CS 3853 Computer Architecture Recitation 2 Answers Fall 2013

Quiz 2a

1. E_new = .7 + .3
   E_old = .7 + .3 × 1.9 = 1.27
   E_old/E_new = 1.27/1 = 1.27
2. by formula:
   speedup =

   1

   ---

   .6 + .4/1.9

   = 1.23377
   
   from scratch:
   E_old = .6 + .4
   E_new = .6 + .4/1.9 = .91111
   E_old/E_new = 1/.81053 = 1.23377
3. CPI = 5 × .8 + 6 × .2 = 5.2

Quiz 2b

1. E_new = .7 + .3
   E_old = .7 + .3 × 1.6 = 1.18
   E_old/E_new = 1.18/1 = 1.18
2. by formula:
   speedup =

   1

   ---

   .6 + .4/1.6

   = 1.17647
   
   from scratch:
   E_old = .6 + .4
   E_new = .6 + .4/1.6 = .85000
   E_old/E_new = 1/.85000 = 1.17647
3. CPI = 4 × .7 + 9 × .3 = 5.5

Problems

1. old time per instruction = 6, new time per instruction = 5 × 1.2 = 6, so speedup = 1.
2. today is 2013, 34 years later. Number of 18-month intervals = 22.67, 30,000 × 2^22.67 = 200 billion.
3. in 31 years the increase was a factor of 100,000. 2^x = 100,000 → x = 16.6, so it doubles every 31/16.6 years = 1.8675 years.
4. You cannot tell from the information given.
5. E_B = E_A (.4 × 1.3 + .3/1.5 + .3) = 1.02 × E_A, so machine A is faster and the speedup is 1.02.
6. 1. 1/(.5 + .5/2) = 1/.75 = 4/3 = 1.33
   2. 1/(.1 + .9/2) = 20/11 = 1.82
   3. We can think of this as having 4 copies of A running while 1 copy of B is running.
      Program A's new exection time is 3/4 of the old, so 4 copies will take 3 units.
      Program B's new execution time is 11/20 of the old, so one copy takes 11/20 units.
      Together they take 3.55 units instead of 5 for a speedup of 5/3.55 = 1.41.