Consider a segmented paging system with the segment table andthe page table with
ID: 3617847 • Letter: C
Question
Consider a segmented paging system with the segment table andthe page table within a segment stored in memory. First part of theaddress is segment number, second part is the page number within asegment and the third part is the offset within a page. [40points]
a) If a memory reference takes 200nano seconds, how long does a segmented page memory referencetake?
b) If we add Translation Lookaside Buffer (TLB), and (i)90%, (ii) 80%, and (iii) 75% of all memory references are found inthe TLB, what is the effective memory reference time in each case?(Assume that finding a segmented page-table entry in the TLBs takeszero time, if the entry is there. Otherwise memory tables must besearched to determine the physical address. )
Explanation / Answer
http://codex.cs.yale.edu/avi/os-book/OS8/os8c/index.html Slides: http://codex.cs.yale.edu/avi/os-book/OS8/os8c/slide-dir/index.htmlb) Your teacher is making it hard. He is asking for EAT(effective access time). The name is a little confusing like that.Use the slides above and these formulas.
EAT =(1 – a) x memory access time + a ( 2*memoryaccess time + page fault overhead + swap page out + swap pagein + restart overhead) Page Fault (PF) Rate, a: 0<= a <=1
Use these formulas, but if you don't have one of these in thequestion, ignore it. Some things are not always needed. Swappingpages refers to reading/writing from/to the disk.
Those equations are from memory, but I just remembered: eat=(1-a)mem + a(2mem + pf +po +pi +rs)=mem+a(mem + pf+po +pi +rs) But, mem should be the memory access time + something else (Ithink access to the TLB aka cache) that you ALWAYS have as a cost.Anyways, you will find the formula in the slides just likethat.
To further explain it: You first check the TLB to see if it(the address) was recently used and then either read it in memoryor search for it in memory (sometimes a teacher/book will ignorethis part, so you can too). If it still is not in memory, find iton the disk, write out a page to make room (when needed), write inthe new page, then restart the failed instruction and check memoryagain. I think it is without checking the TLB again, but notsure.
b) Your teacher is making it hard. He is asking for EAT(effective access time). The name is a little confusing like that.Use the slides above and these formulas.
EAT =(1 – a) x memory access time + a ( 2*memoryaccess time + page fault overhead + swap page out + swap pagein + restart overhead) Page Fault (PF) Rate, a: 0<= a <=1
Use these formulas, but if you don't have one of these in thequestion, ignore it. Some things are not always needed. Swappingpages refers to reading/writing from/to the disk.
Those equations are from memory, but I just remembered: eat=(1-a)mem + a(2mem + pf +po +pi +rs)=mem+a(mem + pf+po +pi +rs) But, mem should be the memory access time + something else (Ithink access to the TLB aka cache) that you ALWAYS have as a cost.Anyways, you will find the formula in the slides just likethat.
To further explain it: You first check the TLB to see if it(the address) was recently used and then either read it in memoryor search for it in memory (sometimes a teacher/book will ignorethis part, so you can too). If it still is not in memory, find iton the disk, write out a page to make room (when needed), write inthe new page, then restart the failed instruction and check memoryagain. I think it is without checking the TLB again, but notsure.
EAT =(1 – a) x memory access time + a ( 2*memoryaccess time + page fault overhead + swap page out + swap pagein + restart overhead) Page Fault (PF) Rate, a: 0<= a <=1
Use these formulas, but if you don't have one of these in thequestion, ignore it. Some things are not always needed. Swappingpages refers to reading/writing from/to the disk.
Those equations are from memory, but I just remembered: eat=(1-a)mem + a(2mem + pf +po +pi +rs)=mem+a(mem + pf+po +pi +rs) But, mem should be the memory access time + something else (Ithink access to the TLB aka cache) that you ALWAYS have as a cost.Anyways, you will find the formula in the slides just likethat.
To further explain it: You first check the TLB to see if it(the address) was recently used and then either read it in memoryor search for it in memory (sometimes a teacher/book will ignorethis part, so you can too). If it still is not in memory, find iton the disk, write out a page to make room (when needed), write inthe new page, then restart the failed instruction and check memoryagain. I think it is without checking the TLB again, but notsure. EAT =(1 – a) x memory access time + a ( 2*memoryaccess time + page fault overhead + swap page out + swap pagein + restart overhead) Page Fault (PF) Rate, a: 0<= a <=1
Use these formulas, but if you don't have one of these in thequestion, ignore it. Some things are not always needed. Swappingpages refers to reading/writing from/to the disk.
Those equations are from memory, but I just remembered: eat=(1-a)mem + a(2mem + pf +po +pi +rs)=mem+a(mem + pf+po +pi +rs) But, mem should be the memory access time + something else (Ithink access to the TLB aka cache) that you ALWAYS have as a cost.Anyways, you will find the formula in the slides just likethat.
To further explain it: You first check the TLB to see if it(the address) was recently used and then either read it in memoryor search for it in memory (sometimes a teacher/book will ignorethis part, so you can too). If it still is not in memory, find iton the disk, write out a page to make room (when needed), write inthe new page, then restart the failed instruction and check memoryagain. I think it is without checking the TLB again, but notsure.