Assignment 11

Q2) A memory system has a memory access time of 250 nanoseconds and a page fault service time of 50 milliseconds.

b) If there are no page faults, the effective access time is equal to the memory access time. Suppose that even with page faults, we do not want the effective access time to degrade by more than 10 %. What should the page fault rate be to achieve this?

p < 25 / 49999750 p < 5 * 10-7

p * 49999750 < 25p * 49999750 < 275 – 250250+p * 49999750 < 275250 +p(– 250 + 50000000) < 275250 – 250 p + 50000000 p < 275

(1-p)*250+p*50000000 < 275(1-p) * memory access time + p*page fault service time < ( 250 ns + 25 ns )

We wantEAT < (memory access time + 10% of memory access time)

A page has 3 cases

Page in TLB

Page in PT&

v=1 (in Memory)

Page in PT&

v=0 (not in Memory)(page fault)

TLB hit ratio * (TLB lookup time + memory access time)

(1- TLB hit ratio – page fault rate) * (TLB lookup time + PT lookup time + memory access time)

page fault rate *(TLB lookup time + PT lookup time + swap page out time + swap page in time + memory access time)

+

+

EAT = Q3

Assignment 12

Block number = address/block size

Byte offset = address – address at the beginning of the block = address – (block number*block size)

Q2

Maximum number of bytes addressed by n direct pointers is= Number of direct pointers * BlockSize

Maximum number of bytes addressed by a single indirect pointer is= NumberOfEntriesInOneBlock * BlockSize= (BlockSize / PointerSize) * BlockSize

Maximum number of bytes addressed by a double indirect pointer is= NumberOfEntriesInOneBlock 2 * BlockSize= (BlockSize / PointerSize)2 * BlockSize

Maximum number of bytes addressed by a triple indirect pointer is= NumberOfEntriesInOneBlock 3 * BlockSize = (BlockSize / PointerSize)3 * BlockSize

K M G T 210 B = KB220 B = MB230 B = GB240 B = TB

Q3

1

2

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.13

14

15

16

direct

indirect

double

triple

i-node

Block = 8K

4 bytes

8KB/4B= 2K entry(pointer) = 2000 entry

8KB/4B= 2K entry(pointer)=2000 entry

8KB/4B= 2K entry(pointer)= 2000 entry

104KB

16MB

32 GB

64 TB

1

2

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.13

14

15

16

direct

indirect

double

triple

i-node

Block = 8K

4 bytes

104KB 13 * 8KB = 104 KB

1

2

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.

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.13

14

15

16

direct

indirect

double

triple

i-node

Block = 8K

4 bytes

8KB/4B= 2K entry(pointer) = 2000 entry

16MB

2000 * 8 KB = 16000 KB covert to MB= 16 MB

1

2

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.

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.13

14

15

16

direct

indirect

double

triple

i-node

Block = 8K

4 bytes 8KB/4B= 2K entry(pointer)=2000 entry

32 GB

2000 2000*2000 * 8Kentry entry 4,000,000 * 8K 32,000,000 KB convert to GB by /1,000,000 32 GB

1

2

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.

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.13

14

15

16

direct

indirect

double

triple

i-node

Block = 8K

4 bytes

8KB/4B= 2K entry(pointer)= 2000 entry

64 TB

2000 2000*2000entry entry = 4,000,0000 4,000,000*2000 * 8K entry entry 8,000,000,000 * 8K = 64,000,000,000 KB convert to TB by /1000,000,000 = 64 TB

Maximum number of bytes addressed by n direct pointers is= Number of direct pointers * BlockSize

Maximum number of bytes addressed by a single indirect pointer is= NumberOfEntriesInOneBlock * BlockSize= (BlockSize / PointerSize) * BlockSize

K M G T 210 B = KB220 B = MB230 B = GB240 B = TB

= 13* 8 KB

= 104 KB

= (8 KB / 4 B )* 8 KB= 2 K * 8 KB

= 2000 * 8 KB= 16000 KB= 16 MB

Maximum number of bytes addressed by a double indirect pointer is= NumberOfEntriesInOneBlock 2 * BlockSize= (BlockSize / PointerSize)2 * BlockSize

Maximum number of bytes addressed by a triple indirect pointer is= NumberOfEntriesInOneBlock 3 * BlockSize = (BlockSize / PointerSize)3 * BlockSize

K M G T 210 B = KB220 B = MB230 B = GB240 B = TB= ( 8 KB / 4 B)2 * 8 KB

= ( 23 . 210 / 22)2 * (23 . 210 B) = ( 211 )2 * 213 B = 222 * 213 B= 22 * 220 * 23*210 B= 25 * 230 B= 32 GB

= ( 8 KB / 4 B)3 * 8 KB

= ( 23 . 210 / 22)3 * (23 . 210 B) = ( 211 )3 * 213 B = 233 * 213 B= 23 * 230 * 23*210 B= 26 * 240 B= 64 TB

Maximum file size is 64 TB + 32 GB + 16 MB + 104 KB