Aspire Faculty ID #10760 · Topic: NIMCET 2021 · Just now
NIMCET 2021

Consider a computer system with speed of 106  instructions per second. A program P, having 2n2  steps is run on this system, where n is the input size. If n = 10000, what is the execution time for P?

Solution

$\text{Speed of computer} = 10^6 \text{per second} $
$\text{For n} =10000=10^4$ 
$$Time = \frac{\text{No of tasks}}{\text{Speed  of  computer}}$$ 
$$Time = \frac{2n^2}{10^6}$$ 
$$= \frac{2\times (10^4)^2}{10^6}$$
$$= \frac{2\times 10^8}{10^6}$$
$$=2 \times 10^2$$ 
$$= 200sec$$

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Manmohan dwivedi
Manmohan dwivedi , Nimcet
Commented May 21, 2022
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Shivam Gupta
Shivam Gupta , Aspire
Commented May 02, 2023
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