I have a historical question. PGP has been around since 1991-ish. At that time, there were still some 8086 kicking around. How long would it take generate a 1024 bit (or whatever the 1991 standard was) asymmetric key on something like an 8086?

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    I think the 6800 was definitely considered old-hat by then. Do you mean 68000 (one extra zero, but a completely different architecture) Commented Jan 13, 2021 at 20:57
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    AFAIR, in the early days of PGP (early 1990s), 512 bit for an asymmetric key was considered very secure, and 384 bit was the typical key size.
    – Leo B.
    Commented Jan 13, 2021 at 21:39
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    @bdegnan — The Motorola 6800 was a very early (1974) 8-bit processor. The Motorola 68000 was a 16-bit processor from 1979 that drove early workstations, the original Macintosh, and later the Atari ST and Amiga computers. It's unrelated to the 6800 except in name. There's roughly an order of magnitude difference in complexity and performance between them. The Apple IIgs, on the other hand, used the WDC 65816, a 16-bit-derivative of the 6502, which, in turn is arguably a derivative of the 6800. Commented Jan 13, 2021 at 22:44
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    As for your question: I don't have an answer, but PGP distributions for DOS, the Amiga or the Atari ST are readily available (e.g. at the.earth.li/pub/pgp/OLD). You could grab an emulator, set the speed to the relevant machines, and try it yourself. Commented Jan 13, 2021 at 22:50
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    Found a rather detailed timeline.
    – Raffzahn
    Commented Jan 14, 2021 at 1:27

1 Answer 1


Generating a 1024-bit PGP key with PGP 2.6.3ia on a (double-speed) HP 200LX (80186 cpu) takes about fifteen minutes.


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