ith the following parameters: Assu b: block size given number of words S: number

Question

ith the following parameters: Assu b: block size given number of words S: number of sets N: number of ways A: number of address bits word: 4 bytes A 16-word cache has the parameters given 2 above. Assume that a program running on this computer executes the followin 0x74 OxAO 0x78 0x38C OxAC 0x84 0x88 0x8C 0x7C 0x34 0x38 0x13C 0x388 Ox18C Assuming least recently used (LRU) replacement strategy for associative caches, show the placement of the instruction addresses in cache for each of the following cache design 3.A.) A direct mapped cache with b-1 word 3.B.) A fully associative cache with b-2 words 3.?.) A two-way set associative cache with b-2 words 3.D.) A direct mapped cache with b-4 words me a cache w g sequence of lw addresses in the exact sequence given:

Explanation / Answer

Solution:

3. A)

Block 0

0xA0

Block 1

Block 2

Block 3

Block 4

(0x74)=> (0x84)=> 0x34

Block 5

Block 6

Block 7

Block 8

(0x78)=> (0x88)=>(0x38)=>0x388

Block 9

Block 10

Block 11

Block 12

(0x38C)=> (0xAC)=>(0x8C)=>(0x7C)=>(0x13C)=>0x18C

Block 13

Block 14

Block 15

3. B)

In fully set associate the words can go to anywhere in the cache.

The entire cache will act as a single set.

The mapping can be generated as given below.

The replacement will happen only id entire cache will full.

Block 0

0x74

Block 1

0xA0

Block 2

0x78

Block 3

0x38C

Block 4

0xAC

Block 5

0x84

Block 6

0x88

Block 7

0x8C

Block 8

0x7C

Block 9

0x34

Block 10

0x38

Block 11

0x13C

Block 12

0x388

Block 13

0x18C

Block 14

Block 15

3. C)

Here the last 8 bits of the address will be taken and it will be mapped to either of the sets based on the first 3 bits among this.

Since 16 words cache, there will be 8 sets.

Set 0

0x34=>0x13C

1

0x38

Set 1

1

Set 2

1

Set 3

0x74=>0x7C

1

0x74

Set 4

0x38C=>0x18C

1

0x388

Set 5

0xA0

1

0xAC

Set 6

0x88

1

0x8C

Set 7

1

3. D)

The directed mapped cache with block size 4 words can be populated as given below.

Block 0

0x74=>0x88

0xA0=>0x34

0x78=>0x38

0x84=>0x388

Block 1

Block 2

0x38C=>0x13C

0xAC=>0x18C

0x8C

0x7C

Block 3

Block 0

0xA0

Block 1

Block 2

Block 3

Block 4

(0x74)=> (0x84)=> 0x34

Block 5

Block 6

Block 7

Block 8

(0x78)=> (0x88)=>(0x38)=>0x388

Block 9

Block 10

Block 11

Block 12

(0x38C)=> (0xAC)=>(0x8C)=>(0x7C)=>(0x13C)=>0x18C

Block 13

Block 14

Block 15

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