COSC 3330/6308Solutions to theThird Problem Set

Jehan-François PârisNovember 2012

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.

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)

Answer

For the direct mapping organization Entry tag is page number Page number is 64 – log2 4096 = 52 bits

Last log2 64 = 6 bits of tag are given by the

entry index Minimum size of tag is 52 – 6 = 46 bits

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 log2 32 = 5 bits of tag are given by the

entry index Minimum size of tag is 52 – 5 = 47 bits

Part B

Describe at the bit level how the hardware should access each of these caches?

Answer

For the direct mapping organizationTLB has 64 linesUse log2 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

Answer

For the two-way associative organizationTLB has 32 lines and 2 entries per lineUse log2 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

Part C

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

Answer

For the direct mapping organizationEntry that was not in the TLB replaces old

TLB entry whose page number had samelog2 64 = 6 least significant bits

Answer

For the two-way set associative organizationEntry that was not in the TLB will replace one on

the two old TLB entries whose page number had same log2 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

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)

Answer

Only one percent of misses are associativity misses

Should use simpler, faster direct mapping organization

Second problem

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

Part A

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

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

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)

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

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?

Simplest Answer

Probability a given disk fails during the five years5 0.05 = 0.25

Probability a given disk fails and it leads to a data loss0.25 6.85 10-5 = 1.71 10-5

Probability any of the 200 disk fails and it leads to a data loss1.71 10-5 200 = 3.42 10-3

More rigorous answer (I)

Probability a given disk fails during the five years5 0.05 = 0.25

Probability a given disk fails and it leads to a data loss0.25 6.85 10-5 = 1.71 10-5

Probability a pair experiences a data loss2 1.71 10-5 = 3.42 10-5

Probability a pair experiences no data loss1 - 3.42 10-5 = 0.999965753

More rigorous answer (II)

Probability array experiences no data loss0.999965753100 = 0.996581141

Probability of a data loss1 – 0.996581141 = 3.42 10-3

How is that possible?

For very small values of x

(1 – x)n 1 – nx

Third problem

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

Part A

What is the mean time between failures of this server? (5 points)

Answer

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

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)

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 = 0.995MTTF = 0.95 daysMTTR = 0.05 days = 1.2 hours

Fourth problem

A RAID level 6 array has Eight data blocks (b0 to b7)

Two parity blocks p and q

per stripe.

Part A

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

Answer

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

8/(8 + 2) = 0.8 or 80 percent

Part B

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

Answer

What is the best way to update block b5 and its two

parity blocks? (10 points) Read old block b5, old parity block p and

old parity block q Compute

new p = old p old b5 new d

new q = old q old b5 new d

Write new b5, new p and new q

Give full credit to people who do not have the formulas

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)

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

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)

Answer

The question is whether it would be really help Must know the

Peak Memory BW Arithmetic Intensity

of program If Peak Memory BWArithmetic Intensity is

less than the Peak Floating-Point Performanceupgrading the CPU will not help

Another very good answer

If the CPU has multiple cores,check whether the program can use the multiple cores in parallel