Question

Consider an error-free 64 kbps satellite channel used to send 512 byte data frames in one direction, with very short acknowledgements coming back the other way. What window size will keep the channel full?

Answer 1

The answer is "2".

Step-by-step explanation:

In the given question some information is missing, that is "The propagation time for satellite to earth" which is "270 milliseconds" so, the description to this question can be defined as follows:

Given values:

Bandwidth = 64 kbps

Data frames = 512 bytes

Propagation Time (t​​​​​​p ) =270 ms

Change Bandwidth kbps to bps:

1 kb= 1024 bytes

calculated bandwidth= 64 kbps = 64×1024 bps = 65536 bps

1 bytes = 8 bits

512 bytes = 512 × 8 = 4096 bits

Frame length = 4096 bits

Formula

Transmission time (T​​​​​​t) = Frame length /Bandwidth

Window size = 1+2a

where a = Propagation time/Transmission time

Calculate Transmission time:

Transmission time (T​​​​​​t) = 4096 / 65536

Transmission time (T​​​​​​t)= 625 m.sec

Calculate Window size:

Window size = 1+2(270/625)

Window size = 1+2(0.432)

Window size = 1+0.864

Window size = 1.864

Window size = 2