Question

Assume that, in a stop-and-wait arq system, the bandwidth of the line is 1 mbps, and 1 bit takes 20 ms to make a round trip. What is the bandwidth-delay product

Answer 1

The bandwidth-delay product is 20,000 bits

Explanation:

Given that,

Bandwidth of the line =  1 mbps

1 bit takes 20 ms to make a round trip

The length of system data frames = 1000 bits

bandwidth-delay product = $1 \times 10^6 \times 20 \times 10^{-3}$

= $20,000$ bits

During the time it takes the data to go from the sender to the recipient and then back again, the device will send 20,000 bits.

The unit, however, only sends 1,000 bits. We can say that the use of the link is only 1000/20,000, or 5%. For this reason, the use of Stop-and-Wait ARQ is wasting the link's power for a connection with high bandwidth or long delay.

Hence, the bandwidth-delay product is 20,000 bits

Answer 2

Answer:

(i)The bandwidth-delay product is (1 × 10^6

) × (20 × 10^-3) = 20,000 bits. (4)

The system can send 20,000 bits during the time it takes for the data to go from the

sender to the receiver and the acknowledgment to come back.

(ii) However, the system sends only 1,000 bits. (4)

We can say that the link utilization is only 1,000/20,000, or 5 percent