2006-03-10 Exam Solutions

Problem 1.
 0 00 00 | --- | --- |   0
 0 00 01 |  0  | 1/4 |  1/4
 0 00 10 |  0  | 2/4 |  2/4
 0 00 11 |  0  | 3/4 |  3/4
 0 01 00 |  0  | 4/4 |  4/4
 0 01 01 |  0  | 5/4 |  5/4
 0 01 10 |  0  | 6/4 |  6/4
 0 01 11 |  0  | 7/4 |  7/4
 0 10 00 |  1  | 4/4 |  8/4
 0 10 01 |  1  | 5/4 | 10/4
 0 10 10 |  1  | 6/4 | 12/4
 0 10 11 |  1  | 7/4 | 14/4
 0 11 00 | --- | --- | +infinity

Problem 2.
 number        decimal      binary
 Zero       |    0       |  00 0000           
 n/a        |   -1       |  11 1111
 n/a        |    5       |  00 0101
 n/a        |  -10       |  11 0110
 n/a        |   26       |  01 1010
 n/a        |  -26       |  10 0110
 TMax       |   31       |  01 1111
 TMin       |  -32       |  10 0000
 TMax+TMax  |   -2       |  11 1110
 TMin+TMin  |    0       |  00 0000
 TMin+1     |  -31       |  10 0001
 TMin-1     |   31(Tmax) |  01 1111
 TMax+1     |  -32(Tmin) |  10 0000
 -TMax      |  -31       |  10 0001
 -TMin      |  -32(Tmin) |  10 0000

Problem 3.
M=9   N=15

Problem 4.
int foo(int x, int y)
{
  int i, result=0;
  for (i = x; i > y; i--) {
    result += i;
  }
  return result;
}

Problem 5.
 buf[0] = 0x64636261
 buf[1] = 0x68676665
 buf[2] = 0x08040069     
 ebp    = 0x68676665
 eip    = 0x08040069

Problem 6.
Version Measured CPE   Theoretical CPE
 A1      4.00           4/1 = 4.00
 A2      2.67           8/3 = 2.67
 A3      1.67           4/3 = 1.33
 A4      1.67           4/3 = 1.33
 A5      2.67           8/3 = 2.67

Problem 7.
cache	m	   C	 B	  E	  S	 t	s	b
1.	32 	1024	 4	  4 	 64	24	6	2
2.	32 	1024 	 4	256 	  1	30	0	2
3.	32 	1024 	 8 	  1 	128	22	7	3
4.	32 	1024 	 8 	128 	  1	29	0	3
5.	32 	1024 	32 	  1 	 32	22	5	5
6.	32 	1024 	32 	  4 	  8 	24	3	5