Solve the following exercises in the textbook (68000 Family Assembly Language):

1. Chapter 1:

     Exercise:  16

2. Chapter 5:

     Exercises:  12, 16

3. Chapter 6:

     Exercises:  2, 5, 8

4. Chapter 7:

     Exercises:  1, 7

5. Chapter 8:

     Exercises:  6, 7

6. Chapter 9:

     Exercise:  5

7. Chapter 10:

     Exercises:  5, 6

8. The following 68000 program, presented in class on 12-1-99, reverses the contents of a string that ends with 0 and stored in memory starting at address label String:

   
           LEA      String,A1   Point to the beginning of the string
   Loop1   TST.B    (A1)+       Move to the end of the string
           BNE      Loop1       Until the EOS is encountered
           SUBA     #1,A1       Back up to the EOS
           LEA      String,A0   Make A0 point to the beginning
   Loop2   MOVE.B   -(A1),D0    Save the bottom byte
           CMPA.L   A1,A0       If A0 has reached or passed A1
           BHS      Exit        Then the string is reversed
           MOVE.B   (A0),(A1)   Move the top to the bottom byte
           MOVE.B   D0,(A0)+    Move the previously saved bottom byte to the top byte
           BRA      Loop2       Loop back for another byte
   Exit    STOP     #$2700
   
   

   Assuming a string length of 50000 bytes, calculate the average number of cycles per instruction (CPI) for the above code, and calculate the program execution time in seconds when run on a 68000 CPU running at a clock rate of 16 MHz.