Explanation:
A.)
we have two machines M1 and M2
cpi stands for clocks per instruction.
to get cpi for machine 1:
= we multiply frequencies with their corresponding M1 cycles and add everything up
50/100 x 1 = 0.5
20/100 x 2 = 0.4
30/100 x 3 = 0.9
CPI for M1 = 0.5 + 0.4 + 0.9 = 1.8
We find CPI for machine 2
we use the same formula we used for 1 above
50/100 x 2 = 1
20/100 x 3 = 0.6
30/100 x 4 = 1.2
CPI for m2 = 1 + 0.6 + 1.2 = 2.8
B.)
CPU execution time for m1 and m2
this is calculated by using the formula;
I * CPI/clock cycle time
execution time for A:
= I * 1.8/60X10⁶
= I x 30 nsec
execution time b:
I x 2.8/80x10⁶
= I x 35 nsec
Answer:
You are most likely to automatically encode information about the sequence of your day's events.
Answer:
Achieved Instruction per cycle rate is 
Solution:
As per the question:
Execution time, t = 1.4 s
No. of instructions being executed, n = 1.8 billion =
Clock speed of CPU, f = 3.2 GHz = 
Now, to calculate the actual issue rate achieved:
CPI (Cycle per Instruction) = 
CPI = 
Instruction per cycle is given as the reciprocal of CPI:
Instruction per cycle = 
Answer:
Explanation:
The system will be deadlock free if the below two conditions holds :
Proof below:
Suppose N = Summation of all Need(i), A = Addition of all Allocation(i), M = Addition of all Max(i). Use contradiction to prove.
Suppose this system isn't deadlock free. If a deadlock state exists, then A = m due to the fact that there's only one kind of resource and resources can be requested and released only one at a time.
Condition B, N + A equals M < m + n. Equals N + m < m + n. And we get N < n. It means that at least one process i that Need(i) = 0.
Condition A, Pi can let out at least 1 resource. So there will be n-1 processes sharing m resources now, Condition a and b still hold. In respect to the argument, No process will wait forever or permanently, so there's no deadlock.
