Chapter 1, Problem 1.9.2E

Explanation of Solution

Number of processors	Arithmetic instructions	L/S instructions	Branch instructions	Execution Time (Seconds)
1	$\text{2}\text{.56}\text{×}{\text{10}}^{\text{9}}$	$1.28\text{×}{\text{10}}^{\text{9}}$	$\text{2}\text{.56}\text{×}{\text{10}}^{\text{9}}$	9.6
2	$1.83\text{×}{\text{10}}^{\text{9}}$	$9.14\text{×}{\text{10}}^8$	$\text{2}\text{.56}\text{×}{\text{10}}^{\text{9}}$	7.04
4	$9.12\text{×}{\text{10}}^8$	$4.57\text{×}{\text{10}}^8$	$\text{2}\text{.56}\text{×}{\text{10}}^{\text{9}}$	3.84
8	$4.57\text{×}{\text{10}}^8$	$\text{2}\text{.29}\text{×}{\text{10}}^8$	$\text{2}\text{.56}\text{×}{\text{10}}^{\text{9}}$	2.24

The arithmetic instructions are parallelized to run over multiple cores as follows

$\text{Arithmatic}\text{instructions}\text{=}\frac{\text{Number}\text{of}\text{instructions}}{\text{0}\text{.7}\text{×}\text{Number}\text{of}\text{processors}}$

Substitute, “$\text{2}\text{.56}\text{×}{\text{10}}^{\text{9}}$” for “number of instructions” and “2” for “number of processors” in the above formula

$\begin{array}{c}\text{Arithmatic}{\text{instructions}}_{(2)}\text{=}\frac{\text{2}\text{.56}\text{×}{\text{10}}^{\text{9}}}{\text{0}\text{.7}\text{×}\text{2}}\\ \text{=}1.83\text{×}{\text{10}}^{\text{9}}\end{array}$

Thus, the number of arithmetic instructions with 2 processors is $1.83\text{×}{\text{10}}^{\text{9}}$.

Substitute, “$\text{2}\text{.56}\text{×}{\text{10}}^{\text{9}}$” for “number of instructions” and “4” for “number of processors” in the above formula

$\begin{array}{c}\text{Arithmatic}{\text{instructions}}_{(4)}\text{=}\frac{\text{2}\text{.56}\text{×}{\text{10}}^{\text{9}}}{\text{0}\text{.7}\text{×}4}\\ \text{=}9.12\text{×}{\text{10}}^8\end{array}$

Thus, the number of arithmetic instructions with 4 processors is $9.12\text{×}{\text{10}}^8$.

Substitute, “$\text{2}\text{.56}\text{×}{\text{10}}^{\text{9}}$” for “number of instructions” and “8” for “number of processors” in the above formula

$\begin{array}{c}\text{Arithmatic}{\text{instructions}}_{(8)}\text{=}\frac{\text{2}\text{.56}\text{×}{\text{10}}^{\text{9}}}{\text{0}\text{.7}\text{×}8}\\ \text{=}4.57\text{×}{\text{10}}^8\end{array}$

Thus, the number of arithmetic instructions with 8 processors is $4.57\text{×}{\text{10}}^8$.

The L/S instructions are parallelized to run over multiple cores as follows

$\text{L/S}\text{instructions}\text{=}\frac{\text{Number}\text{of}\text{instructions}}{\text{0}\text{.7}\text{×}\text{Number}\text{of}\text{processors}}$

Substitute, “$1.28\text{×}{\text{10}}^{\text{9}}$” for “number of instructions” and “2” for “number of processors” in the above formula

$\begin{array}{c}\text{L/S}{\text{instructions}}_{(2)}\text{=}\frac{1.28\text{×}{\text{10}}^{\text{9}}}{\text{0}\text{.7}\text{×}\text{2}}\\ \text{=}9.14\text{×}{\text{10}}^8\end{array}$

Thus, the number of L/S instructions with 2 processors is $9.14\text{×}{\text{10}}^8$.

Substitute, “$1.28\text{×}{\text{10}}^{\text{9}}$” for “number of instructions” and “4” for “number of processors” in the above formula

$\begin{array}{c}\text{L/S}{\text{instructions}}_{(4)}\text{=}\frac{1.28\text{×}{\text{10}}^{\text{9}}}{\text{0}\text{.7}\text{×}4}\\ \text{=}4.57\text{×}{\text{10}}^8\end{array}$

Thus, the number of L/S instructions with 4 processors is $4.57\text{×}{\text{10}}^8$.

Substitute, “$1.28\text{×}{\text{10}}^{\text{9}}$” for “number of instructions” and “8” for “number of processors” in the above formula

$\begin{array}{c}\text{L/S}{\text{instructions}}_{(8)}\text{=}\frac{1.28\text{×}{\text{10}}^{\text{9}}}{\text{0}\text{.7}\text{×}8}\\ \text{=}2.29\text{×}{\text{10}}^8\end{array}$

Thus, the number of L/S instructions with 8 processors is $2.29\text{×}{\text{10}}^8$.

The total execution time of the program can be calculated using the following formula

$\text{Execution}\text{time}\text{=}\frac{\text{CPI}\text{×}\text{Number}\text{of}\text{instructions}}{\text{Clock}\text{rate}}$

Substitute, “2” for “CPI of arithmetic instructions”, “12” for “CPI of L/S instructions”, “5” for “CPI of branch instructions”,, “$2.56\times {10}^9$” for “number of arithmetic instructions”, “$1.28\times {10}^9$” for “number of L/S instructions” “$0.25\times {10}^9$”for “number of branch instructions” and “$2\times {10}^9$” for “clock rate”

$\begin{array}{c}\text{Execution}{\text{time}}_{(1)}\text{=}\frac{\text{(2×}\text{2}\text{.56}\text{×}{\text{10}}^{\text{9}}\text{)}\text{+}\text{(12×}\text{1}\text{.28}\text{×}{\text{10}}^{\text{9}}\text{)}\text{+}\text{(5×}\text{0}\text{.25}\text{×}{\text{10}}^{\text{9}}\text{)}}{\text{2}\text{×}{\text{10}}^{\text{9}}}\\ \text{=}10.88\end{array}$

Thus, the total execution time of the program with 1 processor is “10.88 seconds”.

Substitute, “2” for “CPI of arithmetic instructions”, “12” for “CPI of L/S instructions”, “5” for “CPI of branch instructions”,, “$1.83\times {10}^9$” for “number of arithmetic instructions”, “$0.91\times {10}^9$” for “number of L/S instructions” “$0.25\times {10}^9$”for “number of branch instructions” and “$2\times {10}^9$” for “clock rate”

$\begin{array}{c}\text{Execution}{\text{time}}_{(2)}\text{=}\frac{\text{(2×}1.83\text{×}{\text{10}}^{\text{9}}\text{)}\text{+}\text{(12×}0.91\text{×}{\text{10}}^{\text{9}}\text{)}\text{+}\text{(5×}\text{0}\text{.25}\text{×}{\text{10}}^{\text{9}}\text{)}}{\text{2}\text{×}{\text{10}}^{\text{9}}}\\ \text{=}7.95\end{array}$

Thus, the total execution time of the program with 1 processor is “7.95 seconds”.

Substitute, “2” for “CPI of arithmetic instructions”, “12” for “CPI of L/S instructions”, “5” for “CPI of branch instructions”,, “$0.91\times {10}^9$” for “number of arithmetic instructions”, “$0.45\times {10}^9$” for “number of L/S instructions” “$0.25\times {10}^9$”for “number of branch instructions” and “$2\times {10}^9$” for “clock rate”

$\begin{array}{c}\text{Execution}{\text{time}}_{(4)}\text{=}\frac{\text{(2×}0.91\text{×}{\text{10}}^{\text{9}}\text{)}\text{+}\text{(12×}0.45\text{×}{\text{10}}^{\text{9}}\text{)}\text{+}\text{(5×}\text{0}\text{.25}\text{×}{\text{10}}^{\text{9}}\text{)}}{\text{2}\text{×}{\text{10}}^{\text{9}}}\\ \text{=}4.29\end{array}$

Thus, the total execution time of the program with 1 processor is “4.29 seconds”.

Substitute, “2” for “CPI of arithmetic instructions”, “12” for “CPI of L/S instructions”, “5” for “CPI of branch instructions”,, “$0.45\times {10}^9$” for “number of arithmetic instructions”, “$0.22\times {10}^9$” for “number of L/S instructions” “$0.25\times {10}^9$”for “number of branch instructions” and “$2\times {10}^9$” for “clock rate”

$\begin{array}{c}\text{Execution}{\text{time}}_{(8)}\text{=}\frac{\text{(2×}0.45\text{×}{\text{10}}^{\text{9}}\text{)}\text{+}\text{(12×}0.22\text{×}{\text{10}}^{\text{9}}\text{)}\text{+}\text{(5×}\text{0}\text{.25}\text{×}{\text{10}}^{\text{9}}\text{)}}{\text{2}\text{×}{\text{10}}^{\text{9}}}\\ \text{=}2.47\end{array}$

Thus, the total execution time of the program with 1 processor is “2.47 seconds”.