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The address sequence generated by tracing a particular program executing in a pure demand paging system with 100 bytes per page is
0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0410.
Suppose that the memory can store only one page and if x is the address which causes a page fault then the bytes from addresses x to x + 99 are loaded on to the memory.
0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0410.
Suppose that the memory can store only one page and if x is the address which causes a page fault then the bytes from addresses x to x + 99 are loaded on to the memory.
Q. How many page faults will occur ?
- a)0
- b)4
- c)7
- d)8
Correct answer is option 'C'. Can you explain this answer?
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The address sequence generated by tracing a particular program executi...
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Given:
- Demand paging system with 100 bytes per page
- Memory can store only one page
To find:
- Number of page faults that will occur
Address Sequence:
0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0410
Solution:
To determine the number of page faults, we need to keep track of the pages currently in memory. Let's analyze the given address sequence and identify the page faults.
Address 0100:
- Since the memory is initially empty, a page fault occurs and the page containing addresses 0100-0199 is loaded into memory.
Address 0200:
- The page containing addresses 0200-0299 is not in memory, so another page fault occurs and this page is loaded.
Address 0430:
- The page containing addresses 0400-0499 is not in memory, so a page fault occurs and this page is loaded.
Address 0499:
- The page containing addresses 0400-0499 is already in memory, so no page fault occurs.
Address 0510:
- The page containing addresses 0500-0599 is not in memory, so a page fault occurs and this page is loaded.
Address 0530:
- The page containing addresses 0500-0599 is already in memory, so no page fault occurs.
Address 0560:
- The page containing addresses 0500-0599 is already in memory, so no page fault occurs.
Address 0120:
- The page containing addresses 0100-0199 is already in memory, so no page fault occurs.
Address 0220:
- The page containing addresses 0200-0299 is already in memory, so no page fault occurs.
Address 0240:
- The page containing addresses 0200-0299 is already in memory, so no page fault occurs.
Address 0260:
- The page containing addresses 0200-0299 is already in memory, so no page fault occurs.
Address 0320:
- The page containing addresses 0300-0399 is not in memory, so a page fault occurs and this page is loaded.
Address 0410:
- The page containing addresses 0400-0499 is already in memory, so no page fault occurs.
Summary:
In total, there are 7 page faults that occur while executing the given program in the pure demand paging system. Therefore, the correct answer is option C.
- Demand paging system with 100 bytes per page
- Memory can store only one page
To find:
- Number of page faults that will occur
Address Sequence:
0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0410
Solution:
To determine the number of page faults, we need to keep track of the pages currently in memory. Let's analyze the given address sequence and identify the page faults.
Address 0100:
- Since the memory is initially empty, a page fault occurs and the page containing addresses 0100-0199 is loaded into memory.
Address 0200:
- The page containing addresses 0200-0299 is not in memory, so another page fault occurs and this page is loaded.
Address 0430:
- The page containing addresses 0400-0499 is not in memory, so a page fault occurs and this page is loaded.
Address 0499:
- The page containing addresses 0400-0499 is already in memory, so no page fault occurs.
Address 0510:
- The page containing addresses 0500-0599 is not in memory, so a page fault occurs and this page is loaded.
Address 0530:
- The page containing addresses 0500-0599 is already in memory, so no page fault occurs.
Address 0560:
- The page containing addresses 0500-0599 is already in memory, so no page fault occurs.
Address 0120:
- The page containing addresses 0100-0199 is already in memory, so no page fault occurs.
Address 0220:
- The page containing addresses 0200-0299 is already in memory, so no page fault occurs.
Address 0240:
- The page containing addresses 0200-0299 is already in memory, so no page fault occurs.
Address 0260:
- The page containing addresses 0200-0299 is already in memory, so no page fault occurs.
Address 0320:
- The page containing addresses 0300-0399 is not in memory, so a page fault occurs and this page is loaded.
Address 0410:
- The page containing addresses 0400-0499 is already in memory, so no page fault occurs.
Summary:
In total, there are 7 page faults that occur while executing the given program in the pure demand paging system. Therefore, the correct answer is option C.
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The address sequence generated by tracing a particular program executing in a pure demand paging system with 100 bytes per page is0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0410.Suppose that the memory can store only one page and if x is the address which causes a page fault then the bytes from addresses x to x + 99 are loaded on to the memory.Q. How many page faults will occur ?a)0b)4c)7d)8Correct answer is option 'C'. Can you explain this answer?
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The address sequence generated by tracing a particular program executing in a pure demand paging system with 100 bytes per page is0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0410.Suppose that the memory can store only one page and if x is the address which causes a page fault then the bytes from addresses x to x + 99 are loaded on to the memory.Q. How many page faults will occur ?a)0b)4c)7d)8Correct answer is option 'C'. Can you explain this answer? for Computer Science Engineering (CSE) 2025 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about The address sequence generated by tracing a particular program executing in a pure demand paging system with 100 bytes per page is0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0410.Suppose that the memory can store only one page and if x is the address which causes a page fault then the bytes from addresses x to x + 99 are loaded on to the memory.Q. How many page faults will occur ?a)0b)4c)7d)8Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The address sequence generated by tracing a particular program executing in a pure demand paging system with 100 bytes per page is0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0410.Suppose that the memory can store only one page and if x is the address which causes a page fault then the bytes from addresses x to x + 99 are loaded on to the memory.Q. How many page faults will occur ?a)0b)4c)7d)8Correct answer is option 'C'. Can you explain this answer?.
The address sequence generated by tracing a particular program executing in a pure demand paging system with 100 bytes per page is0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0410.Suppose that the memory can store only one page and if x is the address which causes a page fault then the bytes from addresses x to x + 99 are loaded on to the memory.Q. How many page faults will occur ?a)0b)4c)7d)8Correct answer is option 'C'. Can you explain this answer? for Computer Science Engineering (CSE) 2025 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about The address sequence generated by tracing a particular program executing in a pure demand paging system with 100 bytes per page is0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0410.Suppose that the memory can store only one page and if x is the address which causes a page fault then the bytes from addresses x to x + 99 are loaded on to the memory.Q. How many page faults will occur ?a)0b)4c)7d)8Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The address sequence generated by tracing a particular program executing in a pure demand paging system with 100 bytes per page is0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0410.Suppose that the memory can store only one page and if x is the address which causes a page fault then the bytes from addresses x to x + 99 are loaded on to the memory.Q. How many page faults will occur ?a)0b)4c)7d)8Correct answer is option 'C'. Can you explain this answer?.
Solutions for The address sequence generated by tracing a particular program executing in a pure demand paging system with 100 bytes per page is0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0410.Suppose that the memory can store only one page and if x is the address which causes a page fault then the bytes from addresses x to x + 99 are loaded on to the memory.Q. How many page faults will occur ?a)0b)4c)7d)8Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Computer Science Engineering (CSE).
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