Computer Networks Summer 2013

Transcription

1 Computer Networks Summer 2013 Problem 1 Homework 2 Due Date: 6/17/2013 Two hosts A and B are each connected to a switch S via 10-Mbps links as shown in Figure 1. The propagation delay on each link is 20 s. S is a store-and-forward device; it begins retransmitting a received packet 35 s after it has finished receiving it. A S B Figure 1 Network Configuration Calculate the total time required to transmit 10,000 bits from A to B. a) As a single packet b) As two 5000-bit packets sent one right after the other a) Per-link transmit delay is 10 4 bits / 10 7 bits/sec = 1000 s. Total transmission delay, D, can be computed as D = = 2075 s. b) The diagram showing the transmission of the two 5000-bit packets from A to B is depicted in Figure 2. The following events take place, during the lifetime of the transmission: 1. T = 0 A starts sending packet 1 2. T = 500 A finishes sending packet 1, and starts sending packet 2 3. T = 520 Packet 1 finishes arriving at S 4. T = 555 Packet 1 departs from S to B 5. T = 1000 A finishes sending packet 2 6. T = 1020 Packet 2 arrives at S 7. T = 1055 Packet 2 departs from S to B 8. T = 1075 Bit 1 of packet 2 arrives at B 9. T = 1575 Last bit of packet 2 arrives at B

2 0 A S B Packet Packet 2 Packet 1 Packet Time Time Time Figure 2 Time Diagram of Transmission Event The total transmission delay, D, can be expressed algebraically as follows: D = 3 T P F 35 = 1575 s, where T=500 s is the transmission time of a packet, P=20 s is the propagation delay, and F=35 s is store-and-forward delay This is an improvement over the first method. The improvement is due to the pipelining effect enabled by packet switching.

3 Problem 2 Consider sending a file of F = M L bits over a path of Q links. Each link transmits at a rate of R bps. The network is lightly loaded so that there are no queuing delays. A form of packet switching is used, where the M L bits are divided into M packets, each packet with L bits. We further assume that propagation delay is negligible. a) Suppose the network is a packet-switched virtual-circuit network. Denote the VC set-up time by t s seconds. Suppose to each packet the sending layers add a total of h bits of header. How long does it take to send the file from source to destination? b) Suppose the network is a packet-switched datagram network, and a connectionless service is used. Now suppose each packet has 2h bits of header. How long does it take to send the file? c) Repeat (b), but assume message switching is used (i.e., 2h bits are added to the message, and the message is not segmented). d) Finally, suppose that the network is a circuit switched network. Further, suppose that the transmission rate of the circuit between source and destination is R bps. Assuming t s set-up time and h bits of header appended to the entire file, how long does it take to send the file? a) The time to transmit one packet onto a link is (L + h) / R. The time to deliver the packet over Q links is Q(L + h)/r. Thus the total latency is T= t s + M(L+h)/R +( L + h)(q-1)/r. M(L+h)/R is the time it the transmit the entire file between over the first hop, (L+h)(Q-1)/R is the time it takes to transmit the last packet over (Q-1) hops, and t s is the set up time (It is assume that t s the time it takes to establish a circuit and close the circuit). b) Packet switched datagram service does not require a VC setup. Therefore, the total latency is T = M(L+2h)/R +( L +2 h)(q-1)/r. c) Message switched datagram service does not require a VC setup. Therefore, the total latency is T = (F+2h)/R +( F+2h)(Q-1)/R. (We assume that the file can be contained in one message. Each message contains 2h header). d) The network is assumed to be light loaded, with no queuing delay at the link. Therefore, there are no store-and-forward delays at the links. Consequently, the total delay is T = t s + (h +F)/R.

4 Problem 3 The sender wants to transmit the message to the receiver. The sender uses the CRC polynomial x for error detection. a) Use polynomial long division to determine the message that should be transmitted. b) Suppose the leftmost bit of the message is inverted due to noise on the transmission link. What is the result of the receiver s CRC calculation? How does the receiver know that an error has occurred? a) The sender appends 000 to the original message The result is M= It then divides M by 1001 to obtain a reminder r=011. The sender than appends 011 to the original message and transmits T= b) After noise inverts the first bit, the received message R= The receiver divides R by 1001 and obtains a quotient of and a remainder of 10. Since the remainder is not equal to 0, the receiver concludes that an error has occurred.

5 Problem 4 The sender wants to transmit the message to the receiver. The sender uses the CRC polynomial x 8 + x 2 + x for error detection. a) Use polynomial long division to determine the message that should be transmitted. b) Suppose the leftmost bit of the message is inverted due to noise on the transmission link. What is the result of the receiver s CRC calculation? How does the receiver know that an error has occurred? a) The senders appends 8 0 s to the message M= The sender then divides the result by (x 8 + x 2 + x 1 + 1). The remainder is R = The sender transmits appends the remainder R to the original message, M, resulting in message T = The sender transmits the message T to the receiver. b) Due to noise inverting the first bit of the transmitted message T, the received message is R= The receiver divides the received message R by (x 8 + x 2 + x 1 + 1) to obtain a remainder of The receiver notices that the reminder is not equal to 0 and concludes that an error has occurred.

6 Problem 5 Consider the use of 1000-bit frames on a 1-Mbps satellite channel with 270-ms delay. We define the throughput, T, of a channel as follows as the number of bits transmitted per second. We assume that the channel is error free and the processing time of frames and acknowledgments is negligible. What is the maximum throughput (bits per second) for the following flow control schemes? 1. Stop-and-wait flow control? 2. Sliding window flow control with a window size of 7? 3. Sliding window flow control with a window size of 127? 4. Sliding window flow control with a window size of 255? Let TF represent the time to transmit a frame, UL, represent the link utilization and W denote the window size. UL = (W TF )/2D, where D is the one way satellite delay. TF = 1000/1o 6 = 10-3 sec; D= 540 msec. 1. Stop-and-wait flow control? W=1: UL = 10-3 /( )= Sliding window flow control with a window size of 7? W=7; UL = 7(10-3 /( ))= Sliding window flow control with a window size of 127? W=127; UL = 127(10-3 /( ))= Sliding window flow control with a window size of 255? W=255; UL = 255(10-3 /( ))= 0.47 Note:the solution treats the TF is negligible comparing to 2D, if you count the UL = (W TF )/(2D+ TF) is also acceptable.