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Computers and Technology, 22.05.2021 06:10 cooltrey777
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Computers and Technology, 23.06.2019 01:30, solikhalifeoy3j1r
In deadlock avoidance using banker’s algorithm, what would be the consequence(s) of: (i) a process declaring its maximum need as maximum possible for each resource. in other words, if a resource a has 5 instances, then each process declares its maximum need as 5. (ii) a process declaring its minimum needs as maximum needs. for example, a process may need 2-5 instances of resource a. but it declares its maximum need as 2.
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Computers and Technology, 23.06.2019 06:30, Knownothing
When early motion pictures played in movie theaters, they were often accompanied by live organ or piano music. which of the following are the most likely reasons that this happened? (select all that apply). the music was provided to distract audience members from the loud sounds made when filmstrips were changed. the music accompanied the movies because the movies were silent and audiences were used to hearing music during plays in theaters. the music usually was played before, and sometimes after the movie, as an alternative form of entertainment. the music viewers to interpret the dramatic action in the films.
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Computers and Technology, 23.06.2019 17:30, Annlee23
When making changes to optimize part of a processor, it is often the case that speeding up one type of instruction comes at the cost of slowing down something else. for example, if we put in a complicated fast floating-point unit, that takes space, and something might have to be moved farther away from the middle to accommodate it, adding an extra cycle in delay to reach that unit. the basic amdahl's law equation does not take into account this trade-off. a. if the new fast floating-point unit speeds up floating-point operations by, on average, 2ă—, and floating-point operations take 20% of the original program's execution time, what is the overall speedup (ignoring the penalty to any other instructions)? b. now assume that speeding up the floating-point unit slowed down data cache accesses, resulting in a 1.5ă— slowdown (or 2/3 speedup). data cache accesses consume 10% of the execution time. what is the overall speedup now? c. after implementing the new floating-point operations, what percentage of execution time is spent on floating-point operations? what percentage is spent on data cache accesses?
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