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Marat540 [252]
3 years ago
7

Memory Question!

Engineering
1 answer:
mafiozo [28]3 years ago
7 0

Answer:

Demand Paging

The basic idea behind demand paging is that when a process is swapped in, its pages are not swapped in all at once. Rather they are swapped in only when the process needs them. ( on demand. ) This is termed a lazy swapper, although a pager is a more accurate term.

Figure 9.4 - Transfer of a paged memory to contiguous disk space

9.2.1 Basic Concepts

The basic idea behind paging is that when a process is swapped in, the pager only loads into memory those pages that it expects the process to need ( right away. )

Pages that are not loaded into memory are marked as invalid in the page table, using the invalid bit. ( The rest of the page table entry may either be blank or contain information about where to find the swapped-out page on the hard drive. )

If the process only ever accesses pages that are loaded in memory ( memory resident pages ), then the process runs exactly as if all the pages were loaded in to memory.

Figure 9.5 - Page table when some pages are not in main memory.

On the other hand, if a page is needed that was not originally loaded up, then a page fault trap is generated, which must be handled in a series of steps:

The memory address requested is first checked, to make sure it was a valid memory request.

If the reference was invalid, the process is terminated. Otherwise, the page must be paged in.

A free frame is located, possibly from a free-frame list.

A disk operation is scheduled to bring in the necessary page from disk. ( This will usually block the process on an I/O wait, allowing some other process to use the CPU in the meantime. )

When the I/O operation is complete, the process's page table is updated with the new frame number, and the invalid bit is changed to indicate that this is now a valid page reference.

The instruction that caused the page fault must now be restarted from the beginning, ( as soon as this process gets another turn on the CPU. )

Figure 9.6 - Steps in handling a page fault

In an extreme case, NO pages are swapped in for a process until they are requested by page faults. This is known as pure demand paging.

In theory each instruction could generate multiple page faults. In practice this is very rare, due to locality of reference, covered in section 9.6.1.

The hardware necessary to support virtual memory is the same as for paging and swapping: A page table and secondary memory. ( Swap space, whose allocation is discussed in chapter 12. )

A crucial part of the process is that the instruction must be restarted from scratch once the desired page has been made available in memory. For most simple instructions this is not a major difficulty. However there are some architectures that allow a single instruction to modify a fairly large block of data, ( which may span a page boundary ), and if some of the data gets modified before the page fault occurs, this could cause problems. One solution is to access both ends of the block before executing the instruction, guaranteeing that the necessary pages get paged in before the instruction begins.

9.2.2 Performance of Demand Paging

Obviously there is some slowdown and performance hit whenever a page fault occurs and the system has to go get it from memory, but just how big a hit is it exactly?

There are many steps that occur when servicing a page fault ( see book for full details ), and some of the steps are optional or variable. But just for the sake of discussion, suppose that a normal memory access requires 200 nanoseconds, and that servicing a page fault takes 8 milliseconds. ( 8,000,000 nanoseconds, or 40,000 times a normal memory access. ) With a page fault rate of p, ( on a scale from 0 to 1 ), the effective access time is now:

( 1 - p ) * ( 200 ) + p * 8000000

= 200 + 7,999,800 * p

which clearly depends heavily on p! Even if only one access in 1000 causes a page fault, the effective access time drops from 200 nanoseconds to 8.2 microseconds, a slowdown of a factor of 40 times. In order to keep the slowdown less than 10%, the page fault rate must be less than 0.0000025, or one in 399,990 accesses.

A subtlety is that swap space is faster to access than the regular file system, because it does not have to go through the whole directory structure. For this reason some systems will transfer an entire process from the file system to swap space before starting up the process, so that future paging all occurs from the ( relatively ) faster swap space.

Some systems use demand paging directly from the file system for binary code ( which never changes and hence does not have to be stored on a page operation ), and to reserve the swap space for data segments that must be stored. This approach is used by both Solaris and BSD Unix

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The answer is A. Immediately inform her colleague
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1). Mention any four operations that requires airlines. 2). Explain how airflow is applicable to the above mentioned operations.
vladimir2022 [97]

Answer:

Following are the answer to this question:

Explanation:

1)

Following the four operations in the airlines:

Landside operations:

In Airlines, the airports are divided into areas on the countryside and on the airside, in which landside region is available to the public, although strictly controlled access to the airside zone. Its area covers all areas of the airport across the aircraft, including parts of the buildings which can only be reached by customers and employees.

Airside Operations:

It's also committed to ensuring which air operations military exercises Ballarat airfields are safe and secure. It includes the provision of parking and flight escort services to itinerant and automates. Organizing operational response to incidents, accidents, or emergencies at the airport.

Billing and invoicing Operations:

This requires several steps, each of which must be performed with absolute accuracy to ensure that perhaps the airport operator is adequately paid for supplying passengers with all the services and infrastructure. After this, the receipts want to be produced and sent to customers on the airline.

Information management:

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2)

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4 0
3 years ago
The elevation of the end of the steel beam supported by a concrete floor is adjusted by means of the steel wedges E and F. The b
Wewaii [24]

Answer:

a) P ≥ 22.164 Kips

b) Q = 5.4 Kips

Explanation:

GIven

W = 18 Kips

μ₁ = 0.30

μ₂ = 0.60

a) P = ?

We get F₁  and F₂ as follows:

F₁ = μ₁*W = 0.30*18 Kips = 5.4 Kips

F₂ = μ₂*Nef = 0.6*Nef

Then, we apply

∑Fy = 0   (+↑)

Nef*Cos 12º -  F₂*Sin 12º = W

⇒   Nef*Cos 12º -  (0.6*Nef)*Sin 12º = 18

⇒   Nef = 21.09 Kips

Wedge moves if

P ≥ F₁ + F₂*Cos 12º + Nef*Sin 12º

⇒  P ≥ 5.4 Kips + 0.6*21.09 Kips*Cos 12º + 21.09 Kips*Sin 12º

⇒  P ≥ 22.164 Kips

b) For the static equilibrium of base plate

Q = F₁ = 5.4 Kips

We can see the pic shown in order to understand the question.

7 0
3 years ago
Read 2 more answers
Give me some examples of fragile structures.
Anvisha [2.4K]

Answer:

i don't know if this help tell me if i am wrong

Explanation:

Gravity is the force that pulls all elements of matter together. Matter refers to things you can physically touch. The more matter there is, the greater the amount of gravity or force. This means that the Earth or other planets have a great deal of pull and that everything on Earth is pulled back to Earth.

Some examples of the force of gravity include:

The force that holds the gases in the sun.

The force that causes a ball you throw in the air to come down again.

The force that causes a car to coast downhill even when you aren't stepping on the gas.

The force that causes a glass you drop to fall to the floor.

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3 years ago
The steel bracket is used to connect the ends of two cables. if the allowable normal stress for the steel is sallow = 30 ksi, de
garri49 [273]

The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb

<h3>The static equilibrium is given as:</h3>

F = P (Normal force)

Formula for moment at section

M = P(4 + 1.5/2)

= 4.75p

Solve for the cross sectional area

Area = \frac{\pi d^{2} }{4}

d = 1.5

\frac{\pi *1.5^{2} }{4}

= 1.767 inches²

<h3>Solve for inertia</h3>

\frac{\pi *0.75^4}{4}

= 0.2485inches⁴

Solve for the tensile force from here

\frac{F}{A} +\frac{Mc}{I}

30x10³ = \frac{P}{1.767} +\frac{4.75p*0.75}{0.2485} \\\\

30000 = 14.902 p

divide through by 14.902

2013.15 = P

The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb

Read more on tensile force here: brainly.com/question/25748369

4 0
2 years ago
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