Get instant access to our step-by-step TCP IP Protocol Suite solutions manual. Our solution manuals are written by Chegg experts so you can be assured of the highest quality! Complete Solutions Manual. TCP/IP Protocol Suite. Local Area Networks. Business Data Communications. Fourth Edition. DeAnza College with. Sophia Chung Fegan. Higher Education. Boston Burr Ridge, IL Dubuque, A Madison,.

Exercises TCP/IP Networking With Solutions • 1. Exercises TCP/IP Networking With Solutions Jean-Yves Le Boudec Fall 2009 1 Module 1: TCP/IP Architecture Exercise 1.9 1. Consider the transparency ”Nagle’s Algorithm: Example”.

Assume that the packet at line 4 is lost in the network. Give a possible continuation of the message chart. Solution: In the hypothesis that the override timer is bigger than the retransmission timer, we have repacketization: at the retransmission timeout the lost bits are retransmitted, together with the bits arrived in the meantime. In this case Nagle’s algorithm does not come into play.

Assume Nagle’s algorithm is disabled for a given connection. Is it possible that some data written by the application is still delayed?

Prove your answer. 1 file: 22 1 • Solution: Due to the sliding window mechanism, whenever the available window does not allow the application to send all the data that it produces, all the remaining data is buffered and therefore de- layed. Exercise 1.10 Quiz 1. True false When a multiport repeater has some bits to send on a half-duplex Ethernet interface, it should first wait until the channel is idle. Solution: False.

The repeater simply repeats bits, even if this causes a collision. True false When a bridge sends a packet towards the final destination over a full duplex Ethernet interface, it should put as destination MAC address the MAC address of the next hop. Solution: False; the bridge does not modify MAC addresses. True false When a bridge has a packet ready to send on a full-duplex Ethernet port, it listens to the medium and waits until the medium is idle. Solution: False, there is no CSMA/CD over full duplex Ethernet.

True false Bridges are said to be “multiprotocol” because a bridged network works independently of network layer protocols such as IPv4 or IPv6. Solution: True.

True false A bridge is an intermediate system for layer 2. Solution: True. True false Assume host A sends an IP packet to host B via bridge X, and assume all three systems are on the same bridged network. Then the destination MAC address in the packet sent by A is the MAC address of X.

Solution: False. Bridges on Ethernet are transparent. True false On a full duplex Ethernet link, there is no CSMA/CD protocol. Solution: True.

A full duplex Ethernet link uses Ethernet physical layer but is not a shared medium link. True false With an Ethernet switch, there is one collision domain per port. Solution: True. True false A multiport repeater separates collision domains. Solution: False. True false When a bridge has a packet ready to send on a half-duplex Ethernet port, it listens to the medium and waits until the medium is idle.

2 • Solution: True, the bridge executes CSMA/CD on all half-duplex ports. True false In a bridged LAN with more than one bridge and with redundant paths, packet sequence is not guaranteed. Solution: False.

Packet sequence is guaranteed by the spanning tree algorithm, which reduces the active topology to a tree. True false Assume hosts A and B are on the same bridged LAN, with one bridge X. When host A sends a packet to host B, the source MAC address is that of A, and the destination MAC address is that of the bridge Solution: False; Bridges are transparent. The destination address is normally that of B; it may also be the broadcast address, or a multicast address. True false A router is an intermediate system for layer 3. Solution: True. True false Ethernet bridges do not use IP addresses when deciding where to send a packet.

Solution: True. Bridges do not look at layer 3 information and are therefore said to be multiprotocol 15. True false If an IP host A receives an IP packet with TTL=255, then A can conclude that the source of the packet is on-link. Solution: True. True false If host A at EPFL wants to send an IP packet to host B at ETHZ, and if A’s ARP cache is empty, then A sends an ARP request in order to determine the IP address of the next hop router.

Solution: False. The ARP request is to find the MAC address of the next hop router. Acdc Rock Band Torrent Wii Super.

True false Assume A and B are two IPv4 hosts, and that the hosts are on Ethernet. If A and B have the same network mask and the same network prefix, then when A sends a packet to B, the packet still contains an IP destination address, equal to the IP address of B. Solution: True. True false When an IP router between two Ethernet segments forwards an IP packet, it does not modify the destination IP address. Solution: true. True false Assume that host A has an IP packet to send to host B, and that the two hosts are on two Ethernet segments separated by a bridge BR. Assume the ARP table at A is empty.

Host A will send an ARP packet in order to find the MAC address of the bridge BR. Solution: False. The bridge is not visible to A. The ARP is to find the MAC address of B. True false Assume A and B are two IPv4 hosts, and that the hosts are on Ethernet. If A and B have the same network mask and the same network prefix; if A has no entry in its ARP, then 3 • before sending a packet to B, A sends an ARP request with target IP address = IP address of B.

Solution: True. Comment: if proxy ARP is used, a proxy ARP server may respond with another MAC address than that of B 21.

True false The route indicated by traceroute may not be the real one because paral- lel paths may exist in the Internet. Solution: True. True false In an intranet with more than one router, packet sequence is guaranteed by means of the TTL field.

Solution: False. Packet sequence is not guaranteed with IP.

True false When an IP router between two Ethernet segments forwards an IP packet, it does not modify the destination MAC address. Solution: false. True false Assume A and B are two IPv4 hosts on the EPFL network. Assume that host A is configured by error with a network mask equal to 255.255.0.0.

When A sends a packet to another EPFL host B, if the ARP cache at A is empty, then A will send an ARP packet in order to find the MAC address of B. Solution: True. This is not the normal configuration, but it will works because in such cases the default router for A will use proxy ARP and respond with its own MAC address 25. True false If there are some errors in the routing tables at some routers, then, with IPv4, it is possible that a packet loops for ever. Solution: False. The packet is discarded when the TTL fields becomes 1. True false When a router sends a packet towards the final destination over a full duplex Ethernet interface, it should put as destination MAC address the MAC address of the next hop.

Solution: True. True false The subnet mask is used by a host or a router in order to know whether it belongs to the same subnet as a machine identified by some IP address.

Solution: True. True false When an application receives a block of data from TCP, the application knows that the data was sent as one message by the source. Solution: False. True false Assume host A sends data to host B using TCP. In some cases, it may happen that two blocks of data generated by the application at A are grouped by TCP into one single IP datagram. Solution: True. TCP does its own packetization.

True false Assume host A sends data to host B using a TCP socket. If A writes three blocks of data into the TCP socket, then there will be three packets sent to B. Solution: False. TCP does its own packetization. There may be any number of packets, depending on how much data is written by B.

True false It is possible for a UDP source A to send data to a destination process P1 on host B1, using source port a and destination port b, and at the same time send (different) data to an- other destination process P2 on a different host B2, still using the same source port a and destination port b. Solution: True. True false With TCP, the goal of silly window syndrome avoidance is to avoid that out of sequence data is delivered to the application. Solution: False. True false When an application receives data from UDP, the application knows that the data was sent as one message by the source. Solution: True.

True false Assume host A sends data to host B using UDP. In some cases, it may happen that two blocks of data generated by the application at A are grouped by UDP into one single IP datagram.

Solution: False. True false With a sliding window protocol and for a constant round trip time, increasing the window size increases the throughput if there is no loss, up to a certain limit. Solution: True.

True false With a sliding window protocol, the window size is the maximum amount of unacknowledged data that can be sent by the source. Solution: True. True false Assume host A sends one block of data to host B using UDP. In some cases, it may happen that the blocks of data generated by the application at A is fragmented by the IP layer at A into several IP packets. Solution: True. 2 Module 2: Dynamic Routing Exercise 2.1 1. Why do bridges have to build a spanning tree whereas routers do not?

5 • Solution: Bridges have to build a spanning tree because they forward packets according to MAC addresses which are not structured and they do not detect frames that loop. Routers do not have to build a spanning tree since they forward packets according to IP addresses which are structured and eventually discard packets that loop. What happens to packets if there is a routing loop with bridges?

With routers? Solution: Packets loop indefinitely if there is a routing loop with bridges. Packets will eventually be discarded if there is a routing loop with routers because of the TTL field.

Is it possible for a link-state algorithm to use the Bellman-Ford algorithm? Why or why not? Solution: The link-state algorithm can use the Bellman-Ford algorithm (static version) for computing the shortest path to all other nodes since the Bellman-Ford algorithm requires only a partial view of the network and the link-state algorithm provides a complete topology view of the network.

⋆ Exercise 2.3 Consider the network in Figure 1. It represents a small corporate network. The IP addresses are shown explicitly; M1 to M15 mean MAC addresses. B1, B2 and B3 are bridges; R1, R2 and R3 are routers. B1 R1 B2 R2 B3 R3 Internet H C1 D1 C2 D2 C3 D3 (DNS) 192.168.1.18 192.168.1.19 192.168.1.17 192.168.1.X 192.168.1.35 192.168.1.36 192.168.1.33 192.168.1.51 192.168.1.52 192.168.1.49 9.23.31.1 9.1.2.2 145.238.10.78 S M1 M2 M3 M4 M8 M12 M11M7 M5 M6 M9 M13 M14 M15 M10 www.plinn.ws 1 192.168.1.Y 9.23.31.2 Figure 1: A small corporate network (exercise 2.3). D3 is the DNS server for this network. The machines C1, D1, C2, D2, and C3 are configured with DNS 6 • server address = 192.168.1.52.

The network is connected to the Internet only by means of a web proxy (the machine H is an application layer gateway). All interfaces that have IP addresses of the form 192.168.x.y are configured with netmask = 255.255.255.240. The default gateways are configured as follows • at C1 and D1: 192.168.1.17 • at C2 and D2: 192.168.1.33 • at C3 and D3: 192.168.1.49 1.

Give a possible value for the X in the IP address of the interface M4 of router R1 (i. Gharelu Nuskhe In Hindi Pdf Free Download. e. Give a possible value for the address marked 192.168.1.X on the figure). Justify your answer. Same question for the Y in the IP address of the interface M8 of router R2.

Solution: The M4 interface must belong to the same subnet as C2 and D2. Since the mask is over 28 bits, X must lie in the interval [33, 46] (the host parts b0000 [X = 32] and b1111 [X = 47] are not possible). The value must also not be already allocated. A possible value is 34.

Similarly, Y must lie in the interval [49, 62]; a possible value is 50. We assume that R1, R2 and R3 are manually configured, i.e. They do not run any routing protocol. Put in the table below the routing table entries that need to be written in these three routers.

Give only the entries for destination prefixes that are not on-link with this router. Solution: (Manual Configu- ration) Destination prefix Destination mask Next hop R1 0.0.0.0 0.0.0.0 192.168.1.33 R2 192.168.1.16 255.255.255.240 192.168.1.34 9.23.31.0 255.255.255.240 192.168.1.49 R3 0.0.0.0 0.0.0.0 192.168.1.50 3. The user at host C1 uses a web browser to connect to the server www.plinn.ws, which is on the ma- chine marked S on the figure. As a result, the web browser at C1 sends a DNS query to determine the IP address that corresponds to the DNS name www.plinn.ws. A packet sniffer placed at the location labelled 1 on the figure reads the DNS query and its answer.

In the table below, mark the values of the fields that are read in these two packets. Solution: The port p needs to be one of the non-reserved ports. 7 • MAC header IP header Transport Protocol header Packet Source MAC address Destination MAC ad- dress Source IP address Destination IP ad- dress Protocol Source Port Destination Port Query from C1 to DNS server M4 M7 192.168.1.18 192.168.1.52 UDP p>1024 53 Response from DNS server to C1 M7 M4 192.168.1.52 192.168.1.18 UDP 53 p 8 • 4.

The web browser at C1 has now received the response from the DNS server and sends an HTTP query. Same question as before for the packets that contain the HTTP query sent by C1 and for the resulting response. MAC header IP header Transport Protocol header Packet Source MAC address Destination MAC ad- dress Source IP address Destination IP ad- dress Protocol Source Port Destination Port HTTP Re- quest from from C1 M4 M7 192.168.1.18 9.23.31.1 TCP p>1024 80 Response to C1 M7 M4 9.23.31.1 192.168.1.18 TCP 80 p 5. Assume that we change (by mistake) the netmask for the interface M1 of host C1. The new mask value is 255.255.255.0.

Will C1 continue to work normally? Justify your answer. Solution: C1 will continue to work normally if router R1 functions also as an ARP proxy.

Otherwise not, because no one will answer to C1’s ARP requests for the machines in 192.168.1/24, which C1 sees as machines on its local subnet. Instead of manual configuration as in question 2, routers R1 R2 and R3 use now RIP. After RIP has converged, what are the routing tables at each router? Give only the entries for destination prefixes that are not on-link with this router.

Solution: (RIP, Fig- ure 1) Destination prefix Destination mask Next hop R1 192.168.1.48 255.255.255.240 192.168.1.33 9.23.31.0 255.255.255.240 192.168.1.33 R2 192.168.1.16 255.255.255.240 192.168.1.34 9.23.31.1 255.255.255.240 192.168.1.49 R3 192.168.1.32 255.255.255.240 192.168.1.50 192.168.1.16 255.255.255.240 192.168.1.50 10 • B1 R1 B2 R2 B3 R3 C1 D1 C2 D2 C3 D3 (DNS) 192.168.1.18 192.168.1.19 192.168.1.17 192.168.1.X 192.168.1.35 192.168.1.36 192.168.1.33 192.168.1.51 192.168.1.52 192.168.1.49 M1 M2 M3 M4 M8 M12 M11M7 M5 M6 M9 M10 192.168.1.Y???? Figure 2: The second network (exercise 2.3). We now pull the wire between M12 and M13; then we change the IP address of the interface at M12 and connect M12 to bridge B1; the resulting new configuration is in Figure 2. What IP address and netmask should we give to M12? Solution: An address in subnet 192.168.1.16/28, for example 192.168.1.20. The netmask should be 255.255.255.240 (i.e.

The prefix is 28 bits) Explain what RIP does immediately after the re-connection? Solution: After discovering that R3 is a neighbor, R1 gets from R3 a route to 192.168.1.48/28 but it is not better than the existing one so there is no change. Symmetrically, R3 gets from R1 a route to 192.168.1.32/28, but it is not better than the existing one, so again there is no change in the routing table.

Since R3 is now on-link with 192.168.1.16/28, R3 sends to R2 a route to 192.168.1.16/28, but again it is not better than the one that R2 already has. Finally, the entries in R1 and R2 to 9.23.31.x will timeout but this normally does not cause a message to be sent.

In the following table, write the routing tables after RIP has stabilized. (As before, give only the entries for destination prefixes that are not on-link with this router.) 11 • (RIP, Fig- ure 2) Destination prefix Destination mask Next hop R1 192.168.1.48 255.255.255.240 192.168.1.33 R2 192.168.1.16 255.255.255.240 192.168.1.34 R3 192.168.1.32 255.255.255.240 192.168.1.50 12 • B1 R1 B2 R2 B3 R3 C1 D1 C2 D2 C3 D3 (DNS) 192.168.1.18 192.168.1.19 192.168.1.17 192.168.1.35 192.168.1.36 192.168.1.33 192.168.1.51 192.168.1.52 192.168.1.49 M1 M2 M3 M4 M8 M12 M11M7 M5 M6 M9 M10 13 2 Figure 3: The third network (exercise 2.3). We reconfigure the network as shown in Figure 3. The interfaces at M4, M8 and M12 are not used. We change the network mask to 255.255.255.0 on all systems, the IP addresses remain the same.

We do a ping from C1 to C2, C2 to C3 and C3 to C1. Packet sniffers are placed at locations labeled 1, 2 and 3 on the figure. In the table below, mark the values of the fields that are read in the ping packets corresponding to each of the ping exchanges if the packet is visible at this location. Consider only the ping packets themselves, not the replies. 13 • MAC header IP header Sniffing Lo- cation Ping Packet Source MAC address Destination MAC address Source IP address Destination IP address Protocol 1 C1 → C2 C2 → C3 M5 M9 192.168.1.35 192.168.1.51 ICMP C3 → C1 M9 M1 192.168.1.51 192.168.1.18 ICMP 2 C1 → C2 C2 → C3 C3 → C1 3 C1 → C2 M1 M5 192.168.1.18 192.168.1.35 ICMP C2 → C3 C3 → C1 M9 M1 192.168.1.51 192.168.1.18 ICMP Solution: All the hosts are on the same network: 192.168.1.0/24. One of the links must be disabled by the spanning tree, so at one of the observation points we should see nothing. We assume it is the link between B1 and B3, so nothing is observed at point 2.