Given scenario: The given performance parameters of virtual machine is as follows: Base CPI =1.5 Privileged O/S accesses per 10000 instructions = 120 Performance impact to guest O/S= 15 cycles Performance impact to VMM = 175 cycles I/O Access per 10000 instructions = 30 I/O access time = 1100 cycles

Question

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Answer 1

Question

Chapter 5, Problem 5.15.1E

Program Plan Intro

Given scenario:

The given performance parameters of virtual machine is as follows:

Base CPI =1.5

Privileged O/S accesses per 10000 instructions = 120

Performance impact to guest O/S= 15 cycles

Performance impact to VMM = 175 cycles

I/O Access per 10000 instructions = 30

I/O access time = 1100 cycles

Expert Solution & Answer

Explanation of Solution

Calculation of CPI:

CPI = Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS +Performance impact to Trap to VMM)=1.5+12010000×(15 cycles + 175 cycles)=1.5+12010000×(190)=1.5+0.012×(190)=1.5+2.28=3.78

Calculation of CPI if the VMM performance impact double:

CPI = Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS + Double of the Performance impact to Trap to VMM)=1.5+12010000×(15 cycles + 350 cycles)=1.5+12010000×(365)=1.5+0.012×(365)=1.5+4.38=5.88

Calculation of CPI if the VMM performance cut in half:

CPI = Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS + Half of the Performance impact to Trap to VMM)=1.5+12010000×(15 cycles + 87.5 cycles)=1.5+12010000×(102.5 )=1.5+0.012×(102.5 )=1.5+1.23=2.73

Longest possible penalty to trap to the VMM:

To obtain a 10% performance degradation, use the following formula:

{1.1× Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS )}={(O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS )+n}1.1×(1.5+12010000×(15 cycles))=(1.5+12010000×((15 cycles)+n))1.1×(1.5+12010000×(15 ))=1.512×((15 cycles)+n)1.1×(1.5+0.18)=1.512×((15 cycles)+n)1.1×1.68=22.68+(1.512n)1.848-22.68=1.512n20.832=1.512nn=13.78 Computer Organization and Design MIPS Edition, Fifth Edition: The Hardware/Software Interface (The Morgan Kaufmann Series in Computer Architecture and Design), Chapter 5, Problem 5.15.1E

Therefore, 13 cycles is the longest possible penalty to trap to the VMM.

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Answer 2

Question

Chapter 5, Problem 5.15.1E

Program Plan Intro

Given scenario:

The given performance parameters of virtual machine is as follows:

Base CPI =1.5

Privileged O/S accesses per 10000 instructions = 120

Performance impact to guest O/S= 15 cycles

Performance impact to VMM = 175 cycles

I/O Access per 10000 instructions = 30

I/O access time = 1100 cycles

Expert Solution & Answer

Explanation of Solution

Calculation of CPI:

CPI = Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS +Performance impact to Trap to VMM)=1.5+12010000×(15 cycles + 175 cycles)=1.5+12010000×(190)=1.5+0.012×(190)=1.5+2.28=3.78

Calculation of CPI if the VMM performance impact double:

CPI = Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS + Double of the Performance impact to Trap to VMM)=1.5+12010000×(15 cycles + 350 cycles)=1.5+12010000×(365)=1.5+0.012×(365)=1.5+4.38=5.88

Calculation of CPI if the VMM performance cut in half:

CPI = Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS + Half of the Performance impact to Trap to VMM)=1.5+12010000×(15 cycles + 87.5 cycles)=1.5+12010000×(102.5 )=1.5+0.012×(102.5 )=1.5+1.23=2.73

Longest possible penalty to trap to the VMM:

To obtain a 10% performance degradation, use the following formula:

{1.1× Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS )}={(O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS )+n}1.1×(1.5+12010000×(15 cycles))=(1.5+12010000×((15 cycles)+n))1.1×(1.5+12010000×(15 ))=1.512×((15 cycles)+n)1.1×(1.5+0.18)=1.512×((15 cycles)+n)1.1×1.68=22.68+(1.512n)1.848-22.68=1.512n20.832=1.512nn=13.78 Computer Organization and Design MIPS Edition, Fifth Edition: The Hardware/Software Interface (The Morgan Kaufmann Series in Computer Architecture and Design), Chapter 5, Problem 5.15.1E

Therefore, 13 cycles is the longest possible penalty to trap to the VMM.

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Answer 3

Question

Chapter 5, Problem 5.15.1E

Program Plan Intro

Given scenario:

The given performance parameters of virtual machine is as follows:

Base CPI =1.5

Privileged O/S accesses per 10000 instructions = 120

Performance impact to guest O/S= 15 cycles

Performance impact to VMM = 175 cycles

I/O Access per 10000 instructions = 30

I/O access time = 1100 cycles

Expert Solution & Answer

Explanation of Solution

Calculation of CPI:

CPI = Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS +Performance impact to Trap to VMM)=1.5+12010000×(15 cycles + 175 cycles)=1.5+12010000×(190)=1.5+0.012×(190)=1.5+2.28=3.78

Calculation of CPI if the VMM performance impact double:

CPI = Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS + Double of the Performance impact to Trap to VMM)=1.5+12010000×(15 cycles + 350 cycles)=1.5+12010000×(365)=1.5+0.012×(365)=1.5+4.38=5.88

Calculation of CPI if the VMM performance cut in half:

CPI = Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS + Half of the Performance impact to Trap to VMM)=1.5+12010000×(15 cycles + 87.5 cycles)=1.5+12010000×(102.5 )=1.5+0.012×(102.5 )=1.5+1.23=2.73

Longest possible penalty to trap to the VMM:

To obtain a 10% performance degradation, use the following formula:

{1.1× Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS )}={(O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS )+n}1.1×(1.5+12010000×(15 cycles))=(1.5+12010000×((15 cycles)+n))1.1×(1.5+12010000×(15 ))=1.512×((15 cycles)+n)1.1×(1.5+0.18)=1.512×((15 cycles)+n)1.1×1.68=22.68+(1.512n)1.848-22.68=1.512n20.832=1.512nn=13.78 Computer Organization and Design MIPS Edition, Fifth Edition: The Hardware/Software Interface (The Morgan Kaufmann Series in Computer Architecture and Design), Chapter 5, Problem 5.15.1E

Therefore, 13 cycles is the longest possible penalty to trap to the VMM.

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Answer 4

Question

Chapter 5, Problem 5.15.1E

Program Plan Intro

Given scenario:

The given performance parameters of virtual machine is as follows:

Base CPI =1.5

Privileged O/S accesses per 10000 instructions = 120

Performance impact to guest O/S= 15 cycles

Performance impact to VMM = 175 cycles

I/O Access per 10000 instructions = 30

I/O access time = 1100 cycles

Expert Solution & Answer

Explanation of Solution

Calculation of CPI:

CPI = Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS +Performance impact to Trap to VMM)=1.5+12010000×(15 cycles + 175 cycles)=1.5+12010000×(190)=1.5+0.012×(190)=1.5+2.28=3.78

Calculation of CPI if the VMM performance impact double:

CPI = Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS + Double of the Performance impact to Trap to VMM)=1.5+12010000×(15 cycles + 350 cycles)=1.5+12010000×(365)=1.5+0.012×(365)=1.5+4.38=5.88

Calculation of CPI if the VMM performance cut in half:

CPI = Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS + Half of the Performance impact to Trap to VMM)=1.5+12010000×(15 cycles + 87.5 cycles)=1.5+12010000×(102.5 )=1.5+0.012×(102.5 )=1.5+1.23=2.73

Longest possible penalty to trap to the VMM:

To obtain a 10% performance degradation, use the following formula:

{1.1× Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS )}={(O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS )+n}1.1×(1.5+12010000×(15 cycles))=(1.5+12010000×((15 cycles)+n))1.1×(1.5+12010000×(15 ))=1.512×((15 cycles)+n)1.1×(1.5+0.18)=1.512×((15 cycles)+n)1.1×1.68=22.68+(1.512n)1.848-22.68=1.512n20.832=1.512nn=13.78 Computer Organization and Design MIPS Edition, Fifth Edition: The Hardware/Software Interface (The Morgan Kaufmann Series in Computer Architecture and Design), Chapter 5, Problem 5.15.1E

Therefore, 13 cycles is the longest possible penalty to trap to the VMM.

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Answer 5

Chapter 5, Problem 5.15.1E

Program Plan Intro

Given scenario:

The given performance parameters of virtual machine is as follows:

Base CPI =1.5

Privileged O/S accesses per 10000 instructions = 120

Performance impact to guest O/S= 15 cycles

Performance impact to VMM = 175 cycles

I/O Access per 10000 instructions = 30

I/O access time = 1100 cycles

Expert Solution & Answer

Explanation of Solution

Calculation of CPI:

CPI = Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS +Performance impact to Trap to VMM)=1.5+12010000×(15 cycles + 175 cycles)=1.5+12010000×(190)=1.5+0.012×(190)=1.5+2.28=3.78

Calculation of CPI if the VMM performance impact double:

CPI = Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS + Double of the Performance impact to Trap to VMM)=1.5+12010000×(15 cycles + 350 cycles)=1.5+12010000×(365)=1.5+0.012×(365)=1.5+4.38=5.88

Calculation of CPI if the VMM performance cut in half:

CPI = Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS + Half of the Performance impact to Trap to VMM)=1.5+12010000×(15 cycles + 87.5 cycles)=1.5+12010000×(102.5 )=1.5+0.012×(102.5 )=1.5+1.23=2.73

Longest possible penalty to trap to the VMM:

To obtain a 10% performance degradation, use the following formula:

{1.1× Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS )}={(O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS )+n}1.1×(1.5+12010000×(15 cycles))=(1.5+12010000×((15 cycles)+n))1.1×(1.5+12010000×(15 ))=1.512×((15 cycles)+n)1.1×(1.5+0.18)=1.512×((15 cycles)+n)1.1×1.68=22.68+(1.512n)1.848-22.68=1.512n20.832=1.512nn=13.78 Computer Organization and Design MIPS Edition, Fifth Edition: The Hardware/Software Interface (The Morgan Kaufmann Series in Computer Architecture and Design), Chapter 5, Problem 5.15.1E

Therefore, 13 cycles is the longest possible penalty to trap to the VMM.

Answer 6

Chapter 5, Problem 5.15.1E

Program Plan Intro

Given scenario:

The given performance parameters of virtual machine is as follows:

Base CPI =1.5

Privileged O/S accesses per 10000 instructions = 120

Performance impact to guest O/S= 15 cycles

Performance impact to VMM = 175 cycles

I/O Access per 10000 instructions = 30

I/O access time = 1100 cycles

Expert Solution & Answer

Explanation of Solution

Calculation of CPI:

CPI = Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS +Performance impact to Trap to VMM)=1.5+12010000×(15 cycles + 175 cycles)=1.5+12010000×(190)=1.5+0.012×(190)=1.5+2.28=3.78

Calculation of CPI if the VMM performance impact double:

CPI = Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS + Double of the Performance impact to Trap to VMM)=1.5+12010000×(15 cycles + 350 cycles)=1.5+12010000×(365)=1.5+0.012×(365)=1.5+4.38=5.88

Calculation of CPI if the VMM performance cut in half:

CPI = Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS + Half of the Performance impact to Trap to VMM)=1.5+12010000×(15 cycles + 87.5 cycles)=1.5+12010000×(102.5 )=1.5+0.012×(102.5 )=1.5+1.23=2.73

Longest possible penalty to trap to the VMM:

To obtain a 10% performance degradation, use the following formula:

{1.1× Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS )}={(O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS )+n}1.1×(1.5+12010000×(15 cycles))=(1.5+12010000×((15 cycles)+n))1.1×(1.5+12010000×(15 ))=1.512×((15 cycles)+n)1.1×(1.5+0.18)=1.512×((15 cycles)+n)1.1×1.68=22.68+(1.512n)1.848-22.68=1.512n20.832=1.512nn=13.78 Computer Organization and Design MIPS Edition, Fifth Edition: The Hardware/Software Interface (The Morgan Kaufmann Series in Computer Architecture and Design), Chapter 5, Problem 5.15.1E

Therefore, 13 cycles is the longest possible penalty to trap to the VMM.

Answer 7

Chapter 5, Problem 5.15.1E

Program Plan Intro

Given scenario:

The given performance parameters of virtual machine is as follows:

Base CPI =1.5

Privileged O/S accesses per 10000 instructions = 120

Performance impact to guest O/S= 15 cycles

Performance impact to VMM = 175 cycles

I/O Access per 10000 instructions = 30

I/O access time = 1100 cycles

Answer 8

Expert Solution & Answer

Explanation of Solution

Calculation of CPI:

CPI = Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS +Performance impact to Trap to VMM)=1.5+12010000×(15 cycles + 175 cycles)=1.5+12010000×(190)=1.5+0.012×(190)=1.5+2.28=3.78

Calculation of CPI if the VMM performance impact double:

CPI = Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS + Double of the Performance impact to Trap to VMM)=1.5+12010000×(15 cycles + 350 cycles)=1.5+12010000×(365)=1.5+0.012×(365)=1.5+4.38=5.88

Calculation of CPI if the VMM performance cut in half:

CPI = Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS + Half of the Performance impact to Trap to VMM)=1.5+12010000×(15 cycles + 87.5 cycles)=1.5+12010000×(102.5 )=1.5+0.012×(102.5 )=1.5+1.23=2.73

Longest possible penalty to trap to the VMM:

To obtain a 10% performance degradation, use the following formula:

{1.1× Base CPI + (O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS )}={(O/S Accesses per 10000 instructions)10000×(Performance impact of Guest OS )+n}1.1×(1.5+12010000×(15 cycles))=(1.5+12010000×((15 cycles)+n))1.1×(1.5+12010000×(15 ))=1.512×((15 cycles)+n)1.1×(1.5+0.18)=1.512×((15 cycles)+n)1.1×1.68=22.68+(1.512n)1.848-22.68=1.512n20.832=1.512nn=13.78 Computer Organization and Design MIPS Edition, Fifth Edition: The Hardware/Software Interface (The Morgan Kaufmann Series in Computer Architecture and Design), Chapter 5, Problem 5.15.1E

Therefore, 13 cycles is the longest possible penalty to trap to the VMM.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Ch. 5 - Prob. 5.1.1E Ch. 5 - Prob. 5.1.2E Ch. 5 - Prob. 5.1.3E Ch. 5 - Prob. 5.1.4E Ch. 5 - Prob. 5.1.5E Ch. 5 - Prob. 5.1.6E Ch. 5 - Prob. 5.2.1E Ch. 5 - Prob. 5.2.2E Ch. 5 - Prob. 5.2.3E Ch. 5 - Prob. 5.2.4E

Explanation of Solution

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Chapter 5 Solutions