CS 282 Winter 2003

Solutions for Assignment 4

Do addition on the following unsigned numbers. Identify which (if any) of the operations result in overflow. Show a check of your work by converting to decimal and redoing our calculations.

(a) 100110 (38)

+ 001000 (8)

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101110 ( 46)

No overflow !

(b) 1000101 (69)

+ 0111100 (60)

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10000001 ( 1, if you consider only 7 lower bits ).

Hence, there is an overflow!

Do addition on the following two’s complement numbers. Identify which (if any) of the operations result in overflow for 8-bit words. Show a check of your work by converting to decimal and redoing our calculations.

(a) 01110110 ( 118 )

+ 01100011 ( 99 )

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11011001 (217) Hence, there is no overflow !!

(b) 11000101 ( 1 + 4 + 64 – 128 = -59 )

+ 00111100 ( 60 )

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100000001 ( 1, if you consider only 8 lower bits )

As we have studied before, we observe that there is no overflow, when two numbers of opposite signs are added. We do have a carry, but that can be ignored and the lower 8 bits still give the right answer.

Do the following two’s complement subtraction. Identify which (if any) of the operations result in overflow for 8-bit words. Show a check of your work by converting to decimal and redoing our calculations.

(a) 10110101

- 00110011

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We can convert this subtraction into an addition operation as follows:

X – Y = X + (-Y) = X + ( Additive inverse of Y )

In this case, X = 10110101 an Y = 00110011.

Additve inverse of Y = two’s complement of Y in this case = 11001101

Hence the new operation is

10110101 ( 5 + 16 + 32 – 128 = -75)

+ 11001101 ( 5 + 8 + 64 – 128 = -51 )

----------------

110000010 (-126, if you consider only the lower 8 bits.)

There is no overflow. We do have a carry, but that can be ignored and the lower 8 bits still give the right answer.

(b) 10110110

- 01110100

----------------

This is similar to above problem.

In this case, X = 10110110 and Y = 01110100.

Additve inverse of Y = two’s complement of Y in this case = 10001100

Hence the new operation is

10110110 ( 6 + 16 + 32 – 128 = -74)

+ 10001100 ( 12 – 128 = -116)

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101000010 ( +66, if you consider only the lower 8 bits ).

This is clearly an overflow. As per the rules we have studied before, by adding two negative numbers, we got a positive result. And that means there is an overflow.

Do the following multiplications in two’s complement representation. Convert the decimal numbers to two’s complement representation first, and then show the multiplication. Check that your answer is correct.

Treat them as 4 bit numbers and perform sign extension before multiplying.

(a) 7 x 5

This is same as (0111 x 0101 ) in two’s complement representation.

Since both are positive numbers, sign extension is not even needed.

0111

* 0101

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0111

0000

0111

------------------

0100011 ( 35 )

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There is no overflow and the result is within 2 * n = 8 bits, where operands are each of n = 4 bits size.!

(b) -6 x 4

This is same as (1010 x 0100) in two’s complement representation.

After sign extension, we have, ( 11111010 x 00000100 )

11111010

* 00000100

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00000000

0000000

111010

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xxxxxxxx11101000 ( 8 + 32 + 64 – 128 = -24 )

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There is no overflow and the result is within 2 * n = 8 lower bits, where operands are each of n = 4 bits size.! The higher 8 bits can be ignored.

5. Consider the 8-bit quantity: 10110010

(a) Perform a logical left shift operation on the above number.

(b) Perform an arithmetic right shift on the above number.

(a) 01100100

(b) 11011001

6. Consider the 8-bit quantity: 11000110

(a) Perform a left rotate operation on the above number.

(b) Perform a right rotate operation on the above number.

(a) 10001101

(b) 01100011