Business, 15.02.2020 01:52 rschaum08

Assume your computer is able to complete 4 double floating-point operations per cycle when operands are in registers and it takes an additional delay of 100 cycles to access any operands that are not in registers. The clock frequency of your computer is 2 Ghz.

Answers: 3

Show answers

Answers

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++) {

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%.

Assume your computer is able to complete 4 double floating-point operations per cycle when operands

Answer from: Quest

Can you take a photo so i can see the table

Answer from: Quest

answer; the above statement/question is ( true);

Another question on Business

Business, 22.06.2019 04:00

Assume that the following conditions exist: a. all banks are fully loaned up- there are no excess reserves, and desired excess reserves are always zero. b. the money multiplier is 5 . c. the planned investment schedule is such that at a 4 percent rate of interest, investment =$1450 billion. at 5 percent, investment is $1420 billion. d. the investment multiplier is 3 . e.. the initial equilibrium level of real gdp is $12 trillion. f. the equilibrium rate of interest is 4 percent now the fed engages in contractionary monetary policy. it sells $1 billion worth of bonds, which reduces the money supply, which in turn raises the market rate of interest by 1 percentage point. calculate the decrease in money supply after fed's sale of bonds: $nothing billion.

Answers: 2

Answer

Business, 22.06.2019 09:00

Afood worker has just rinsed a dish after cleaning it.what should he do next?

Answers: 2

Answer

Business, 22.06.2019 09:30

Which are the best examples of costs that should be considered when creating a project budget?

Answers: 2

Answer

Business, 22.06.2019 09:30

The 39 percent and 38 percent tax rates both represent what is called a tax "bubble." suppose the government wanted to lower the upper threshold of the 39 percent marginal tax bracket from $335,000 to $208,000. what would the new 39 percent bubble rate have to be? (do not round intermediate calculations. enter your answer as a percent rounded to 2 decimal places,e.g., 32.16.)

Answers: 3

Answer

You know the right answer?

Assume your computer is able to complete 4 double floating-point operations per cycle when operands...

Questions