You are configuring a wireless network with two wireless access points. Both access points connect to the same wired network. Yo
1 answer:
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Explanation:
Code with comments:
#include <stdio.h>
#include <math.h>
/* math library is required to use pow() function
First we need to create a function to calculate the Monthly Payment
It would take three inputs Rate, Years, and Principle amount
Then we converted the Rate and Months according to problem statement
pow() function is used to find (1+R)^M
Then finaly this function returns the value of Monthly Payment */
float MonthlyPay(float R, int M, int P)
{
R=R/1200;
M=M*12;
float po=pow((1+R),M);
float PayM=(R+(R/(po-1)))*P;
return PayM;
}
/* In the main() function we get 3 inputs from the user Years, Rate and Principle amount
then we call the MonthlyPay function and pass it 3 inputs that we got from the user
then we calculated the total Payment by multiplying the Monthly Pay and total number of months
then we calculated the interest Amount by subtracting the Principle Amount from total Pay
Testing is performed and output result is given at the end and also attached in the image */
int main(void)
{
float Rate;
int Year, Amount;
printf("Enter Years: ");
scanf("%d",&Year);
printf("Enter Rate: ");
scanf("%f",&Rate);
printf("Enter Principle Amount: ");
scanf("%d",&Amount);
float Pay = MonthlyPay(Rate, Year, Amount);
printf ("Monthly Payment is: %f \n", Pay);
float totalPay=Pay*Year*12;
printf ("Total Payment is: %f \n", totalPay);
float interest=totalPay-Amount;
printf ("Interest Expense is: %f \n", interest);
return 0;
}
Output:
Enter Years: 5
Enter Rate: 7
Enter Principle Amount: 12200
Monthly Payment is: 241.572767
Total Payment is: 14494.367188
Interest Expense is: 2294.367188
Answer:
The answer to this question can be described as follows:
Explanation:
Given data:
Performance of the CPU:
The Fastest Factor Fraction of Work:
Current Feature Speedup:
1.5
CPU upgrade=6000
Disk activity:
The quickest part is the proportion of the work performed:
Current Feature Speedup:
Disk upgrade=8000
System speedup formula:
Finding the CPU activity and disk activity by above formula:
CPU activity:
Disk activity:
CPU:
Formula for CPU upgrade:
DISK:
Formula for DISK upgrade:
equation (3) and (4),
Thus, for the least money the CPU alternative is the best performance upgrade.
b)
From (3) and (4) result,
The disc choice is therefore the best choice for a quicker system if you ever don't care about the cost.
c)
The break-event point for the upgrades:
=4702.19 x-0.5
= -2351.095
From (2) and (3))
Therefore, when you pay the sum for disc upgrades, all is equal $ -2351.095