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Question

Q. Consider a 10 mpbs Ethernet lan that has attach to 2.8km long coaxial cable given that transmission speed  is 2.8*10^8 m/s .The packet size is 128 Byte and out of which 30 Byte is overhead .Find the effective transmission rate and max rate at which that network can  send data.

Comments

Effective Transmission Rate = 4.183Mbps

Maximum Transmission Rate = 12.8Mbps

Are these figure up to what you are expecting?

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please explain in detail

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I am gettong 6.4029 Mbps as effective data speed

and max as 8.363 Mbps

is it correc??

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@Shubhanshu Yup its correct i am also getting same 6.405mbps and 8366kbps

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can you explain

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can u plz provide detail solution.

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@kallu singh

@arch

Calculate

TT = DataSize/ BW

TT = 128 Bytes / 10^7 bits/sec

TT = 1024bits / 10^7 bits/sec

TT = 102.4 micro sec

Now, Calculate, PT

PT = len / Velocity

PT = (2.8 * 10^3 meter )/ (2.8 * 10^8 meter / sec)

PT = 10 micro sec

Now, Calculate Throughput (max rate)

Throughput = TT/(TT + 2PT)

You will get  

Throughput = 8.363 Mbps

Now in this we want effective transmission rate which means the rate at which the actual data is being transmitted,

it is given in the question that out of 128 B, 30 B is overhead, which means in this 98 Bytes is of data which is 76.56 % of total (128 Bytes)

So, simply multiply 0.7656 to max throughput = 8.363

We get effective transmission rate 6.4027 Mbps.

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i got it thank u

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i think utilization of ethernet is 1/(1+6.44a)

but you take 1/(1+2a).

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that is considered in case of collision but here their is no collision.

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means when in question collision is mentioned then we us 6.44?

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i have read in some comment about 1 more utilization formula that is 1/(1+5.44a) when we use this formula in ethernet?

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I also in dilemma, see this:- https://gateoverflow.in/169104/csma-cd-with-and-without-collision

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u asked same query which i want to clear but in that no one cleared anything.