Transmission time has to be L/B= 256/10000000 right? Please explain

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+3 votes

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%

(A) 70% (B) 90% (C) 80% (D) 60%

+5 votes

Best answer

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%

0

Why frame size converted to bits where as bandwidth is in mega'bytes' .

Transmission time has to be L/B= 256/10000000 right? Please explain

Transmission time has to be L/B= 256/10000000 right? Please explain

+1

@chandra sal

It is given in question " The data rate is 10 Mbps " where Mbps means mega bits per second . So Banwidth is given in Mega "bits " , thats why frame size is converted to bits .

Where as MBPS is Mega Bytes per second.

--------------

" Transmission time has to be L/B= 256/10000000 " , No this is wrong , L = 256 * 8 = 2048 bits

and B = 10 Mbps = 10 * 10^{6} bps = 10^{7 }bps

so L/B =( 2048 / 10^{7}) * 10^{6} microseconds [ 1 seconds = 10^{6} microseconds ]

= 2048 / 10 microseconds

= 204.8 microseconds

Hope you understand it .

+2

Can anyone help me out that how C = Number of Contention Intervals = (1-1/N)^{1-N} - 1 ?

what i know is that A=N*p* (1-p)^{1-N} and C= 1/A

where N= no. of station,p=probability

for max utilization p=1/N ,so C= (1-1/N)^{1-N }

0

To analyzing CSMA/CD performance we use the concept of a contention interval.

For two stations contending for the media , contention interval is the time it takes to resolve their conflict.

see this lecture note, --> http://www.piya.ee.engr.tu.ac.th/course/le517/lecture_10.pdf

C = Number of Contention Intervals = (1-1/N)^{1-N} - 1 is given there at slide # 7 ..

0

@Diamond

Why C = Number of Contention Intervals = (1-1/N)^{1-N} - 1 , is correct means why we need to subtract 1 here i am giving you another reason .

suppose C = Number of Contention Intervals = (1-1/N)^{1-N}

and Number of stations, N = 128 is given ,

so C = (1-1/N)^{1-N}

= { 1- (1/128) }^{-127}

= (0.992) ^{-127}

= 2.773

hence U = (100 * 204.8) / ( 204.8 + 51.2 * 2.773)

= 20480 / (204.8 + 141.9776 )

= 20480 / 346.7776

= 59.058

~ 59% and it is not matching with any options, yes it close to option D which is 60%

but if we consider -1 then answer come directly 70% which match with option A..

so from here we should get the idea that (-1) should consider in number of contention interval formula !

+1

@Bikram

To match options we consider -1???

Please help me out logic behind it because in tannenbaum A=(1-1/N)^N -1

C=1/A

So, C=(1-1/N)^1-N

To match options we consider -1???

Please help me out logic behind it because in tannenbaum A=(1-1/N)^N -1

C=1/A

So, C=(1-1/N)^1-N

0

@diamond

matching choice is my secondary reason which i just got out ..but before that did you check this link http://www.piya.ee.engr.tu.ac.th/course/le517/lecture_10.pdf slide number 7 ?

Book follow for making this slide is Stallings and Farauzan ... see yourself http://www.piya.ee.engr.tu.ac.th/course/le517/course_syllabus.pdf

most probably Stallings have this formula ...you can check that book.

0

@diamond

I am now confirm that to calculate C = Number of Contention Intervals

we have to use this formula C= (1-1/N)^{1-N} - 1 , here -1 is needed ..

see another example problem, https://gateoverflow.in/111277/local-area-network , in this problem if we take C = 2.7 [ by not taking that -1 from (1-1/N)^{1-N} ]

then, U = (100 * 409.6) / ( 409.6 + 51.2* 2.7)

= 40960 / (409.6 + 138.24)

= 40960 / 547.84

= 74.766

now see the options given 18%, 82%, 90% and NONE .. so **no option is matching **if we don't use "-1 " in that formula .. :)

so now you understand we have to use this formula , C= (1-1/N)^{1-N }- 1 ..

+1

here C= (1-1/N)^{1-N} - 1 is correct

because C gives us the (no. of stations dont got the hold of channel)/(no. of stations which got)

A=(1-1/N)^(N -1)

ie A gives us the success probabilty

so when we calculate C= (1/A) - 1

we will get the (no. of stations which donot got the hold of channel)/(no. of stations which got)= (1-A)/A

hope this hepled

+1 vote

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 * 10^{6})

= 204.8 μs

Given time to acquire channel = 51.2 μs

And propogation delay = 127 * 500 / (3 * 10^{8})

= 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..**

0

I think in default assumption of these calculations we do not consider the contention period unless and until mentioned..

Also due to contention efficiency will decrease and here maximum theoretical efficiency is asked..

Also due to contention efficiency will decrease and here maximum theoretical efficiency is asked..

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