That would depend on the rounding method. For instance, years ago (I haven't checked lately) the rounding methods used by both Lotus 123 and Excel were statistically inaccurate.
Hmm, unless MS an Lotus made some dorky implementation of a simple function I can't see it happening. Mathematically speaking rounding is the best choice; for example given a fair bit that can be 0 or 1 with equal probability (1/2), the best choice IS the mean value i.e. 0.5 the same applies to rounding.
About the 1.44MB; a floppy has 1,474,560b unformatted and 1,457,644b when formatted in FAT. The calculation is done using either 1440/1024~1.40 or 1457644/(1024)^2~1.39 which is the real space.
Take a look here:
Taken from obscure DOS documentation
Code: Select all
PSS ID Number: Q121839
Article last modified on 03-06-1996
3.x 4.x 5.x 6.00 6.20 6.21 6.22 | 3.10 3.11 95
MS-DOS | WINDOWS
---------------------------------------------------------------------
The information in this article applies to:
- Microsoft MS-DOS operating system versions 3.x, 4.x, 5.x, 6.0,
6.2, 6.21, 6.22
- Microsoft Windows operating system versions 3.1, 3.11
- Microsoft Windows for Workgroups versions 3.1, 3.11
- Microsoft Windows 95
---------------------------------------------------------------------
SUMMARY
=======
The 1.44-megabyte (MB) value associated with the 3.5-inch disk format
does
not represent the actual size or free space of these disks. Although its
size has been popularly called 1.44 MB, the correct size is actually
1.40
MB.
MORE INFORMATION
================
The correct size is determined by multiplying the number of tracks,
sides,
sectors per track, and 512 bytes per sector, then subtracting the bytes
required to format the disk, and then dividing this figure by 1024. For
a
"1.44-MB" 3.5-inch floppy disk, there are
80 tracks
18 sectors per track
512 bytes per sector
2 sides
Multiplying the above gives you 1,474,560 bytes. This is the unformatted
size.
To determine the number of bytes formatting requires, you need to know
how
many bytes are used for the boot sector, file allocation table (FAT),
and
root directory.
There is 1 sector used for the boot sector, which is 512 bytes; 18
sectors for the two FATs (9 sectors each), which is 9216 bytes (512 *
18 = 9216); and 14 sectors for the root directory, which is 7168
bytes.
NOTE: There are two ways to arrive at the 7168 number:
224 entries * 32 bytes per entry = 7168 bytes
-or-
512 bytes per sector (14 * 512 = 7168 bytes)
Adding these figures gives you 16,896 bytes.
Subtracting the amount used for formatting from the total unformatted
size
gives you 1,457,664. (1,474,560 - 16,896 = 1,457,664 bytes)
Dividing the above figure by 1024 bytes generates 1440. (1,474,560 /
1024 = 1440 KB)
To convert to megabytes, divide by 1024. (1440 KB / 1024 = 1.406 MB)
This formula works for 1.2-MB disks as well. The only variable is the
number of sectors, which is 15, for the calculations with 1.2-MB disks.
From the calculations shown above, we can see that the 3.5-inch disk
considered to have 1.44 MB free disk space actually has 1.40 MB, and the
5.25-inch disk considered to have 1.2 MB actually has 1.17 MB.
The misunderstanding comes from the incorrect calculation below:
1440 KB / 1000 = 1.44 MB
The calculation should be:
1440 KB / 1024 = 1.40 MB
There are 1024 bytes in a kilobyte, not 1000.
Note that in Windows 95, the properties for a blank, formatted 3.5-inch
1.44-MB disk show that there are 1.38 MB of free disk space.
Cheers.
) someone agrees that rounding is preferable because it is a FACT (not just an opinion) that rounding is more accurate than truncating generally.
