Danielle Gardner

Fundamentals of Computer Science

Dr Kihlstrom

26 September 2005

 

Chapter 5 Exc. #2, 9, 15, 16

Chapter 6 #4, 6, 11, 15, 18

 

2. a. 1 million bytes=

             20 bits, because 2^2= 1, 048, 576 which is close to 1 million.

b. 10 million bytes=

            24 bits, because 2^24- 16, 777, 216 which is close to 10 million

c. 100 million bytes=

            27  bits, because 2^27= 134, 217, 728 which is close to one hundred million bytes.

d. 1 billion bytes=

            30 bits, because 2^30= 1, 073, 741, 824 which is close to 1 billion.

 

9.

            You would need a 32 by 5 multiplexor circuit in order to select exactly one of those 20 operations.

            Need 20 but 2^4=16, which is not enough  to cover 20 operations.

            2^5= 32 which will cover all 20 operations.

            It will be a 5 input

 

15.  a. For an op code of 6, you would take 2^6= 64 distinct operations.

       b. For  Address-18 you would take 2^18= 262, 144 which equals the maximum memory size in the machine.

       c.  For each operation, you would add 6+18+18= 42 bits.

            Since there are 8 bits in a byte, then you would divide the 42/8= 5.25

Answer: 6 bytes are required for each operation.

 

16.

            a.

                        0          LOAD                         202

                        1          SUBTRACT               203

                        2          ADD                           204

                        3          STORE                       200

           

            b.         0          LOAD                         201

                        1          ADD                           202

                        2          STORE                       201

                        3          LOAD                         203

                        4          ADD                           204

                        5          STORE                       203

                        6          LOAD                         201

                        7          SUBTRACT               203

                        8          STORE                       200

 

c.                 0              COMPARE                200, 202

                    1              JUMPLT                     4

                    2              LOAD                         203

                    3              STORE                       202

                    4              JUMP                         7

                    5              LOAD                         204

                    6              STORE                       202

                    7                                                              Next instruction begin new

 

d.                 0              COMPARE                203, 204

                    1              JUMPGT                    9

                    2              LOAD                         203

                    3              ADD                           201

                    4              ADD                           204

                    5              STORE                       203

                    6              LOAD                         204

                    7              ADD                           200

                    8              STORE                       204

                    9                                                              Next instruction begin new

 

Chapter 6 #4, 6, 11, 15, 18

                   

4. R:13

    60: 472

     61: -1

A. R:472         60:472             61:-1

B. R:472          60:472             61:-1

C. R:944          60:472             61: -1

D. R:944         60:472             61:-1

E. R:944          60:472             61: 50

F. R:944          60:472             61:50

 

 

 

6. .

DATA16387= 14 Bits

Since the 16, 387 is 3 over 2^14 if we execute the .data16387 it would spill over from the 12 bit address into 4 bits of op code which would lead to machine language instruction like

0010 000000000011 which would remind the machine that it should clear the decimal number 3, which is represented in the .data16387.

 

11. The Statement CLEAR SUM on line 2 is necessary because if you do not clear the sum, you might have a sum in there already from a previous operation and that number could get mixed up with the new operation. The Statement LOAD ZERO on line 4 is also necessary because Register R could have already something in it which would affect the operation.

 

15. Counting out the order of operations:

AGAIN- 3

ANS- 8

X-9

ONE-10

 

19. Some of the drawbacks in using password to limit access to a computer system would be just general forgetfulness of the user, death or a mental incapability or severe mental limitation. The alternative safeguards might be using a thumb print access guarantee so that the user is guaranteed access unless their fingers have been decapitated. In addition, you could have a password generated from the userŐs blood type, something that will never change and thus register the password. These alternative safeguards would be appropriate in applications where lives are at stake or if a threat to humanity is posted and can be stopped if someone could access the password.