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sineoko [7]
3 years ago
9

Consider a multiprocessor system and a multithreaded program written using the many-to-many threading model. Let the number of u

ser-level threads in the program be more than the number of processors in the system. Discuss the performance implications of the following scenarios.
The number of kernel threads allocated to the program is less than the number of processing cores
The number of kernel threads allocated to the program is equal to the number of processing cores
The number of kernel threads allocated to the program is greater than the number of processing cores but less than the number of user-level threads
Engineering
1 answer:
Montano1993 [528]3 years ago
4 0

Answer:

At the point when the quantity of bit strings is not exactly the quantity of processors, at that point a portion of the processors would stay inert since the scheduler maps just part strings to processors and not client level strings to processors. At the point when the quantity of part strings is actually equivalent to the quantity of processors, at that point it is conceivable that the entirety of the processors may be used all the while. Be that as it may, when a part string obstructs inside the portion (because of a page flaw or while summoning framework calls), the comparing processor would stay inert. When there are more portion strings than processors, a blocked piece string could be swapped out for another bit string that is prepared to execute, in this way expanding the use of the multiprocessor system.When the quantity of part strings is not exactly the quantity of processors, at that point a portion of the processors would stay inert since the scheduler maps just bit strings to processors and not client level strings to processors. At the point when the quantity of bit strings is actually equivalent to the quantity of processors, at that point it is conceivable that the entirety of the processors may be used at the same time. Be that as it may, when a part string hinders inside the piece (because of a page flaw or while summoning framework calls), the relating processor would stay inert. When there are more portion strings than processors, a blocked piece string could be swapped out for another bit string that is prepared to execute, along these lines expanding the usage of the multiprocessor framework.

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Marina CMI [18]

Answer:

D the closest

Explanation:

i looked it up and 8t says 200

5 0
3 years ago
Assume a function requires 20 lines of machine code and will be called 10 times in the main program. You can choose to implement
Volgvan

Answer:

"Macro Instruction"

Explanation:

A macro definition is a rule or pattern that specifies how a certain input sequence should be mapped to a replacement output sequence according to a defined procedure. The mapping process that instantiates a macro use into a specific sequence is known as macro expansion.

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Explanation:!!

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3 years ago
Read 2 more answers
Shows a closed tank holding air and oil to which is connected a U-tube mercury manometer and a pressure gage. Determine the read
damaskus [11]

Answer:

P_2-P_1=27209h

Explanation:

For pressure gage we can determine this by saying:

The closed tank with oil and air has a pressure of P₁ and the pressure of oil at a certain height in the U-tube on mercury is p₁gh₁. The pressure of mercury on the air in pressure gauge is p₂gh₂. The pressure of the gage is P₂.

P_1+p_1gh_1=p_2_gh_2+P_2

We want to work out P₁-P₂: Heights aren't given so we can solve it in terms of height: assuming h₁=h₂=h

P_1-P_2=p_1gh_1-p_2gh_2=(55)\cdot{32.2}h-845\cdot{32.2}h

P_2-P_1=27209h

3 0
4 years ago
The Stefan-Boltzmann law can be employed to estimate the rate of radiation of energy H from a surface, as in
Mazyrski [523]

Explanation:

A.

H = Aeσ^4

Using the stefan Boltzmann law

When we differentiate

dH/dT = 4AeσT³

dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³

= 8.4085

Exact error = 8.4085x20

= 168.17

H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴

= 1366.376watts

B.

Verifying values

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴

= 1542.468

H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴

= 1205.8104

Error = 1542.468-1205.8104/2

= 168.329

ΔT = 40

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴

= 1735.05

H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴

= 1735.05-1059.83/2

= 675.22/2

= 337.61

5 0
3 years ago
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