Answer:
Means no matter how many processors you use, speed up never increase from 10 times.
Explanation:
If a problem of size W has a serial component Ws,then performance using parallelism:
Using Amdahl's Law:
Tp = (W - Ws )/ N + Ws
Here, Ws = .1,
W - Ws = .9
Performance Tp = (.9 / N) + .1
---------------------------------------------------------
Speed Up = 1 / ( (.9 / N) + .1)
If N -> infinity, Speed Up <= 10
Means no matter how many processors you use, speed up never increase from 10 times.
Answer/Explanation:
It is best to use Performance Monitor and Process counters to observe performance.
Cheers
Real time system means that the system is subjected to real time, i.e., response should be guaranteed within a specified timing constraint or system should meet the specified deadline. For example: flight control system, real time monitors etc.
Normal or random variations that are considered part of operating the system at its current capability are <u> c. common cause variations.</u>
Explanation:
Common cause variation is fluctuation caused by unknown factors resulting in a steady but random distribution of output around the average of the data.
Common-cause variation is the natural or expected variation in a process.
Common-cause variation is characterised by:
- Phenomena constantly active within the system
- Variation predictable probabilistically
- Irregular variation within a historical experience base
It is a measure of the process potential, or how well the process can perform when special cause variation removed.
Common cause variation arises from external sources that are not inherent in the process and is where statistical quality control methods are most useful.
Statistical process control charts are used when trying to monitor and control 5- and 6-sigma quality levels.
The first thing we are going to do is find the equation of motion:
ωf = ωi + αt
θ = ωi*t + 1/2αt^2
Where:
ωf = final angular velocity
ωi = initial angular velocity
α = Angular acceleration
θ = Revolutions.
t = time.
We have then:
ωf = (7200) * ((2 * pi) / 60) = 753.60 rad / s
ωi = 0
α = 190 rad / s2
Clearing t:
753.60 = 0 + 190*t
t = 753.60 / 190
t = 3.97 s
Then, replacing the time:
θ1 = 0 + (1/2) * (190) * (3.97) ^ 2
θ1 = 1494.51 rad
For (10-3.97) s:
θ2 = ωf * t
θ2 = (753.60 rad / s) * (10-3.97) s
θ2 = 4544,208 rad
Number of final revolutions:
θ1 + θ2 = (1494.51 rad + 4544.208 rad) * (180 / π)
θ1 + θ2 = 961.57 rev
Answer:
the disk has made 961.57 rev 10.0 s after it starts up