1. COSC 3330/6308 Solutions to the Third Problem Set Jehan-François Pâris November 2012

2. First problem You are to design a TLB for a virtual memory system with 64-bit addresses and 4-kilobyte pages. It should have with 64 entries with each entry mapping one virtual page into one physical page. You are to consider two possible TLB organizations, namely One using direct mapping Another using two-way set associativity.

3. Part A What would be the minimum sizes in bits of the entry tags for each of these two cache organizations? (2×5 easy points)

4. Answer  For the direct mapping organization Entry tag is page number Page number is 64 – log 2 4096 = 52 bits Last log 2 64 = 6 bits of tag are given by the entry index Minimum size of tag is 52 – 6 = 46 bits

5. Answer  For the two-way set associative organization Entry tag is page number TLB has 32 lines each with two entries Page number is still 52 bits Last log 2 32 = 5 bits of tag are given by the entry index Minimum size of tag is 52 – 5 = 47 bits

6. Part B Describe at the bit level how the hardware should access each of these caches?

7. Answer For the direct mapping organization  TLB has 64 lines  Use log 2 64 = 6 least significant bits of page number as index  If page number matches the tag : Get page frame number and bits else : Declare a TLB miss

8. Answer For the two-way associative organization  TLB has 32 lines and 2 entries per line  Use log 2 32 = 5 least significant bits of page number as index  If page number matches tag of one of the two entries : Get page frame number and bits else : Declare a TLB miss

9. Part C How will each of the two TLBs handle collisions? (2 + 8 points for a detailed solution)

10. Answer For the direct mapping organization  Entry that was not in the TLB replaces old TLB entry whose page number had same log 2 64 = 6 least significant bits

11. Answer For the two-way set associative organization  Entry that was not in the TLB will replace one on the two old TLB entries whose page number had same log 2 32 = 5 least significant bits  Should select Least Recently Used entry  Will add to cache an extra bit set to Zero when first entry of line is accessed One when second entry is accessed

12. Part D Measurements on the direct mapping cache indicate that  80 percent of the misses are compulsory misses  19 percent of them are capacity misses  Remaining misses are all associativity misses. What conclusions can you draw? (5 points)

13. Answer  Only one percent of misses are associativity misses  Should use simpler, faster direct mapping organization

14. Second problem A given make of disks has a failure rate of 5 percent per year.

15. Part A What is the mean time to failure of these disks? (5 easy points)

16. Answer What is the mean time to failure of these disks? (5 easy points)  Failure rate is 5 percent per year  One failure every 20 years  MTTF is 20 years

17. Part B Assuming that all data are mirrored on two disks, and that it takes twelve hours to replace a failed disk and restore its contents, what is the probability that a single disk failure will lead to a data loss? (5 points)

18. Answer Assuming that all data are mirrored on two disks, and that it takes twelve hours to replace a failed disk and restore its contents, what is the probability that a single disk failure will lead to a data loss? (5 points)  Probability of second failure while first disk gets repaired is ½  1/365  0.05 = 6.85 10 -5

19. Part C Consider a disk farm consisting of 100 pairs of mirrored disks for a total of 200 disks. What would be the probability of a data loss over a period of five years?

20. Simplest Answer Probability a given disk fails during the five years 5  0.05 = 0.25 Probability a given disk fails and it leads to a data loss 0.25  6.85 10 -5 = 1.71 10 -5 Probability any of the 200 disk fails and it leads to a data loss 1.71 10 -5  200 = 3.42 10 -3

21. More rigorous answer (I) Probability a given disk fails during the five years 5  0.05 = 0.25 Probability a given disk fails and it leads to a data loss 0.25  6.85 10 -5 = 1.71 10 -5 Probability a pair experiences a data loss 2  1.71 10 -5 = 3.42 10 -5 Probability a pair experiences no data loss 1 - 3.42 10 -5 = 0.999965753

22. More rigorous answer (II) Probability array experiences no data loss 0.999965753 100 = 0.996581141 Probability of a data loss 1 – 0.996581141 = 3.42 10 -3

23. How is that possible? For very small values of x (1 – x) n  1 – nx

24. Third problem A file server crashes on the average once every ten days.

25. Part A What is the mean time between failures of this server? (5 points)

26. Answer What is the mean time between failures of this server? (5 points)  One failure every ten days  MTBF is ten days

27. Part B What is the maximum mean time to repair that we can tolerate if we want to achieve an average availability of 99.5 percent? (5 points)

28. Answer What is the maximum mean time to repair that we can tolerate if we want to achieve an average availability of 99.5 percent? (5 points)  We have MTTF + MTTR = 10 days MTTF/(MTTF + MTTR) = 0.995  MTTF/10 = 0.995 MTTF = 0.95 days MTTR = 0.05 days = 1.2 hours

29. Fourth problem A RAID level 6 array has  Eight data blocks (b 0 to b 7 )  Two parity blocks p and q per stripe.

30. Part A How much of the total disk space is used by data blocks? (5 easy points)

31. Answer How much of the total disk space is used by data blocks? (5 easy points)  8/(8 + 2) = 0.8 or 80 percent

32. Part B What is the best way to update block b5 and its two parity blocks? (10 points)

33. Answer What is the best way to update block b 5 and its two parity blocks? (10 points) Read old block b 5, old parity block p and old parity block q Compute new p = old p  old b 5  new d new q = old q  old b 5  new d Write new b 5, new p and new q Give full credit to people who do not have the formulas

34. Fifth problem A task is distributed along 128 processors but the maximum observed speedup is only 50. People suspect that this low figure is due to a single processor. What is the share of the workload that is executed by that processor? (5 points)

35. Answer A task is distributed along 128 processors but the maximum observed speedup is only 50. People suspect that this low figure is due to a single processor. What is the share of the workload that is executed by that processor? (5 points)  1/50 of the workload

36. Sixth problem People in your research group are unhappy with the speed of the computer they use to run a huge program. Somebody suggests upgrading the CPU of the computer. What do we need to know before deciding to do so? (5 points)

37. Answer The question is whether it would be really help Must know the  Peak Memory BW  Arithmetic Intensity of program If Peak Memory BW  Arithmetic Intensity is less than the Peak Floating-Point Performance upgrading the CPU will not help

38. Another very good answer If the CPU has multiple cores, check whether the program can use the multiple cores in parallel