Henry, a graphic artist, wants to create posters. Which software should Henry use for this purpose?

Additional resources required for the projects

Added time necessary for the project

Clarification:

In any project management scenario, there will naturally be unexpected changes and additional needs, hence to successfully complete a project, one must allocate more time and resources. It is advisable that, based on the project specifics, the end user should maintain a sufficient buffer to accommodate any variations in human resources and the extra time necessary for project completion.

When planning the project, a consideration of extra time per each task is essential.

Every task within project management is categorized under distinct scopes of work.

Response:

C++ code provided below with suitable annotations

Clarification:

pattern.cpp

#include<iostream>

using namespace std;

void printCross(int n)

{

int i,j,k;

if(n%2) //odd number of lines

{

for(int i=n;i>=1;i--)

{

for(int j=n;j>=1;j--)

{

if(j==i || j==(n-i+1))

cout<<j;

else

cout<<" ";

}

cout<<"\n";

}

}

else //even number of lines

{

for(int i=1;i<=n;i++)

{

for(int j=1;j<=n;j++)

{

if(j==i || j==(n-i+1))

{

cout<<" "<<j<<" ";

}

else

cout<<" ";

}

cout<<"\n";

}

}

void printForwardSlash(int n)

{

if(n%2)

{

for(int i=n;i>=1;i--)

{

for(int j=n;j>=1;j--)

{

if(j==n-i+1)

{

cout<<j;

}

else

cout<<" ";

}

cout<<"\n";

}

}

else

{

for(int i=1;i<=n;i++)

{

for(int j=1;j<=n;j++)

{

if(j==(n-i+1))

{

cout<<j;

}

else

cout<<" ";

}

cout<<"\n";

}

}

}

void printBackwardSlash(int n)

{

if(n%2) // odd number of lines

{

for(int i=n;i>=1;i--)

{

for(int j=n;j>=1;j--)

{

if(j==i)

{

cout<<j;

}

else

cout<<" ";

}

cout<<"\n";

}

}

else //even number of lines

{

for(int i=1;i<=n;i++)

{

for(int j=1;j<=n;j++)

{

if(j==i)

{

cout<<j;

}

else

cout<<" ";

}

cout<<"\n";

}

}

}

int main()

{

int num;

char ch;

cout<<"Create a numberes shape that can be sized."<<endl;

cout<<"Input an integer [1,50] and a character [x,b,f]."<<endl;

cin>>num>>ch;

if(ch=='x' || ch=='X')

printCross(num);

else if(ch=='f' || ch=='F')

printForwardSlash(num);

else if(ch=='b' || ch=='B')

printBackwardSlash(num);

else

cout<<"\nWrong input"<<endl;

return 0;

}

The provided question is lacking details. It can be retrieved from search engines. However, please see the full question below:

Question

It points out a mistake in using a part of the performance equation as a measure of performance. For example, examine these two processors. P1 operates at a clock frequency of 4 GHz, has an average CPI of 0.9, and needs to execute 5.0E9 instructions. P2 runs at 3 GHz, with an average CPI of 0.75, needing to execute 1.0E9 instructions. 1. A common misunderstanding is assuming that the processor with the highest clock rate has the best performance. Determine if this holds true for P1 and P2. 2. Another misconception is that the processor with the greater number of executed instructions necessarily has a longer CPU time. If processor P1 processes 1.0E9 instructions and both processors have unchanged CPI values, calculate how many instructions P2 can complete in the same duration that P1 uses to execute 1.0E9 instructions. 3. A frequent error is to use MIPS (millions of instructions per second) to evaluate the performance of different processors, believing that the one with the highest MIPS is the best. Verify whether this applies to P1 and P2. 4. MFLOPS (millions of floating-point operations per second) is another common metric, defined as MFLOPS = No. FP operations / (execution time x 1E6), but it suffers from the same issues as MIPS. Assuming 40% of the instructions executed on both P1 and P2 are floating-point instructions, calculate the MFLOPS values for the programs.

Answer:

(1) We apply the following formula:

                                         CPU time = number of instructions x CPI / Clock rate

By substituting 1 GHz = 10⁹ Hz, we find:

CPU time₁ = 5 x 10⁹ x 0.9 / 4 GHz

              = 4.5 x 10⁹ / 4 x 10⁹ Hz = 1.125 s

and,

CPU time₂ = 1 x 10⁹ x 0.75 / 3 GHz

                    = 0.75 x 10⁹ / 3 x 10⁹ Hz = 0.25 s

This shows that P2 is significantly faster than P1 because CPU₂ is shorter than CPU₁

(2)

Determine the CPU time of P1 using (*)

CPU time₁ = 1 x 10⁹ x 0.9 / 4 GHz

                  = 0.9 x 10⁹ / 4 x 10⁹ Hz = 0.225 s

Next, we need the count of instructions₂ so that CPU time₂ = 0.225 s, applying (*) with clock rate₂ = 3 GHz and CPI₂ = 0.75

Thus, instruction count₂ x 0.75 / 3 GHz = 0.225 s

Consequently, instruction count₂ = 0.225 x 3 x 10⁹ / 0.75 = 9 x 10⁸

Thus, P1 can handle more instructions than P2 within the same time frame.

(3)

We remember that:

MIPS = Clock rate / CPI x 10⁶

 So, MIPS₁ = 4 GHz / 0.9 x 10⁶ = 4 x 10⁹ Hz / 0.9 x 10⁶ = 4444

MIPS₂ = 3 GHz / 0.75 x 10⁶ = 3 x 10⁹ / 0.75 x 10⁶ = 4000

This indicates that P1 has a higher MIPS

(4)

 Now we note that:

MFLOPS = FLOPS Instructions / time x 10⁶

              = 0.4 x instructions / time x 10⁶ = 0.4 MIPS

Accordingly,

                    MFLOPS₁ = 1777.6

                    MFLOPS₂ = 1600

Again, P1 boasts a greater MFLOPS