CS 3853 Computer Architecture Recitation 1 Answers Fall 2013

Quiz 1a

  1. EB/EA = 1.3., EA = 5, so EB = 1.3 × 5 = 6.5 seconds.
  2. EB/EA = 1.3., EB = 5, so EA = 5 / 1.3 = 3.85 seconds.
  3. EB/EA = 1.3., EC/EB = 1.4, so EC/EA = 1.3 × 1.4 = 1.82 and A is 82% faster than C.

Quiz 1b

  1. EB/EA = 1.4., EA = 10, so EB = 1.4 × 10 = 14 seconds.
  2. EB/EA = 1.4., EB = 10, so EA = 10 / 1.4 = 7.14 seconds.
  3. EB/EA = 1.4., EC/EB = 1.5, so EC/EA = 1.4 × 1.5 = 2.10 and A is 110% faster than C.

Problems

  1. (1.2)5 = 2.4883, so the improvement is 148.83%.
  2. EB/EA = 1.6 and EC/EA = 1.8, so EC/EB = 1.8/1.6 = 1.125, so B is 12.5% faster than C.
  3. (1 +
    n
    100
    )30 = 10000, so 30 log10 (1 +
    n
    100
    ) = 4, so log10 (1 +
    n
    100
    ) = .1333, and 1 +
    n
    100
    = 10.1333 = 1.3593, giving n = .3593.
    Each year the performance increase is 35.93%.
  4. wafer diameter = 300 mm, wafer radius = 150 mm, die area = 150 mm2, sqrt(2 × die area) = 17.32
    dies per wafer =
    π × (wafer diameter/2)2
    die area
     - 
    π × wafer diameter
    √2 × die area
    =
    π × (150)2
    150
     - 
    π × 300
    17.32
    = 416.82, so about 416 dies per wafer.
  5. $5000/416 = $12.02, but with a 50% yield, each good one costs $24.04.
    1. There are 8766 hours in an average year so the probablility that a disk will die in a year is 8,766/1,000,000 = .008776, so for 1000 disks, between 8 and 9 are likely to die in the first year.
    2. All of them, since disk drives do not last 20 years.