Assume the clock rates for both the machines is 300GHz a) Find CPU time for prog

Question

Assume the clock rates for both the machines is 300GHz

a)   Find CPU time for program 1 on both machines.  

b)   Find the CPI for program 1 on both machines.  

c)   Which machine is faster and by how much?

d) Find the MIPS for program 1 on both machines.

Now change the Clock rate of machine 2 to 500 GHz, machine clock rate remains same 300 GHz.

a)   Find CPU time for program 1 on both machines.  

b)   Find the CPI for program 1 on both machines.      

c)   Which machine is faster and by how much?                   

d)   Find the MIPS for program 1 on both machines.

After changing the Clock rate of machine 2 to 500 GHz what difference you find write your opinion about machine2.   

1 Computer architects want to compare the performance of two different computers: M1 and M2. The following measurements have been made on these computers: Machine 1 Machine 2 Programl executes 3 different Program l executes 3 different Each takes 2,3 and 4 clocks cyclesEach takes 4,5 and 6 clocks cycles Instructions 3millions load instructions, 5millions ALU instruction and lmillions store instructions. instructions 3million load instructons,5 million ALU instruction and 4millions store instructions

Explanation / Answer

Hi,
CPU time is given by t= (Number of instructions * CPI)/Clock Rate
so, lets first calculate CPI which is clock cycles per instruction

Machine 1:
total instructions= 3+5+1= 9 million
multiplying them with respective clock cycles, 3*2+5*3+1*4= 25 million cycles/9 million instructions
CPI= 2.77
now, CPU time=(9 * 106 * 2.77)/300 * 109 = 83.1 micro sec
now, MIPS = Instruction count/execution time * 106 = 9*106 / 83.1 * 10-6 *106 = 108303.24
Machine 2:
total instructions= 3+5+4= 12 million
multiplying them with respective clock cycles, 3*4+5*5+4*6= 61 million cycles/12million instructions
CPI= 5.083
now, CPU time=(12 * 106 *5.083)/300 * 109 =203.32 micro sec.
now, MIPS = Instruction count/execution time * 106 = 12*106 / 203.1 * 10-6 *106 = 59084.19
Hence Machine 1 is almost three times faster than machine 1
Second part, changing clock rate of machine 2, machine values remain same,
Machine 2:
total instructions= 3+5+4= 12 million
multiplying them with respective clock cycles, 3*4+5*5+4*6= 61 million cycles/12million instructions
CPI= 5.083
now, CPU time=(12 * 106 *5.083)/500 * 109 =121.992 micro sec.
now, MIPS = Instruction count/execution time * 106 = 12*106 / 121.99 * 10-6 *106 = 98368.71
by bumping the clock rate, there is a significant change in the execution time
Although the comparison is a little flawed becuase we are given different number of instructions to be run on both machines, but i think the objective is to help you understand the calculation of values

Thumbs up if this was helpful, otherwise let me know in comments

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