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3 years ago

When is it appropriate to utilize the nat network connection type?

Computers and Technology

1 answer:

lbvjy [14]3 years ago

6 0

Whenever the VM does not need to be access at a known address by other network nodes.

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3. is the system of rules that govern the ordering of values? culture a. Ethics b. organization c. protocol d. value system 3 .

Norma-Jean [14]

Answer:

ethics

Explanation:

The set of principles that guides the ordering of values is known as ethics. Scenario, challenge, or event in which a person has to choose between various morally correct or bad behaviors. The moral principles and norms that govern conduct in the business world. The rules or norms that regulate a specific individual or a group's behavior. It is also the ethical ideas and standards that society accepts as right or good as opposed to immoral or undesirable.

7 0

2 years ago

Computer system with a 32-bit logical address and 4k-byte page size. assume that each entry of a page table consists of 4bytes.

Annette [7]

The number of bits in the physical address is 26 bits. The number of entries in a page table is $\mathbf{2^{20}}$ entries. The size of the page table in a one-level paging scheme is 4MB.

<h3>
What is paging in Operating System?</h3>

Paging is a storage method used in operating systems to recover activities as pages from secondary storage and place them in primary memory. The basic purpose of pagination is to separate each procedure into pages.

We are given the following parameters:

32-bit logical address
Page size = 4 KB = $\mathbf{2^{12} \ bytes }$
Size of each Page table entry = 4 bytes

Suppose the system supports physical memory size = 64 MB = $\mathbf{2^{26} \ bytes}$

Thus, the number of bits in the physical address is computed as:

= $\mathbf{log_2 \{Physical-memory-size\}}$

$=\mathbf{log_2(2^{26})}}$

= 26 bits

The number of entries in a page table = logical address space size/page size

The number of entries in a page table $\mathbf{= \dfrac{2^{32}}{2^{12}}}$

$\mathbf{=2^{20}}$ entries

In a one-level paging scheme, the size of the table is:

= entire no. of page entries × page table size

= $\mathbf{2^{20}\times 4 \ bytes}$

= 4 MB

suppose that this system supports up to 2^30 bytes of physical memory.

The size of the page table will be the same as 4 MB due to the fact that the number of entries, as well as, the page table entry size is the same.

Since the size of the page table surpasses that of a single page. A page cannot include a whole page table. Therefore, the page table must be broken into parts to fit onto numerous pages, and an additional level of the page table is required to access this page table.

This is called the Multi-Level Paging system.

Therefore, we can conclude that the number of bits in the physical address is 26 bits, the number of entries in a page table is $\mathbf{2^{20}}$ entries, and the size of the page table in a one-level paging scheme is 4MB.

Learn more about Paging in Operating System here:

brainly.com/question/17004314

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4 0

2 years ago

ANYBODY WANNA PLAY FoRnITe

ValentinkaMS [17]

Answer:

No but my baby brother can play with u

6 0

3 years ago

Read 2 more answers