Chapter 11, Problem 3E

Explanation of Solution

Verification of the ratio being consistent when compared with the other system:

• System performance is considered as one of main factor of a processor.
• It is used to determine the speed a problem can be solved.
• It is also used to determine the factors such as number of problems that can be allocated at particular amount of time and also the number of problems that can be handled by the processor.
• The relative performance between the two systems is measured and expected by the run time of the program of the individual.

Given:

The information about the system A and System c are shown in the below table:

Program	Execution time System A(sec)	Execution time System B(sec)	Execution time System C(sec)
V	150	200	75
W	200	250	150
X	250	175	200
Y	400	800	500
Z	1000	1200	1100

Consider there are n programs and each programs are considered to have their own runtime on each systems.

Program	Execution time System A(sec)	Execution time System B(sec)	Execution time System C(sec)
V	150	200	75
W	200	250	150
X	250	175	200
Y	400	800	500
Z	1000	1200	1100

System A:

calculate the arithmetic mean for the system A:

$\begin{array}{c}\text{A=}\left(\frac{150+200+250+400+1000}{5}\right)\\ =400\end{array}$

System B:

calculate the arithmetic mean for the system B:

$\begin{array}{c}\text{A=}\left(\frac{\text{200+250+175+800+1200}}{\text{5}}\right)\\ \text{=525}\end{array}$

System C:

calculate the arithmetic mean for the system C:

$\begin{array}{c}\text{A=}\left(\frac{75+150+200+500+1100}{\text{5}}\right)\\ =405\end{array}$

Calculating the ratio of the geometic mean obtained for the various systems are:

$\begin{array}{c}\frac{\text{AM\_A}}{\text{AM\_B}}\text{=}\frac{\text{400}}{525}\\ \text{=0}\text{.762}\end{array}$

$\begin{array}{c}\frac{\text{AM\_B}}{\text{AM\_C}}\text{=}\frac{\text{525}}{\text{405}}\\ \text{=1}\text{.296}\end{array}$

$\begin{array}{c}\frac{\text{AM\_A}}{\text{AM\_C}}\text{=}\frac{400}{405}\\ \text{=0}\text{.987}\end{array}$

Consider there are n programs and each programs are considered to have their own runtime on each systems.

The geometric mean of the one system’s runtime is obtained by normalizing it with the another system.

The process of normalization is carried out by taking the products of the ratio of the run time and by taking the nth root of the product.

The below tables illustrates the how the system B and system C is being normalized with that of the system A.

Program	Execution time System A(sec)	Normalized  to A	Execution time System B(sec)	Normalized  to B	Execution time System C(sec)	Normalized to C
V	150	1	200	0.75	75	2
W	200	1	250	0.8	150	0.75
X	250	1	175	1.428	200	0.8
Y	400	1	800	0.5	500	1.25
Z	5000	1	1200	0.83	1100	0.22

Calculating geometic Mean:

The formula to calculate the geometric mean:

$\text{G=}{\left(\text{x1×x2×x3×}\mathrm{...}\text{×xn}\right)}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{n}$}\right.}$

System A:

calculate the geometric mean for the system A:

$\begin{array}{c}\text{G=}{\left(\text{1×1×1×1×1}\right)}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{5}$}\right.}\\ =1\end{array}$

System B:

calculate the geometric mean for the system B:

$\begin{array}{c}\text{G=}{\left(\text{0}\text{.7×0}\text{.8×1}\text{.428×0}\text{.5×0}\text{.83}\right)}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}\\ ={\left(0.3555\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}\\ =0.813\end{array}$

System C:

calculate the geometric mean for the system C:

$\begin{array}{c}\text{G=}{\left(\text{2}\times \text{0}\text{.75}\times \text{0}\text{.8}\times \text{1}\text{.25}\times \text{0}\text{.22}\right)}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}\\ ={\left(0.33\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}\\ =0.8011\end{array}$

Calculating the ratio of the geometic mean obtained for the various systems are:

$\begin{array}{c}\frac{\text{GM\_A}}{\text{GM\_B}}\text{=}\frac{\text{1}}{\text{0}\text{.813}}\\ \text{=1}\text{.23}\end{array}$

$\begin{array}{c}\frac{\text{GM\_B}}{\text{GM\_C}}\text{=}\frac{\text{0}\text{.813}}{\text{0}\text{.801}}\\ \text{=1}\text{.01}\end{array}$

$\begin{array}{c}\frac{\text{GM\_A}}{\text{GM\_C}}\text{=}\frac{\text{1}}{\text{0}\text{.801}}\\ \text{=1}\text{.24}\end{array}$

From the comparison of the  arithmetic and geometric mean that is obtained with the every pair of system, the consistency is found only with the sytem B because the ratio of runtine in the system A and system C  is less than 1.

Likewise, The relative performance compariosn seems to be difficult to compared with that of the system A or B or with the system B or C.