Given Information : A 4-way set associative cache is given: Index Tag V Bytes 0-3 Tag V Bytes 0-3 Tag V Bytes 0-3 Tag V Bytes 0-3 0 F0 1 ED 32 0A A2 8A 1 BF 80 1D FC 14 1 EF 09 86 2A BC 0 25 44 6F 1A 1 BC 0 03 3E CD 38 A0 0 16 7B ED 5A BC 1 8E 4C DF 18 E4 1 FB B7 12 02 2 BC 1 54 9E 1E FA B6 1 DC 81 B2 14 00 0 B6 1F 7B 44 74 0 10 F5 B8 2E 3 BE 0 2F 7E 3D A8 C0 1 27 95 A4 74 C4 0 07 11 6B D8 BC 0 C7 B7 AF C2 4 7E 1 32 21 1C 2C 8A 1 22 C2 DC 34 BC 1 BA DD 37 D8 DC 0 E7 A2 39 BA 5 98 0 A9 76 2B EE 54 0 BC 91 D5 92 98 1 80 BA 9B F6 BC 1 48 16 81 0A 6 38 0 5D 4D F7 DA BC 1 69 C2 8C 74 8A 1 A8 CE 7F DA 38 1 FA 93 EB 48 7 8A 1 04 2A 32 6A 9E 0 B1 86 56 0E CC 1 96 30 47 F2 BC 1 F8 1D 42 30 The number of physical address bits is represented by “m”. The size of cache represented is by “C”. The block of size in bytes is represented as “B”. Associativity is represented as “E”. The number of cache sets is represented by “S”. The number of tag bits is represented by “t”. The number of block offset bits is represented as “b”.

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Answer 1

Question

Chapter 6, Problem 6.33HW

Program Plan Intro

Given Information:

A 4-way set associative cache is given:

Index Tag V Bytes 0-3 Tag V Bytes 0-3 Tag V Bytes 0-3 Tag V Bytes 0-3
0 F0 1 ED 32 0A A2 8A 1 BF 80 1D FC 14 1 EF 09 86 2A BC 0 25 44 6F 1A
1 BC 0 03 3E CD 38 A0 0 16 7B ED 5A BC 1 8E 4C DF 18 E4 1 FB B7 12 02
2 BC 1 54 9E 1E FA B6 1 DC 81 B2 14 00 0 B6 1F 7B 44 74 0 10 F5 B8 2E
3 BE 0 2F 7E 3D A8 C0 1 27 95 A4 74 C4 0 07 11 6B D8 BC 0 C7 B7 AF C2
4 7E 1 32 21 1C 2C 8A 1 22 C2 DC 34 BC 1 BA DD 37 D8 DC 0 E7 A2 39 BA
5 98 0 A9 76 2B EE 54 0 BC 91 D5 92 98 1 80 BA 9B F6 BC 1 48 16 81 0A
6 38 0 5D 4D F7 DA BC 1 69 C2 8C 74 8A 1 A8 CE 7F DA 38 1
FA 93

EB 48
7 8A 1 04 2A 32 6A 9E 0 B1 86 56 0E CC 1 96 30 47 F2 BC 1 F8 1D 42 30

The number of physical address bits is represented by “m”.

The size of cache represented is by “C”.

The block of size in bytes is represented as “B”.

Associativity is represented as “E”.

The number of cache sets is represented by “S”.

The number of tag bits is represented by “t”.

The number of block offset bits is represented as “b”.

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Considering the TM example of binary sum ( see attached)do the step-by-step of execution for the binary numbers 1101 and 11. Feel free to use the Formal Language Editor Tool to execute it; Write it down the current state of the tape (including the head position) and indicate the current state of the TM at each step.

I need help on inculding additonal code where I can can do the opposite code of MatLab, where the function of t that I enter becomes the result of F(t), in other words, turning the time-domain f(t) into the frequency-domain function F(s):

I need help with the TM computation step-by-step execution for the binary numbers 1101 and 11. Formal Language Editor Tool can be used to execute it; Write it down the current state of the tape (including the head position) and indicate the current state of the TM at each step;

Answer 2

Question

Chapter 6, Problem 6.33HW

Program Plan Intro

Given Information:

A 4-way set associative cache is given:

Index Tag V Bytes 0-3 Tag V Bytes 0-3 Tag V Bytes 0-3 Tag V Bytes 0-3
0 F0 1 ED 32 0A A2 8A 1 BF 80 1D FC 14 1 EF 09 86 2A BC 0 25 44 6F 1A
1 BC 0 03 3E CD 38 A0 0 16 7B ED 5A BC 1 8E 4C DF 18 E4 1 FB B7 12 02
2 BC 1 54 9E 1E FA B6 1 DC 81 B2 14 00 0 B6 1F 7B 44 74 0 10 F5 B8 2E
3 BE 0 2F 7E 3D A8 C0 1 27 95 A4 74 C4 0 07 11 6B D8 BC 0 C7 B7 AF C2
4 7E 1 32 21 1C 2C 8A 1 22 C2 DC 34 BC 1 BA DD 37 D8 DC 0 E7 A2 39 BA
5 98 0 A9 76 2B EE 54 0 BC 91 D5 92 98 1 80 BA 9B F6 BC 1 48 16 81 0A
6 38 0 5D 4D F7 DA BC 1 69 C2 8C 74 8A 1 A8 CE 7F DA 38 1
FA 93

EB 48
7 8A 1 04 2A 32 6A 9E 0 B1 86 56 0E CC 1 96 30 47 F2 BC 1 F8 1D 42 30

The number of physical address bits is represented by “m”.

The size of cache represented is by “C”.

The block of size in bytes is represented as “B”.

Associativity is represented as “E”.

The number of cache sets is represented by “S”.

The number of tag bits is represented by “t”.

The number of block offset bits is represented as “b”.

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Answer 3

Question

Chapter 6, Problem 6.33HW

Program Plan Intro

Given Information:

A 4-way set associative cache is given:

Index Tag V Bytes 0-3 Tag V Bytes 0-3 Tag V Bytes 0-3 Tag V Bytes 0-3
0 F0 1 ED 32 0A A2 8A 1 BF 80 1D FC 14 1 EF 09 86 2A BC 0 25 44 6F 1A
1 BC 0 03 3E CD 38 A0 0 16 7B ED 5A BC 1 8E 4C DF 18 E4 1 FB B7 12 02
2 BC 1 54 9E 1E FA B6 1 DC 81 B2 14 00 0 B6 1F 7B 44 74 0 10 F5 B8 2E
3 BE 0 2F 7E 3D A8 C0 1 27 95 A4 74 C4 0 07 11 6B D8 BC 0 C7 B7 AF C2
4 7E 1 32 21 1C 2C 8A 1 22 C2 DC 34 BC 1 BA DD 37 D8 DC 0 E7 A2 39 BA
5 98 0 A9 76 2B EE 54 0 BC 91 D5 92 98 1 80 BA 9B F6 BC 1 48 16 81 0A
6 38 0 5D 4D F7 DA BC 1 69 C2 8C 74 8A 1 A8 CE 7F DA 38 1
FA 93

EB 48
7 8A 1 04 2A 32 6A 9E 0 B1 86 56 0E CC 1 96 30 47 F2 BC 1 F8 1D 42 30

The number of physical address bits is represented by “m”.

The size of cache represented is by “C”.

The block of size in bytes is represented as “B”.

Associativity is represented as “E”.

The number of cache sets is represented by “S”.

The number of tag bits is represented by “t”.

The number of block offset bits is represented as “b”.

Expert Solution & Answer

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Answer 4

Question

Chapter 6, Problem 6.33HW

Program Plan Intro

Given Information:

A 4-way set associative cache is given:

Index Tag V Bytes 0-3 Tag V Bytes 0-3 Tag V Bytes 0-3 Tag V Bytes 0-3
0 F0 1 ED 32 0A A2 8A 1 BF 80 1D FC 14 1 EF 09 86 2A BC 0 25 44 6F 1A
1 BC 0 03 3E CD 38 A0 0 16 7B ED 5A BC 1 8E 4C DF 18 E4 1 FB B7 12 02
2 BC 1 54 9E 1E FA B6 1 DC 81 B2 14 00 0 B6 1F 7B 44 74 0 10 F5 B8 2E
3 BE 0 2F 7E 3D A8 C0 1 27 95 A4 74 C4 0 07 11 6B D8 BC 0 C7 B7 AF C2
4 7E 1 32 21 1C 2C 8A 1 22 C2 DC 34 BC 1 BA DD 37 D8 DC 0 E7 A2 39 BA
5 98 0 A9 76 2B EE 54 0 BC 91 D5 92 98 1 80 BA 9B F6 BC 1 48 16 81 0A
6 38 0 5D 4D F7 DA BC 1 69 C2 8C 74 8A 1 A8 CE 7F DA 38 1
FA 93

EB 48
7 8A 1 04 2A 32 6A 9E 0 B1 86 56 0E CC 1 96 30 47 F2 BC 1 F8 1D 42 30

The number of physical address bits is represented by “m”.

The size of cache represented is by “C”.

The block of size in bytes is represented as “B”.

Associativity is represented as “E”.

The number of cache sets is represented by “S”.

The number of tag bits is represented by “t”.

The number of block offset bits is represented as “b”.

Expert Solution & Answer

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Answer 5

Chapter 6, Problem 6.33HW

Program Plan Intro

Given Information:

A 4-way set associative cache is given:

Index Tag V Bytes 0-3 Tag V Bytes 0-3 Tag V Bytes 0-3 Tag V Bytes 0-3
0 F0 1 ED 32 0A A2 8A 1 BF 80 1D FC 14 1 EF 09 86 2A BC 0 25 44 6F 1A
1 BC 0 03 3E CD 38 A0 0 16 7B ED 5A BC 1 8E 4C DF 18 E4 1 FB B7 12 02
2 BC 1 54 9E 1E FA B6 1 DC 81 B2 14 00 0 B6 1F 7B 44 74 0 10 F5 B8 2E
3 BE 0 2F 7E 3D A8 C0 1 27 95 A4 74 C4 0 07 11 6B D8 BC 0 C7 B7 AF C2
4 7E 1 32 21 1C 2C 8A 1 22 C2 DC 34 BC 1 BA DD 37 D8 DC 0 E7 A2 39 BA
5 98 0 A9 76 2B EE 54 0 BC 91 D5 92 98 1 80 BA 9B F6 BC 1 48 16 81 0A
6 38 0 5D 4D F7 DA BC 1 69 C2 8C 74 8A 1 A8 CE 7F DA 38 1
FA 93

EB 48
7 8A 1 04 2A 32 6A 9E 0 B1 86 56 0E CC 1 96 30 47 F2 BC 1 F8 1D 42 30

The number of physical address bits is represented by “m”.

The size of cache represented is by “C”.

The block of size in bytes is represented as “B”.

Associativity is represented as “E”.

The number of cache sets is represented by “S”.

The number of tag bits is represented by “t”.

The number of block offset bits is represented as “b”.

Answer 6

Chapter 6, Problem 6.33HW

Program Plan Intro

Given Information:

A 4-way set associative cache is given:

Index Tag V Bytes 0-3 Tag V Bytes 0-3 Tag V Bytes 0-3 Tag V Bytes 0-3
0 F0 1 ED 32 0A A2 8A 1 BF 80 1D FC 14 1 EF 09 86 2A BC 0 25 44 6F 1A
1 BC 0 03 3E CD 38 A0 0 16 7B ED 5A BC 1 8E 4C DF 18 E4 1 FB B7 12 02
2 BC 1 54 9E 1E FA B6 1 DC 81 B2 14 00 0 B6 1F 7B 44 74 0 10 F5 B8 2E
3 BE 0 2F 7E 3D A8 C0 1 27 95 A4 74 C4 0 07 11 6B D8 BC 0 C7 B7 AF C2
4 7E 1 32 21 1C 2C 8A 1 22 C2 DC 34 BC 1 BA DD 37 D8 DC 0 E7 A2 39 BA
5 98 0 A9 76 2B EE 54 0 BC 91 D5 92 98 1 80 BA 9B F6 BC 1 48 16 81 0A
6 38 0 5D 4D F7 DA BC 1 69 C2 8C 74 8A 1 A8 CE 7F DA 38 1
FA 93

EB 48
7 8A 1 04 2A 32 6A 9E 0 B1 86 56 0E CC 1 96 30 47 F2 BC 1 F8 1D 42 30

The number of physical address bits is represented by “m”.

The size of cache represented is by “C”.

The block of size in bytes is represented as “B”.

Associativity is represented as “E”.

The number of cache sets is represented by “S”.

The number of tag bits is represented by “t”.

The number of block offset bits is represented as “b”.

Answer 7

Chapter 6, Problem 6.33HW

Program Plan Intro

Given Information:

A 4-way set associative cache is given:

Index Tag V Bytes 0-3 Tag V Bytes 0-3 Tag V Bytes 0-3 Tag V Bytes 0-3
0 F0 1 ED 32 0A A2 8A 1 BF 80 1D FC 14 1 EF 09 86 2A BC 0 25 44 6F 1A
1 BC 0 03 3E CD 38 A0 0 16 7B ED 5A BC 1 8E 4C DF 18 E4 1 FB B7 12 02
2 BC 1 54 9E 1E FA B6 1 DC 81 B2 14 00 0 B6 1F 7B 44 74 0 10 F5 B8 2E
3 BE 0 2F 7E 3D A8 C0 1 27 95 A4 74 C4 0 07 11 6B D8 BC 0 C7 B7 AF C2
4 7E 1 32 21 1C 2C 8A 1 22 C2 DC 34 BC 1 BA DD 37 D8 DC 0 E7 A2 39 BA
5 98 0 A9 76 2B EE 54 0 BC 91 D5 92 98 1 80 BA 9B F6 BC 1 48 16 81 0A
6 38 0 5D 4D F7 DA BC 1 69 C2 8C 74 8A 1 A8 CE 7F DA 38 1
FA 93

EB 48
7 8A 1 04 2A 32 6A 9E 0 B1 86 56 0E CC 1 96 30 47 F2 BC 1 F8 1D 42 30

The number of physical address bits is represented by “m”.

The size of cache represented is by “C”.

The block of size in bytes is represented as “B”.

Associativity is represented as “E”.

The number of cache sets is represented by “S”.

The number of tag bits is represented by “t”.

The number of block offset bits is represented as “b”.

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 6.1 - Prob. 6.1PP Ch. 6.1 - Prob. 6.2PP Ch. 6.1 - Prob. 6.3PP Ch. 6.1 - Prob. 6.4PP Ch. 6.1 - Prob. 6.5PP Ch. 6.1 - Prob. 6.6PP Ch. 6.2 - Prob. 6.7PP Ch. 6.2 - Prob. 6.8PP Ch. 6.4 - Prob. 6.9PP Ch. 6.4 - Prob. 6.10PP

Solution Summary: The author explains that a 4-way set associative cache is given. The number of physical address bits is represented by "m".

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Chapter 6 Solutions

Index	Tag	V	Bytes 0-3	Tag	V	Bytes 0-3	Tag	V	Bytes 0-3	Tag	V	Bytes 0-3
0	F0	1	ED 32 0A A2	8A	1	BF 80 1D FC	14	1	EF 09 86 2A	BC	0	25 44 6F 1A
1	BC	0	03 3E CD 38	A0	0	16 7B ED 5A	BC	1	8E 4C DF 18	E4	1	FB B7 12 02
2	BC	1	54 9E 1E FA	B6	1	DC 81 B2 14	00	0	B6 1F 7B 44	74	0	10 F5 B8 2E
3	BE	0	2F 7E 3D A8	C0	1	27 95 A4 74	C4	0	07 11 6B D8	BC	0	C7 B7 AF C2
4	7E	1	32 21 1C 2C	8A	1	22 C2 DC 34	BC	1	BA DD 37 D8	DC	0	E7 A2 39 BA
5	98	0	A9 76 2B EE	54	0	BC 91 D5 92	98	1	80 BA 9B F6	BC	1	48 16 81 0A
6	38	0	5D 4D F7 DA	BC	1	69 C2 8C 74	8A	1	A8 CE 7F DA	38	1	FA 93 EB 48
7	8A	1	04 2A 32 6A	9E	0	B1 86 56 0E	CC	1	96 30 47 F2	BC	1	F8 1D 42 30