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Airida [17]
2 years ago
13

Designers are comparing performance between a multi-core processor equipped with 16 in-order cores and a dual-core superscalar p

rocessor. The dual-core superscalar processor has two identical Out- of-Order cores. Each core is a 4-issue superscalar. The multi-core processor has 16 in-order processor cores. Each in-order core of the multi-core processor can achieve about one third the performance of the OoO superscalar core. A benchmark program K includes both sequential codes that can only be executed on one CPU core and parallel codes that can be distributed to multiple CPU cores. For program K, execution of the sequential codes takes 30% of the total execution time.
Required:
Which processor would deliver faster performance for benchmark program K and show your calculations?
Computers and Technology
1 answer:
Nataly [62]2 years ago
5 0

Explanation:

it will be k because it process faster an it comsume less elecricity

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Give short answers.
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Answer:

16

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2 years ago
Read 2 more answers
DNS forwarding can be set up so that if the DNS server that receives the forwarded request cannot resolve the name, then the ser
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Nonexclusive forwarding

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8 0
2 years ago
The Taylor series expansion for ax is: Write a MATLAB program that determines ax using the Taylor series expansion. The program
zalisa [80]

Answer: (a). 11.3137

(b). 22.849

Explanation:

Provided below is a step by step analysis to solving this problem

(a)

clc;close all;clear all;

a=2;x=3.5;

E=10;n=0;k=1;sn1=0;

while E >0.000001

cn=((log(a))^n)*(x^n)/factorial(n);

sn=sn1+cn;

E=abs((sn-sn1)/sn1);

sn1=sn;

n=n+1;

k=k+1;

end

fprintf('2^3.5 from tailor series=%6.4f after adding n=%d terms\n',sn,n);

2^3.5 from tailor series=11.3137 after adding n=15 terms

disp('2^3.5 using calculator =11.3137085');

Command window:

2^3.5 from tailor series=11.3137 after adding n=15 terms

2^3.5 using calculator =11.3137085

(b)

clc;close all;clear all;

a=6.3;x=1.7;

E=10;n=0;k=1;sn1=0;

while E >0.000001

cn=((log(a))^n)*(x^n)/factorial(n);

sn=sn1+cn;

E=abs((sn-sn1)/sn1);

sn1=sn;

n=n+1;

k=k+1;

end

fprintf('6.3^1.7 from tailor series=%6.4f after adding n=%d terms\n',sn,n);

disp('6.3^1.7 using calculator =22.84961748');

Command window:

6.3^1.7 from tailor series=22.8496 after adding n=16 terms

6.3^1.7 using calculator =22.84961748

cheers i hope this helped !!!

7 0
3 years ago
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