File Size Equivalents | ||||||

Term | Long name | Scientific | Computer | Actual | Translation | Roughly |

1 KB | one kilobyte | 2^{10} bytes
| 1024 bytes | 1,024 bytes | kilo=thousand | 1,000 |

1 MB | one megabyte | 2^{20} bytes
| 1024 KB | 1,048,576 bytes | mega=million | 1,000,000 |

1 GB | one gigabyte | 2^{30} bytes
| 1024 MB | 1,073,741,824 bytes | giga=billion | 1,000,000,000 |

Why is a kilo 1024 instead of an even 1000 in computer lingo? The reason is that computer numbers are binary - representing the two states of on and off. When 2 (the two possible states) is multiplied by itself over and over again, the closest one can get to 1000 is 1024.

While the correct computer shorthand values are based on 1024, we often just casually refer to the numbers in even thousands because the math more easily fits our layman's base-10 math training. However, the larger the number gets, the farther off our layman's version slips.

How does image size affect storage space? By adding up the file sizes of all of your hosted images, you can determine how much storage space is needed. If auction images are nicely compressed to an average of 50K bytes each, then 100 images would require 5,000KB of storage, which is equivalent to 5MB of storage.

Conversely, you can divide the average size of your photos into the storage space allocated by your host. So if a host offered 50MB (50,000KB) of storage, and your photos averaged about 50KB each, then you could load approximately 1000 images onto the host.

Bandwidth requirement is harder to determine in advance, but it can be estimated if one already has auctions running.

**Average Views.** Check the traffic counter in a number of different auctions to see how many views a one-week auction attracts.
Get the average number by adding together all the traffic counts and dividing by the number of auctions in the sampling.
eBay's counters only log unique visitors, so double that averaged number to account for repeat traffic.
Triple the number (or more) if the items are popular with active bidding and lots of watchers.

As an example, let's say the average number of unique hits turned out to be 30 visitors. We might double that figure and use 60 views for each auction during the 7-day auction period.

**Average Image size.** How many photos are in those auctions? How big are they?
Add the file sizes for all the auction photos in all the listings and divide the total by the number of listings.
That will give the total image load per auction for the 7-day period.
The value can be in bytes or KB - just remember which was used when adding up the image sizes.

**Weekly load per auction.** Multiply the average number of views by the average image size and that will give you a bandwidth load for one auction for the week.

**Total Weekly load.** Multiply the weekly load per auction by the number of auctions you usually run per week and that will give you the total image load for the week.

**Accounting Period.** How does the image host determine bandwidth limits? Is it daily or monthly?
Divide the total image load for the week by 7 to get your daily usage, or multiply by 4 to get a montly usage.
That value can be used to determine what level of bandwidth usage should be purchased.
The numbers can be run in reverse to determine how many photos per auction are feasible,
given the stated bandwidth limits assigned to a specific service level.

courtesy of shipscript

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