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inysia [295]
3 years ago
11

An enhancement to a computer accelerates some mode of execution by a factor of 10. The enhanced mode is used 50% of the time, me

asured as a percentage of the execution time when the enhanced mode is in use. Recall that Amdahl's Law depends on the fraction of the original unenhanced execution time that could make use of enhanced mode. Thus the 50% measurement cannot be used directly to compute speedup with Amdahl's Law. I What is the speedup obtained from fast mode? I What percentage of the original execution time has been converted to fast mode?
Computers and Technology
1 answer:
Ann [662]3 years ago
5 0

Answer:

The answer is "5.5 and 90.90%"

Explanation:

For point 1:

To calculate the speed from fast mode, its run time without enhancement should be worked out. Designers are aware of which two selves are implicated throughout the accelerated project planning: the empty (50 \%) and the increased stages (50 \%).

Although not enhanced, this would take almost as long (50 \%) and for the combine to give phase; even so, the increased phase would've been 10 times longer, as well as 500 \%. Thus the corresponding total speed without enhancement is =50\% + 500\% = 550\%.

Its overall speed is:

=\frac{\text{Accelerated runtime}}{ \text{accelerated  runtime}} = \frac{550 \%}{ 100 \%}= 5.5

For point 2:

We re-connect these figures in Amdahl's Law throughout order to identify how long it would take for both the initial implementation:

\text{Vectorized fraction}= \frac{\text{Overall velocity}\times \text{Accelerated acceleration}-\text{Accelerated acceleration}}{\text{Overall acceleration} \times \text{Accelerated acceleration}-\text{Overall acceleration}}

                              = \frac{5.5  \times 10 -10 }{ 5.5 \times 10 - 5.5}\\\\=\frac{45}{49.5}\\\\=0.9090\\\\=90.90\%

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A computer retail store has 15 personal computers in stock. A buyer wants to purchase 3 of them. Unknown to either the retail st
Dima020 [189]

Answer:

a. 1365 ways

b. Probability = 0.4096

c. Probability = 0.5904

Explanation:

Given

PCs = 15

Purchase = 3

Solving (a): Ways to select 4 computers out of 15, we make use of Combination formula as follows;

^nC_r = \frac{n!}{(n-r)!r!}

Where n = 15\ and\ r = 4

^{15}C_4 = \frac{15!}{(15-4)!4!}

^{15}C_4 = \frac{15!}{11!4!}

^{15}C_4 = \frac{15 * 14 * 13 * 12 * 11!}{11! * 4 * 3 * 2 * 1}

^{15}C_4 = \frac{15 * 14 * 13 * 12}{4 * 3 * 2 * 1}

^{15}C_4 = \frac{32760}{24}

^{15}C_4 = 1365

<em>Hence, there are 1365 ways </em>

Solving (b): The probability that exactly 1 will be defective (from the selected 4)

First, we calculate the probability of a PC being defective (p) and probability of a PC not being defective (q)

<em>From the given parameters; 3 out of 15 is detective;</em>

So;

p = 3/15

p = 0.2

q = 1 - p

q = 1 - 0.2

q = 0.8

Solving further using binomial;

(p + q)^n = p^n + ^nC_1p^{n-1}q + ^nC_2p^{n-2}q^2 + .....+q^n

Where n = 4

For the probability that exactly 1 out of 4 will be defective, we make use of

Probability =  ^nC_3pq^3

Substitute 4 for n, 0.2 for p and 0.8 for q

Probability =  ^4C_3 * 0.2 * 0.8^3

Probability =  \frac{4!}{3!1!} * 0.2 * 0.8^3

Probability = 4 * 0.2 * 0.8^3

Probability = 0.4096

Solving (c): Probability that at least one is defective;

In probability, opposite probability sums to 1;

Hence;

<em>Probability that at least one is defective + Probability that at none is defective = 1</em>

Probability that none is defective is calculated as thus;

Probability =  q^n

Substitute 4 for n and 0.8 for q

Probability =  0.8^4

Probability = 0.4096

Substitute 0.4096 for Probability that at none is defective

Probability that at least one is defective + 0.4096= 1

Collect Like Terms

Probability = 1 - 0.4096

Probability = 0.5904

8 0
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