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In a stop and wait ARQ, Bandwidth Delay product is 20,000 bits. Given the frame size is 1000 bits. What is the percentage of utilization of link if we have a protocol that can send 15 frames without waiting for acknowledgement?

### 4 Comments

Is it 75% utilisation ?
Yes, btw what is bandwidth delay product?
w\o waiting for a ack then it will work like Go-back-n protocol

@Nisha Bharti it is BW * Round trip delay

## 2 Answers

Bandwidth delay product = $Bw*RTT$; $Bw$ = bandwidth, $RTT = Routed \ Time = 2*Tp$

Here it’s given : $Bw*RTT = 20000\ bits$, and the protocol is sending $15$ frames of size $1000bits$ = $15000bits$

Hence, link utilization or channel utilization or bandwidth utilization or efficiency = (Bits \ successfully \ sent)/ (Bandwidth  \ delay \ product)*100\% = 15000/20000*100 = 75\%$### 2 Comments Never heard of this formula for efficiency:$\eta = \frac{Bits Successfully Sent}{Bandwidth Delay Product}*100$%. Could you please give a little insight into the same? @Devanshu Sachdev Bandwidth Delay Product = Bandwidth * Round Trip Time. This Bandwidth Delay Product is equivalent to the maximum amount of data on the network circuit at any given time. Now you know efficiency = Useful Bytes/ Total Bytes sent. So from there we can derive the formula…$\eta = \frac{Bits Successfully Sent}{Bandwidth Delay Product}*100\$

In a stop-and-wait ARQ, the bandwidth-delay product determines the maximum number of unacknowledged frames that can be in transit at any given time. In this case, the bandwidth-delay product is 20,000 bits, which means that at any given time, there can be up to 20,000 bits worth of data in transit.

If the frame size is 1000 bits, and the protocol allows 15 frames to be sent without waiting for acknowledgement, then the total number of bits in transit would be 15 * 1000 = 15,000 bits.

To calculate the percentage of utilization of the link, we can divide the number of bits in transit by the bandwidth-delay product and multiply by 100 to express the result as a percentage:

utilization = (15,000 bits / 20,000 bits) * 100% = 75%

Therefore, the percentage of utilization of the link would be 75%.

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