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

Explanation of Solution

Calculation of CPI:

$\begin{array}{l}\text{CPI = Base CPI + }\frac{\left(\text{O/S Accesses per 10000 instructions}\right)}{10000}\times \left(\begin{array}{l}\text{Performance impact of Guest OS }\\ \text{+Performance impact }\\ \text{to Trap to VMM}\end{array}\right)\\ =\text{1}\text{.5+}\frac{120}{10000}\times \left(15\text{ cycles + 175 cycles}\right)\\ =1.5+\frac{120}{10000}\times \left(190\right)\\ =1.5+0.012\times \left(190\right)\\ =1.5+2.28\\ =3.78\end{array}$

Calculation of CPI if the VMM performance impact double:

$\begin{array}{l}\text{CPI = Base CPI + }\frac{\left(\text{O/S Accesses per 10000 instructions}\right)}{\text{10000}}\text{×}\left(\begin{array}{l}\text{Performance impact of Guest OS }\\ \text{+ Double of the Performance impact }\\ \text{to Trap to VMM}\end{array}\right)\\ \text{=1}\text{.5+}\frac{\text{120}}{\text{10000}}\text{×}\left(\text{15 cycles + 350 cycles}\right)\\ \text{=1}\text{.5+}\frac{\text{120}}{\text{10000}}\text{×}\left(\text{365}\right)\\ \text{=1}\text{.5+0}\text{.012×}\left(\text{365}\right)\\ \text{=1}\text{.5+4}\text{.38}\\ \text{=5}\text{.88}\end{array}$

Calculation of CPI if the VMM performance cut in half:

$\begin{array}{l}\text{CPI = Base CPI + }\frac{\left(\text{O/S Accesses per 10000 instructions}\right)}{\text{10000}}\text{×}\left(\begin{array}{l}\text{Performance impact of Guest OS + }\\ \text{Half of the Performance impact to }\\ \text{Trap to VMM}\end{array}\right)\\ \text{=1}\text{.5+}\frac{\text{120}}{\text{10000}}\text{×}\left(\text{15 cycles + 87}\text{.5 cycles}\right)\\ \text{=1}\text{.5+}\frac{\text{120}}{\text{10000}}\text{×}\left(\text{102}\text{.5 }\right)\\ \text{=1}\text{.5+0}\text{.012×}\left(\text{102}\text{.5 }\right)\\ \text{=1}\text{.5+1}\text{.23}\\ \text{=2}\text{.73}\end{array}$

Longest possible penalty to trap to the VMM:

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

$\begin{array}{l}\{\begin{array}{l}\text{1}\text{.1× Base CPI + }\frac{\left(\text{O/S Accesses per 10000 instructions}\right)}{\text{10000}}\text{×}\\ \left(\text{Performance impact of Guest OS }\right)\end{array}\}\text{=}\left\{\begin{array}{l}\frac{\left(\text{O/S Accesses per 10000 instructions}\right)}{\text{10000}}\text{×}\\ \left(\text{Performance impact of Guest OS }\right)\text{+n}\end{array}\right\}\\ \text{1}\text{.1×}\left(\text{1}\text{.5+}\frac{\text{120}}{\text{10000}}\text{×}\left(\text{15 cycles}\right)\right)\text{=}\left(\text{1}\text{.5+}\frac{\text{120}}{\text{10000}}\text{×}\left(\left(\text{15 cycles}\right)\text{+n}\right)\right)\\ \text{1}\text{.1×}\left(\text{1}\text{.5+}\frac{\text{120}}{\text{10000}}\text{×}\left(\text{15 }\right)\right)\text{=1}\text{.512×}\left(\left(\text{15 cycles}\right)\text{+n}\right)\\ \text{1}\text{.1×}\left(\text{1}\text{.5+0}\text{.18}\right)\text{=1}\text{.512×}\left(\left(\text{15 cycles}\right)\text{+n}\right)\\ \text{1}\text{.1×1}\text{.68=22}\text{.68+}\left(\text{1}\text{.512n}\right)\\ \text{1}\text{.848-22}\text{.68=1}\text{.512n}\\ \text{20}\text{.832=1}\text{.512n}\\ \text{n=13}\text{.78}\end{array}$

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