Homework 4

1. Page 55 1.4

   Assume a color display using 8 bits for each color per pixel and a 
   frame size of 1280 x 1024.  
   
   a. What is the minimum size in bytes of the fram buffer to store a frame?
   
   1280 x 1024 pixels x 3 colors / pixel x 8 bits/ color = 31,457,280 bits.
        31,457,280 bits x 1 byte/8 bits = 3,932,160 bytes.
        4 MB
   
   b. How long would it take, at a minimum, for a frame to be sent over
   a 100 Mb/sec network
   
       31,457,280 bits x 1 second/100x106 bits = .31 seconds.
      

2. 1.5 page 55

    
   Consider three different processors P1, P2, and P3 executing the 
   same instruction set. 
       P1 has a 3.0 GHz clock and a CPI of 1.5
       P2 has a 2.5 GHz clock and a CPI of 1.0
       P3 has a 4.0 GHz clock and a CPI of 2.2
   
      (a) Which process has the highest performance expressed in 
          instructions per second.
   
       1 instruction/ C cycles x  n cycles/ 1 second = n/c instructions/second
   
       P1:   3.0/1.5 x 109 instructions = 2.0 Ginstructions
       P2:   2.5/1.0 x 109 instructions = 2.6 Ginstructions
       P3:   4.0/2.2 x 109 instructions = 1.8 Ginstructions
   
       P2 has a higher instruction per second value so it is faster, assuming
       all three are running the same program with the same instruction mix.
   
       (b) If the processors each execute a program in 10 seconds, find the 
          number of cycles and instructions.
   
        i instructions/ sec x 10 seconds = 10i instructions
   
        10i instructions x c cycles/instruction = 10i × c cycles.
   
        P1: 2.0x109 x 10  = 20x109 instructions
        P2: 2.6x109 x 10  = 26x109 instructions
        P3: 1.8x109 x 10  = 18x109 instructions
   
        P1: 20x109 x 1.5 = 30 x 109cycles
        P2: 26x109 x 1.0 = 26 x 109cycles
        P3: 18x109 x 2.2 = 40 x 109cycles
      

3. 1.6

         P1: .1 x 1 + .2 x 2 + .5 x 3 + .2 x 3 = 2.6 CPI
         P2: .1 x 2 + .2 x 2 + .5 x 2 + .2 x 2 = 2.0 
   
         1.0 x 106 instructions x c cycles/instruction = c x 106 cycles.
   
         P1: 2.6 x 106 cycles
         P2: 2.0 x 106 cycles
      

4. 1.12

   
     Start off by computing the time for each to perform the computation.
   
     i instructions x c cycles/instruction x 1 second/ k x 109 cycles
        = ic/k  x 10-9 seconds
   
        P1: i = 5x109               P2:   i = 1.0x109
            k = 4  (GHz)                  k = 3  (GHz)
   	 c = 0.9                       c = 0.75
   
   	 5 x 0.9/4 = 1.125 sec          1x0.75/3 =.25 sec
   
   	 P2 is clearly faster due to the low instruction count and other
   	 factors being nearly the same.
   
        (a) in spite  of P1 having the faster clock, it is not the faster machine
            for this program
   
        (b) 1.0 x 109 instructions x 0.9 cycles/instruction x 1 second/4.0x109 cycles 
             = 1x0.9/4 seconds
   	  =  0.225 seconds
   
   	 0.225 seconds = n x109instructions x 0.75 cycles/instruction x 1 second/3x109 cycles
   	 0.225 seconds = nx0.75/3 
   	 n = 0.225/0.75 x 3
   	   = .9
   	 It would require 9x108 instructions.
   
   
        (c)  MIPS
                 Assume the conditions of the original problem.
   
   	  k x 109 instructions / s seconds x 1 million instructions/ 106 instructions
   	    = k/s x 103 MIPS
   
   	         = 5/1.125 x 103 MIPS
   		 = 4.4x103 MIPS
   
   	  P1:  5/1.125 x 103 MIPS = 4.4x 103 MIPS
             P2: 1.0/.25 x 103 = 4 x 103 MIPS.
   
   	  P2 has a lower mips rating, but is faster.
   
       (d) flops is the SAME computation with only 40% of the instructions.
             P1: 1.76 x 103 Mflops
   	  P2: 1.6 x 103 Mflops.