Sliding Window protocols Summary With Questions

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 (W_s) = 1
2. Receiver window size (W_r) = 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 = T_p / T_t

Go Back N –

1. Sender window size W_s = N
2. Receiver window size W_r = 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 = T_p / T_t

Selective Repeat –

1. Sender window size W_s = N
2. Receiver window size W_r = 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 = T_p / T_t

Practice Questions –

• Example-1. In Stop and wait protocol every 4^th 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 T_p = 49.5ms & T_t = 1ms. What is the Efficiency of the protocol and Throughput given Bandwidth = 1000 bps?
• Explanation – Efficiency = N/(1+2a), N = 10 (given), a = T_p/T_t = 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 $ retransmission 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 W_s = 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 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 W_s = 1 and W_r = 1 For GBN, W_s = 2^K-1 and W_r = 1 For SR, W_s = 2^K-1 and W_r = 2^(K-1)