Sliding-Window-Protocol
Question 1 |
- The time taken for processing the data frame by the receiver is negligible.
- The time taken for processing the acknowledgement frame by the sender is negligible.
- The sender has an infinite number of frames available for transmission.
- The size of the data frame is 2,000 bits and the size of the acknowledgment frame is 10 bits.
- The link data rate in each direction is 1 Mbps (=106bits per second).
- One way propagation delay of the link is 100 milliseconds.
51 |
Tt(packet) = L / B.W => 2000 bits / 10^6 bps = 2 x 10^-3 sec = 2 millisec
Tt(Ack) = L / B.W. => 10 bits / 10^6 bps = 10^-5 sec = 10^-2 millisec = 0.01 millisec
Tp = 100 millisec
Total time = Tt(packet) + 2 x Tp + Tt(Ack)
=> 2 + 2 x 100 + 0.01 = 202.01 millisec
Efficiency = 50 % = ½
Efficiency = Useful Time / Total time
½ = n x Tt / Total time
=> 2 x n x Tt = Total time
=>2 x n x 2 = 202.01
=> n = 202.01 / 4 => 50.50
For minimum, we have to take ceil, Hence size of window = 51
Question 2 |
Host A is sending data to host B over a full duplex link. A and B are using the sliding window protocol for flow control. The send and receive window sizes are 5 packets each. Data packets (sent only from A to B) are all 1000 bytes long and the transmission time for such a packet is 50 µs. Acknowledgement packets (sent only from B to A) are very small and require negligible transmission time. The propagation delay over the link is 200 µs. What is the maximum achievable throughput in this communication?
7.69 × 106 bps
| |
11.11 × 106 bps | |
12.33 × 106 bps | |
15.00 × 106 bps |
Transmission rate , Tt = L / B.W
Therefore, B.W. = L / Tt = 1000 bytes/ 50 μs = 8000 bits / 50 μs=160 Mbps
Efficiency = N / 1 + 2a, where a = Tp / Tt
Efficiency = 5 * 50 / (50+400) = 250/450 = 5/9
Maximum achievable throughput = Efficiency * B.W = (5/9)*160 Mbps = 88.88 Mbps = = 11.11 x 106 bytes per second
*Actual option should be in bytes per second.
Question 3 |
SYN bit = 1, SEQ number = X+1, ACK bit = 0, ACK number = Y, FIN bit = 0 | |
SYN bit = 0, SEQ number = X+1, ACK bit = 0, ACK number = Y, FIN bit = 1 | |
SYN bit = 1, SEQ number = Y, ACK bit = 1, ACK number = X+1, FIN bit = 0 | |
SYN bit = 1, SEQ number = Y, ACK bit = 1, ACK number = X, FIN bit = 0 |
Q will send the SYN bit = 1 to the connection establishment.
Q Seq number will be Y different from X
ACK bit = 1 because sending the ACK
ACK number = X+1 (Next seq number id)
FIN bit = 0 (Because establishing the connection)
Question 4 |
135 |
1 frames takes = Tt = L / B.w. => 1000 / 10^6 = 1 millisec
1000 frame Tt = 1000 x 1 millisec = 1 sec
In 1 sec, 1000 frames sends, which is 1 millisec per frame.
So, G = 1
Efficiency of Pure Aloha (η) = G x e-2G
where G = Number of requests per time slot willing to transmit.
e = Mathematical constant approximately equal to 2.718
So, η = 1 x 2.718(-2 x 1) = 0.1353
Therefore, In 1 sec1000 frames = 0.1353 x 1000 = 135.3(closest integer) =>135
Throughput => 135
Question 5 |
Suppose that it takes 1 unit of time to transmit a packet (of fixed size) on a communication link. The link layer uses a window flow control protocol with a window size of N packets. Each packet causes an ack or a nak to be generated by the receiver, and ack/nak transmission times are negligible. Further, the round trip time on the link is equal to N units. Consider time i > N. If only acks have been received till time i(no naks), then the goodput evaluated at the transmitter at time i(in packets per unit time) is
1 – N/i | |
i/(N + i) | |
1 | |
1 – e(i/N) |
So, successful delivery of packet can be assured if ack has been received for it.
So till time 'i' we would have transmitted 'i' packets but only (i - N) can be acknowledged as minimum time for a packet to get acknowledged is N (since RTT is N which is equal to the window size, there is no waiting for the sender).
So, successfully delivered packets = (i - N)
Time for transmission = i
Goodput = Successfully delivered data/Time
= (i - N)/i
= 1 - N/i