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:
Used the command syntax; awk -F":" '{ print "username: " $<number location> "\t\tuid:" $<number location> }' <target folder>
Explanation:
Linux operating system is a fast open-source computer platform for programmers and network administrators. Its system is arranged in a hierarchical tree structure with the root represented as "/" (for absolute path).
The passwd is a folder in the Linux OS that holds the login details of all users in the system network. The 'awk' is one of the commands used to get information from a file in a folder. It prints out the result by specifying the location of the values (like the username and user id) as a variable (with prefix '$') and then the target folder.
Like when a seed drops on the ground, and a bee come to take some honey. It sticks to the bee and where the bee goes, the bee will rub it off and the seed will start to grow.
