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Homework #1 solutions, due Wednesday, November 18 1. Complete the following table: binary 011100 111100 110011.10011 101010101010.1 1100100 1101111 100010.11 11011.0011 1010101111001101 0001000100010001.0001 100010001000.1 111010111010.11 2. Complete the following table: decimal 8 bit 2’s complement (hex) 7B 40 81 D5 FF AB 20 F1 overflow overflow overflow 00 8 bit 1’s complement (hex) 7B 40 80 D4 FD AA 19 F0 overflow overflow overflow FF 16 bit 2’s complement (hex) 007B 0040 FF81 FFD5 FFFF FFAB 0020 FFF1 ABCD 8765 ABCE 0000 16 bit 1’s complement (hex) 007B 0040 FF80 FFD4 FFFD FFAA 0019 FFF0 ABCC 8764 ABCD FFFF Decimal 28 60 51.59375 2730.5 100 111 34.75 27.1875 43981 4369.0625 2184.5 3770.75 hexadecimal 1C 3C 33.98 AAA.8 64 6F 22.C 1D.3 ABCD 1111.1 888.8 EBA.C

123 64 -127 -43 -1 -84 25 -15 -21555 -30874 -21554 -0

3. Find the following differences using 8-bit 2’s complement arithmetic, show your steps !!! a. 123 – 127 = 01111011 - 01111111 = 01111011 + 10000001 = 11111100 = -4 b. -123 – 127 = 10000101 - 01111111 = 10000101 + 10000001 = 00000110 (overflow) c. 64 – 43 = 01000000 - 00101011 = 01000000 + 11010101 = 00010101 = 21 d. 43 – 43 = 00101011 - 00101011 = 00101011 + 11010101 = 00000000 = 0 e. 127 – 123 = 01111111 - 01111011 = 01111111 + 10000101 = 00000100 = 4

4. Complete the following table: IEEE single-precision format (hex) 41100000H 3E200000H BE200000H 40C40000H 42E48000H 3F880000H 00800000H C7F00000H Decimal 9 5/32 -5/32 6.125 114.25 1.0625 0.0 -122880.0

5. Perform the following operations using IEEE single precision floating point arithmetic and express the normalized result in hexadecimal. Describe each step of the calculation procedure clearly. a. 125.25 - 75.5 125.25 = (1.11110101) × 26 to represent it in IEEE 754 format sign bit = 0 exponent code 10000101 mantissa code 1111010 10000000 00000000 75.5 = (1.0010111)× 26 to represent it in IEEE 754 format sign bit=0 exponent code 10000101 mantissa code 0010111 00000000 00000000 their exponents...