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What is the maximum theoretical percent utilization in a 802.3 LAN (that uses 10 base 5 cable) with 128 stations connected to five 500 meter segments. The data rate is 10 Mbps. Assume the frame size is 256 bytes and standard slot time is 51.2 μsec.

(A) 70%  (B) 90%  (C) 80%  (D) 60%

2 Answers

Best answer
5 votes
5 votes

In IEEE 802.3 Ethernet, 
 Utilization U= 100 * (F/R) / (F/R + T*C)
where , U= Percentage of Utilization
R = Transmission Rate
F = Number of bits in a frame
T = Slot Time
C = Number of Contention Intervals = (1-1/N)1-N - 1
N = Number stations

Frame size = (256 *8 ) bits = 2048 bits

Data rate is 10 Mbps

F/R = 2048 bits/10 Mbps = 204.8 µsec

     Number of stations, N = 128

C = (1-1/N)1-N – 1

= { 1- (1/128) }-127 - 1 

= (0.992) -127  - 1 

= 2.7 - 1 

= 1.7

 U = (100 * 204.8) / ( 204.8 + 51.2* 1.7) 

= 20480 / (204.8 + 87.07)

= 20480 / 291.84

= 70.17 

Hence, channel utilization is 70%.

so answer is option A , 70%

1 votes
1 votes

First we find how much time we need to transfer 256 bytes..

Given bandwidth  =  10 Mbps

So time taken to transfer 256 bytes   =   (256 * 8) / (10 * 106)

                                                      =   204.8 μs

Given time to acquire channel           =  51.2 μs

And propogation delay                      =  127 * 500 / (3 * 108)

                                                      =   211.67 μs

So link utilisation                             =   Transmission time(including slot time prior to transmission) / Total time

                                                      =   54.7 %

So going by options it is coming out to be D)..Hence D) should be correct answer..

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