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
No .
Sorry , but I never try them ,
Answer:
Interval Recording
Explanation:
According to my research on different data collection methods, I can say that based on the information provided within the question this method of data collection is called Interval Recording. This refers to the process of collecting different pieces of data from different parts of the same interview or study. Which is what Mrs. Golden is doing.
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Answer:
In Email's Subject Line “Re:” most commonly understood as “Reply/Response” to an Email. “Re: ” is not taken as abbreviation of Regards/Reference. For Reference, Abbreviation is “Ref.”Explanation:
