CS273

Homework 2 Solution

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For all the problems that involve conversions to and from decimal, use the multiplication method to convert to decimal, and the division method to convert from decimal. Also, when going from two's complement binary form to decimal, be sure to use `+' and `-' with decimal number.

From the Book

from page 45, 46, and 48

• 1-1 (20 points)
  (a) (+23) + (-120)

  Begin: +23

  Old	Radix	New	Digit
  23	2	11	1
  11	2	5	1
  5	2	2	1
  2	2	1	0
  1	2	0	1

  Second: Convert to two's complement form: 00010111_2C

  Next: -120

  Old	Radix	New	Digit
  120	2	60	0
  60	2	30	0
  30	2	15	0
  15	2	7	1
  7	2	3	1
  3	2	1	1
  1	2	0	1

  Fourth: Convert to two's complement form (use algorithm because started negative):
  01111000 -> 10000111 + 00000001 -> 10001000_2C

  Last: Add the two together:

            00010111
           +10001000
           ---------
            10011111 

  answer: 10011111_2C with no overflow; with negative bit set
  

  (b) (-20) + (-32)

  Begin: -20

  Old	Radix	New	Digit
  20	2	10	0
  10	2	5	0
  5	2	2	1
  2	2	1	0
  1	2	0	1

  Second: Convert to two's complement form (use algorithm because started negative):
  00010100 -> 11101011 + 00000001 -> 11101100_2C

  Next: -32

  Old	Radix	New	Digit
  32	2	16	0
  16	2	8	0
  8	2	4	0
  4	2	2	0
  2	2	1	0
  1	2	0	1

  Fourth: Convert to two's complement form (use algorithm because started negative):
  00100000 -> 11011111 + 00000001 -> 11100000_2C

  Last: Add the two together:

            11101100
           +11100000
           ---------
            11001100 

  answer: 11001100_2C with no overflow; with carry-out; with negative bit set
  

  (c) (-44) + (+44)

  Begin: -44

  Old	Radix	New	Digit
  44	2	22	0
  22	2	11	0
  11	2	5	1
  5	2	2	1
  2	2	1	0
  1	2	0	1

  Second: Convert to two's complement form (use algorithm because started negative):
  00101100 -> 11010011 + 00000001 -> 11010100_2C

  Next: +44

  Old	Radix	New	Digit
  44	2	22	0
  22	2	11	0
  11	2	5	1
  5	2	2	1
  2	2	1	0
  1	2	0	1

  Fourth: Convert to two's complement form: 00101100_2C

  Last: Add the two together:

            11010100
           +00101100
           ---------
            00000000 

  answer: 00000000_2C with no overflow; with carry-out; with zero bit set
  

  (e) (+47) - (+92)

  Begin: +47

  Old	Radix	New	Digit
  47	2	23	1
  23	2	11	1
  11	2	5	1
  5	2	2	1
  2	2	1	0
  1	2	0	1

  Second: Convert to two's complement form: 00101111_2C

  Next: -92

  Old	Radix	New	Digit
  92	2	46	0
  46	2	23	0
  23	2	11	1
  11	2	5	1
  5	2	2	1
  2	2	1	0
  1	2	0	1

  Fourth: Convert to two's complement form (use algorithm because started negative):
  01011100 -> 10100011 + 00000001 -> 10100100_2C

  Last: Add the two together:

            00101111
           +10100100
           ---------
            11010011 

  answer: 11010011_2C with no overflow; with negative bit set
  

  (i) (-56) + (-72)

  Begin: -56

  Old	Radix	New	Digit
  56	2	28	0
  28	2	14	0
  14	2	7	0
  7	2	3	1
  3	2	1	1
  1	2	0	1

  Second: Convert to two's complement form (use algorithm because started negative):
  00111000 -> 11000111 + 00000001 -> 11001000_2C

  Next: -72

  Old	Radix	New	Digit
  72	2	36	0
  36	2	18	0
  18	2	9	0
  9	2	4	1
  4	2	2	0
  2	2	1	0
  1	2	0	1

  Fourth: Convert to two's complement form (use algorithm because started negative):
  01001000-> 10110111 + 00000001 -> 10111000_2C

  Last: Add the two together:

            11001000
           +10111000
           ---------
            10000000 

  answer: 10000000_2C with no overflow; with a carry-out; with negative bit set
  

• 1-2 (4 points) (a) NOTE: There is two parts to this question, take the two's complement binary number 1111101_2C

1) to decimal:

First: convert from 2C form: 00000010 + 00000001 -> 00000011₂

Second: use mult. method table:


Old	Digit	New	Radix
0	0	0	2
0	0	0	2
0	0	0	2
0	0	0	2
0	0	0	2
0	0	0	2
0	0	0	2
0	1	1	2
2	1	3

Third: remember that the 2C form has a 1 in the MSB, final answer: -3₁₀

2) go from decimal to hex:

First: use division table to get digits:

Old	Radix	New	Digit
3	16	0	3

Second: Convert to two's complement form because starting with negative number:

	    0     3   
	   f-0   f-3
	   ---   ---
	    f     c
	  + 0     1
	   ---   ---
	    f     d 

answer: fd₁₆

• 1-11 (2 points)

Begin: Largest 16 bit Positive number in 2C: 0111111111111111

Second: convert to hexadecimal:

	   0111  1111  1111  1111
	     7     f     f     f 

answer: 7fff₁₆

Next: Convert from hexadecimal to decimal:

Old	Digit	New	Radix
0	7	7	16
112	f	127	16
2032	f	2047	16
32752	f	32767

answer: Began with positive 2C binary number so: +32767₁₀

• 1-12 (2 points) NOTE: Give the smallest two's complement number that is affected by the change in sign rule. For 8 bits, we defined this to be -127₁₀ in class.

Begin: Largest 16 bit Negative number in 2C: 1000000000000001

Second: convert to hexadecimal:

	   1000  0000  0000  0001
	     8     0     0     1 

answer: 8001₁₆

Next: Convert from hexadecimal to decimal:

First: convert from 2C form: 8001

	    8     0     0     1
	   f-8   f-0   f-0   f-1
	   ---   ---   ---   ---
	    7     f     f     e
	  + 0     0     0     1
	   ---   ---   ---   ---
	    7     f     f     f 

Old	Digit	New	Radix
0	7	7	16
112	f	127	16
2032	f	2047	16
32752	f	32767

answer: Began with negative 2C binary number so: -32767₁₀

• 1-46 (2 points)
  As learned in class, the "fetch" refers to the first part of the instruction cycle. That is to do an instruction fetch from memory. To get data back from memory, we do a read.

Not From the Book

• (10 points) Using your Lab1 program, MODIFY the program to subtract +10₁₀ from the B accumulator just before the ``bra'' instruction. Now answer the following questions using either simulator:
  
  • What opcode is shown for the subtraction instruction?
    answer: c0
  • What is in the A and B accumulators on cycle 0?
    answer: A - 0 B - 0
  • What is in the A and B accumulators on cycle 2?
    answer: A - 0 B - 24
  • What cycle are we on just after the execution of the subtraction instruction?
    answer: cycle 6
  • What is in the A and B accumulators on cycle 9?
    answer: A - 0 B - 2C
  • What is in the PC on cycle 9?
    answer: $f806

Last modified: Sun Sep 23 20:41:32 MDT 2001