The following problems concern the measurement of network performance. a. How lo
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The following problems concern the measurement of network performance. a. How long is a bit in copper wire, where the speed of propagation is 2.3 x 10*m/s? b. Suppose a 10-Gbps point-to-point link is being set up between the Earth and a new lunar colony. The distance from the moon to the Earth is approximately 385,000 km, and the data travels over the link at the speed of light-3 x 108 m/s. Calculate the minimum round-trip-time (RTT) for the link. c. Using the RTT as the delay, calculate the delay x bandwidth product for the link. d. What is the significance of the delay x bandwidth product computed in part c? e. A camera on the lunar base takes pictures of the Earth and saves them in digital format to a disk. Suppose Mission Control on Earth wishes to download the most current image, which is 40 MB. What is the minimum amount of time that will elapse between when the request for the data goes out and the transfer is finished?Explanation / Answer
Propagation=distance / speed in propagation
i.e. delay=distance / speed of propagation
distance = delay x speed of propagation
= 1 x 10-10 s x 2.3 x 108m/s (10 Gps is the bandwidth so 10 bits per 10-9 seconds)
= 2.3 x 10-2 m
= 0.023 meters
= distance / speed of light *2
= 2 * 385000000 m / 3 x 108 m/s
= 2.57 s, minimum RTT as the delay.
Delay * bandwidth
= RTT/2 * 10 x 109 bps
= 1.283 * 10 x 109 bits
= 12.83 Gbs
The Bandwidth*Delay Product, or BDP for short determines the amount of data that can be in transit in the network. BDP is a very important concept in a Window based protocol such as TCP/ sliding window protocol. It plays an vital role in high-speed / high-latency networks, like broadband internet connections.
40 MB => 40*8 Mb
Bandwidth is 10 Gbps
= 320 x 106 / 10 x 109
= 320 / 10000
= 0.032 seconds
Now, total time is transmitting time + delay
So, 0.032 + 1.283 seconds
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