TCP
May 3, 2024Question 10717 – GATE 2007-IT
May 3, 2024Sliding-Window-Protocol
| Question 7 |
Consider a network connected two systems located 8000 kilometers apart. The bandwidth of the network is 500 × 106 bits per second. The propagation speed of the media is 4 × 106 meters per second. It is needed to design a Go-Back-N sliding window protocol for this network. The average packet size is 107 bits. The network is to be used to its full capacity. Assume that processing delays at nodes are negligible. Then the minimum size in bits of the sequence number field has to be ___________.
| 8 | |
| 7 | |
| 6 | |
| 5 |
Question 7 Explanation:
<img src="https://solutionsadda.in/wp-content/uploads/2020/01/g3.jpg"
η=100%
n = ?
∴a=Tp/Tn =2/0.02=100
Given Protocol, Go-back-N protocol, So η = w/(1+2a) where w = 2n-1
100/100 = w/(1+2a) ⇒ w = 1+2a
⇒ 2(n-1) = 1+2(100)
⇒ 2n – 1 = 201
⇒ 2n = 202 ⇒ 2n = 28
⇒ n = 8
η=100%
n = ?
∴a=Tp/Tn =2/0.02=100
Given Protocol, Go-back-N protocol, So η = w/(1+2a) where w = 2n-1
100/100 = w/(1+2a) ⇒ w = 1+2a
⇒ 2(n-1) = 1+2(100)
⇒ 2n – 1 = 201
⇒ 2n = 202 ⇒ 2n = 28
⇒ n = 8
Correct Answer: A
Question 7 Explanation:
<img src="https://solutionsadda.in/wp-content/uploads/2020/01/g3.jpg"
η=100%
n = ?
∴a=Tp/Tn =2/0.02=100
Given Protocol, Go-back-N protocol, So η = w/(1+2a) where w = 2n-1
100/100 = w/(1+2a) ⇒ w = 1+2a
⇒ 2(n-1) = 1+2(100)
⇒ 2n – 1 = 201
⇒ 2n = 202 ⇒ 2n = 28
⇒ n = 8
η=100%
n = ?
∴a=Tp/Tn =2/0.02=100
Given Protocol, Go-back-N protocol, So η = w/(1+2a) where w = 2n-1
100/100 = w/(1+2a) ⇒ w = 1+2a
⇒ 2(n-1) = 1+2(100)
⇒ 2n – 1 = 201
⇒ 2n = 202 ⇒ 2n = 28
⇒ n = 8
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