CSC 362 Homework Assignment #1
CSC 362 Homework Assignment #1 answer key
1) Convert 10110011 to decimal assuming the number is stored in each of the following representations:
a. Unsigned magnitude: 128 + 32 + 16 + 2 + 1 = 179
b. Signed magnitude: -(32 + 16 + 2 + 1) = -51
c. One’s complement: negative, flip all bits ( 01001100 = 64 + 8 + 4 = 76, so -76
d. Two’s complement: negative, flip all bits and add 1 ( 01001101 = 64 + 8 + 4 + 1 = 77, so -77
2) Convert 1101100110101000 to decimal assuming that the number is stored in
a. Signed magnitude: -(16384 + 4096 + 2048 + 256 + 128 + 32 + 8) = -22952
b. Two’s complement: negative, flip all bits and add 1 ( 0010011001011000 = 8192 + 1024 + 512 + 64 + 16 + 8 = 9816, so -9816
3) Do the following conversions
a. -25 to 8-bit 2’s complement: 25 = 16 + 8 + 1 = 00011001, for -25, flip all bits and add 1, so -25 = 11100111
b. -1678 to 16-bit 1’s complement: 1678 = 1024 + 512 + 128 + 8 + 4 + 2 = 0000011010001110, for -1678, flip all bits, so -1678 = 1111100101110001
c. -17235 to 16-bit signed magnitude: 17235 = 16384 + 512 + 256 + 64 + 16 + 2 + 1 = 0100001101010011, for -17235, add leading sign bit of 1, so -17235 = 1100001101010011
d. 1111000011111101 from 1’s complement to decimal: negative, flip all bits for positive: 0000111100000010 = 2048 + 1024 + 512 + 256 + 2 = 3842, so -3842
e. 1101101100111000 from 2’s complement to decimal: negative, flip all bits and add 1: 0010010011001000 = 8192 + 1024 + 128 + 64 + 8 = 9416, so -9416
4)
a. 01011110000111 to decimal: sign bit = 0 (positive), exponent = 10111, mantissa = 10000111, or +.10000111*210111, the exponent is in excess 16, so it is really 7 which gives us +.10000111*27 = 1000011.1 = 64 + 2 + 1 + .5 = 67.5.
b. -3.15625 to binary: sign bit = 1, convert 3.15625 to binary = 2 + 1 + 1/8 + 1/32 = 11.00101. Normalize by shifting the decimal point before the first 1 gives us .11001010 * 22. Converting the exponent into excess 16 gives us 10010, so our exponent is 10010 and our mantissa is 11001010. The number is then 11001011001010, or 1 10010 11001010 for readability.
c. 10111011011000 to decimal: sign bit = 1 (negative), exponent = 011011, mantissa = .11011000 = -(.11011 * 201110). The exponent, 01110 is in excess 16, which is really -2. So our value is –(.11011*2-2) = -.0011011 = -(1/8 + 1/16 + 1/64 + 1/128) = -.2109375.
d. 9.8 to binary = 8 + 1 + 1/2 + 1/4 + 1/32 + 1/64 + 1/256 + 1/2048 + … Unfortunately, we will have run out of precision as we only get 8 bits for the mantissa. So 9.8 = 8 + 1 + 1/2 + 1/4 = 1001.1100. We cannot store the rest of the fractional portion. 1001.1100 = .10011100 * 24 = .10011100 * 210100, so our floating point value = 01010010011100 or 0 10100 10011100 for readability. We have lost .05 in precision!
5)
a. 11010 * 10111 (unsigned)
|C |A |Q |M = 10111 |
|0 |00000 |11010 |Q0 = 0, right shift |
|0 |00000 |01101 |Q0 = 1, C||A = A + M |
|0 |10111 |01101 |Right shift |
|0 |01011 |10110 |Q0 = 0, right shift |
|0 |00101 |11011 |Q0 = 1, C||A = A + M |
|0 |11101 |11011 |Right shift |
|0 |01110 |01101 |Q0 = 1, C||A = A + M |
|1 |00101 |01101 |Right shift |
|0 |10010 |10110 |Done |
11010 * 10111 = 1001010110 (26 * 23 = 598)
b. 10110 * 00101 c. 10101 * 10011
|A |Q |Q-1 |M=00101, -M=11011 |
|00000 |10110 |0 |Q0Q-1=00, arithmetic right shift |
|00000 |01011 |0 |Q0Q-1=10, A = A – M |
|11011 |01011 |0 |Arithmetic right shift |
|11101 |10101 |1 |Q0Q-1=11, arithmetic right shift |
|11110 |11010 |1 |Q0Q-1=01, A = A + M |
|00011 |11010 |1 |Arithmetic right shift |
|00001 |11101 |0 |Q0Q-1=10, A = A – M |
|11100 |11101 |0 |Arithmetic right shift |
|11110 |01110 |1 |Done |
|A |Q |Q-1 |M=10011, -M=01101 |
|00000 |10101 |0 |Q0Q-1=10, A = A – M |
|01101 |10101 |0 |Arithmetic right shift |
|00110 |11010 |1 |Q0Q-1=01, A = A + M |
|11001 |11010 |1 |Arithmetic right shift |
|11100 |11101 |0 |Q0Q-1=10, A = A - M |
|01001 |11101 |0 |Arithmetic right shift |
|00100 |11110 |1 |Q0Q-1=01, A = A + M |
|10111 |11110 |1 |Arithmetic right shift |
|11011 |11111 |0 |Q0Q-1=10, A = A – M |
|01000 |11111 |0 |Arithmetic right shift |
|00100 |01111 |1 |Done |
10110 * 00101 = 1111001110 (-10 * 5 = -50)
10101 * 10011 = 0010001111 (-11 * -13 = 143)
d. 110011 / 000101 (unsigned)
|A |Q |M = 000101 |
|000000 |110011 |Left shift |
|000001 |10011_ |A < M, Q0 = 0 |
|000001 |100110 |Left shift |
|000011 |00110_ |A < M, Q0 = 0 |
|000011 |001100 |Left shift |
|000110 |01100_ |A >= M, Q0 = 1, A = A – M |
|000001 |011001 |Left shift |
|000010 |11001_ |A < M, Q0 = 0 |
|000010 |110010 |Left shift |
|000101 |10010_ |A >= M, Q0 = 1, A = A – M |
|000000 |100101 |Left shift |
|000001 |00101_ |A < M, Q0 = 0 |
|000001 |001010 |Done |
110011 / 000101 = 001010 with a remainder of 000001 (51 / 5 = 10 with a remainder of 1)
e. 111101 / 000011 (unsigned)
|A |Q |M = 000011 |
|000000 |111101 |Left shift |
|000001 |11101_ |A < M, Q0 = 0 |
|000001 |111010 |Left shift |
|000011 |11010_ |A >= M, Q0 = 1, A = A – M |
|000000 |110101 |Left shift |
|000001 |10101_ |A < M, Q0 = 0 |
|000001 |101010 |Left shift |
|000011 |01010_ |A >= M, Q0 = 1, A = A – M |
|000000 |010101 |Left shift |
|000000 |10101_ |A < M, Q0 = 0 |
|000000 |101010 |Left shift |
|000001 |01010_ |A < M, Q0 = 0 |
|000001 |010100 |Done |
111101 / 000011 = 010100 with a remainder of 000001 (61 / 3 = 20 with a remainder of 1)
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