Question

Could someone help with just part a please ?

You are hired to design a reliable byte stream protocol that uses a sliding window (like TCP). This protocol will run over a 1 -Gbps network. The RTT of the network is 100 ms. and the maximum segment lifetime is 30 seconds. (a) How many bits would you include in the AdvertisedWindow and SequenceNum fields of your protocol header? (b) How would you determine the numbers given above, and which values might be less certain?

Answer 1

Given,

Bandwidth = 1 Gbps = 10⁹ bps = (10⁹ /8) bytes per second

Round Trip Time = RTT= 100ms = 0.1 sec

The window size (in bytes) should be equal to = RTT x Bandwidth

Window size = (10⁹ / 8) × 0.1 = bytes = 12500000 Bytes

Number of bits needed for window size =

log2 (12500000)) = 24

Therefore, we need to 24 bits for advertisedwindow (which allows a maximum window size of 16777216 bytes).

In 60 seconds,

Number of bytes that can be transmitted = (10⁹ /8 ) x 60 = 7500000000 bytes.

All of those bytes must have unique sequence number.

Number of bits needed for sequence numbers =

log2 (7500000000) = 33

Therefore, 33 bits are required for sequence number field.