**Memory Management and Page Table Size Calculation****Problem Statement:**Given a system with a 32-bit logical address and paging used for memory management, the page size is 4 KB (4096 bytes). If 4 bytes would be needed to store one entry of the page table, then what is the page table size for a process with a 4 MB (4096 x 4096 bytes) address space?**Options:**a. 1 KB b. 8 KB c. 16 KB d. 4 KB **Explanation:**In this problem, you need to calculate the page table size for a process with a given address space. The size of each page is 4 KB, and the size required to store one page table entry is 4 bytes.**Steps to Calculate:**1. **Address Space and Page Size:** - The process has an address space of 4 MB. - The page size is 4 KB.2. **Number of Pages Required:** - Total address space = 4 MB = 4096 KB - Each page is of size 4 KB. - Therefore, the number of pages = Total address space / Page size = 4096 KB / 4 KB = 1024 pages.3. **Page Table Size:** - Each entry in the page table requires 4 bytes. - Number of entries in the page table is equal to the number of pages, which is 1024. - Therefore, page table size = Number of pages * Size of each entry = 1024 * 4 bytes = 4096 bytes = 4 KB.**Correct Answer:**d. 4 KB

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Description: **Memory Management and Page Table Size Calculation****Problem Statement:**Given a system with a 32-bit logical address and paging used for memory management, the page size is 4 KB (4096 bytes). If 4 bytes would be needed to store one entry of the p

A logical address is a virtual address that is generated by the CPU and used by a program or process…

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Given a system with 32 bits logical address and paging is used for memory management, the page size is 4 KB (4096 bytes); If 4 Bytes would be needed to store one entry of the page table, then what is the page table size for a process of 4 MB (4096x496 bytes) address space? O a. 1 KB O b. 8 KB O C. 16 KB O d. 4 KB

Given a system with 32 bits logical address and paging is used for memory management, the page size is 4 KB (4096 bytes); If 4 Bytes would be needed to store one entry of the page table, then what is the page table size for a process of 4 MB (4096x496 bytes) address space? O a. 1 KB O b. 8 KB O C. 16 KB O d. 4 KB

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Chapter1: Introduction

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Question

**Memory Management and Page Table Size Calculation**

**Problem Statement:**

Given a system with a 32-bit logical address and paging used for memory management, the page size is 4 KB (4096 bytes). If 4 bytes would be needed to store one entry of the page table, then what is the page table size for a process with a 4 MB (4096 x 4096 bytes) address space?

**Options:**

a. 1 KB
b. 8 KB
c. 16 KB
d. 4 KB

**Explanation:**

In this problem, you need to calculate the page table size for a process with a given address space. The size of each page is 4 KB, and the size required to store one page table entry is 4 bytes.

**Steps to Calculate:**

1. **Address Space and Page Size:**
- The process has an address space of 4 MB.
- The page size is 4 KB.

2. **Number of Pages Required:**
- Total address space = 4 MB = 4096 KB
- Each page is of size 4 KB.
- Therefore, the number of pages = Total address space / Page size = 4096 KB / 4 KB = 1024 pages.

3. **Page Table Size:**
- Each entry in the page table requires 4 bytes.
- Number of entries in the page table is equal to the number of pages, which is 1024.
- Therefore, page table size = Number of pages * Size of each entry = 1024 * 4 bytes = 4096 bytes = 4 KB.

**Correct Answer:**
d. 4 KB

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