Question 15135 – DSSSB TGT 2021
May 2, 2024Question 8207 – Computer-Networks
May 2, 2024Question 8157 – Computer-Networks
Consider a 128 × 103 bits/ second satellite communication link with one way propagation delay of 150 milliseconds. Selective retransmission (repeat) protocol is used on this link to send data with a frame size of 1 kilobyte. Neglect the transmission time of acknowledgement. The minimum number of bits required for the sequence number field to achieve 100% utilization is __________.
Correct Answer: A
Question 41 Explanation:
To achieve 100% efficiency, the number of frames that we should send N = 1 + 2 * a
a = Tp / Tt where Tp is propagation delay, and Tt is transmission delay.
Given, B = 128 kbps, Tp = 150 msec,
L = 1 KB = 1 * 8 * 210 bits
Tt = L / B ⇒ 1 * 8 * 210 bits / 128 * 103 bps ⇒ 0.064 sec = 64 msec
So, a = 150 msec / 64 msec = 2.343
Efficiency (η) = 100 % ⇒ 1 = N/ 1 + 2 * a
So, N = 1 + 2 * a ⇒ 1 + 2 * 2.343 = 5.686
No. of sequence numbers requires in SR is 2*N = 2 *5.686 = 11.375
a = Tp / Tt where Tp is propagation delay, and Tt is transmission delay.
Given, B = 128 kbps, Tp = 150 msec,
L = 1 KB = 1 * 8 * 210 bits
Tt = L / B ⇒ 1 * 8 * 210 bits / 128 * 103 bps ⇒ 0.064 sec = 64 msec
So, a = 150 msec / 64 msec = 2.343
Efficiency (η) = 100 % ⇒ 1 = N/ 1 + 2 * a
So, N = 1 + 2 * a ⇒ 1 + 2 * 2.343 = 5.686
No. of sequence numbers requires in SR is 2*N = 2 *5.686 = 11.375
Minimum No. of bits required in the sequence number = [ log2 (11.375) ] = 4
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