計(jì)算機(jī)組成與設(shè)計(jì)第五版答案

1、Solutions 3S-4 Chapter 3 Solutions3.13 62?12Step Action Multiplicand Product/Multiplier0 Initial Vals 110 010 000 000 001 0101 lsb=0, no op 110 010 000 000 001 010Rshift Product 110 010 000 000 000 1012 Prod=Prod+Mcand

2、110 010 110 010 000 101Rshift Mplier 110 010 011 001 000 0103 lsb=0, no op 110 010 011 001 000 010Rshift Mplier 110 010 001 100 100 0014 Prod=Prod+Mcand 110 010 111 110 100 001Rshift Mplier 110 010 011 111 010 0005 lsb=0

3、, no op 110 010 011 111 010 000Rshift Mplier 110 010 001 111 101 0006 lsb=0, no op 110 010 001 111 101 000Rshift Mplier 110 010 000 111 110 1003.14 For hardware, it takes 1 cycle to do the add, 1 cycle to do the shift,

4、 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

5、 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.(3?8)?4tu ? 96 time units for hardware(5?8)?4tu ? 160 time units for software3.15 It take

6、s B time units to get through an adder, and there will be A ? 1 adders. Word is 8 bits wide, requiring 7 adders. 7?4tu ? 28 time units.3.16 It takes B time units to get through an adder, and the adders are arranged i

7、n a tree structure. It will require log2(A) levels. 8 bit wide word requires 7 adders in 3 levels. 3?4tu ? 12 time units.3.17 0x33 ? 0x55 ? 0x10EF. 0x33 ? 51, and 51 ? 32?16?2?1. We can shift0x55 left5 places (0xAA0)

8、, then add 0x55 shifted left4 places (0x550), then add 0x55 shifted leftonce (0xAA), then add 0x55. 0xAA0?0x550?0xAA?0x55 ? 0x10EF. 3 shifts, 3 adds.(Could also use 0x55, which is 64?16?4?1, and shift0x33 left6 times,