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ValentinkaMS [17]
3 years ago
6

A marketing strategy that involves a firm using different marketing mix activities to help consumers perceive the product as bei

ng different and better than competing products is referred to as __________.
a. product repositioning
b. points of difference
c. product positioning
d. product differentiation
e. market differentiation
Business
1 answer:
svet-max [94.6K]3 years ago
4 0

Answer:

It is referred to as product differentiation.

Explanation:

Product differentiation is a strategic type of marketing in which a firm uses campaigns and promotions to highlight features that make its product unique as well as the benefits of using the product or service.

This kind of marketing differentiate the firm's product or services from those of competitors and makes consumer perceive such differentiated product or service as better than other similar competing products.

Explanation:

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

<em>For 1st algorithm</em><em>: 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%.</em>

<em>For 2nd algorithm:</em><em>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%.</em>

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

<u>1st Algorithm</u>

<em>/*dgemm0: simple ijk version triple loop</em>

<em>algorithm*/</em>

<em>for (i=0; i<n; i++)</em>

<em>for (j=0; j<n; j++)</em>

<em>for (k=0; k<n; k++)</em>

<em>c[i*n+j] += a[i*n+k] * b[k*n+j];</em>

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\%

<em>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%.</em>

<u>2nd Algorithm</u>

<em>/*dgemm1: simple ijk version triple loop</em>

<em>algorithm with register reuse*/</em>

<em>for (i=0; i<n; i++)</em>

<em>for (j=0; j<n; j++) {</em>

<em>register double r = c[i*n+j];</em>

<em>for (k=0; k<n; k++)</em>

<em>r += a[i*n+k] * b[k*n+j];</em>

<em>c[i*n+j] = r;</em>

<em>}</em>

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\%

<em>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%.</em>

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