1 / 2

Answers of Exercise 4

Answers of Exercise 4. 1. Prove that the total number of links in a point-to-point network with N computers is (N 2 -N)/2 Proof: 1 st computer 2 nd computer … (N-1)th computer Nth computer

nico
Télécharger la présentation

Answers of Exercise 4

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 4 1. Prove that the total number of links in a point-to-point network with N computers is (N2-N)/2 Proof: 1st computer 2nd computer … (N-1)th computer Nth computer links N-1 N-2 1 0 Total = (N-1) + (N-2) + … +2+1= (N2 - N)/2 2. For each of bus, star and ring topologies, discuss the consequences if a station has broken. Answer: Ring topology can’t work if a station is broken since it can not pass data to the next. No affect to bus and star topologies. 3. In most technologies, a sending station can choose the amount of data in a frame, but the frame header is a fixed size. Calculate the percentage of bits in a frame devoted to the header, trailer and preamble for the largest and smallest Ethernet frame. Answer: Ethernet frame Preamble + Header + trailer = 8+6+6+2+4 = 26 Bytes For the largest frame: 26/(26+1500) = 1.7% For the smallest frame: 26/(26+46) = 36.1%

  2. Answers of Exercise 4 4. Assume a one megabyte file must be send from one computer to another (both are connected to a same 10Base-T Ethernet). What is the minimum time to send the file across the network? What is the minimum time across a Fast Ethernet? Across a gigabit Ethernet? Answer: 1 Mbytes  1024X1024=1048576bytes The maximum data can be sent in one Ethernet packet is 1500 bytes. 1048576bytes/1500 = 698 (1500_byte_packets) + 1 (576_byte_packet) Therefore at least 699 packets are needed to send the 1 Mbytes file. The length of preamble, header and trailer in an Ethernet packet is 26 bytes. Total bits to be sent for the file are 1048576 + 699x26 = 1066650 bytes = 8.533 X 106 bits 10Base-T: 0.853 sec, Fast Ethernet: 0.0853 sec, Gigabit Ethernet: 0.00853 sec 5. Give the definitions and examples of unicast, broadcast and multicast, respectively. Answer: Unicast: sending data to a single destination. Examples: modem, telephone, accessing a Web site, sending an email to an individual Broadcast: sending data to all destinations. Examples: radio and TV broadcast, email from network administrator to all users Multicast: sending data to partial destinations. Examples: news group, audiovisual conferencing

More Related