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R1. What are some of the possible services that a link-layer protocol can offer to the network layer? Which of these link-layer services have corresponding services in IP? In TCP? Answer : Framing: there is also framing in IP and TCP; link access; reliable delivery: there is also reliable delivery in TCP; flow control: there is also flow control in TCP; error detection: there is also error detection in IP and TCP; error correction; Half/full duplex: TCP is also full duplex. R2. If all the links in the Internet were to provide reliable delivery service, would the TCP reliable delivery service be redundant? Why or why not? Answer : Although each link guarantees that an IP datagram sent over the link will be received at the other end of the link without errors, it is not guaranteed that IP datagrams will arrive at the ultimate destination in the proper order. With IP, datagrams in the same TCP connection can take different routes in the network, and therefore arrive out of order. TCP is still needed to provide the receiving end of the application the byte stream in the correct order. Also, IP can lose packets due to routing loops or equipment failures. R3. In section 5.3, we listed four desirable characteristics of a broadcast channel. Which of these characteristic does slotted ALOHA have ? Which of these characteristic does token passing have? Answer : Slotted Aloha: 1, 2 and 4 (slotted ALOHA is only partially decentralized, since it requires the clocks in all nodes to be synchronized). Token ring: 1, 2, 3, 4. R4. Suppose two nodes start to transmit at the same time a packet of length L over a broadcast channel of rate R. Denote the propagation delay between the two nodes an dprop. Will there be a collision if dprop< L/R? why or why not? Answer :There will be a collision in the sense that while a node is transmitting it will start to receive a packet from the other node. R7. Suppose nodes A, B, and C each attach to the same broadcast LAN (through their adapters). If A sends thousands of IP datagrams to B with each encapsulating frame addressed to the MAC address of B, will C’s adapter process these frames? If so, will C’s adapter pass the IP datagrams in these frames to the network layer C? how would your answers change if A sends frames with the MAC broadcast address? Answer :C’s adapter will process the frames, but the adapter will not pass the datagrams up the protocol stack. If the LAN broadcast address is used, then C’s adapter will both process the frames and pass the datagrams up the protocol stack. R8.How big is the MAC address space? The IPv4 address space? The IPv6 address space? Answer : 248 MAC addresses; 232 IPv4 addresses; 2128 IPv6 addresses. R9. Why is an ARP query sent within a broadcast frame? Why is an ARP response sent within a frame with a specific destination MAC address? Answer : An ARP query is sent in a broadcast frame because the querying host does not which adapter address corresponds to the IP address in question. For the response, the sending node knows the adapter address to which the response should be sent, so there is no need to send a broadcast frame (which would have to be processed by all the other nodes on the LAN). P1. Suppose the information content of a packet is the bit pattern 1110110010001010 and an even parity scheme is being used. What would the value of the field containing the parity bits be for the case of a twodimensional parity scheme? Your answer should be such that a minimum-length checksum field is used. Answer: The rightmost column and bottom row are for parity bits. 11101 11000 10001 10100 00000 P2. Suppose the information portion of a packet (D in Figure 5.4 ) contains 10 bytes consisting of the 8-bit unsigned binary representation of the integers 0 through 9. Compute the Internet checksum for this data. Answer: P3. Consider the previous problem, but instead of containing the binary of the numbers 0 through 9 suppose these 10 bytes contain a. The binary representation of the numbers 1 through 10. b. The ASCll representation of the letters A through J (uppercase). c. The ASCll representation of the letters a through j (lowercase). Compute the Internet checksum for this data. Answer: P6 .Consider the 4-bit generator, G, shown in Figure 5.8,and suppose that D has the value a. 10010001 b. 10100011 c. 01010101 What is the value of R? Answer: • If we divide 1001 into 10010001000 we get 10000001, with a remainder of R = 001. b) If we divide 1001 into we get 10100011000 we get 10110101, with a remainder of R = 101. c) If we divide 1001 into 01010101000 we get 010111101, with a remainder of R =110. P12. Consider three LANs interconnect by two routers, as shown in Figure 5.38. C A E B F Subnet 1 D Subnet 2 Subnet 3 a. b. c. d. e. Redraw the diagram to include adapters. Assign IP addresses to all the interfaces. For Subnet 1 use addresses of the form 111.111.111.xxx; for Subnet 2 uses addresses of the form 122.122.122.xxx; and for Subnet 3 use addresses of the form 133.133.1333.xxx. Assign MAC addresses to all of the adapters. Consider sending an IP datagram from Host A to Host F. Suppose all of the ARP tables are up to date. Enumerate all the steps, as done for the single-router example in Section5.4.2. Repeat (d), now assuming that the ARP table in the sending host is empty (and the other tables are up to date). a), b), c) See figure below. e) ARP in A must now determine the LAN address of 111.111.111.002. Host A sends out an ARP query packet within a broadcast Ethernet frame. The first router receives the query packet and sends to Host A an ARP response packet. This ARP response packet is carried by an Ethernet frame with Ethernet destination address 00-00-00-00-00-00. P14. Recall that with the CSMA/CD protocol, the adapter waits K*512 bits times after a collision, where K is drawn randomly. For K = 100, how long does the adapter wait until returning to Step 2 for a 1 Mbps Ethernet? For a 10Mbps Ethernet? Answer : Wait for 51,200 bit times. For 10 Mbps, this wait is For 100 Mbps, the wait is 512 μ sec. P15. Suppose nodes A and B are on the same 10 Mbps Ethernet bus, and the propagation delay between the two nodes is 225 bit times. Suppose A and B send frames at the same time, the frames collide, and then A and B choose different values of K in the CSMA/CD algorithm. Assuming no other nodes are active, can the retransmissions from A and B collide? For our purposes, it suffices to work out the following example. Suppose A and B begin transmission at t = 0 bit times. They both detect collisions at t = 225 bit times. They finish transmitting a jam signal at t = 225+48=273 bit times. Suppose KA =0 and KB = 1. At what time does B schedule its retransmission? At what time does A begin transmission? (Note :The nodes must wait for an idle channel after returning to Step 2 – see protocol. ) At what time does A’s signal reach B? does B refrain from transmitting at its scheduled time. Answer : P16. Suppose nodes A and B are on the same 10Mbps Ethernet bus, and the propagation delay between the two nodes is 225 bit times. Suppose node A begins transmitting a frame and, before it finishes, node B begins transmitting a frame. Can A finish transmitting before it detects that B has transmitted? Why or why not? If the answer is yes, then A incorrectly believes that its frame was successfully transmitted without a collision. Hint: Suppose at time t = 0 bit times, A begins to transmitting a frame. In the worst case, A transmits a minimum-sized frame of 512+64 bit times. So A would finish transmitting the frame at t = 512 +64 bit times. Thus ,the answer is no, if B’s signal reaches A before bit time t= 512+64 bits. In the worst case, when does B’s signal reach A? Answer : At t = 0 A transmits. At t = 576 , A would finish transmitting. In the worst case, B begins transmitting at time t = 224 . At time t = 224 + 225 = 449 B 's first bit arrives at A . Because 449 < 576 , A aborts before completing the transmission of the packet, as it is supposed to do. Thus A cannot finish transmitting before it detects that B transmitted. This implies that if A does not detect the presence of a host, then no other host begins transmitting while A is transmitting. P19. Suppose two nodes, A and B, are attached to opposite ends of a 900 m cable, and that they each have one frame of 1,000 bits (including all headers and preambles) to send to each other. Both nodes attempt to transmit at time t= 0. suppose there are four repeaters between A and B, each inserting a 20-bit delay. Assume the transmission rate is 10 Mbps, and CSMA/CD with backoff intervals of multiples of 512 bits is used. After the first collision, A draws K =0 and B draws K =1in the exponential backoff protocol. Ignore the jam signal and the 96-bit time delay. a. What is the one-way propagation delay (include repeater delays) between A and B in seconds? Assume that the signal propagation speed is 2*108 m/sec . b. At what time ( in seconds) is A’s packet completely delivered at B ? c. Now suppose that only A has a packet and that the repeaters are replaced with switches. Suppose that each switch has a 20-bit processing delay in addition to a storeand-forward delay. At what time, in seconds ,is A’s packet delivered at B? Answer: P21.Suppose now that the leftmost router in Figure 5.38 is replaced by a switch. Hosts A, B, C, and D and the right router are all star-connected into this switch. Give the source and destination MAC address in the frame encapsulating this IP datagram as the frame is transmitted (i) from A to the switch, (ii) from the switch to the right router, (iii) from the right router to E. Also give the source and the destination IP addresses in the IP datagram encapsulated within the frame at each of these points in time. C A E B F Subnet 1 D Subnet 2 Subnet 3 Answer: P23. Consider Figure 5.26. suppose that all links are 10 Mbps. What is the maximum total aggregate throughput that can be achieved among the 14 end systems in this network? Why? 1Gbps 1Gbps 1Gbps 100Mbps (fiber) 100Mbps (fiber) Mixture of 10 Mbps, 100Mbps,1Gbps, Cat 5 cable 100Mbps (fiber) Electrical Engineering Computer Engineering Computer Science Answer: If all the 14 nodes send out data at the maximum possible rate of 10 Mbps, a total aggregate throughput of 14*10 = 140 Mbps is possible P24. Suppose the three departmental switches in Figure 5.26 are replaced by hubs. All links are 10 Mbps. What is the maximum total aggregate throughput that can be achieved among the 14 end system in this network? Why? 1Gbps 1Gbps 1Gbps 100Mbps (fiber) 100Mbps (fiber) Mixture of 10 Mbps, 100Mbps,1Gbps, Cat 5 cable 100Mbps (fiber) Electrical Engineering Computer Engineering Computer Science Answer: Each departmental hub is a single collision domain that can have a maximum throughput of 10 Mbps. The links connecting the web server and the mail server has a maximum throughput of 10 Mbps. Hence, if the three collision domains and the web server and mail server send out data at their maximum possible rates of 10 Mbps each, a maximum total aggregate throughput of 50 Mbps can be achieved among the 14 end systems.