Difference between efficency and throughput?

Question

I am confused with this silly doubt.. someone please help me.

To calculate efficiency of sliding window protocol we use formula

Efficiency =  1/1+2a ------Formula1

i understood how this formula derived by series of steps. And the we calculate throughput by using formula

Throughput= Efficiency * Bandwidth ------Formula2

So if i re-write throughput formula as Efficiency = Throughput / Bandwidth.

So how this efficiency differ in formula1 and formula2? And why we are multiplying Efficiency * Bandwidth to get throughput?

P:S I am not trying to mug up formulas but trying to understand how they are derived and how each term have its significance

Comments

Throughput is the no. of bits that a receiver can accept in total time duration (i.e. transmission time(t_x) + 2*propagation delay(t_p)).

For Stop and Wait, the receiver can accept only one frame in the total duration. One frame is of L bits.

Then Throughput= L/(t_x + 2t_p) = L/t_x(1+2a)

t_x=L/B => L/t_x=B  [B is bandwidth]

So, Throughput=B/(1+2a)

Efficiency(or Channel Utilization) is the fraction of time the sender is using the channel out of the total time duration. The sender is using the channel for t_x time.

So, Efficiency= t_x/(t_x + 2t_p) = 1/(1+2a)

Throughput=B* Efficiency

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Efficiency is just , for how much time a sender is actually using the link , this is also known as Link utilization .

Now the channel capacity or ,  the maximum data rate that the channel supports is given by Bandwidth .

Throughput is what is the maximum Data Rate at which a sender can send data through the channel .

In sliding window protocols , it is just Size of data that can be sent per round trip time(RTT) .

 Now since , according to your question how throughput is Efficiency x Bandwidth.

 

Let's try to prove it ,

Let's take the Stop and Wait Protocol.

Efficiency in noiseless channel = 1/(1+2a)

Let X = Efficiency x Bandwidth

1/(1+2a) * Bw

since a = Propagation delay / Transmission Delay.

Let's write it that way ,

(Transmission delay / (Transmission delay + 2*Propagation delay))*Bw

Transmission delay = Size of frame / Bw.

Doing simple calculation it is seen that ,

X = Size of frame / (Transmission delay + 2*Propagation Delay)

 

Now Transmission Delay + 2* Propagation Delay is RTT , if ACK size is considered as negligible .

 

Therefore , X= Size of frame / RTT.

 

Which is nothing but Throughput .

 

Please point if i'm wrong