My laptop has 16GiB - Revision history

Simson at 15:28, 2 May 2018

2018-05-02T15:28:44Z

← Older revision		Revision as of 08:28, 2 May 2018
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	~~My MacBook has 17~~,~~179,869,184 bytes~~ of ~~RAM, and not 16,000,000,000~~. ~~Recall that 17,179,869,184 = 16GiB = 16 x 2~~<sup>30</sup>, which is ~~the correct measurement under IEC 60027-2~~. ~~16GB would be 16,000,000,000 = 16 x 10~~<sup>9</sup>.		RAM is measured in GiB, which is measured as even powers of 1024. (A GiB is 1024<sup>3</sup>)

			Disk and network bandwidth is measured in GB, which is even powers of 1000. (A GB is 1000<sup>3</sup>).

	The differences between 1 KiB and 1 KB is not significant, but when you get up to the Gibibytes, the differences are indeed significant!		The differences between 1 KiB and 1 KB is not significant, but when you get up to the Gibibytes, the differences are indeed significant!

			In the 1980s the powers-of-1024 prefixes did not exist. When Bill Gates gave his famous quote "640 kilobytes ought to be enough for anybody," he didn't mean 640,000 bytes: he meant 640 * 1024 = 655,360 bytes. What is an extra 15,360 bytes between friends? But as memories went to megabytes and then gigabytes, the difference became meaningful. So in the 1990s, the IEC created the new suffixes which are based on powers of 2, rather than powers of 10. However, they are only used for systems that are organized with binary addresses, which basically means memory. Hard drives are peculiar: the sectors are even powers of two, but the number of sectors is arbitrary, so they are typically described in powers of ten.

			My MacBook has 17,179,869,184 bytes of RAM, and not 16,000,000,000. Note that 17,179,869,184 = 16GiB = 16 x 2<sup>30</sup>, which is the correct measurement under IEC 60027-2. 16GB would be 16,000,000,000 = 16 x 10<sup>9</sup>.

	For more information, see https://physics.nist.gov/cuu/Units/binary.html		For more information, see https://physics.nist.gov/cuu/Units/binary.html

Simson at 14:41, 27 April 2018

2018-04-27T14:41:14Z

← Older revision		Revision as of 07:41, 27 April 2018
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	My MacBook has 17,179,869,184 bytes of RAM, and not 16,000,000,000. Recall that 17,179,869,184 = 16GiB = 16 x 2^30, which is the correct measurement under IEC 60027-2. 16GB would be 16,000,000,000 = 16 x 10^9.		My MacBook has 17,179,869,184 bytes of RAM, and not 16,000,000,000. Recall that 17,179,869,184 = 16GiB = 16 x 2<sup>30</sup>, which is the correct measurement under IEC 60027-2. 16GB would be 16,000,000,000 = 16 x 10<sup>9</sup>.

	The differences between 1 KiB and 1 KB is not significant, but when you get up to the Gibibytes, the differences are indeed significant!		The differences between 1 KiB and 1 KB is not significant, but when you get up to the Gibibytes, the differences are indeed significant!

	For more information, see https://physics.nist.gov/cuu/Units/binary.html		For more information, see https://physics.nist.gov/cuu/Units/binary.html

Simson: Created page with "My MacBook has 17,179,869,184 bytes of RAM, and not 16,000,000,000. Recall that 17,179,869,184 = 16GiB = 16 x 2^30, which is the correct measurement under IEC 60027-2. 16GB..."

2018-04-27T14:40:41Z

Created page with "My MacBook has 17,179,869,184 bytes of RAM, and not 16,000,000,000. Recall that 17,179,869,184 = 16GiB = 16 x 2^30, which is the correct measurement under IEC 60027-2. 16GB..."

New page

My MacBook has 17,179,869,184 bytes of RAM, and not 16,000,000,000. Recall that 17,179,869,184 = 16GiB = 16 x 2^30, which is the correct measurement under IEC 60027-2. 16GB would be 16,000,000,000 = 16 x 10^9.

The differences between 1 KiB and 1 KB is not significant, but when you get up to the Gibibytes, the differences are indeed significant!

For more information, see https://physics.nist.gov/cuu/Units/binary.html