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Re:How much would an iPhone have cost in 1991? by fortepianissimo on Friday February 07, 2014 @01:23PM (#7197796)
Care to detail your calculation?
Re:How much would an iPhone have cost in 1991? by prr on Friday February 07, 2014 @01:40PM (#7197912)
$3 560 000 divided by 2, 12 times :-)

The effect of Moore's "law" is that processing power doubles approximately every 2 years, while the cost halves.

If 1 "unit" of processing currently costs $1; in 2 years time, 2 units will cost $1. Or that 1 unit costs 0.50.

Re:How much would an iPhone have cost in 1991? by T-hawk on Friday February 07, 2014 @02:20PM (#7198146)
Moore's law is that the density of discrete components (transistors, storage cells) increases. Cost and speed are knock-on effects. It just so happens that density does drive CPUs to be faster for the same production cost and storage to be either more capacious or physically smaller. But some aspects of computing performance don't benefit from density, like memory response time for example.
Re:How much would an iPhone have cost in 1991? by Galactor963 on Friday February 07, 2014 @01:30PM (#7197837)
Divided by two for each cycle (each doubling of transistors). 3560000/2^12 = 869.140625
Re:How much would an iPhone have cost in 1991? by baddox on Friday February 07, 2014 @01:31PM (#7197849)
Just cut 3.56 million in half 12 times.

    3,560,000 * 0.5 ^ 12 ≈ 869.14
Re:How much would an iPhone have cost in 1991? by lutusp on Friday February 07, 2014 @01:42PM (#7197919)
I offer this notational suggestion, just to avoid either a floating-point or fractional constant:

3,560,000 * 2 ^ -12 ≈ 869.14

Based on: 1/2 ^ 12 = 2 ^ -12

Re:How much would an iPhone have cost in 1991? by baddox on Friday February 07, 2014 @01:49PM (#7197960)
That's much less clear if someone isn't aware of how negative exponents are defined. That's why I chose this notation, and why I put into words "cut in half 12 times."
Re:How much would an iPhone have cost in 1991? by lutusp on Friday February 07, 2014 @01:58PM (#7198014)
> That's much less clear if someone isn't aware of how negative exponents are defined.

Yes, I've noticed that in other contexts. It's too bad because it makes the representation less clear, more complex than necessary. I first became aware of this when I tried to say that gravitational force declines as r^-2, to numerous protests.

Oh, well -- a tempest in a teapot, as they say.