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GarryVolchara [31]
3 years ago
7

If coal mining produces a negative externality because it leads to environmental damage, then, at the market equilibrium, the:__

______. a. price of coal will be higher than the socially optimal price b. quantity of coal produced will be less than the socially optimal quantity. c. quantity of coal produced will be greater than the socially optimal quantity d. supply curve will lie to the left of the regulated supply curve.
Business
1 answer:
Aneli [31]3 years ago
3 0

Answer: c. quantity of coal produced will be greater than the socially optimal quantity

Explanation:

The Socially Optimal quantity is the level of production that takes into account the negative externality being produced by coal and weighs it against the marginal benefit it brings.

If coal produces the negative externality of polluting the environment here then the socially optimal quantity of coal will be less than the market equilibrium because the marker equilibrium would not have taken into account the negative externality and will lead to more coal being produced than should be.

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$300 utilities bill and immediately paid it. general journal entry to record this transaction will include a:
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4 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:

<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>

8 0
3 years ago
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