Project ForgeComputer Networks
OSI/TCP-IP Layers & Routing
GATE 2007
Expert
The distance between two stations M and N is L kilometres. All frames are K bits long. The propagation delay per kilometre is t seconds. Let R bits/second be the channel capacity. Assuming that processing delay is negligible, the minimum number of bits for the sequence number field in a frame for maximum utilization, when the sliding window protocol is used, is:
- A⌈log₂((2LtR+2K)/K)⌉
- B⌈log₂(2LtR/K)⌉
- ⌈log₂((2LtR+K)/K)⌉
- D⌈log₂((2LtR+K)/(2K))⌉
Explanation
For 100% utilization, the window size must be at least 1+2a, where a=Tprop/Ttrans=(Lt)/(K/R)=LtR/K. Window size = 1+2LtR/K = (K+2LtR)/K. The number of sequence-number bits needed is ⌈log₂(window size)⌉ = ⌈log₂((2LtR+K)/K)⌉.
Source: GATE CS 2007 (2007)
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