320 bits

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no they are actually correct

we have to consider distance from a to satellite and from satellite to b

then tp = 40msec

and when we put efficiency as 1/2

1/2 = 62 / (1 + (2*40*10^3)/L)

after solving

L = 650.40 bits
therefore L = 81.3 Bytes

correct me if wrong

but in there answers they have multiplied it by 100 dont know why !!

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almost similar question https://gateoverflow.in/3375/gate2008-it-64 with proper explanation

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1/2 = 62 / (1 + (2*40*10^3)/L)  -> 1/2 = 63 / (1 + (2*40*10^3)/L)

63 is the sender window size,which should be used in the formulae right?

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they also did the same thing
good explanation there

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Dont trust made easy answers.

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@balchandar reddy san yea then ans becomes 79.68
thank you for pointing it out 

actually in made easy they took 62 that's why i mistakenly used 62