Answer:
3.1 ns ; 1.25 ; 3.097
Explanation:
Given :
IF, 3 ns;
ID, 2.5 ns;
EX, 2 ns;
MEM, 3 ns;
WB, 1.5 ns.
Use 0.1 ns for the pipelineregisterdelay
maximum time required for MEM = 3 ns
Pipeline register delay = 0.1 ns.
Clock cycled time of the pipelined machine= maximum time required + delay
3ns+0.1 ns = 3.1 ns
2.) for stall after every 4 instruction :
CPI of new machine :
(1 + (1 /4)) = 1 + 0.25 = 1.25
3.)
The speedup of pipelined machine over the single-cycle machine is given by :
Average time per instruction of single cycle ÷ average time per instruction of pipelined
Clock time of original machine = 12ns
Ideal CP1 = 1
CPI of new machine = 1.25
Clock period = 3.1 ns
(12 * 1) / (1.25 * 3.1) = 12 / 3.875
= 3.097
D. Speed up will equal the number of stages in the machine
Frequent cleaning with an air duster can would do the job
A hybrid network is any computer network that uses more than one type of connecting technology or topology. For example, a home network that uses both Wi-Fi and Ethernet cables to connect computers is a hybrid.
Here's the best answer I can give you, but bear with me.
The second option is incorrect because a class method must have a class identifier not an object identifier. What makes myObject an object identifier is the fact that it was created as an instance of the class MyClass in the constructor in line 5 (MyClass myObject= new MyClass(12.4,20);
The answer here should be MyClass.method2(20); Methods must have a set of parentheses, even if it has nothing inside. The first answer has a class identifier but the SOME_VALUE acts much like the Integer.MAX_VALUE; code which stores a constant value and does not actually perform tasks like most methods.
In short, the answer should be the last one but I hope my explanation cleared some things up for you, even if it was a bit more concept heavy than the question probably intended.
Explanation:
in fourth generation microprocessor was developed.
may it helped u
