1.)
<span>((i <= n) && (a[i] == 0)) || (((i >= n) && (a[i-1] == 0))) </span>
<span>The expression will be true IF the first part is true, or if the first part is false and the second part is true. This is because || uses "short circuit" evaluation. If the first term is true, then the second term is *never even evaluated*. </span>
<span>For || the expression is true if *either* part is true, and for && the expression is true only if *both* parts are true. </span>
<span>a.) (i <= n) || (i >= n) </span>
<span>This means that either, or both, of these terms is true. This isn't sufficient to make the original term true. </span>
<span>b.) (a[i] == 0) && (a[i-1] == 0) </span>
<span>This means that both of these terms are true. We substitute. </span>
<span>((i <= n) && true) || (((i >= n) && true)) </span>
<span>Remember that && is true only if both parts are true. So if you have x && true, then the truth depends entirely on x. Thus x && true is the same as just x. The above predicate reduces to: </span>
<span>(i <= n) || (i >= n) </span>
<span>This is clearly always true. </span>
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:
I would say D hope this helps
Machine language is the language understood by a computer. It is very difficult to understand, but it is the only thing that the computer can work with. All programs and programming languages eventually generate or run programs in machine language
