Computer Organization and Design Chapter 3 Solution

3 Solutions Solution 3.1 3.1.1 a. 3716 b. 6041 3.1.2 a. 3716 b. 1467 3.1.3 a. 1660 1660 b. 2165 –117 3.1.4 a. 6374 b. 753 3.1.5 a. 7504 (–3504) b. 7777 (–3777) 3.1.6 a. 111000100000 b. 100011110101 The attraction is that each octal digit contains one of 8 different characters (0–7). Since with 3 binary bits you can represent 8 different patterns, in octal each digit requires exactly 3 binary bits. You can write down the conversion directly. S2 Chapter 3 Solutions Solution 3.2 3.2.1 a. 7B75 b. 6D95 3.2.2 a. 7B75 b. 6D95 3.2.3 a. 5190 5190 b. 9312 9312 3.2.4 a. 8CA4 b. 5730 3.2.5 a. FA00 b. 5730 3.2.6 a. 1100001101010010 b. 0101111011010100 The attraction is that each hex digit contains one of 16 different characters (0–9, A–E). Since with 4 binary bits you can represent 16 different patterns, in hex each digit requires exactly 4 binary bits. And bytes are by definition 8 bits long, so two hex digits are all that are required to represent the contents of 1 byte. Chapter 3 Solutions Solution 3.3 3.3.1 a. Underflow (–39) b. Neither (63) 3.3.2 a. Overflow (result = –215, which does not fit into an SM 8-bit format) b. Neither (65) 3.3.3 a. Neither (39) b. Overflow (result = –179, which does not fit into an SM 8-bit format) 3.3.4 a. 15 – 117 = –102 b. –105 – 42 = –128 (–147) 3.3.5 a. 15 + 117 = 127 (132) b. –105 + 42 = –63 3.3.6 a. 15 + 139 = 154 b. 151 + 214 = 255 (365) S3 S4 Chapter 3 Solutions Solution 3.4 3.4.1 a. 62 × 12 Step Action 0 Initial Vals 1 lsb = 0, no op 2 3 4 5 6 Multiplier 001 010 001 010 Multiplicand 000 000 110 010 000 000 110 010 Lshift Mcand 001 010 000 001 100 100 Rshift Mplier 000 101 000 001 100 100 Prod = Prod + Mcand 000 101 000 001 100 100 Lshift Mcand 000 101 000 011 001 000 Rshift Mplier 000 010 000 011 001 000 lsb = 0, no op 000 010 000 011 001 000 Lshift Mcand 000 010 000 110 010 000 Rshift Mplier 000 001 000 110 010 000 Prod = Prod + Mcand 000 001 000 110 010 000 Lshift Mcand 000 001 001 100 100 000 Rshift Mplier 000 000 001 100 100 000 lsb = 0, no op 000 000 001 100 100 000 Lshift Mcand 000 000 011 001 000 000 Rshift Mplier 000 000 011 001 000 000 lsb = 0, no op 000 000 110 010 000 000 Product 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 001 100 100 000 001 100 100 000 001 100 100 000 001 100 100 000 001 100 100 000 001 100 100 000 111 110 100 000 111 110 100 000 111 110 100 000 111 110 100 000 111 110 100 000 111 110 100 000 111 110 100 Lshift Mcand 000 000 110 010 000 000 000 111 110 100 Rshift Mplier 000 000 110 010 000 000 000 111 110 100 b. 35 × 26 Step 0 Action Initial Vals lsb = 0, no op 1 2 Multiplier 010 110 010 110 Multiplicand 000 000 011 101 000 000 011 101 Lshift Mcand 010 110 000 000 111 010 Rshift Mplier 001 011 000 000 111 010 Prod = Prod + Mcand 001 011 000 000 111 010 Product 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 111 010 Lshift Mcand 001 011 000 001 110 100 000 000 111 010 Rshift Mplier 000 101 000 001 110 100 000 000 111 010 Chapter 3 Solutions Step 3 4 5 6 Action Prod = Prod + Mcand Multiplier 000 101 Multiplicand 000 001 110 100 Lshift Mcand 000 101 000 011 101 000 Rshift Mplier 000 010 000 011 101 000 lsb = 0, no op 000 010 000 011 101 000 Lshift Mcand 000 010 000 111 010 000 Rshift Mplier 000 001 000 111 010 000 Prod = Prod + Mcand 000 001 000 111 010 000 Lshift Mcand 000 001 001 110 100 000 Rshift Mplier 000 000 001 110 100 000 lsb = 0, no op 000 000 001 110 100 000 Product 000 010 101 110 000 010 101 110 000 010 101 110 000 010 101 110 000 010 101 110 000 010 101 110 001 001 111 110 001 001 111 110 001 001 111 110 001 001 111 110 Lshift Mcand 000 000 011 101 000 000 001 001 111 110 Rshift Mplier 000 000 011 101 000 000 001 001 111 110 3.4.2 a. 62 × 12 Step Action 0 Initial Vals 1 lsb = 0, no op 2 Prod = Prod + Mcand 3 lsb = 0, no op 4 Prod = Prod + Mcand 5 lsb = 0, no op 6 lsb = 0, no op Rshift Product Rshift Mplier Rshift Mplier Rshift Mplier Rshift Mplier Rshift Mplier Multiplicand 110 010 110 010 110 010 110 010 110 010 110 010 110 010 110 010 110 010 110 010 110 010 110 010 110 010 Product/Multiplier 000 000 001 010 000 000 001 010 000 000 000 101 110 010 000 101 011 001 000 010 011 001 000 010 001 100 100 001 111 110 100 001 011 111 010 000 011 111 010 000 001 111 101 000 001 111 101 000 000 111 110 100 S5 S6 Chapter 3 Solutions b. 35 × 26 Step Action 0 Initial Vals 1 lsb = 0, no op 2 Prod = Prod + Mcand 3 Prod = Prod + Mcand 4 lsb = 0, no op 5 Prod = Prod + Mcand 6 lsb = 0, no op Multiplicand 011 101 011 101 Rshift Mplier 011 101 011 101 Rshift Product 011 101 011 101 Rshift Mplier 011 101 011 101 Rshift Mplier 011 101 011 101 Rshift Mplier 011 101 011 101 Rshift Mplier 011 101 Product/Multiplier 000 000 010 110 000 000 010 110 000 000 001 011 011 101 001 011 001 110 100 101 101 011 100 101 010 101 110 010 010 101 110 010 001 010 111 001 100 111 111 001 010 011 111 100 010 011 111 100 001 001 111 110 3.4.3 No solution provided 3.4.4 a. 41 × 33 = 4033 Step 0 Action Initial Values Mplier 011 011 Multiplicand 000 000 100 001 Product 000 000 000 000 Sign 0 Multiplier.sign XOR Multiplicand.sign (0 XOR 1) Make positive 1 2 3 Prod = Prod + Mcand 1 011 011 011 011 000 000 000 001 000 000 000 001 000 000 000 000 000 000 000 001 Lshift Mcand 011 011 000 000 000 010 000 000 000 001 Rshift Mplier 001 101 000 000 000 010 000 000 000 001 Prod = Prod + Mcand 001 101 000 000 000 010 000 000 000 011 Lshift Mcand 001 101 000 000 000 100 000 000 000 011 Rshift Mplier 000 110 000 000 000 100 000 000 000 011 lsb = 0, no op 000 110 000 000 000 100 000 000 000 011 1 1 1 1 1 1 1 1 Lshift Mcand 000 110 000 000 001 000 000 000 000 011 1 Rshift Mplier 000 011 000 000 001 000 000 000 000 011 1 S7 Chapter 3 Solutions Step 4 5 6 7 Action Prod = Prod + Mcand Mplier Multiplicand 000 011 000 000 001 000 Product 000 000 001 011 Lshift Mcand 000 011 000 000 010 000 000 000 001 011 Rshift Mplier 000 001 000 000 010 000 000 000 001 011 Prod = Prod + Mcand 000 001 000 000 010 000 000 000 011 011 Lshift Mcand 000 001 000 000 100 000 000 000 011 011 Rshift Mplier 000 000 000 000 100 000 000 000 011 011 lsb = 0, no op 000 000 000 000 100 000 000 000 011 011 Lshift Mcand 000 000 000 001 000 000 000 000 011 011 Rshift Mplier 000 000 000 001 000 000 000 000 011 011 Prod msb = sign 000 000 000 001 000 000 100 000 011 011 Sign 1 1 1 1 1 1 1 1 1 1 b. 60 × 26 = 4540 Step 0 Action Initial Values Mplier Multiplicand 010 110 000 000 110 000 Product 000 000 000 000 Sign 0 Multiplier.sign XOR Multiplicand.sign (0 XOR 1) Make positive 1 2 3 4 5 lsb = 0, no op 1 010 110 010 110 000 000 010 000 000 000 010 000 000 000 000 000 000 000 000 000 Lshift Mcand 010 110 000 000 100 000 000 000 000 000 Rshift Mplier 001 011 000 000 100 000 000 000 000 000 Prod = Prod + Mcand 001 011 000 000 100 000 000 000 100 000 Lshift Mcand 001 011 000 001 000 000 000 000 100 000 Rshift Mplier 000 101 000 001 000 000 000 000 100 000 Prod = Prod + Mcand 000 101 000 001 000 000 000 001 100 000 Lshift Mcand 000 101 000 010 000 000 000 001 100 000 Rshift Mplier 000 010 000 010 000 000 000 001 100 000 lsb = 0, no op 000 010 000 010 000 000 000 001 100 000 Lshift Mcand 000 010 000 100 000 000 000 001 100 000 Rshift Mplier 000 001 000 100 000 000 000 001 100 000 Prod = Prod + Mcand 000 001 000 100 000 000 000 101 100 000 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Lshift Mcand 000 001 001 000 000 000 000 101 100 000 1 Rshift Mplier 000 000 001 000 000 000 000 101 100 000 1 S8 Chapter 3 Solutions Step 6 7 Action lsb = 0, no op Mplier Multiplicand 000 000 001 000 000 000 Product 000 101 100 000 Lshift Mcand 000 000 010 000 000 000 000 101 100 000 Rshift Mplier 000 000 010 000 000 000 000 101 100 000 Prod msb = sign 000 000 010 000 000 000 100 101 100 000 Sign 1 1 1 1 3.4.5 a. 41 × 33 = −37 × 33 = −1505 (6273) Step Action 0 Initial Vals 1 Prod = Prod + Mcand 2 Prod = Prod + Mcand 3 lsb = 0, no op 4 Prod = Prod + Mcand 5 Prod = Prod + Mcand 6 lsb = 0, no op Multiplicand 100 001 100 001 Rshift Mplier 100 001 100 001 Rshift Product 100 001 100 001 Rshift Mplier 100 001 100 001 Rshift Mplier 100 001 100 001 Rshift Mplier 100 001 100 001 Rshift Mplier 100 001 Product/Multiplier 0 000 000 011 011 1 100 001 011 011 1 110 000 101 101 1 010 001 101 101 1 101 000 110 110 1 101 000 110 110 1 110 100 011 011 1 010 101 011 011 1 101 010 101 101 1 001 011 101 101 1 100 101 110 110 1 100 101 110 110 1 110 010 111 011 b. 60 × 26 = −20 × 26 = −540 (7240) Step Action Multiplicand Initial Vals 1 lsb = 0, no op 2 Prod = Prod + Mcand Rshift Product 110 000 1 111 000 000 101 3 Prod = Prod + Mcand 110 000 1 101 000 000 101 4 lsb = 0, no op Rshift Mplier Rshift Mplier Rshift Mplier 110 000 Product/Multiplier 0 110 000 110 000 110 000 110 000 110 000 110 000 0 000 000 010 110 0 000 000 010 110 0 000 000 001 011 1 110 000 001 011 1 110 100 000 010 1 110 100 000 010 1 111 010 000 001 Chapter 3 Solutions Step Action 5 Multiplicand Prod = Prod + Mcand Rshift Mplier 6 lsb = 0, no op Rshift Mplier 110 000 110 000 110 000 110 000 Product/Multiplier 1 101 010 000 001 1 110 101 000 000 1 110 101 000 000 1 111 010 100 000 3.4.6 No solution provided Solution 3.5 3.5.1 For hardware, it takes 1 cycle to do the add, 1 cycle to do the shift, and 1 cycle to decide if we are done. So the loop takes (3 × A) cycles, with each cycle being B time units long. For a software implementation, it takes 1 cycle to decide what to add, 1 cycle to do the add, 1 cycle to do each shift, and 1 cycle to decide if we are done. So the loop takes (5 × A) cycles, with each cycle being B time units long. a. (3 × 8) × 4tu = 96 time units for hardware (5 × 8) × 4tu = 160 time units for software b. (3 × 64) × 8tu = 1536 time units for hardware (5 × 64) × 8tu = 2560 time units for software 3.5.2 It takes B time units to get through an adder, and there will be A – 1 adders. a. Word is 8 bits wide, requiring 7 adders. 7 × 4tu = 28 time units. b. Word is 64 bits wide, requiring 63 adders. 63 × 8tu = 504 time units. 3.5.3 It takes B time units to get through an adder, and the adders are arranged in a tree structure. It will require log2(A) levels. a. 8-bit wide word requires 7 adders in 3 levels. 3 × 4tu = 12 time units. b. 64-bit word requires 63 adders in 6 levels. 6 × 8tu = 48 time units. Solution 3.6 3.6.1 a. 0x33 × 0x55 = 0x10EF. 0x33 = 51, and 51 = 32 + 16 + 2 + 1. We can shift 0x55 left 5 places (0xAA0), then add 0x55 shifted left 4 places (0x550), then add 0x55 shifted left once (0xAA), then add 0x55. 0xAA0 + 0x550 + 0xAA + 0x55 = 0x10EF. 3 shifts, 3 adds. (Could also use 0x55, which is 64 + 16 + 4 + 1, and shift 0x33 left 6 times, add to it 0x33 shifted left 4 times, add to that 0x33 shifted left 2 times, and add to that 0x33. Same number of shifts and adds.) S9 S10 Chapter 3 Solutions b. 0x8A × 0xED = 0x7FC2 0x8A = 128 + 8 + 2, 0xED = 128 + 64 + 32 + 8 + 4 + 1. Best way is to shift 0xED left 7 places (0x7680), then add to that 0xED shifted left 3 places (0x768), and then add 0xED shifted left 1 place (0x1DA). 3 shifts, 2 adds. 3.6.2 a. 0x33 × 0x55 = 0x10EF. 0x33 = 51, and 51 = 32 + 16 + 2 + 1. We can shift 0x55 left 5 places (0xAA0), then add 0x55 shifted left 4 places (0x550), then add 0x55 shifted left once (0xAA), then add 0x55. 0xAA0 + 0x550 + 0xAA + 0x55 = 0x10EF. 3 shifts, 3 adds. (Could also use 0x55, which is 64 + 16 + 4 + 1, and shift 0x33 left 6 times, add to it 0x33 shifted left 4 times, add to that 0x33 shifted left 2 times, and add to that 0x33. Same number of shifts and adds.) b. 0x8A × 0xED = –0x0A × –0x6D = 0x442 0x0A = 8 + 2, 0x6D = 64 + 32 + 8 + 4 + 1. Best way is to shift 0x6D left 3 places (0x368), then add to that 0x6D shifted left 1 place (0xDA). 2 shifts, 1 add. 3.6.3 No solution provided 3.6.4 Quoting the Wikipedia entry directly: Booth’s algorithm involves repeatedly adding one of two predetermined values A and S to a product P, then performing a rightward arithmetic shift on P. Let x and y be the multiplicand and multiplier, respectively; and let x and y represent the number of bits in x and y. 1. Determine the values of A and S, and the initial value of P. All of these numbers should have a length equal to (x + y + 1). a. A: Fill the most signifi cant (leftmost) bits with the value of x. Fill the remaining (y + 1) bits with zeros. b. S: Fill the most significant bits with the value of (−x) in two’s complement notation. Fill the remaining (y + 1) bits with zeros. c. P: Fill the most signifi cant x bits with zeros. To the right of this, append the value of y. Fill the least signifi cant (rightmost) bit with a zero. 2. Determine the two least significant (rightmost) bits of P. a. If they are 01, fi nd the value of P + A. Ignore any overflow. b. If they are 10, find the value of P + S. Ignore any overflow. c. If they are 00 or 11, do nothing. Use P directly in the next step. 3. Arithmetically shift the value obtained in the previous step by a single place to the right. Let P now equal this new value. 4. Repeat steps 2 and 3 until they have been done y times. 5. Drop the least signifi cant (rightmost) bit from P. This is the product of x and y. Chapter 3 Solutions 3.6.5 a. 0xF6 × 0x7F = −0xA × 0x7F = −10 × 127 = −1270 = 0xFB0A Action Multiplicand Product/Multiplier Initial Vals 1111 0110 0000 0000 0111 1111 0 10, subtract shift 1111 0110 1111 0110 0000 1010 0111 1111 0 0000 0101 0011 1111 1 11, nop shift 1111 0110 1111 0110 0000 0101 0011 1111 1 0000 0010 1001 1111 1 11, nop 1111 0110 0000 0010 1001 1111 1 shift 1111 0110 0000 0001 0100 1111 1 11, nop shift 1111 0110 1111 0110 0000 0001 0100 1111 1 0000 0000 1010 0111 1 11, nop shift 1111 0110 1111 0110 0000 0000 1010 0111 1 0000 0000 0101 0011 1 11, nop shift 1111 0110 1111 0110 0000 0000 0101 0011 1 0000 0000 0010 1001 1 11, nop shift 1111 0110 1111 0110 0000 0000 0010 1001 1 0000 0000 0001 0100 1 01, add shift 1111 0110 1111 0110 1111 0110 0001 0100 1 1111 1011 0000 1010 0 Multiplicand Product/Multiplier b. 0x08 × 0x55 = 0x2A8 Action Initial Vals 0000 1000 0000 0000 0101 0101 0 10, subtract shift 0000 1000 0000 1000 1111 1000 0101 0101 0 1111 1100 0010 1010 1 01, add shift 0000 1000 0000 1000 0000 0100 0010 1010 1 0000 0010 0001 0101 0 10, subtract shift 0000 1000 0000 1000 1111 1010 0001 0101 0 1111 1101 0000 1010 1 01, add shift 0000 1000 0000 1000 0000 0101 0000 1010 1 0000 0010 1000 0101 0 10, subtract shift 0000 1000 0000 1000 1111 1010 1000 0101 0 1111 1101 0100 0010 1 01, add 0000 1000 0000 0101 0100 0010 1 shift 0000 1000 0000 0010 1010 0001 1 S11 S12 Chapter 3 Solutions Multiplicand Product/Multiplier 10, subtract shift Action 0000 1000 0000 1000 1111 1010 1010 0001 0 1111 1101 0101 0000 1 01, add shift 0000 1000 0000 1000 0000 0101 0101 0000 1 0000 0010 1010 1000 1 3.6.6 No solution provided Solution 3.7 3.7.1 a. 74/21 = 3 remainder 9 Step Action Quotient 1 Rem = Rem – Div 000 000 010 001 000 000 101 111 111 100 Rem < 0, R + D, Q<< 000 000 010 001 000 000 000 000 111 100 Rshift Div 001 000 100 000 111 000 011 100 Rem < 0, R + D, Q<< 000 000 001 000 100 000 000 000 111 100 000 000 111 100 000 100 010 000 111 100 101 100 Rem < 0, R + D, Q<< 000 000 000 100 010 000 000 000 111 100 000 000 000 010 001 000 000 000 111 100 Rem = Rem – Div 000 000 000 010 001 000 111 110 110 100 Rem < 0, R + D, Q<< 000 000 000 010 001 000 000 000 111 100 000 000 000 001 000 100 000 000 111 100 Rem = Rem – Div 000 000 000 001 000 100 111 111 111 000 Rem < 0, R + D, Q<< 000 000 000 001 000 100 000 000 111 100 Rem = Rem – Div Rem > 0, Q << 1 7 000 100 010 000 000 000 Rshift Div 6 000 000 Rem = Rem – Div Rshift Div 5 000 000 111 100 000 000 Rshift Div 4 001 000 100 000 000 000 111 100 Rem = Rem – Div Rshift Div 3 000 000 010 001 000 000 Remainder Initial Vals 2 000 000 Divisor 0 000 000 000 000 000 001 000 000 100 010 000 000 100 010 000 000 100 010 000 000 111 100 000 000 011 010 000 000 011 010 Rshift Div 000 001 000 000 010 001 000 000 011 010 Rem = Rem – Div 000 001 000 000 010 001 000 000 001 001 Rem > 0, Q << 1 Rshift Div 000 011 000 011 000 000 010 001 000 000 001 000 000 000 001 001 000 000 001 001 Chapter 3 Solutions b. 76/52 = 1 remainder 24 Step Action Quotient Remainder Initial Vals 1 Rem = Rem – Div 000 000 101 010 000 000 101 001 000 010 Rem < 0, R + D, Q<< 000 000 101 010 000 000 000 000 111 110 Rshift Div 2 101 011 111 110 Rem < 0, R + D, Q<< 000 000 010 101 000 000 000 000 111 110 001 010 100 000 000 000 111 110 000 000 001 010 100 000 110 110 011 110 Rem < 0, R + D, Q<< 000 000 001 010 100 000 000 000 111 110 000 000 000 101 010 000 000 000 111 110 Rem = Rem – Div 000 000 000 101 010 000 111 011 101 110 Rem < 0, R + D, Q<< 000 000 000 101 010 000 000 000 111 110 000 000 000 010 101 000 000 000 111 110 Rem = Rem – Div 000 000 000 010 101 000 111 110 010 110 Rem < 0, R + D, Q<< 000 000 000 010 101 000 000 000 111 110 000 000 000 001 010 100 000 000 111 110 Rem = Rem – Div 000 000 000 001 010 100 111 111 101 101 Rem < 0, R = D, Q<< 000 000 000 001 010 100 000 000 111 110 Rshift Div 7 000 000 Rem = Rem – Div Rshift Div 6 000 000 111 110 010 101 000 000 Rshift Div 5 010 101 000 000 000 000 111 110 000 000 Rshift Div 4 000 000 101 010 000 000 Rem = Rem – Div Rshift Div 3 000 000 Divisor 0 000 000 000 000 101 010 000 000 111 110 Rem = Rem – Div 000 000 000 000 101 010 000 000 010 100 Rem > 0, Q << 1 000 001 000 000 101 010 000 000 010 100 Rshift Div 000 001 000 000 010 101 000 000 010 100 3.7.2 In these solutions a 1 or a 0 was added to the quotient if the remainder was greater than or equal to 0. However, an equally valid solution is to shift in a 1 or 0, but if you do this you must do a compensating right shift of the remainder (only the remainder, not the entire remainder/quotient combination) after the last step. a. 74/21 = 3 remainder 11 Step Action Divisor Remainder/Quotient 0 Initial Vals 010 001 1 R<< 010 001 000 000 111 100 000 001 111 000 Rem = Rem – Div 010 001 111 000 111 000 Rem < 0, R + D 010 001 000 001 111 000 S13 S14 Chapter 3 Solutions Step 2 3 4 5 Action Remainder/Quotient 010 001 000 011 110 000 Rem = Rem – Div 010 001 110 010 110 000 Rem < 0, R + D 010 001 000 011 110 000 R<< 010 001 000 111 100 000 Rem = Rem – Div 010 001 110 110 110 000 Rem < 0, R + D 010 001 000 111 100 000 R<< 010 001 001 111 000 000 Rem = Rem – Div 010 001 111 110 000 000 Rem < 0, R + D 010 001 001 111 000 000 R<< 010 001 011 110 000 000 Rem = Rem – Div 010 001 111 110 000 000 Rem > 0, R0 = 1 6 Divisor R<< 010 001 001 101 000 001 R<< 010 001 011 010 000 010 Rem = Rem – Div 010 001 001 001 000 010 Rem > 0, R0 = 1 010 001 001 001 000 011 b. 76/52 = 1 remainder 24 Step Action Divisor Remainder/Quotient 0 Initial Vals 101 010 000 000 111 110 1 R<< 101 010 000 001 111 100 2 3 4 Rem = Rem – Div 101 010 101 001 111 100 Rem < 0, R + D 101 010 000 001 111 100 101 010 000 011 111 000 R<< Rem = Rem – Div 101 010 100 111 111 000 Rem < 0, R + D 101 010 000 011 111 000 101 010 000 111 110 000 R<< Rem = Rem – Div 101 010 100 011 110 000 Rem < 0, R + D 101 010 000 111 110 000 101 010 001 111 100 000 R<< Rem = Rem – Div 101 010 100 101 100 000 Rem < 0, R + D 101 010 001 111 100 000 Chapter 3 Solutions Step 5 6 Action Divisor Remainder/Quotient R<< 101 010 011 111 000 000 Rem = Rem – Div 101 010 110 101 000 000 Rem < 0, R + D 101 010 011 111 000 000 R<< 101 010 111 110 000 000 Rem = Rem – Div 101 010 010 100 000 000 Rem > 0, R0 = 1 101 010 010 100 000 001 3.7.3 No solution provided 3.7.4 a. 72/07 = 3 remainder 5: Dividend negative Sign of Quotient = (Sign bit of Divisor) XOR (Sign bit of Dividend) = negative Sign of Remainder = Sign of Dividend = negative Step Action Quotient 1 Rem = Rem – Div 000 000 000 111 000 000 111 001 011 010 Rem < 0, R + D, Q<< 000 000 000 111 000 000 000 000 011 010 Rshift Div 000 011 100 000 111 100 111 010 Rem < 0, R + D, Q<< 000 000 000 011 100 000 000 000 011 010 000 001 110 000 000 000 011 010 000 000 000 001 110 000 111 110 101 010 Rem < 0, R + D, Q<< 000 000 000 001 110 000 000 000 011 010 000 000 000 000 111 000 000 000 011 010 Rem = Rem – Div 000 000 000 000 111 000 111 111 100 010 Rem < 0, R + D, Q<< 000 000 000 000 111 000 000 000 011 010 000 000 000 000 011 100 000 000 011 010 Rem = Rem – Div 000 000 000 000 011 100 111 111 111 110 Rem < 0, R + D, Q<< 000 000 000 000 011 100 000 000 011 010 Rshift Div 6 000 000 Rem = Rem – Div Rshift Div 5 000 000 011 010 000 000 Rshift Div 4 000 011 100 000 000 000 011 010 Rem = Rem – Div Rshift Div 3 000 000 000 111 000 000 Remainder Initial Vals 2 000 000 Divisor 0 000 000 000 000 001 110 000 000 011 010 Rem = Rem – Div 000 000 000 000 001 110 000 000 001 100 Rem > 0, Q << 1 000 001 000 000 001 110 000 000 001 100 Rshift Div 000 001 000 000 000 111 000 000 001 100 S15 S16 Chapter 3 Solutions Step 7 Action Divisor Remainder 000 001 000 000 000 111 000 000 000 101 Rem < 0, Q << 1 000 011 000 000 000 111 000 000 000 101 Rshift Div 8 Quotient Rem = Rem – Div Set sign bits 000 011 100 011 000 000 000 011 000 000 000 011 000 000 000 101 100 000 000 101 b. 75/44 = 7 remainder 1: Dividend negative Sign of Quotient = (Sign bit of Divisor) XOR (Sign bit of Dividend) = positive Sign of Remainder = Sign of Dividend = negative Step Action Quotient Divisor Remainder 0 Initial Vals 000 000 000 100 000 000 000 000 011 101 1 Rem = Rem – Div 000 000 000 100 000 000 111 100 011 101 000 000 000 100 000 000 000 000 011 101 Rem < 0, R + D, Q<< Rshift Div 2 Rem = Rem – Div Rem < 0, R + D, Q<< Rshift Div 3 Rem = Rem – Div Rem < 0, R + D, Q<< Rshift Div 4 Rem = Rem – Div Rem < 0, R + D, Q<< Rshift Div 5 000 010 000 000 111 110 011 101 000 000 000 010 000 000 000 000 011 101 000 000 000 001 000 000 000 000 011 101 000 000 000 001 000 000 111 111 011 101 000 000 000 001 000 000 000 000 011 101 000 000 000 000 100 000 000 000 011 101 000 000 000 000 100 000 111 111 111 101 000 000 000 000 100 000 000 000 011 101 000 000 000 000 010 000 000 000 011 101 000 000 010 000 000 000 001 101 Rem > 0, Q << 1 000 001 000 000 010 000 000 000 001 101 000 001 000 000 001 000 000 000 001 101 Rem = Rem – Div 000 001 000 000 001 000 000 000 000 101 Rem > 0, Q << 1 000 011 000 000 001 000 000 000 000 101 000 011 000 000 000 100 000 000 000 101 Rem = Rem – Div 000 011 000 000 000 100 000 000 000 001 Rem > 0, Q << 1 000 111 000 000 000 100 000 000 000 001 Rshift Div 8 000 000 011 101 000 000 000 000 Rshift Div 7 000 010 000 000 Rem = Rem – Div Rshift Div 6 000 000 Set sign bits 000 111 000 111 000 000 000 010 000 000 000 010 000 000 000 001 100 000 000 001 Chapter 3 Solutions 3.7.5 a. 72/07 = 3 remainder 5: Dividend negative Sign of Quotient = (Sign bit of Divisor) XOR (Sign bit of Dividend) = negative Sign of Remainder = Sign of Dividend = negative Step Action Divisor Remainder/Quotient 0 Initial Vals 000 111 1 R<< 000 111 000 000 110 100 Rem = Rem – Div 000 111 111 001 110 100 Rem < 0, R + D 000 111 000 000 110 100 2 3 4 5 R<< 000 111 000 001 101 000 Rem = Rem – Div 000 111 111 010 101 000 Rem < 0, R + D 000 111 000 001 101 000 R<< 000 111 000 011 010 000 Rem = Rem – Div 000 111 111 100 010 000 Rem < 0, R + D 000 111 000 011 010 000 R<< 000 111 000 110 100 000 Rem = Rem – Div 000 111 111 111 100 000 Rem < 0, R + D 000 111 000 110 100 000 R<< 000 111 001 101 000 000 Rem = Rem – Div 000 111 000 110 000 000 Rem > 0, R0 = 1 6 7 000 000 011 010 000 111 000 110 000 001 R<< 000 111 001 100 000 010 Rem = Rem – Div 000 111 000 101 000 010 Rem > 0, R0 = 1 000 111 000 101 000 011 Adjust signs 000 111 100 101 100 011 (Q = –3, Rem = –5) b. 75/44 = 7 remainder 1: Dividend negative Sign of Quotient = (Sign bit of Divisor) XOR (Sign bit of Dividend) = positive Sign of Remainder = Sign of Dividend = negative Step Action Divisor Remainder/Quotient 0 Initial Vals 000 100 1 R<< 000 100 000 000 011 101 000 000 111 010 Rem = Rem – Div 000 100 111 100 111 010 Rem < 0, R + D 000 100 000 000 111 010 S17 S18 Chapter 3 Solutions Step 2 3 4 Action 000 001 110 100 Rem = Rem – Div 000 100 111 101 110 100 Rem < 0, R + D 000 100 000 001 110 100 R<< 000 100 000 011 101 000 Rem = Rem – Div 000 100 111 111 101 000 Rem < 0, R + D 000 100 000 011 101 000 R<< 000 100 000 111 010 000 Rem = Rem – Div 000 100 000 011 010 000 000 011 010 001 000 100 000 110 100 010 Rem = Rem – Div 000 100 000 010 100 010 000 100 000 010 100 011 R<< 000 100 000 101 000 110 Rem = Rem – Div 000 100 000 001 000 110 Rem > 0, R0 = 1 7 000 100 R<< Rem > 0,R0 = 1 6 Remainder/Quotient 000 100 Rem > 0, R0 = 1 5 Divisor R<< Adjust signs 000 100 000 100 000 001 000 111 100 001 000 111 (Q = 7, Rem = –1) 3.7.6 No solution provided Solution 3.8 3.8.1 In these solutions a 1 will be shifted into the quotient and a compensating right shift of the remainder will be performed. This is the alternate approach mentioned in Solution Solution 3.7.2: In these solutions a 1 or a 0 was added to the quotient if the remainder was greater than or equal to 0.. a. 26/05 = 4 remainder 2 Step 0 1 Action Remainder/Quotient 000 101 R<< 000 101 000 000 101 100 Rem = Rem – Div 000 101 111 011 101 100 Rem < 0, Q << 0, Addnext Rem = Rem + Div 2 Divisor Initial Vals Rem < 0, Q << 0, Addnext Rem = Rem + Div 000 101 000 101 000 101 000 101 000 000 010 110 110 111 011 000 111 100 011 000 111 000 110 000 111 101 110 000 Chapter 3 Solutions Step 3 Action Rem < 0, Q << 0, Addnext Rem = Rem + Div 4 Rem > = 0, Q << 1, Sub 5 Rem < 0, Q << 0, Add 6 Rem < 0, Q << 0, Add 7 Rem < 0, Rem = Rem + Div Rem = Rem – Div Rem = Rem + Div Rem = Rem + Div Shift Rem >> 1 Divisor Remainder/Quotient 000 101 111 011 100 000 000 101 000 101 000 101 000 101 000 101 000 101 000 101 000 101 000 101 000 000 100 000 000 001 000 001 111 100 000 001 111 000 000 010 111 101 000 010 111 010 000 100 111 111 000 100 000 100 000 100 000 010 000 100 (Q = 4, Rem = 2) b. 37/15 = 2 remainder 5 Step 0 1 Action Divisor Remainder/Quotient Initial Vals 001 101 R<< 001 101 000 000 111 110 Rem = Rem – Div 001 101 110 011 111 110 Rem < 0, Q << 0, Addnext Rem = Rem + Div 2 Rem < 0, Q << 0, Addnext 3 Rem < 0, Q << 0, Addnext 4 Rem < 0, Q << 0, Addnext 5 Rem > 0, Q << 1, Subnext 6 Rem < 0, Q << 0, Addnext 7 Rem < 0, Rem = Rem + Div Rem = Rem + Div Rem = Rem + Div Rem = Rem + Div Rem = Rem – Div Rem = Rem + Div Shift Rem >> 1 3.8.2 No solution provided 3.8.3 No solution provided 001 101 001 101 001 101 001 101 001 101 001 101 001 101 001 101 001 101 001 101 001 101 001 101 001 101 001 101 000 000 011 111 100 111 111 100 110 100 111 100 101 001 111 000 110 110 111 000 101 101 110 000 111 010 110 000 110 101 100 000 000 010 100 000 000 101 000 001 111 000 000 001 110 000 000 010 111 101 000 010 001 010 000 010 000 101 000 010 (Q = 2, Rem = 5) S19 S20 Chapter 3 Solutions 3.8.4 a. 27/6 = 3 remainder 5 Step Action 0 Initial Vals 1 Temp = Rem – Div Quotient 000000 000000 Temp < 0, Q << 0 000000 Rshift Div 2 000000 Temp = Rem – Div 000000 Temp < 0, Q << 0 000000 Rshift Div 3 000000 Temp = Rem – Div 000000 Temp < 0, Q << 0 000000 Rshift Div 4 000000 Temp = Rem – Div 000000 Temp < 0, Q << 0 000000 Rshift Div 5 000000 Temp = Rem – Div 000000 Temp < 0, Q << 0 000000 Rshift Div 6 000000 Temp = Rem – Div 000000 T > 0, Q << 1, R = T 000001 Rshift Div 7 000001 Temp = Rem – Div 000001 T > 0, Q << 1, R = T 000011 Rshift Div 000011 Temp 000000 000000 111010 010111 111010 010111 111010 010111 111101 010111 111101 010111 111101 010111 111111 110111 111111 110111 Divisor 000110 000000 000110 000000 000110 000000 000011 000000 000011 000000 000011 000000 000001 100000 000001 100000 000001 100000 Remainder 000000 010111 000000 010111 000000 010111 000000 010111 000000 010111 000000 010111 000000 010111 000000 010111 000000 010111 111111 110111 000000 110000 000000 010111 111111 100111 000000 110000 000000 010111 111111 100111 111111 100111 111111 111111 111111 111111 000000 110000 000000 011000 000000 011000 000000 011000 111111 111111 000000 001100 000000 001011 000000 001100 000000 001011 000000 001100 000000 010111 000000 010111 000000 010111 000000 010111 000000 010111 000000 010111 000000 001011 000000 001011 000000 000110 000000 001011 000000 000101 000000 000110 000000 001011 000000 000101 000000 000101 000000 000110 000000 000011 000000 000101 000000 000101 b. 54/12 = 4 remainder 4 Step Action 0 Initial Vals 1 Temp = Rem – Div Temp < 0, Q << 0 2 Quotient 000000 000000 000000 Temp 000000 000000 110110 101100 110110 101100 Divisor 001010 000000 001010 000000 001010 000000 Remainder 000000 101100 000000 101100 000000 101100 Rshift Div 000000 110110 101100 000101 000000 000000 101100 Temp = Rem – Div 000000 111011 101100 000101 000000 000000 101100 Temp < 0, Q << 0 Rshift Div 000000 000000 111011 101100 111011 101100 000101 000000 000010 100000 000000 101100 000000 101100 Chapter 3 Solutions Step Action 3 000000 101100 Temp < 0, Q << 0 000000 111110 001100 000010 100000 000000 101100 000000 111110 001100 000001 010000 000000 101100 Temp = Rem – Div 000000 111111 011100 000001 010000 000000 101100 Temp < 0, Q << 0 000000 111111 011100 000001 010000 000000 101100 000000 111111 011100 000000 101000 000000 101100 Temp = Rem – Div 000000 000000 000100 000000 101000 000000 101100 T > 0, Q << 1, R = T 000001 000000 000100 000000 101000 000000 000100 000001 000000 000100 000000 010100 000000 000100 Temp = Rem – Div 000001 111111 110000 000000 010100 000000 000100 Temp < 0, Q << 0 000010 111111 110000 000000 010100 000000 000100 Rshift Div 7 Remainder 000010 100000 Rshift Div 6 Divisor 111110 001100 Rshift Div 5 Temp 000000 Rshift Div 4 Quotient Temp = Rem – Div 000010 111111 110000 000000 001010 000000 000100 Temp = Rem – Div 000010 111111 111010 000000 001010 000000 000100 Temp < 0, Q << 0 000100 111111 111010 000000 001010 000000 000100 Rshift Div 000100 111111 111010 000000 000101 3.8.5 No solution provided 3.8.6 No solution provided Solution 3.9 3.9.1 No solution provided 3.9.2 No solution provided 3.9.3 No solution provided Solution 3.10 3.10.1 a. 201326592 201326592 b. –1000144896 3294822400 3.10.2 a. jal 0x00000000 b. lwc1 $3,0($3) 000000 000100 S21 S22 Chapter 3 Solutions 3.10.3 a. 0x0C000000 = 0000 1100 0000 0000 0000 0000 0000 0000 = 0 0001 1000 0000 0000 0000 0000 0000 000 sign is positive exp = 0x18 = 24 – 127 = –103 there is a hidden 1 mantissa = 0 answer = 1.0 x 2 –103 b. 0xC4630000 = 1100 0100 0110 0011 0000 0000 0000 0000 = 1 1000 1000 1100 0110 0000 0000 0000 000 sign is negative exp = 0x88 = 136 – 127 = 9 there is a hidden 1 –1 + 6 x 16 –2 mantissa = 0xC60000 = 12 x 16 = .75 +.0234375 9 answer = –1.7734375 x 2 3.10.4 a. 63.25 x 10 0 = 111111.01 x 2 0 normalize, move binary point 5 to the left 1.1111101 x 2 5 sign = positive, exp = 127 + 5 = 132 Final bit pattern: 0 1000 0100 1111 1010 0000 0000 0000 000 = 0100 0010 0111 1101 0000 0000 0000 0000 = 0x427D0000 b. 0 146987.40625 x 10 = 100011111000101011.011010 x 2 normalize, move binary point 17 to the left 17 1.00011111000101011011010 x 2 sign = positive, exp = 127 + 17 = 144 Final bit pattern: 0 1001 0000 0001 1111 0001 0101 1011 010 = 0100 1000 0000 1111 1000 1010 1101 1010 = 0x480F8ADA 0 3.10.5 a. 63.25 x 10 0 = 111111.01 x 2 0 normalize, move binary point 5 to the left 1.1111101 x 2 5 sign = positive, exp = 1023 + 5 = 1028 Final bit pattern: 0 100 0000 0100 1111 1010 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 0x404FA00000000000 b. 0 0 = 100011111000101011.011010 x 2 146987.40625 x 10 normalize, move binary point 17 to the left 17 1.00011111000101011011010 x 2 sign = positive, exp = 1023 + 17 = 1040 Final bit pattern: 0 100 0001 0000 0001 1111 0001 0101 1011 0100 0000 0000 0000 0000 0000 0000 0000 = 0x4101F15B40000000 Chapter 3 Solutions 3.10.6 a. 63.25 × 100 = 111111.01 × 20 = 3F.40 × 160 move hex point 2 to the left .3F40 × 162 sign = positive, exp = 64 + 2 Final bit pattern: 01000010001111110100000000000000 b. 146987.40625 × 100 = 10 0011 1110 0010 1011.011010 × 20 = 23E2B.68 × 160 move hex point 5 to the left .00100011111000101011011010 × 165 sign = positive, exp = 64 + 5 = 69 Final bit pattern: 01000101001000111110001010110110 Solution 3.11 3.11.1 a. –1.5625 × 10–1 = –.15625 × 100 = –.00101 × 20 move the binary point 2 to the right = –.101 × 2–2 exponent = –2, mantissa = –.101000000000000000000000 answer: 111111111110101100000000000000000000 b. 9.356875 × 102 = 935.6875 × 100 = 0x3A7.B × 160 = 1110100111.1011 × 20 move the binary point 10 to the left = .11101001111011 × 210 exponent = +10, mantissa = +.11101001111011 answer: 000000001010011101001111011000000000 3.11.2 a. –1.5625 × 10–1 = –.15625 × 100 = –.00101 × 20 move the binary point 3 to the right, = –1.01 × 2–3 exponent = –3 = –3 + 16 = 13, mantissa = –.0100000000 answer: 1011010100000000 b. 9.356875 × 102 = 935.6875 × 100 = 0x3A7.B × 160 = 1110100111.1011 × 20 move the binary point 9 to the left = 1.1101001111011 × 29 exponent = +9 = 9 + 16 = 25, mantissa = +.1101001111011 answer: 0110011101001111 S23 S24 Chapter 3 Solutions 3.11.3 a. –1.5625 × 10–1 = –.15625 × 100 = –.00101 × 20 move the binary point 2 to the right = –.101 × 2–2 exponent = –2, mantissa = –.1010000000000000000000000000 answer: 10110000000000000000000000000101 b. 9.356875 × 102 = 935.6875 × 100 = 0x3A7.B × 160 = 1110100111.1011 × 20 move the binary point 10 to the left = .11101001111011 × 210 exponent = +10, mantissa = +.11101001111011 answer: 01110100111101100000000000010100 3.11.4 a. 2.6125 × 101 + 4.150390625 × 10–1 2.6125 × 101 = 26.125 = 11010.001 = 1.1010001000 × 24 4.150390625 × 10–1 = .4150390625 = .011010100111 = 1.1010100111 × 2–2 Shift binary point 6 to the left to align exponents, GR 1.1010001000 00 +.0000011010 10 0111 (Guard = 1, Round = 0, Sticky = 1) -------------------1.1010100010 10 In this case the extra bits (G,R,S) are more than half of the least significant bit (0). Thus, the value is rounded up. 1.1010100011 × 24 = 11010.100011 × 20 = 26.546875 = 2.6546875 × 101 b. –4.484375 × 101 + 1.3953125 × 101 –4.484375 × 101 = –44.84375 = –1.0110011011 × 25 1.1953125 × 101 = 11.953125 = 1.0111111010 × 23 Shift binary point 2 to the left and align exponents, GR –1.0110011011 00 0.0101111110 10 (Guard = 1, Round = 0, Sticky = 0) -----------------–1.0000011100 10 In this case, the Guard is 1 and the Round and Sticky bits are zero. This is the “exactly half” case—if the LSB was odd (1) we would add, but since it is even (0) we do nothing. –1.0000011100 × 25 = –100000.11100 × 20 = –32.875 = –3.2875 × 101 Chapter 3 Solutions 3.11.5 No solution provided 3.11.6 No solution provided Solution 3.12 3.12.1 a. –8.0546875 × –1.79931640625 × 10–1 –8.0546875 = –1.0000000111 × 23 –1.79931640625 × 10–1 = –1.0111000010 × 2–3 Exp: –3 + 3 = 0, 0 + 16 = 16 (10000) Signs: both negative, result positive Mantissa: 1.0000000111 × 1.0111000010 -----------00000000000 10000000111 00000000000 00000000000 00000000000 00000000000 10000000111 10000000111 10000000111 00000000000 10000000111 1.01110011000001001110 1.0111001100 00 01001110 Guard = 0, Round = 0, Sticky = 1: NoRnd 1.0111001100 × 20 = 0100000111001100 (1.0111001100 = 1.44921875) –8.0546875 × –.179931640625 = 1.4492931365966796875 Some information was lost because the result did not fit into the available 10-bit field. Answer (only) off by .0000743865966796875. S25 S26 Chapter 3 Solutions b. 8.59375 × 10–2 × 8.125 × 10–1 8.59375 × 10–2 = .0859375 = 1.0110000000 × 2–4 8.125 × 10–1 = .8125 = 1.1010000000 × 2–1 Exp: –4 + –1 = –5, –5 + 16 = 11 (01011) Signs: both positive, result positive Mantissa: 1.0110000000 × 1.1010000000 -----------00000000000 00000000000 00000000000 00000000000 00000000000 00000000000 00000000000 10110000000 00000000000 10110000000 10110000000 1000111100000000000000 Normalize, add one to exponent, negate –1.0001111000 00 000000000 Guard = 0, Round = 0, Sticky = 0: Nornd –1.0001111000 × 2–4 = 1011000001111000 (.00010001111000) = .06982421875 .0859375 × .8125 = .06982421875 In this case the two match exactly, since no information was lost during the shifting. 3.12.2 No solution provided 3.12.3 No solution provided Chapter 3 Solutions 3.12.4 a. 8.625 × 101 / –4.875 × 100 8.625 × 101 = 1.0101100100 × 26 –4.875 = –1.0011100000 × 22 Exponent = 6 – 2 = 4, 4 + 16 = 20 (10100) Signs: one positive, one negative, result negative Mantissa: 1.00011011000100111 10011100000. | 10101100100.0000000000000000 –10011100000. ------------10000100.0000 –1001110.0000 ------------1100110.00000 –100111.00000 ------------1111.0000000 –1001.1100000 ------------101.01000000 –100.11100000 ------------000.011000000000 –.010011100000 ------------.000100100000000 –.000010011100000 ----------------.0000100001000000 –.0000010011100000 ----------------.00000011011000000 –.00000010011100000 -----------------.00000000110000000 1.000110110001001111 Guard = 0, Round = 1, Sticky = 1: No Round, fix sign –1.0001101100 × 24 = 1101000001101100 = 10001.101100 = –17.6875 86.25 / –4.875 = –17.692307692307 Some information was lost because the result did not fit into the available 10-bit field. Answer off by .00480769230. S27 S28 Chapter 3 Solutions b. 1.84375 × 100 / 1.3203125 × 100 1.84375 × 100 = 1.84375 = 1.1101100000 × 20 1.3203125 × 100 = 1.3203125 = 1.0101001000 × 20 Exponent = 0–0 = 0, 0 + 16 = 16 (10000) Signs: both positive, result positive Mantissa: 1.011001010111110 10101001000. | 11101100000.0000000000000000 –10101001000. -----------1000011000.00 – 101010010.00 -----------11000110.000 – 10101001.000 -----------11101.000000 – 10101.001000 -----------111.11100000 – 101.01001000 -----------10.1001100000 – 1.0101001000 -----------1.01000110000 – .10101001000 -----------.100111010000 – .010101001000 -----------.0100100010000 – .0010101001000 -----------.00011110010000 – .00010101001000 -----------.00001001001000 – .000010101001000 1.0110010101 11 110 Guard = 1, Round = 1, Sticky = 1: Round up 1.0110010110 × 20 = 0100000110010110 = 1.0110010110 = 1.396484375 1.84375 / 1.3203125 = 1.3964497041420118343195266 Some information was lost because the result did not fit into the available 10-bit field. Answer off by .000034671. 3.12.5 No solution provided 3.12.6 No solution provided Chapter 3 Solutions Solution 3.13 3.13.1 a. (3.984375 × 10–1 + 3.4375 × 10–1) + 1.771 × 103) 3.984375 × 10–1 = 1.1001100000 × 2–2 3.4375 × 10–1 = 1.0110000000 × 2–2 1.771 × 103 = 1771 = 1.1011101011 × 210 shift binary point of smaller left 12 so exponents match (A) (B) 1.1001100000 +1.0110000000 -------------10.1111100000 Normalize, (A+B) 1.0111110000 × 2–1 (C) +1.1011101011 (A+B) .0000000000 10 111110000 Guard=1, Round=0, Sticky=1 -------------(A+B)+C +1.1011101011 10 1 Round up (A+B)+C =1.1011101100 × 210 = 0110101011101100 = 1772 b. (3.96875 × 100 + 8.46875 × 100) + 2.1921875 × 101 3.96875 × 100 = 1.1111110000 × 21 8.46875 × 100 = 1.0000111100 × 23 2.1921875 × 101 = 1.0101111011 × 24 shift binary point of smaller left 6 so exponents match (A) (B) .0111111100 00 Guard=0, Round=0, Sticky=0 1.0000111100 -------------(A+B) 1.1000111000 No round (A+B) .1100011100 0 Guard=0, Round=0, Sticky=0 (C) +1.0101111011 -------------(A+B)+C10.0010010111Normalize, add 1 to exponent, round to even (A+B)+C = 1.0001001100 × 25 = 0101010001001100 = 34.375 3.13.2 a. 3.984375 × 10–1 + (3.4375 × 10–1 + 1.771 × 103) 3.984375 × 10–1 = 1.1001100000 × 2–2 3.4375 × 10–1 = 1.0110000000 × 2–2 1.771 × 103 = 1771 = 1.1011101011 × 210 shift binary point of smaller left 12 so exponents match (B) .0000000000 01 0110000000 Guard=0, Round=1, Sticky=1 (C) +1.1011101011 -------------(B+C) +1.1011101011 (A) .0000000000 011001100000 -------------A+(B+C) +1.1011101011 No round A+(B+C) +1.1011101011 × 210 = 0110101011101011 = 1771 S29 S30 Chapter 3 Solutions b. 3.96875 × 100 + (8.46875 × 100 + 2.1921875 × 101) 3.96875 × 100 = 1.1111110000 × 21 8.46875 × 100 = 1.0000111100 × 23 2.1921875 × 101 = 1.0101111011 × 24 shift binary point of smaller left 6 so exponents match (B) (C) .1000011110 0 Guard=0, Round=0, Sticky=0 1.0101111011 -------------(B+C) 1.1110011001 No round (A) .0011111110 000 Guard=0, Round=0, Sticky=0 (B+C) 1.1110011001 -------------(A+B)+C10.0010010111Normalize, add 1 to exponent, round to even (A+B)+C = 1.0001001100 × 25 = 0101010001001100 = 34.375 3.13.3 a. No, they are not equal: (A + B) + C = 1772, A + (B + C) = 1771 (steps shown above). Exact: .398437 + .34375 + 1771 = 1771.742187. b. Yes, they are equal: (A + B) + C = 34.375, A + (B + C) = 34.375 (steps shown above). Exact answer is 34.359375. 3.13.4 a. (3.41796875 × 10–3 × 6.34765625 × 10–3) × 1.05625 × 102 (A) 3.41796875 × 10–3 = 1.1100000000 × 2–9 (B) 4.150390625 × 10–3 = 1.0001000000 × 2–8 (C) 1.05625 × 102 = 1.1010011010 × 26 Exp: –9 –8 = –17 Signs: both positive, result positive Mantissa: (A) (B) A×B A×B 1.1100000000 × 1.0001000000 -----------11100000000 11100000000 ---------------------1.11011100000000000000 1.1101110000 00 00000000 Guard = 0, Round = 0, Sticky = 0: No Round 1.1101110000 × 2–17 UNDERFLOW: Cannot represent number Chapter 3 Solutions b. (1.140625 × 102 × –9.135 × 102) × 9.84375 × 10–1 (A) 1.140625 × 102 = 1.1100100001 × 26 (B) –9.135 × 102 = –1.1100100011 × 29 (C) 9.84375 × 10–1 = 1.1111100000 × 2–1 Exp: 6 + 9 = 15 Signs: one positive, one negative - result negative Mantissa: (A) (B) 1.1100100001 × 1.1100100011 -----------11100100001 11100100001 11100100001 11100100001 11100100001 11100100001 ---------------------11.00101110000010000011 Normalize, add 1 to exponent 1.1001011100 00 010000011 Guard=0, Round=0, Sticky=1: No Round A×B –1.1001011100 × 216 OVERFLOW: Cannot represent number S31 S32 Chapter 3 Solutions 3.13.5 a. 3.41796875 × 10–3 × (6.34765625 × 10–3 × 1.05625 × 102) (A) 3.41796875 × 10–3 = 1.1100000000 × 2–9 (B) 4.150390625 × 10–3 = 1.0001000000 × 2–8 (C) 1.05625 × 102 = 1.1010011010 × 26 Exp: –8 + 6 = –2 Signs: both positive, result positive Mantissa: (B) (C) 1.0001000000 × 1.1010011010 -----------10001000000 10001000000 10001000000 10001000000 10001000000 10001000000 ---------------------1.110000001110100000000 1.1100000011 10 100000000 Guard=1, Round=0, Sticky=1: Round B × C 1.1100000100 × 2–2 Exp: –9 –2 = –11 Signs: both positive, result positive Mantissa: (A) (B × C) 1.1100000000 × 1.1100000100 -----------11100000000 11100000000 11100000000 11100000000 ---------------------11.00010001110000000000 Normalize, add 1 to exponent 1.1000100011 10 0000000000 Guard = 1, Round = 0, Sticky = 0: Round to even A × (B × C) 1.1000100100 × 2–10 Chapter 3 Solutions b. 1.140625 × 102 × (–9.135 × 102 × 9.84375 × 10–1) (A) 1.140625 × 102 = 1.1100100001 × 26 (B) –9.135 × 102 = –1.1100100011 × 29 (C) 9.84375 × 10–1 = 1.1111100000 × 2–1 Exp: 9 – 1 = 8 Signs: one negative, one positive - result negative Mantissa: (B) (C) 1.1100100011 × 1.1111100000 -----------11100100011 11100100011 11100100011 11100100011 11100100011 11100100011 ---------------------11.100000110011101 Normalize, add 1 to exponent 1.1100000110 01 1101000000 Guard=0, Round=1, Sticky=1: No Round B × C –1.1100000110 × 29 Exp: 5 + 9 = 14 Signs: one negative, one positive - result negative Mantissa: (A) (B×C) 1.1100100001 × 1.1100000110 -----------11100100001 11100100001 11100100001 11100100001 11100100001 ---------------------11.00100001000111000110 Normalize, add 1 to exponent 1.1001000010 00 111000110 Guard=0, Round=0, Sticky=1: No Round A × (B × C) 1.1001000010 × 215 3.13.6 a. b) No: A × B = 1.1101110000 × 2–17 UNDERFLOW: Cannot represent A × (B × C) = 1.1000100100 × 2–10 A and B are both small, so their product does not fit into the 16-bit floating point format being used. b. e) No: A × (B × C) = –1.1001000010 × 215 A × B = –1.1001011100 × 216 OVERFLOW: Cannot be represented A and B are both large, so their product does not fit into the 16-bit floating point format being used. S33 S34 Chapter 3 Solutions Solution 3.14 3.14.1 a. 1.666015625 × 100 × (1.9760 × 104 – 1.9744 × 104) (A) 1.666015625 × 100 = 1.1010101010 × 20 (B) 1.9760 × 104 = 1.0011010011 × 214 (C) –1.9744 × 104 = –1.0011010010 × 214 Exponents match, no shifting necessary (B) 1.0011010011 (C) –1.0011010010 --------------(B+C) 0.0000000001 × 214 (B+C) 1.0000000000 × 24 Exp: 0 + 4 = 4 Signs: both positive, result positive Mantissa: (A) (B+C) 1.1010101010 × 1.0000000000 -----------11010101010 ---------------------1.10101010100000000000 A×(B+C) 1.1010101010 0000000000 Guard=0, Round=0, Sticky=0: No Round A × (B + C) 1.1010101010 × 24 b. 3.48 × 102 × (6.34765625 × 10–2 – 4.052734375 × 10–2) (A) 3.48 × 102 = 1.0101110000 × 28 (B) 6.34765625 × 10–2 = 1.0000010000 × 2–4 (C) –4.052734375 × 10–2 = 1.0100110000 × 2–5 Shift binary point of smaller left 1 so exponents match (B) 1.0000010000 × 2–4 (C) –.1010011000 0 × 2–4 --------------(B+C) .0101111000 Normalize, subtract 2 from exponent (B + C) 1.0111100000 × 2–6 Exp: 8 – 6 = 2 Signs: both positive, result positive Mantissa: (A) (B+C) 1.0101110000 × 1.0111100000 -----------10101110000 10101110000 10101110000 10101110000 10101110000 A×(B+C) 1.1111111100 10000000000 Guard=1, Round=0, Sticky=0: Round to even A × (B + C) 1.1111111100 × 22 Chapter 3 Solutions 3.14.2 a. 1.666015625 × 100 × (1.9760 × 104 – 1.9744 × 104) (A) 1.666015625 × 100 = 1.1010101010 × 20 (B) 1.9760 × 104 = 1.0011010011 × 214 (C) –1.9744 × 104 = –1.0011010010 × 214 Exp: 0 + 14 = 14 Signs: both positive, result positive Mantissa: (A) (B) 1.1010101010 × 1.0011010011 -----------11010101010 11010101010 11010101010 11010101010 11010101010 11010101010 ---------------------10.0000001001100001111 Normalize, add 1 to exponent A×B 1.0000000100 11 00001111 Guard=1, Round=1, Sticky=1: Round A×B 1.0000000101 × 215 Exp: 0 + 14 = 14 Signs: one negative, one positive, result negative Mantissa: (A) (C) 1.1010101010 × 1.0011010010 -----------11010101010 11010101010 11010101010 11010101010 11010101010 -----------------------10.0000000111110111010 Normalize, add 1 to exponent A×C 1.0000000011 11 101110100 Guard=1, Round=1, Sticky=1: Round A×C –1.0000000100 × 215 A×B 1.0000000101 × 215 A×C –1.0000000100 × 215 ------------A×B+A×C .0000000001 × 215 A × B + A × C 1.0000000000 × 25 S35 S36 Chapter 3 Solutions b. 3.48 × 102 × (6.34765625 × 10–2 – 4.052734375 × 10–2) (A) 3.48 × 102 = 1.0101110000 × 28 (B) 6.34765625 × 10–2 = 1.0000010000 × 2–4 (C) –4.052734375 × 10–2 = 1.0100110000 × 2–5 Exp: 8 – 4 = 4 Signs: both positive, result positive Mantissa: (A) (B) 1.0101110000 × 1.0000010000 -----------10101110000 10101110000 ---------------------1.01100001011100000000 A×B 1.0110000101 11 00000000 Guard=1, Round=1, Sticky=0: Round A × B 1.0110000110 × 24 Exp: 8 – 5 = 3 Signs: one negative, one positive, result negative Mantissa: (A) (C) 1.0101110000 × 1.0100110000 -----------10101110000 10101110000 10101110000 10101110000 -----------------------1.11000011010100000000 A×C 1.1100001101 0100000000 Guard=0, Round=1, Sticky=0: No Round A × C –1.1100001101 × 23 A×B 1.0110000110 × 24 A×C –.1110000110 1 × 24 (Guard=1, Round=0, Sticky=0: Round to even) ------------A×B+A×C .1000000000 × 24 A × B + A × C 1.000000000 × 23 3.14.3 a. b) No: A × (B + C) = 1.1010101010 × 24 = 26.65625, and (A × B) + (A × C) = 1.0000000000 × 25 = 32 Exact: 1.666015625 × (19760 – 19744) = 26.65625 b. e) No: A × B + A × C = 1.0000000000 × 23 = 8, and A × (B + C) = 1.1111111100 × 22 = 7.984375 Exact: 348 × (.0634765625 – .04052734375) = 7.986328125 Chapter 3 Solutions 3.14.4 Answer Sign Exp Exact? a. 1 01111101 00000000000000000000000 – –2 Yes b. 0 01111011 10011001100110011001101 + –4 No 3.14.5 a. b + b + b + b = –1 b × 4 = –1 They are the same b. e + e + e + e + e + e + e + e + e + e = 1.000000000000000000000100 e × 10 = 1.000000000000000000000100 3.14.6 No solution provided Solution 3.15 3.15.1 a. 0101 0101 0101 0101 0101 0101 0x.555555 No b. 0001 1001 1001 1001 1001 1001 .199999 No 3.15.2 a. 0011 0011 0011 0011 0011 0011 .33333 No b. 0001 0000 0000 0000 0000 0000 .100000 Yes 3.15.3 a. 0101 0000 0000 0000 0000 0000 b. 0001 0111 0111 0111 0111 0111 .500000 .177777 Yes No 3.15.4 a. 01010 00000 00000 00000 .A000 Yes b. 00011 00000 00000 00000 .3000 Yes S37