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sesenic [268]
3 years ago
7

The implementation stage of the SDLC _____.

Computers and Technology
1 answer:
Karolina [17]3 years ago
6 0

Answer:

d. involves analyzing end-user business requirements in order to refine project goals

Explanation:

In the software development life cycle, the implementation stage is the stage where the computer programmers begin the actual coding of the application required by the end-user. At this stage, the developer takes into account the needs and specifications of the customer.

For example, if the end-user wants a software application that can track payment records, the researcher at this stage begins using the right software to code programs that can effect this function.

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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

b.

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

#SPJ1

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As a final lesson to her team, Katie shows them a job that has already been preflighted and is now getting ready to be sent to t
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Answer:

Printed

Explanation:

The job was finished and sent to the printer. In the printer, you have to waite to be printed

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