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Homework solutionsch8

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1) The document contains solutions to chapter 8 homework problems involving page tables, page replacement policies, working set size, and memory management with both paging and segmentation. 2) Key concepts covered include virtual to physical address translation using page tables, page replacement algorithms like FIFO and LRU, calculating memory access rates, and addressing with multiple levels of page tables and segmentation. 3) Memory management techniques like page tables, page replacement, and multi-level addressing schemes are analyzed to solve problems involving virtual memory configuration, address translation, and modeling memory access patterns.

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Chapter 8 Homework Solutions 8.1, 8.3, 8.4, 8.6, 8.10, 8.15, 8.17
Problem 8.1 • Page table with page size 1024 a. Describe how, in general, a virtual address generated by the CPU is translated into a physical main memory address. • Split the binary address into a virtual page number (VPN) and an offset • use VPN as an index into the page table • extract the page frame number from the page table entry • concatenate the page frame number with the offset to get physical memory address Virtual Page Number Valid Bit Reference Bit Modify Bit Page Frame Number 0 1 1 0 4 1 1 1 1 7 2 0 0 0 − 3 1 0 0 2 4 0 0 0 − 5 1 0 1 0
Problem 8.1 b. What physical address, if any, correspond to the virtual address: (i) 1052 1052 = 1024+ 28 = 00..01 0000011100 so VPN = 1 and offset = 28 VPN 1 → Frame number 7, so the physical address = 7*1024+28 = 7196 (ii) 2221 2221 = 2 × 1024 + 173, so the VPN = 2, page fault (iii) 5499 5499 = 5 × 1024 + 379 VPN 5 → PFN 0, so physical address = 0 × 1024+379 = 379) Virtual Page Number Valid Bit Reference Bit Modify Bit Page Frame Number 0 1 1 0 4 1 1 1 1 7 2 0 0 0 − 3 1 0 0 2 4 0 0 0 − 5 1 0 1 0
Problem 8.3 a. How much memory space is needed for the user page table of Fig. 8.4? 4 Mbytes Extra credit question: Who is buried in Grant’s tomb?
Problem 8.3 b. Hashed inverted table for above, 20-bit PN → 6-bit hash number. Entry = page number, frame number, chain pointer; 3 overflow entries per table entry. What is the inverted page table size? • Number of rows: 26+3 =256 entries. • Each entry consist of: 20 (page number) + 20 (frame number) + 8 bits (chain index) = 48 bits = 6 bytes. • Total: 256 × 6= 1536 bytes • Alternate possibility: 3 overflow includes initial entry → 128 × 6 = 768
Problem 8.4 • Time is number of ticks since process load time. • Page fault has occurred at time 164. Which page is replaced and why: a. FIFO Page Frame Number (PFN) 3 since it was loaded longest ago at time 20 b. LRU PFN 1 since it was referenced longest ago at time 160 c. Clock Clear R in PFN 3 (oldest loaded), clear R in PFN 2 (next oldest loaded), victim PFN is 0 since R=0 Virtual Page Number Page Frame Time Loaded Time Reference d R Bit M Bit 2 0 60 161 0 1 1 1 130 160 1 0 0 2 26 162 1 0 3 3 20 163 1 1
Problem 8.4 • Time is number of ticks since process load time. • Page fault has occurred at time 164. Which page is replaced and why: d. Optimal, assuming page reference string 4,0,0,0,2,4,2,1,0,3,2 Replace the page in PFN 3 since VPN 3 (in PFN 3) is used furthest in the future Virtual Page Number Page Frame Time Loaded Time Reference d R Bit M Bit 2 0 60 161 0 1 1 1 130 160 1 0 0 2 26 162 1 0 3 3 20 163 1 1
Problem 8.4 e. Given above (just before page fault), assume page reference string 4,0,0,0,2,4,2,1,0,3,2. How many page faults if working set policy with LRU were used with a window size 4 and a fixed allocation? There are 6 faults, indicated by * Virtual Page Number Page Frame Time Loaded Time Reference d R Bit M Bit 2 0 60 161 0 1 1 1 130 160 1 0 0 2 26 162 1 0 3 3 20 163 1 1
Problem 8.6 Process with 8 pages of VM, fixed allocation of 4 page frames. Page trace: 1,0,2,2,1,7,6,7,0,1,2,0,3,0,4,5,1,5,2,4,5,6,7,6,72,4,2,7,3,3,2,3 a. Show the successive pages residing in the four frames using the LRU replacement policy. Compute the hit ratio in main memory. Assume the frames are initially empty. Hit ratio = 16/33 1 0 2 2 1 7 6 7 0 1 2 0 3 0 4 5 1 5 2 4 5 6 7 6 7 2 4 2 7 3 3 2 3 PF1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 4 4 4 4 4 4 4 4 4 4 2 2 2 2 2 2 2 2 PF2 - 0 0 0 0 0 6 6 6 6 2 2 2 2 2 5 5 5 5 5 5 5 5 5 5 5 4 4 4 4 4 4 4 PF3 - - 2 2 2 2 2 2 0 0 0 0 0 0 0 0 0 0 2 2 2 2 7 7 7 7 7 7 7 7 7 7 7 PF4 - - - - - 7 7 7 7 7 7 7 3 3 3 3 1 1 1 1 1 6 6 6 6 6 6 6 6 3 3 3 3 F F F F F F F F F F F F F F F F F
Problem 8.6 Process with 8 pages of VM, fixed allocation of 4 page frames. Page trace: 1,0,2,2,1,7,6,7,0,1,2,0,3,0,4,5,1,5,2,4,5,6,7,6,72,4,2,7,3,3,2,3 a. Show the successive pages residing in the four frames using the FIFO replacement policy. Compute the hit ratio in main memory. Assume the frames are initially empty. Hit ratio = 16/33 1 0 2 2 1 7 6 7 0 1 2 0 3 0 4 5 1 5 2 4 5 6 7 6 7 2 4 2 7 3 3 2 3 PF1 1 1 1 1 1 1 6 6 6 6 6 6 6 6 4 4 4 4 4 4 4 6 6 6 6 6 6 6 6 6 6 2 2 PF2 - 0 0 0 0 0 0 0 0 1 1 1 1 1 1 5 5 5 5 5 5 5 7 7 7 7 7 7 7 7 7 7 7 PF3 - - 2 2 2 2 2 2 2 2 2 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 4 4 4 4 4 4 4 PF4 - - - - - 7 7 7 7 7 7 7 3 3 3 3 3 3 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 F F F F F F F F F F F F F F F F c. Compare the two policy’s effectiveness with respect to this trace. The two policies are equally effective for this particular page trace.
Problem 8.10 • Page size = 4 Kbytes; page table entry = 4 bytes; 64-bit address space; top level page table fits into a single page. How many levels of page tables are needed? • A one-page page table would point to 4K/4 = 1024 = 210 pages, addressing a total of 210 × 212 = 222 bytes. • The address space however is 264 bytes. • Adding a second layer of page tables, the top page table would point to 210 page tables, addressing a total of 232 bytes. • Continuing this process, we can see that 5 levels do not address the full 64 bit address space, so a 6th level is required. • But only 2 bits of the 6th level are required, not the entire 10 bits. • So instead of requiring your virtual addresses be 72 bits long, you could mask out and ignore all but the 2 lowest order bits of the 6th level. • This would give you a 64 bit address. • Your top level page table then would have only 4 entries. • Yet another option is to revise the criteria that the top level page table fit into a single physical page and instead make it fit into 4 pages. • This would save a physical page, which is not much.
Problem 8.15 • Define the mean working set size after the kth reference as and define the missing page probability after the kth reference as where F(t,∆) = 1 if a page fault occurs at virtual time t and is 0 otherwise. • Consider the following page reference sequence: 1 2 3 4 5 2 1 3 3 2 3 4 5 4 5 1 1 3 2 5 a. Draw a diagram similar to that of Figure 8.19 for the reference sequence for the values ∆ = 1,2,3,4,5,6. |),(| 1 )( 1 ∆=∆ ∑= k t k tW k s ),( 1 )( 1 ∆=∆ ∑= k t k tF k m
Problem 8.15
Problem 8.15 • Define the mean working set size after the kth reference as and define the missing page probability after the kth reference as where F(t,∆) = 1 if a page fault occurs at virtual time t and is 0 otherwise. • Consider the following page reference sequence: 1 2 3 4 5 2 1 3 3 2 3 4 5 4 5 1 1 3 2 5 b. Plot s20(∆) as a function of ∆ c. Plot w20(∆) as a function of ∆ |),(| 1 )( 1 ∆=∆ ∑= k t k tW k s ),( 1 )( 1 ∆=∆ ∑= k t k tF k m s20(Δ) is an increasing function of Δ. m20(Δ) is a nonincreasing function of Δ.
Problem 8.17 • A task is divided into four equal sized segments and the system has built an eight-entry page descriptor table for each segment. Thus the system has a combination of paging and segmentation. The page size is 2 Kbytes. a. What is the maximum size of each segment? 8 × 2K = 16K b. What is the maximum logical address space for the task? 16K × 4 = 64K c. What is the maximum physical address space for the system? 232 = 4 GBytes Assume that an element in physical location 00021ABC is accessed by this task. What is the format of the logical address that the task generates for it?
Problem 8.17 • A task is divided into four equal sized segments and the system has built an eight-entry page descriptor table for each segment. Thus the system has a combination of paging and segmentation. The page size is 2 Kbytes. c. Assume that an element in physical location 00021ABC is accessed by this task. What is the format of the logical address that the task generates for it? Segment: 0 1 2 3 0 7 Page descriptor table Main memory (232 bytes) 00021ABC
Problem 8.17 • A task is divided into four equal sized segments and the system has built an eight-entry page descriptor table for each segment. Thus the system has a combination of paging and segmentation. The page size is 2 Kbytes. c. Assume that an element in physical location 00021ABC is accessed by this task. What is the format of the logical address that the task generates for it? (2) (3) (11) Segment Page Offset X Y 2BC

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Homework solutionsch8

  • 1. Chapter 8 Homework Solutions 8.1, 8.3, 8.4, 8.6, 8.10, 8.15, 8.17
  • 2. Problem 8.1 • Page table with page size 1024 a. Describe how, in general, a virtual address generated by the CPU is translated into a physical main memory address. • Split the binary address into a virtual page number (VPN) and an offset • use VPN as an index into the page table • extract the page frame number from the page table entry • concatenate the page frame number with the offset to get physical memory address Virtual Page Number Valid Bit Reference Bit Modify Bit Page Frame Number 0 1 1 0 4 1 1 1 1 7 2 0 0 0 − 3 1 0 0 2 4 0 0 0 − 5 1 0 1 0
  • 3. Problem 8.1 b. What physical address, if any, correspond to the virtual address: (i) 1052 1052 = 1024+ 28 = 00..01 0000011100 so VPN = 1 and offset = 28 VPN 1 → Frame number 7, so the physical address = 7*1024+28 = 7196 (ii) 2221 2221 = 2 × 1024 + 173, so the VPN = 2, page fault (iii) 5499 5499 = 5 × 1024 + 379 VPN 5 → PFN 0, so physical address = 0 × 1024+379 = 379) Virtual Page Number Valid Bit Reference Bit Modify Bit Page Frame Number 0 1 1 0 4 1 1 1 1 7 2 0 0 0 − 3 1 0 0 2 4 0 0 0 − 5 1 0 1 0
  • 4. Problem 8.3 a. How much memory space is needed for the user page table of Fig. 8.4? 4 Mbytes Extra credit question: Who is buried in Grant’s tomb?
  • 5. Problem 8.3 b. Hashed inverted table for above, 20-bit PN → 6-bit hash number. Entry = page number, frame number, chain pointer; 3 overflow entries per table entry. What is the inverted page table size? • Number of rows: 26+3 =256 entries. • Each entry consist of: 20 (page number) + 20 (frame number) + 8 bits (chain index) = 48 bits = 6 bytes. • Total: 256 × 6= 1536 bytes • Alternate possibility: 3 overflow includes initial entry → 128 × 6 = 768
  • 6. Problem 8.4 • Time is number of ticks since process load time. • Page fault has occurred at time 164. Which page is replaced and why: a. FIFO Page Frame Number (PFN) 3 since it was loaded longest ago at time 20 b. LRU PFN 1 since it was referenced longest ago at time 160 c. Clock Clear R in PFN 3 (oldest loaded), clear R in PFN 2 (next oldest loaded), victim PFN is 0 since R=0 Virtual Page Number Page Frame Time Loaded Time Reference d R Bit M Bit 2 0 60 161 0 1 1 1 130 160 1 0 0 2 26 162 1 0 3 3 20 163 1 1
  • 7. Problem 8.4 • Time is number of ticks since process load time. • Page fault has occurred at time 164. Which page is replaced and why: d. Optimal, assuming page reference string 4,0,0,0,2,4,2,1,0,3,2 Replace the page in PFN 3 since VPN 3 (in PFN 3) is used furthest in the future Virtual Page Number Page Frame Time Loaded Time Reference d R Bit M Bit 2 0 60 161 0 1 1 1 130 160 1 0 0 2 26 162 1 0 3 3 20 163 1 1
  • 8. Problem 8.4 e. Given above (just before page fault), assume page reference string 4,0,0,0,2,4,2,1,0,3,2. How many page faults if working set policy with LRU were used with a window size 4 and a fixed allocation? There are 6 faults, indicated by * Virtual Page Number Page Frame Time Loaded Time Reference d R Bit M Bit 2 0 60 161 0 1 1 1 130 160 1 0 0 2 26 162 1 0 3 3 20 163 1 1
  • 9. Problem 8.6 Process with 8 pages of VM, fixed allocation of 4 page frames. Page trace: 1,0,2,2,1,7,6,7,0,1,2,0,3,0,4,5,1,5,2,4,5,6,7,6,72,4,2,7,3,3,2,3 a. Show the successive pages residing in the four frames using the LRU replacement policy. Compute the hit ratio in main memory. Assume the frames are initially empty. Hit ratio = 16/33 1 0 2 2 1 7 6 7 0 1 2 0 3 0 4 5 1 5 2 4 5 6 7 6 7 2 4 2 7 3 3 2 3 PF1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 4 4 4 4 4 4 4 4 4 4 2 2 2 2 2 2 2 2 PF2 - 0 0 0 0 0 6 6 6 6 2 2 2 2 2 5 5 5 5 5 5 5 5 5 5 5 4 4 4 4 4 4 4 PF3 - - 2 2 2 2 2 2 0 0 0 0 0 0 0 0 0 0 2 2 2 2 7 7 7 7 7 7 7 7 7 7 7 PF4 - - - - - 7 7 7 7 7 7 7 3 3 3 3 1 1 1 1 1 6 6 6 6 6 6 6 6 3 3 3 3 F F F F F F F F F F F F F F F F F
  • 10. Problem 8.6 Process with 8 pages of VM, fixed allocation of 4 page frames. Page trace: 1,0,2,2,1,7,6,7,0,1,2,0,3,0,4,5,1,5,2,4,5,6,7,6,72,4,2,7,3,3,2,3 a. Show the successive pages residing in the four frames using the FIFO replacement policy. Compute the hit ratio in main memory. Assume the frames are initially empty. Hit ratio = 16/33 1 0 2 2 1 7 6 7 0 1 2 0 3 0 4 5 1 5 2 4 5 6 7 6 7 2 4 2 7 3 3 2 3 PF1 1 1 1 1 1 1 6 6 6 6 6 6 6 6 4 4 4 4 4 4 4 6 6 6 6 6 6 6 6 6 6 2 2 PF2 - 0 0 0 0 0 0 0 0 1 1 1 1 1 1 5 5 5 5 5 5 5 7 7 7 7 7 7 7 7 7 7 7 PF3 - - 2 2 2 2 2 2 2 2 2 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 4 4 4 4 4 4 4 PF4 - - - - - 7 7 7 7 7 7 7 3 3 3 3 3 3 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 F F F F F F F F F F F F F F F F c. Compare the two policy’s effectiveness with respect to this trace. The two policies are equally effective for this particular page trace.
  • 11. Problem 8.10 • Page size = 4 Kbytes; page table entry = 4 bytes; 64-bit address space; top level page table fits into a single page. How many levels of page tables are needed? • A one-page page table would point to 4K/4 = 1024 = 210 pages, addressing a total of 210 × 212 = 222 bytes. • The address space however is 264 bytes. • Adding a second layer of page tables, the top page table would point to 210 page tables, addressing a total of 232 bytes. • Continuing this process, we can see that 5 levels do not address the full 64 bit address space, so a 6th level is required. • But only 2 bits of the 6th level are required, not the entire 10 bits. • So instead of requiring your virtual addresses be 72 bits long, you could mask out and ignore all but the 2 lowest order bits of the 6th level. • This would give you a 64 bit address. • Your top level page table then would have only 4 entries. • Yet another option is to revise the criteria that the top level page table fit into a single physical page and instead make it fit into 4 pages. • This would save a physical page, which is not much.
  • 12. Problem 8.15 • Define the mean working set size after the kth reference as and define the missing page probability after the kth reference as where F(t,∆) = 1 if a page fault occurs at virtual time t and is 0 otherwise. • Consider the following page reference sequence: 1 2 3 4 5 2 1 3 3 2 3 4 5 4 5 1 1 3 2 5 a. Draw a diagram similar to that of Figure 8.19 for the reference sequence for the values ∆ = 1,2,3,4,5,6. |),(| 1 )( 1 ∆=∆ ∑= k t k tW k s ),( 1 )( 1 ∆=∆ ∑= k t k tF k m
  • 14. Problem 8.15 • Define the mean working set size after the kth reference as and define the missing page probability after the kth reference as where F(t,∆) = 1 if a page fault occurs at virtual time t and is 0 otherwise. • Consider the following page reference sequence: 1 2 3 4 5 2 1 3 3 2 3 4 5 4 5 1 1 3 2 5 b. Plot s20(∆) as a function of ∆ c. Plot w20(∆) as a function of ∆ |),(| 1 )( 1 ∆=∆ ∑= k t k tW k s ),( 1 )( 1 ∆=∆ ∑= k t k tF k m s20(Δ) is an increasing function of Δ. m20(Δ) is a nonincreasing function of Δ.
  • 15. Problem 8.17 • A task is divided into four equal sized segments and the system has built an eight-entry page descriptor table for each segment. Thus the system has a combination of paging and segmentation. The page size is 2 Kbytes. a. What is the maximum size of each segment? 8 × 2K = 16K b. What is the maximum logical address space for the task? 16K × 4 = 64K c. What is the maximum physical address space for the system? 232 = 4 GBytes Assume that an element in physical location 00021ABC is accessed by this task. What is the format of the logical address that the task generates for it?
  • 16. Problem 8.17 • A task is divided into four equal sized segments and the system has built an eight-entry page descriptor table for each segment. Thus the system has a combination of paging and segmentation. The page size is 2 Kbytes. c. Assume that an element in physical location 00021ABC is accessed by this task. What is the format of the logical address that the task generates for it? Segment: 0 1 2 3 0 7 Page descriptor table Main memory (232 bytes) 00021ABC
  • 17. Problem 8.17 • A task is divided into four equal sized segments and the system has built an eight-entry page descriptor table for each segment. Thus the system has a combination of paging and segmentation. The page size is 2 Kbytes. c. Assume that an element in physical location 00021ABC is accessed by this task. What is the format of the logical address that the task generates for it? (2) (3) (11) Segment Page Offset X Y 2BC