+1 vote
131 views
1. 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.
1. 81031 Bytes
2. 83101 Bytes
3. 801301 Bytes
4. 81301 Bytes

recategorized | 131 views
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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

+1

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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