《逻辑与计算机设计基础》(原书第五版)课后习题答案-chapter10_solutions-5th
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CHAPTER 10
© 2016 Pearson Education, Inc.
10-1.
a) Maximum frequency = 1/pipe stage delay = 1/0.8 ns = 1.25 GHz.
b) The latency time = 0.8 ns × 3 = 2.4ns.
c) The maximum throughput is 1 instruction per cycle or 1.25 billion instructions per second.
10-2.*
a) The latency time = 0.5 ns × 8 = 4.0 ns.
b) The maximum throughput is 1 instruction per cycle or 2 billion instructions per second.
c) The time required to execute is 10 instruction + 8 pipe stages -1 = 17 cycles *0.5ns = 8.5ns 10-3.
10-4.
Register Indirect: Load, Store, JMR
Register, Immediate: ADI, SBI, ANI, ORI, XRI, AIU, SIU
Relative: BZ, BNZ, JMP, JML
None: NOP
Register: All instructions not listed above
10-5.
a) Right, SH = 0F = 15 = 0 + 12 + 3
47 lines = 0000 3DF3 CB4A, 35 lines = 0 0003 DF3C, 32 lines = 0000 7BE7
b) Left, SH = 1D = 29 Rt. Rotate = 64 29 = 35 = 32 + 0 + 3
47 lines = 4B4A 0000 0000, 35 lines = 2000 0000, 32 lines = 4000 0000
1
Cycle 1: PC = 10F
Cycle 2: PC-1 = 110, IR = 4418 2F0116
Cycle 3: PC-2 = 110, RW = 1, DA = 01, MD = 0, BS = 0, PS = X, MW = 0, FS = 2, SH = 01, MA = 0, MB =1, CS=1 BUS A = 0000 001F, BUS B = 0000 2F01
Cycle 4: RW = 1, DA = 01, MD = 0, D0 = 0000 2F20, D1 = XXXX XXXX, D2 = 0000 00000
Cycle 5: R1 = 0000 2F20
10-7.
Cycle 1: IF PC = 10F
Cycle 2: DOF PC-1 = 110,IR = 1A61 001D
Cycle 3: EX PC-2 = 110,RW = 1, DA = 06, MD = 2, BS = 0, PS = X, MW = 0, FS =D, SH = 1D, MA = 0, MB = 0 BUS A = 01AB CDEF, BUS B = XXXX XXXX
Cycle 4: RW = 1, DA = 06, MD = 0, D0 = 0000 0000, D1 = XXXX XXXX, D2 = 0000 0000
Cycle 5: R6 = 0000 0000
10-8.
Cycle 1: PC = 10F
Cycle 2: PC-1 = 110, IR = CA71 9400
Cycle 3: PC-2 = 110, RW = 1, DA = 07, MD = 2, BS = 0, PS = 0, MW = 0, FS = 5, MA = 0, MB = 0, CS = 0 BUS A = 0000 F001, BUS B = 0000 000F
Cycle 4: RW = 1, DA = 07 MD = 0, D0 = 0000 EFF2, D1 = XXXX XXXX, D2 = 0000 0000
Cycle 5: R7 = 0000 0000
10-9.
Cycle 1: PC = 10F
Cycle 2: PC-1 = 110, IR = 8A21 635A
Cycle 3: PC-2 = 110, RW = 1, DA = 02, MD = 0, BS = 0, PS = X, MW = 0, FS=5, SH = 1A, MA= 0, MB = 1, CS=0 BUS A = 0A5F BC2B, BUS B = 0000 635A
Cycle 4: RW = 1, DA = 02, MD = 0, D0 = 0AF5 58D1, D1 = XXXX XXXX, D2 = 0000 00000
Cycle 5: R2 = 0AF5 58D1
10-10.+
Answer not given; varies depending on synthesis software used.
10-11.*
MOVA R7, R6
SUB R8, R8, R6 AND R8, R8, R7
MOV A R7, R6 SUB R8, R8, R6 AND R8, R8, R7