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# Exercises

Exercises. Problem 1. The link delay is 270ms. Each data frame is 1000 bits. Data rate is 1Mbps. What is the link efficiency of the Stop&amp;Wait protocol?. Problem 1. The link delay is 270ms. Each data frame is 1000 bits. Data rate is 1Mbps. What is the link efficiency of the Stop&amp;Wait protocol?

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## Exercises

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1. Exercises

2. Problem 1 • The link delay is 270ms. Each data frame is 1000 bits. Data rate is 1Mbps. What is the link efficiency of the Stop&Wait protocol?

3. Problem 1 • The link delay is 270ms. Each data frame is 1000 bits. Data rate is 1Mbps. What is the link efficiency of the Stop&Wait protocol? • Answer. • Transmission delay: 1000/1000000 = 1ms. • The last bit of the frame reaches the receiver at time 271ms. ACK is received at the sender at 541ms. In 541ms, 1ms is used to send data. Link efficiency is 1/541.

4. Problem 2 • Still the Stop&Wait protocol. The link length is 200m. Signal travels at a speed of 200000000m/s. The file is 1000000 bytes. The server divides the file into frames of 500 bytes. Data rate is 10Mbps. How long does it need for the server to send the entire file? (Assuming no frame loss.)

5. Problem 2 • Still the Stop&Wait protocol. The link length is 200m. Signal travels at a speed of 200000000m/s. The file is 1000000 bytes. The server divides the file into frames of 500 bytes. Data rate is 10Mbps. How long does it need for the server to send the entire file? (Assuming no frame loss.) • Answer. • Transmission delay: 500*8/10000000 = 400us. • The propagation delay is 200/200000000 = 1us. • The last bit of the frame reaches the receiver at time 401us. ACK is received at the sender at 402us. Every 402us, a data frame can sent, hence the total time is 402*2000 = 804000us.

6. Problem 3 • Consider a link with propagation delay of 1ms and the transmission delay of 1ms. Assume Stop&Wait is used and assume that every one of four ACKs is lost. Assume the timeout the sender uses is 3ms, defined as the time to wait AFTER the last bit is sent. What is the link efficiency?

7. Problem 3 • Consider a link with propagation delay of 1ms and the transmission delay of 1ms. Assume Stop&Wait is used and assume that every one of four ACKs is lost. Assume the timeout the sender uses is 3ms, defined as the time to wait AFTER the last bit is sent. What is the link efficiency? • Answer. The first three frames are transmitted correctly using 3*3=9ms. At time 9ms (time starts at 0), the sender started to send the fourth frame. Because the ACK was lost, the sender will timeout and retransmit at 13ms. This time it will go through. So the efficiency is 3/13, that is, every 13ms, three frames will go through.

8. Problem 4 • Consider the Go-back-N protocol. Suppose the link delay is d seconds, the frame size is f bits and the data rate is r bit/s. Find the minimum window size such that the link efficiency is 100%, in the case that the frame never gets lost. Assume the ACK frame is very small.

9. Problem 4 • Consider the Go-back-N protocol. Suppose the link delay is d seconds, the frame size is f bits and the data rate is r bit/s. Find the minimum window size such that the link efficiency is 100%, in the case that the frame never gets lost. Assume the ACK frame is very small. • [Answer:] The time to send a frame is f/r. The minimum window size to achieve 100% efficiency should allow the sender keep on sending before getting the first ACK. If the sender sends at time 0, at time d+f/r, the receiver gets the frame and he immediately sends back ACK which will be received by the sender at time d+d+f/r. So the window size should be (2d+f/r)/(f/r) = 1+2dr/f.

10. Problem 5 • An Ethernet has two stations, A and B. A is at one end of the network, B is at the other end of the network. The propagation delay from one end to the other end is 10us. A and B have data frame to send at 0 and 5us, respectively. Each data frame is 40us. What happens at 0, 5, 10, and 15us?

11. Problem 5 • An Ethernet has two stations, A and B. A is at one end of the network, B is at the other end of the network. The propagation delay from one end to the other end is 10us. A and B have data frame to send at 0 and 5us, respectively. Each data frame is 40us. What happens at 0, 5, 10, and 15us? • Answer. At time 0, A will send. At time 5, B will send, because the signal from A has not propagated to B yet. At time 10, B will notice the collision, and will send a jam signal, and stop sending. At time 15, A will notice the collision, and will send a jam signal, and stop sending.

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