1 / 2

Answers of Exercise 10

Answers of Exercise 10. 1. 100 byte data is sent using IP across an Ethernet. Before sent, the data will be first formed an IP datagram and then the datagram will be encapsulated into an Ethernet Frame. Calculate the percentage of headers in sending the 100 byte data. Assume

damali
Download Presentation

Answers of Exercise 10

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Answers of Exercise 10 1. 100 byte data is sent using IP across an Ethernet. Before sent, the data will be first formed an IP datagram and then the datagram will be encapsulated into an Ethernet Frame. Calculate the percentage of headers in sending the 100 byte data. Assume no optional field in both IP and frame header. Answer: As Ethernet frame header length is 26 bytes and IP datagram header is 20 bytes, total length = 26+20+100=146 bytes. Header percentage=46/146=31.5%. 2. Suppose a file of 20 Kbytes to be sent from host H1 to host H2 across three networks as shown in the following figure. How many IP datagrams will be sent from H1? And how many IP datagrams will be received by H2? Assume no datagram loss, duplication and disorder during the transmissions. Answer: As 20k=20x1024=20480=4x4464+1x2624 thus, number of IP datagram sent from H1 is 5 (four 4464 datagrams and one 2624 datagram) Further, 4464=2x1500+1x1464 and 2624=1x1500+1x1124 Therefore, number of IP datagram received by H2 is 4x3+2=14. Note that there is no need of sub-fragmentation and assembly in R2. Toking Ring MTU=4464 Ethernet MTU=1500 FDDI MTU=4352 H1 R1 R2 H2

  2. Answers of Exercise 10 3. Host A sends a message to host B and never receive reply from B. However, host receives an ICMP message with a header in hexadecimal format as the follows 03 01 1A C8 31 00 B7 Give possible reasons that A does not receive reply from B. Answer: In the above ICMP message, type=03 shows destination unreachable error, further,the code=01 shows host unreachable error. Therefore, the reasons is the host has hardware failure, is not connected to network, or powered down. 4. Explain how traceroute utility works. Use the utility in a Windows OS environment to probe the Internet organization web server. The command is tracert www.ietf.org. How many routes have been passed when your packet travel to the web server? Which one is the slowest? Answer: Each IP datagram has a TTL field which specifies the number of routers the packet can pass. If TTL=0, a router will not forward the packet, discard it and send a ICMP message to the host. The traceroute utility uses the TTL and ICMP error report mechanism to detect routers in route. The traceroute application sets TTL=0, 1, 2, …, n, and shows routers as well as time according to corresponding ICMP messages. Note that the information displayed is usually different when using traceroute utility in different sites, computers and time. When I tested it at 1:00PM in June 21, 2000, the number of routers between my computers to the web server at www.ietf.org is 21, and the slowest router is at 206.204.271.97 with name hevacis-01-pos3-0-0.conxion.net. 5. Summarize main features of IPv6 as compared with IPv4. Answer: (1)Large address size - 128bits; (2) Extensible functions because of extensible header; (3) Support for resource allocation (QoS) - audio and video applications; (4) Support for security.

More Related