Solutions for Homework #1

 

  1. (8 pts. Total)
    1. 135910
    2. 11111001112
    3. 11010101011010112   5463510
    4. 11058   24516
  2. (6 pts. Total)
    1. 100001102       (13410)
    2. 112       (310)
    3. 1100001102     (39010)
  3. (4 pts. Total)
    1. 1011102
    2. 1111012
  4. (10 pts. Total)
    1. 2310
    2. -710
    3. -810
    4. –910
    5. 6580910
  5. (4 pts. Total)
    1. 0 to 264-1
    2. –263 to 263-1
  6. (4 pts. Total)
    1. 000001102
    2. 111110112
  7. (3 pts. Total)

  1. (1 pt. Total)  Prof. Dickey teaches 142 durring 11:30 – 12:30 Mondays
  2. (1 pt. Total)  Mausam’s office hours are 11 – 12 Monday.  Nick’s were not announced yet.
  3. (1 pt. Total)  The author of the textbook is a professor at the University of California Berkely
  4. (1 pt. Total)  I hope true!
  5. (1 pt. Total)  The 2nd message was from Alexander Cho.
  6. (1 pt. Total)  The page was: http://www.cs.washington.edu/education/courses/370/98au/admin/Tools/DesignWorks/Tips.html  Tip #7 states ‘ "Z" : High Impedance, The probe is not connected to an input signal. So maybe there's a gap in your circuit wiring or the ouput of a module isn't connected to any inputs.’
  7. (5 pts. Total)  If we assume that when both switches are off, the light is off, and adopt the encoding 0 = off 1 = on, the following truth table can be derived:

A

B

Light

0

0

0

0

1

1

1

0

1

1

1

0

            This is equivalent to an XOR gate, can can be implemented in gates as

                                               

  1. (6 pts. Total)
    1. (x+y)*(x+y’) = x+(y * y’)         by Distributive Law (8D)

x+(y * y’) = x + 0                     by Complementary Law (5D)

x + 0 = x                                  by Identity Law (1)

ð     (x+y)*(x+y’) = x

    1. x*(x+y) = (x+0)*(x+y) by Identity Law (1)

(x+0)*(x+y) = x+(0 * y)           by Distributive Law (8D)

x+(0 * y) = x + 0                      by Null Law (2D)

x + 0 = x                                  by Identity Law (1)

ð     x*(x+y) = x

  1. (4 pts. Total)
    1. ID = 0100 0000

So, Size = 1

2n > 1 è n = 1

META-SIZE = 10

I-Tag = 1010 1000 00

Need to add 6 0’s to the end to make an even number of bytes

I-Tag = 1010 1000 0000 0000 è A80016

    1. ID = 0000 1111 0100 0010 0100 0000

So, Size = 3

2n > 3 è n = 2

META-SIZE = 110

I-Tag = 1101 1000 0111 1010 0001 0010 0000 0

Need to add 3 0’s to the end to make an even number of bytes

I-Tag = 1101 1000 0111 1010 0001 0010 0000 0000 è D87A120016