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Lesson 3 - Choosing WAN Products and ServicesComparing File Transfer TimesThese data rates clearly show which services are faster than others. However, without some first-hand experience with different types of connections, a speed figure alone does not tell us much about the performance a user can expect. Thus, when comparing WAN services, it is often helpful to compare their performance on a common task. For example, assume a bank transfers many files between its two offices in the same city. These files average approximately 5 megabytes (MB) in size, thus a network designer could evaluate various WAN services by calculating the time each one would require to move 5 MB. Bits vs. BytesAs we have seen, network data rates are expressed in Kbps and Mbps. However, file sizes are expressed in KB and MB. Thus, before we can do the simple math to figure a file transfer time, we must first convert the file size into the same units as our data rate. We could use bits; however, kilobits (Kb) is just as useful. Megabytes to KilobitsFirst, let us remember the basic computer numbering system:
We can convert our 5-MB file to Kb: 5 MB * 1,024 KB/MB * 8 bits/byte = 40,960 Kb If a data rate is already expressed in Kbps, then we are ready to calculate. For example, the time necessary for a 56-Kbps connection to transfer our file is: 40,960 Kb / 56 Kbps = 731 seconds or approximately 12 minutes Megabits to KilobitsIf a data rate is measured in Mbps, it is a one-step conversion to Kbps. We can express the bandwidth of a T1 line in Kbps as follows: 1.544 Mbps * 1,024 Kb per megabit (Mb) = 1,581 Kbps Now we can see how fast our T1 line can transfer that 5-MB file, assuming we use all 24 channels for the task: 40,960 Kb / 1,581 Kbps = 25.9 seconds Signaling OverheadIn the last example, we assumed that the entire capacity of a T1 line (1.544 Mbps) would be used to transfer the file. However, as we learned earlier in this unit, some of a T1's bandwidth is used for control signaling. Thus, when comparing WAN services, it is important to know the effective throughput of each type of connection. Even if a technology offers an impressive data rate, we must remember that not all of that bandwidth is available for our use. ATM is a good example of this principle. Each 53-byte ATM cell carries 48 bytes of "real" data. The other 5 bytes are used by the cell header. Thus, roughly 9 percent of ATM's bandwidth is unavailable for real traffic. ActivitiesSee the Activities and Extended Activities section in Unit 7 Lesson 3 in your textbook Introduction to Networking to test what you have learned so far.
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