Question

moore’s law states that processor speeds, or overall processing power for computers will double about every 18 months. likewise, the cost to produce a comparable computer will be cut in half
what does this mean? to simplify the mathematics, let’s assume it takes 2 yrs to double/half instead of 18 months.
example 1. in 1988, the number of transistors in the intel 386 sx microprocessor was 275,000. what was the approximate transistor count of the pentium ii intel microprocessor in 1998?
this is called exponential growth.
(the actual number of transistors in in a pentium ii chip in 1998 was 7,500,000!!)
example 2: a personal computer that cost $3,000 in 2002 would cost about how much now?
this is called exponential decay.
an exponential function is of the form:
y = a(b)^x
initial value growth/decay factor
condition 1: a ≠ 0
condition 2: the base (b) is a positive number other than 1.

Explanation:

Example 1:

Step1: Determine time period

From 1988 to 1998, the time elapsed is \(1998 - 1988 = 10\) years. Since we assume doubling every 2 years, the number of doubling periods \(n=\frac{10}{2}=5\).

Step2: Apply exponential growth formula

The exponential growth formula is \(y = a(b)^x\), where \(a = 275000\) (initial value), \(b = 2\) (growth factor as it doubles), and \(x = 5\) (number of periods). So \(y=275000\times(2)^5\).

Step3: Calculate \(2^5\)

\(2^5 = 32\).

Step4: Find the final value

\(y = 275000\times32 = 8800000\). (Note: The actual value is given as 7,500,000, but this is the approximate using the 2 - year doubling assumption)

Example 2:

Step1: Determine time period (assuming current year is 2024 for example, you can adjust based on actual current year. Let's take current year as 2024)

From 2002 to 2024, the time elapsed is \(2024 - 2002 = 22\) years. Since we assume halving every 2 years, the number of halving periods \(n=\frac{22}{2}=11\).

Step2: Apply exponential decay formula

The exponential decay formula is \(y = a(b)^x\), where \(a = 3000\) (initial value), \(b=\frac{1}{2}\) (decay factor as it halves), and \(x = 11\) (number of periods). So \(y = 3000\times(\frac{1}{2})^{11}\).

Step3: Calculate \((\frac{1}{2})^{11}\)

\((\frac{1}{2})^{11}=\frac{1}{2048}\approx0.000488\).

Step4: Find the final value

\(y = 3000\times0.000488\approx1.46\) (If we take a more recent year, say 2020, time elapsed is \(2020 - 2002 = 18\) years, \(n=\frac{18}{2}=9\), \(y = 3000\times(\frac{1}{2})^9=3000\times\frac{1}{512}\approx5.86\))

Answer:

(for Example 1):
The approximate transistor count in 1998 (using 2 - year doubling) is \(\boldsymbol{8800000}\) (actual is 7,500,000).