Chapter 3 Question 4: Assume A Program Has 510 Bytes

Chapter 3 Question 4assume A Program Has 510 Bytes And Will Be Loa

Given a program that has a size of 510 bytes and is to be loaded into page frames of 256 bytes each, the task is to determine the number of pages needed to store the entire program and to compute the page number and displacement for each byte address where data is stored. Additionally, the assignment involves analyzing page replacement strategies under different memory constraints using the FIFO algorithm and evaluating the impact of increasing available page frames. Furthermore, it encompasses understanding page fault ratios and their implications, as well as scheduling algorithms like FCFS and SJN with specific job parameters. The questions include calculations of page mappings, fault ratios, scheduling order, and times.

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Understanding Memory Management and Paging

Memory management is a crucial aspect of operating systems, tasked with efficiently allocating, managing, and utilizing main memory to support process execution. One fundamental technique is paging, which divides both physical memory and processes into fixed-size blocks called pages and frames, respectively. This mechanism simplifies the allocation process, mitigates external fragmentation, and allows for more flexible memory usage.

Question 1: Calculating the Number of Pages and Address Mappings

The first question involves a program of 510 bytes, loaded into 256-byte page frames. To determine how many pages are necessary, we divide the total program size by the frame size and consider any remaining bytes. Specifically, 510 bytes divided by 256 bytes per page yields:

Number of pages = ceiling(510 / 256) = 3 pages

Thus, the program requires three pages for complete storage. To map each specific byte address to a page number and displacement within that page, a simple method involves dividing the byte address by the page size.

For example:

• Byte 0: Page number = 0, Displacement = 0
• Byte 377: Page number = 377 / 256 = 1 (integer division), Displacement = 377 % 256 = 121

This indicates that byte 377 is stored in page 1 at displacement 121. Repeating this calculation for each byte address helps map all data points systematically.

Question 2: Page Replacement Algorithms and Memory Constraints

Part A: Memory with 3 Page Frames

In a constrained environment with only three page frames, a sequence of page requests (a, c, a, b, a, d, a, c, b, d, e, f) is processed. Using the FIFO page replacement algorithm, we document page faults and the evolution of the frame content. Initially, frames are empty, and each page load is checked against current frames. When a page fault occurs and no empty frame is available, the oldest page (the first loaded) is replaced.

Tracking this sequence reveals a certain failure ratio, which reflects how often page faults happen versus hits. The FIFO policy ensures that the first page loaded is the first replaced, which can cause recurrent faults if the sequence revisits pages that are soon replaced.

Part B: Memory with 4 Page Frames

When the available memory expands to four frames, the same sequence is re-analyzed. With additional space, fewer page faults occur, improving the fault ratio. The FIFO algorithm still replaces the oldest page when needed but now benefits from the extra frame capacity, reducing the number of page replacements and increasing the stability of recently used pages.

Part C: General Observations

This illustrative example demonstrates that increasing available memory (number of page frames) generally decreases page fault rates for a given sequence of page requests. The relationship highlights the importance of adequate memory allocation in improving system performance, reducing costly page faults, and improving overall throughput. However, the benefit has limits, and beyond a certain point, additional frames yield diminishing returns.

Question 3: FIFO with Different Request Sequences

Part A: Limited Memory with 3 Frames

In this scenario, pages requested follow the sequence: a, c, b, d, a, c, e, a, c, b, d, e. Applying FIFO, each page fault is marked with an asterisk (*). During the process, pages are loaded into frames, and when needed, the oldest page is replaced. Faults occur when a page is not in the current frames.

For example, initial requests fill the frames, but subsequent requests for pages already loaded are hits, not faults. Calculating the overall fault and success ratios helps quantify the efficiency of FIFO under this sequence.

Part B: Memory with 4 Frames

Increasing memory to four page frames enhances the page retention capacity, leading to fewer faults. Repeating the analysis with the same sequence, the number of page faults decreases, reflecting improved performance. The ratios clearly demonstrate how additional memory reduces page faults, emphasizing the importance of appropriate resource allocation.

Part C: Broader Implication

This example underscores the impact of memory size on paging efficiency. Larger memories lower the page fault rate for typical workload sequences, leading to faster processing times and better system responsiveness. Nonetheless, optimal memory management must balance hardware costs and system performance needs.

Question 4: Scheduling Algorithms and Job Processing Times

Part A: FCFS Scheduling

Jobs arrive with estimated CPU cycles: A=12 ms, B=2 ms, C=15 ms, D=7 ms, E=3 ms. Assuming simultaneous arrival, the First-Come, First-Served (FCFS) scheduling processes jobs in the order received: A, B, C, D, E. The total time includes the sum of each job's burst times, with cumulative processing times for each job. Calculating total and average turnaround times offers insights into system responsiveness under FCFS.

Part B: SJN Scheduling

Shortest Job Next (SJN) selects the job with the smallest CPU time, reducing average waiting time. For this set, the processing order minimizes total completion time, improving efficiency. Calculations of total processing and average turnaround times illustrate performance gains over FCFS in low-priority or time-sensitive contexts.

Question 5: Scheduling with Multiple Jobs and Arrival Times

Part A: SJN with Detailed Timing

The jobs with arrival times and CPU cycles are processed to determine start and finish times for each. For example, with arrival times specified, the scheduler chooses the shortest available job at each step, accounting for job arrivals and remaining burst times. Faults or delays are identified by comparing scheduled start times and arrival times, assisting in understanding real-time scheduling complexities.

Part B and C: Extended Scheduling Analysis

Repeating the timing analysis with further extensions, increasing number of frames or considering different algorithms, reveals how scheduling strategies impact responsiveness and throughput. The precise start and finish times exemplify the benefits of SJN in reducing average wait times and optimizing resource use. Ultimately, drawing broader conclusions about scheduling efficacy aids in designing better operating system schedulers.

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