Question

A tcp machine is sending windows of 65,535 bytes over a 1gbps channel which has a 10ms one way delay. what is the maximum throughput achievable

Answer 1

Answer:65635bytes are sent in 20msec(RTT)

SO througput= bits/sec

65535*8/20*10^-3

And efficiency=throughput/Bandwidth(10^9)

Explanation:

Answer 2

The maximum throughput achievable is 26.214 Mbps.

Given:

A TCP machine is sending windows of 65,535 bytes over a 1gbps channel which has a 10ms one-way delay.

To Find:

The maximum throughput that can be achievable.

Solution:

There are two methods through which we can solve maximum throughput.

Either by first calculating the line efficiency and multiplying it with bandwidth or,

by simply deriving the number of bits sent per second.

For one way the delay is 10ms. That is for two-way it will be 20 ms.

Let us use the easier method,

We know,

The Maximum Achievable Throughput = Number of bits sent per second

That is,

MAT = 65535 Bytes / 20 msec

= (65535 x 8 bits) / (20 x 10⁻³ sec) [As 1 byte  = 8 bits and 1 ms = 10⁻³ s]

= 26.214 Mbps

Thus the maximum throughput achievable is 26.214 Mbps.

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