Question

The Ethernet is used in a 2.0km LAN, with coaxial cable of 2. 10° m/sec and data rate of 100 Mbps. Answer auestions (7-(121 Byes 6 2 -1500 Destination Source address address Pad Check TypeDat b) Preamble osnationSource (7) An Ethernet frame generally contains b. Sre and Dst addresses d. all of the mentioned a. c. Data (Payload) (8) Consider that a short message of 2 bytes (e.g., OK) is generated at the application layer, with TCP header and IP header included, what is the Ethernet frame size for this message, and is padding needed or not? Select one: a. 64 bytes; YES b. 64 bytes; NO d. 42 bytes; NO c. 42 bytes; YES (9) What is the maximum size of Ethernet frames and what is the maximum frame time, re- spectively? Select one: a. 1500 bytes; 120pusec b. 1500 bytes; 15usec d. 1518 bytes; 15.18pusec c. 1518 bytes; 121.4pusec (10) What is the collision detection time in this Ethernet? Select one-- a. 20 Jisec b. 30 jisec c. 40 Asecd. 50sec (11) What is the maximum Ethernet efficiency when the maximum frames are used? Select one: a, 50% b.60% d, 80% c.70% (12) Given that the maximum Ethernet frame includes a total of 58 bytes overhead, what is the effective data rate? Select one: a. 100 Mbps b. 96 Mbps d. 86 Mbps C. 92 Mbps (13) Assume that stations A and B both have experienced 3 collisions. Which of the following is NOT an option for backoff time before next attempt? Select one: a. 8 slots; b. 6 slots; d. 0 slot; c. 3 slots (14) Which one of the following statements is TRUE about MAC in WiFi networks? Select one: a. CSMA/CD cannot be used because of hidden terminal and exposed terminal problem b. CSMA/CA is adopted in the WiFi MAC c. RTS and CTS are necessary to avoid collisions. d. All of the above

Answer 1

7).From the frame we can see that ethernet contains checksum,source and destination MAC address as well as data(payload).so,option d).is right option.

8).data size at application layer = 2B

it will be send down to transport layer where TCP header of minimum size 20 B is inserted.so,now segment size becomes = 22B

it will be send down to network layer where IP header of minimum size 20 B is inserted.so,now packet size becomes = 42B

it will be send down to data link layer where frame header of size18 B is inserted.so,now frame size becomes = 60B

since,here scaling factor is 8.and 60 is not divisible by 8 so,we need to add some padding bits until its size becomes divisible by 8.so,60 + 4 = 64 B

so,correct option is option a).64 Bytes,YES

9).maximum payload size in ethernet = 1500B

with overhead data (frame headers included) the frame size becomes = 1500B + 18 B = `1518B

so,maximum frame size = 1518 Bytes

maximum frame time = maximum frame size / bandwidth = 1518 Bytes / 100 * 10⁶ bits per second = (1518 * 8 )/ (100 * 10⁶)

= 12144 / 10⁸ = 121.44 microseconds.

so,the correct option is c).1518 Bytes,121.44 microseconds

10).suppose there is a host at one end and there is another host at another end.

then suppose the host left end sends the frame and just when it reaches the right end the host at right end starts sending its frame so,host at left end will detect at time = 2 * propagation time

1st propagation time for frame to go from left to right. and 2nd propagation time for the collision's effect to reach at left end.

propagation time = distance / speed = 2 * 10³ / 2 * 10⁸ = 10 microseconds

therefore,maximum collision detection time = 2 * 10 = 20 microseconds.

so,the correct option is a).20 microseconds.

11).efficiency in ethernet = 1 / (1+6.44*a)

where a = propagation time / transmission time

from above we know that propagation time = 10 microseconds

transmission time = 1518 * 8 / 10⁸ = 121.44 microseconds

a = 10/121.44 = 0.08235

efficiency = 1 / (1+6.44 * 0.08235) = 1 / (1.530334) = 0.6534

in percentage = 65.34 %

matching to nearest option which is b).

so,the correct option is b).60%

12).maximum frame size = 1518 Bytes

overhead size = 58 Bytes

effective data rate = (useful data / total data )*bandwidth = ((1518 - 58 )/1518 )*100 Mbps = 96.17 Mbps

so,the correct option is b).96 Mbps