Answer:
a) the average CPI for machine M1 = 1.6
the average CPI for machine M2 = 2.5
b) M1 implementation is faster.
c) the clock cycles required for both processors.52.6*10^6.
Explanation:
(a)
The average CPI for M1 = 0.6 x 1 + 0.3 x 2 + 0.1 x 4
= 1.6
The average CPI for M2 = 0.6 x 2 + 0.3 x 3 + 0.1 x 4
= 2.5
(b)
The average MIPS rate is calculated as: Clock Rate/ averageCPI x 10^6
Given 80MHz = 80 * 10^6
The average MIPS ratings for M1 = 80 x 10^6 / 1.6 x 10^6
= 50
Given 100MHz = 100 * 10^6
The average MIPS ratings for M2 = 100 x 10^6 / 2.5 x 10^6
= 40
c)
Machine M2 has a smaller MIPS rating
Changing instruction set A from 2 to 1
The CPI will be increased to 1.9 (1*.6+3*.3+4*.1)
and hence MIPS Rating will now be (100/1.9)*10^6 = 52.6*10^6.
Answer:
scores.append(6,2)
Explanation:
This is a complicated question because in theory, scores.insert can also add values, but I am sure that the only line of code that would work is scores.append(6,2)
The awnser is task 6 because 6 plus 8 is 20 and that equals for 6 gallons to go the the teen choice awards
When a formula contains the address of a cell, it is called a c<span>ell reference.
hope this helps!</span>
No longer produced or used; out of date.
