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Determine the number of page table entries (PTES) that are needed for the following combinations of virtual address size (n) and page size (P): n|P=2P Number of PTES 16 4K 16 8K 32 4K 32 8K

Determine the number of page table entries (PTES) that are needed for the following combinations of virtual address size (n) and page size (P): n|P=2P Number of PTES 16 4K 16 8K 32 4K 32 8K

Systems Architecture
Systems Architecture
7th Edition
ISBN:9781305080195
Author:Stephen D. Burd
Publisher:Stephen D. Burd
Chapter11: Operating Systems
Section: Chapter Questions
Problem 30VE
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### Calculating the Number of Page Table Entries (PTEs) Determine the number of page table entries (PTEs) that are needed for the following combinations of virtual address size (n) and page size (P): | Virtual Address Size (n) | Page Size (P = 2^P) | Number of PTEs | |--------------------------|---------------------|----------------| | 16 | 4K | | | 16 | 8K | | | 32 | 4K | | | 32 | 8K | | To calculate the number of page table entries, you need to use the formula: \[ \text{Number of PTEs} = \frac{\text{Virtual Address Space Size}}{\text{Page Size}} \] For a given virtual address space size \(n\) (in bits), you can derive the virtual address space size as \(2^n\). ### Example Calculation Let's take the case where \(n = 16\) (16-bit virtual address size) and the page size \(P = 4K\): 1. **Virtual Address Space Size Calculation**: \[ \text{Virtual Address Space Size} = 2^{16} = 65536 \text{ bytes} \] 2. **Page Size Calculation**: \[ P = 4K = 4096 \text{ bytes} \] 3. **Number of PTEs Calculation**: \[ \text{Number of PTEs} = \frac{65536}{4096} = 16 \] Likewise, you can apply the same steps to compute the number of PTEs for the remaining combinations: - \(n = 16\), \(P = 8K\) - \(n = 32\), \(P = 4K\) - \(n = 32\), \(P = 8K\) Fill in the table with the computed values to complete the analysis.
Transcribed Image Text:### Calculating the Number of Page Table Entries (PTEs) Determine the number of page table entries (PTEs) that are needed for the following combinations of virtual address size (n) and page size (P): | Virtual Address Size (n) | Page Size (P = 2^P) | Number of PTEs | |--------------------------|---------------------|----------------| | 16 | 4K | | | 16 | 8K | | | 32 | 4K | | | 32 | 8K | | To calculate the number of page table entries, you need to use the formula: \[ \text{Number of PTEs} = \frac{\text{Virtual Address Space Size}}{\text{Page Size}} \] For a given virtual address space size \(n\) (in bits), you can derive the virtual address space size as \(2^n\). ### Example Calculation Let's take the case where \(n = 16\) (16-bit virtual address size) and the page size \(P = 4K\): 1. **Virtual Address Space Size Calculation**: \[ \text{Virtual Address Space Size} = 2^{16} = 65536 \text{ bytes} \] 2. **Page Size Calculation**: \[ P = 4K = 4096 \text{ bytes} \] 3. **Number of PTEs Calculation**: \[ \text{Number of PTEs} = \frac{65536}{4096} = 16 \] Likewise, you can apply the same steps to compute the number of PTEs for the remaining combinations: - \(n = 16\), \(P = 8K\) - \(n = 32\), \(P = 4K\) - \(n = 32\), \(P = 8K\) Fill in the table with the computed values to complete the analysis.
## Part C **Task:** Determine the number of page table entries (PTEs) in the third line. **Instructions:** Express your answer as an integer. **Example Response:** 8192 **Feedback:** Incorrect; Try Again --- ## Part D **Task:** Determine the number of page table entries (PTEs) in the fourth line. **Instructions:** Express your answer as an integer. **Example Response:** 8192 **Feedback:** Incorrect; Try Again
Transcribed Image Text:## Part C **Task:** Determine the number of page table entries (PTEs) in the third line. **Instructions:** Express your answer as an integer. **Example Response:** 8192 **Feedback:** Incorrect; Try Again --- ## Part D **Task:** Determine the number of page table entries (PTEs) in the fourth line. **Instructions:** Express your answer as an integer. **Example Response:** 8192 **Feedback:** Incorrect; Try Again
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