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3 years ago

The issue management process has how many stages?

Business

1 answer:

Gennadij [26K]3 years ago

7 0

Answer:

d. Six.

Explanation:

The issue management process is a set of stages and procedure that helps you manage issues as they occur and are part of the operation team and These includes the identification, analysis, evaluation, action, monitor and review stages.

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4 years ago

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Doss [256]

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3 years ago

Read 2 more answers

lotina received an email notice from her university that tuition would be increasing 2.5% for next year, with no changes to fees

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The element of the marketing mix that is used in the given scenario is Price as it has to do with an increase in tuition.

<h3>What is Marketing Mix?</h3>

This refers to the tactics that is used by a company to promote its product or brand in the market.

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2 years ago

Assume your computer is able to complete 4 double floating-point operations per cycle when operands are in registers and it take

svp [43]

Answer:

For 1st algorithm: The total run time is 200.25 s, while that of total waste time is 200 sec and the percentage of the waste time is 99.8%.

For 2nd algorithm:The total run time is 100.35 s, while that of total waste time is 100.1 sec and the percentage of the waste time is 99.75%.

Explanation:

As the complete question is not visible, therefore, the question is searched online and following reference question is obtained.

Following data is given as

Floating point operation time=T/4

Memory Access Time =100T

Frequency =2 GHz

Number of Cycles=1000

1st Algorithm

/*dgemm0: simple ijk version triple loop

algorithm*/

for (i=0; i<n; i++)

for (j=0; j<n; j++)

for (k=0; k<n; k++)

c[i*n+j] += a[i*n+k] * b[k*n+j];

First by rewriting the operation inside the inner loop:

= + ×

Now first A, B and C are loaded into the registers so

$Load \,Time=3 \times Memory \,Access \,Time=3 \times 100\, T =300\, T$

For 2 floating point computations (addition and multiplication)

$Computation\, Time=2 \times Floating\, Time\\Computation\, Time=2 \times \frac{T}{4}\\Computation\, Time=\frac{T}{2}$

Finally, to store and repeat the cycle as N^3 times the time is estimated as

$Store \,Time=Memory\, Access\, Time=100T$

Total Run time is given as

$T_{run}=N^3 \times [T_{load}+T_{comp}+T_{store}]\\T_{run}=1000^3 \times [300T+\frac{T}{2}+100T]\\T_{run}=1000^3 \times [400.5T]\\T_{run}=200.25 s$

Total Wasted time is given as

$T_{waste}=N^3 \times [T_{load}+T_{store}]\\T_{waste}=1000^3 \times [300T+100T]\\T_{waste}=1000^3 \times [400T]\\T_{waste}=200 s$

Percentage of Waste time is given as

$\%age \, waste=\frac{T_{waste}}{T_{run}}\times 100\\\%age \, waste=\frac{200}{200.25}\times 100\\\%age \, waste=99.8\%$

The total run time is 200.25 s, while that of total waste time is 200 sec and the percentage of the waste time is 99.8%.

2nd Algorithm

/*dgemm1: simple ijk version triple loop

algorithm with register reuse*/

for (i=0; i<n; i++)

for (j=0; j<n; j++) {

register double r = c[i*n+j];

for (k=0; k<n; k++)

r += a[i*n+k] * b[k*n+j];

c[i*n+j] = r;

}

Initialize register r with the content of C for N2 Times as given as $Initialization\,Time=N^2 \times Memory \,Access \,Time=N^3 \times 100\, T$

Time for Loading Operands A and B into registers for N3 Times is given as

$Load \,Time=N^3 \times 2 \times Memory \,Access \,Time=N^3\times 2 \times 100\, T =N^3\times 200\, T$

For 2 floating point computations (addition and multiplication)

$Computation\, Time=N^3 \times\frac{T}{2}$

Final Memory update to store result in the register r to the memory for N2 Times

$Store \,Time=Memory\, Access\, Time=N^2 \times 100T$

Total Run time is given as

$T_{run}=N^3 \times [T_{load}+T_{comp}]+N^2 \times [T_{linit}+T_{store}]\\T_{run}=1000^3 \times [200T+\frac{T}{2}]+1000^2 \times [100T+100T]\\T_{run}=1000^3 \times [200.5T]+1000^2 \times [200T]\\T_{run}=100.35 s$

Total Wasted time is given as

$T_{waste}=N^3 \times [T_{load}]+N^2 \times [T_{init}+T_{store}]\\T_{waste}=1000^3 \times [200]+1000^2 \times [100T+100T]\\T_{waste}=1000^3 \times [200T]+1000^2 \times [200T]\\T_{waste}=100.1 s$

Percentage of Waste time is given as

$\%age \, waste=\frac{T_{waste}}{T_{run}}\times 100\\\%age \, waste=\frac{100.1}{100.35}\times 100\\\%age \, waste=99.75\%$

The total run time is 100.35 s, while that of total waste time is 100.1 sec and the percentage of the waste time is 99.75%.

8 0

3 years ago