Relational-Algebra
August 31, 2024Ethernet
August 31, 2024Sliding-Window-Protocol
| Question 8 |
Consider a selective repeat sliding window protocol that uses a frame size of 1 KB to send data on a 1.5 Mbps link with a one-way latency of 50 msec. To achieve a link utilization of 60%, the minimum number of bits required to represent the sequence number field is ________.
| 5 | |
| 6 | |
| 7 | |
| 8 |
Question 8 Explanation:
Given L = 1 KB
B = 1.5 Mbps
Tp = 50 ms
η = 60%
Efficiency formula for SR protocol is
η = W/(1+2a) ⇒ 60/100 = W/(1+2a) (∵a = Tp/Tx)
Tx = L/B = 8×103/1.5×106 = 5.3ms
a = Tp/Tx = 50/5.3 = 500/53 = 9.43
⇒ 60/100 = W/19.86 ⇒ W = 11.9 ≈ 12
⇒ W = 2n-1 = 12 ⇒ 2n = 24 ⇒ 2n = 24 ≈ 25 ⇒ n = 5
B = 1.5 Mbps
Tp = 50 ms
η = 60%
Efficiency formula for SR protocol is
η = W/(1+2a) ⇒ 60/100 = W/(1+2a) (∵a = Tp/Tx)
Tx = L/B = 8×103/1.5×106 = 5.3ms
a = Tp/Tx = 50/5.3 = 500/53 = 9.43
⇒ 60/100 = W/19.86 ⇒ W = 11.9 ≈ 12
⇒ W = 2n-1 = 12 ⇒ 2n = 24 ⇒ 2n = 24 ≈ 25 ⇒ n = 5
Correct Answer: A
Question 8 Explanation:
Given L = 1 KB
B = 1.5 Mbps
Tp = 50 ms
η = 60%
Efficiency formula for SR protocol is
η = W/(1+2a) ⇒ 60/100 = W/(1+2a) (∵a = Tp/Tx)
Tx = L/B = 8×103/1.5×106 = 5.3ms
a = Tp/Tx = 50/5.3 = 500/53 = 9.43
⇒ 60/100 = W/19.86 ⇒ W = 11.9 ≈ 12
⇒ W = 2n-1 = 12 ⇒ 2n = 24 ⇒ 2n = 24 ≈ 25 ⇒ n = 5
B = 1.5 Mbps
Tp = 50 ms
η = 60%
Efficiency formula for SR protocol is
η = W/(1+2a) ⇒ 60/100 = W/(1+2a) (∵a = Tp/Tx)
Tx = L/B = 8×103/1.5×106 = 5.3ms
a = Tp/Tx = 50/5.3 = 500/53 = 9.43
⇒ 60/100 = W/19.86 ⇒ W = 11.9 ≈ 12
⇒ W = 2n-1 = 12 ⇒ 2n = 24 ⇒ 2n = 24 ≈ 25 ⇒ n = 5
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