Sliding Window protocols Summary With Questions

• Difficulty Level : Medium
• Last Updated : 18 Oct, 2021

Prerequisites – Stop & Wait, Go Back N, Selective Repeat

Summary of all the protocols –

Before starting with the questions a quick recap for all the protocols.

Stop and wait –

1. Sender window size (Ws) = 1
2. Receiver window size (Wr) = 1
3. Sequence Number ≥ 1 + 1
4. Uses independent acknowledgement
5. Discards out of order packets
6. Packet Loss ? Retransmit packet after time out
7. Acknowledgement loss ? Resends packet after time out
8. Efficiency = 1/(1+2a) where a = Tp / Tt

Go Back N –

1. Sender window size Ws = N
2. Receiver window size Wr = 1
3. Sequence number ≥ N + 1
4. Can use both cumulative or independent acknowledgement depends on acknowledge timer
5. Discards out of order packets
6. Packet Loss ? Track back N size from the last packet within the window limit to the lost packet and retransmit them
7. Acknowledgement loss ? If not received before timeout the entire window N size is resend
8. Efficiency = N/(1+2a) where a = Tp / Tt

Selective Repeat –

1. Sender window size Ws = N
2. Receiver window size Wr = N
3. Sequence Number ≥ N + N
4. Uses only independent acknowledgement
5. Can Accept out of order packets
6. Packet Loss ? Resend only the lost packet after timeout
7. Acknowledgement loss ? Resend if not receive before timeout
8. Efficiency = N/(1+2a) where a = Tp / Tt

Practice Questions –

• Example-1. In Stop and wait protocol every 4th packet is lost and we need to send total 10 packets so how many transmission it took to send all the packets?
• Explanation –

  1 2 3 4 5 6 7 8 9 10 (Initially)^1 2 3 4 4 5 6 7 8 9 10 (Packet no. 4 retransmitted)              ^1 2 3 4 4 5 6 7 7 8 9 10 (Packet no. 10 retransmitted)^1 2 3 4 4 5 6 7 7 8 9 10 10 (Result)

  So, we retransmitted packet number 4, 7, 10
  Total count = 13

• Example-2. In S&W protocol if Error probability is p and no. of packets to send is ‘n’. How many packets we have to send ?
• Explanation –
  Total retransmissions
  = n*p0+ n*p1+ n*p2 + n*p3 + n*p4 + …
  = n(1 + p + p2 + p3 + p4 + …)
  = n*(1 / (1-p)) using infinite GP sum formula
• Example-3. In GBN sender Window size = 10 and Tp = 49.5ms & Tt = 1ms. What is the Efficiency of the protocol and Throughput given Bandwidth = 1000 bps?
• Explanation –
  Efficiency = N/(1+2a), N = 10 (given), a = Tp/Tt = 49.5
  Efficiency = 10/(1 + 2 * 49.5) = 10/100 = 0.1 or 10%
  Throughput = Efficiency * Bandwidth
  = 0.1 * 1000 = 100
• Example-4. In GB3 if every 5th packet is lost & we need to send 10 packets so how many retransmissions are required ?
• Explanation –

  1 2 3 4 5 6 7  | 8 9 10        ^   $            (packet no. 5 lost)1 2 3 4 5 6 7 5 6 7 8 9 | 10*   ^   $1 2 3 4 5 6 7 5 6 7 8 9 7 8 9 10*   ^  $1 2 3 4 5 6 7 5 6 7 8 9 7 8 9 10 9 10 (count starts from * till ^)(from ^ to $ retranmission is done)

  Note – From the Last packet is window size to lost pocket we resend the entire window.
  Total no. of transmissions = 18

• Example-5. In SR Ws = 5 and we are sending 10 packets where every 5th packet is lost find number of retransmissions?
• Explanation –

  1 2 3 4 5 6 7 8 9 10^1 2 3 4 5 5 6 7 8 9 10^1 2 3 4 5 6 7 8 9 9 10

  We see here there is no role of Window size in SR only the lost packet is resent.
  Total transmissions = 12

• Example-6. If there is K bits sequence no. define require sender window size and receiver window size for S&W, GBN & SR?

  Explanation –
  Given, K bits, For S&W Ws = 1 and Wr = 1
  For GBN, Ws = 2K-1 and Wr = 1
  For SR, Ws = 2K-1 and Wr = 2(K-1)