1. Machine B runs at 2 GHz and has a CPl of 1.3 for a particular program. Machin

Question

1. Machine B runs at 2 GHz and has a CPl of 1.3 for a particular program. Machine C runs at5 GHz and has a CPI of 2.4 for that program, while executing 20% more instructions. Which machine is faster? What is the speedup over the slower machine? 2. (Amdahl's law backwards) You have an optimization that speeds up only the non- memory operations by a factor of 3, If you spend 50% of your execution time waiting for memory after the optimization, what percentage of the original execution time did you wait for memory? 3. Suppose you change the MIPS ISA from a 3-address to a 2-address ISA (destination register is the same as the second ALU operand), and you double the number of registers, how many new R-type instructions can you support? How many fewer I-type instructions can you support, assuming you still want to use 16-bit immediates

Explanation / Answer

Solution:

1)

Suppose machine B instruction count is 1 then machine C's instruction count will be 1.2

Average memory access time(machine B)= CPI*Instruction count/clock frequency= 1.3*1/(2*10^9)= 6.5*10^-10

Average memory access time(machine C)= CPI*Instruction count/clock frequency= 2.4*1.2/(5*10^9)= 5.76*10^-10

we can see here that average memory access time of C is less this means machine C is faster.

please repost others. Thank you.

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.