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A 2-km-long, 10-Mbps CSMA/CD LAN (not 802.3) has a propagation speed of 200 m/microsec.  
Repeaters are not allowed in this system. Data frames are 512 bits long, including 32 bits of header, checksum, and other overhead.
The first bit slot after a successful transmission is reserved for the receiver to capture the channel in order to send a 32-bit acknowledgement frame.
The effective data rate is _____________________ Mbps (correct to 2 decimal places), excluding overhead, assuming that there are no collisions?
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Is it 6.47 Mbps?
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answer given as 4.21 Mbps
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In CDMA/CD when one sender sends others just listen and wait for their chance to send even receiver could not send. The receiver sends acknowledgment after receiving full frame.

Tpropagation = 2km / 2 *10=10 MicroSec 

Transmission = 512 bits / 10Mbps = 51.2 MicroSec

Transmission = 32bit Ack / 10Mbps = 3,2 MicroSec

1.Transmission delay for sender =51.2 MicroSec

2. Propagation delay from sender to receiver  * 2 =20 MicroSec

3. Wait till the time channel has been free =1 propagation Delay =10MicroSec

4. Ack Tramsmission Delay =3,2 MicroSec

5. Propagation delay from receiver to sender  * 2 =20MicroSec

6.Wait till the time channel has been free Because of Ack =10 MicroSec

Sum of all these = 114.4MicroSec

10Mb can be send in 1 sec So 114.4 MicroSec= 1144bits 

Efficiency = Actual amount of data can be sent /  Capacity of Channel for given period.

              = (512-32) / 1144  = 0.4195 

Effective Bandwidth = 0.4195 * 10Mbps =4.195Mbps

http://webhome.csc.uvic.ca/~wkui/Courses/networks/midterm-F05.pdf

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@Bikram Sir. Thnku. I understand the approach used in the link you provided. My confusion is should we always use that method for calculating in case of CSMA/CD ? Here is a below link which uses standard value by calculating number of collisions and approximating to 6.44

Page No - 17

http://www.nlcpr.com/MUNengineering/uploadedfiles/6876Chapter6_Presentation_Part2.pdf

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@yash

where to use which formula , it's depends upon the question itself . 

that link u shared says " Time slotted in 2 * tprop slots during contention periods ( see slide no - 15 ) "

average contention period = 2 * tprop

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Here CSMA/CD is use and from the question

T(transmission)=512/10*10^6=51.2micro sec

T(propagation)=10micro sec

CSMA/CD efficiency=1/1+6.44a where a=T(p)/T(t)

                    

here we want effeftive bandwidth=efficiency * bandwidth=0.4429*10=4.42
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