Solutions to some of the midterm problems:

1. a)

   int b = 0;
   int a[4000];
   for(int i = 0; i!= 4000; i = i + 8) {
       b = b + a[i];
   }

1. c)

   li   r20, 0
   li   r19, 16000
   li   r15, 0
   nop
   nop 
start:
   ld   r1, 0(r20)
   nop
   nop
   nop
   add  r15, r15, r1
   addi r20, r20, 32
   nop
   nop
   nop
   sub  r3, r19, r20
   nop
   nop
   nop
   bnez r3, start

1. d)
   there are many ways to do it.. but most likely you will need to have
something like this:

   ld r1,  0(r20)
   ld r10, 32(r20)
   ld r11, 64(r20)
   ld r12, 92(r20)
   addi r20, r20, 128

2.

a) CPI  = 0.7 * 1 + 0.1 * 2 + 0.2 * 3 = 1.5
b) CPI2 = 0.1 * 1 + 0.1 * 2 + 0.2 * 0.8 * 1 + 0.2 * 0.2 * 3 = 1.18
c) time old / time new = ( 1.5  / (2 * 10^9) ) / ( 1.18 / (1.5 * 10^9)) = 0.9533
---> the original processor is faster by about 4%.