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tankabanditka [31]
3 years ago
13

Your Java program will be reading input from a file name strInput.txt. Each record contains String firstname String lastName Str

ing strSalary (which will be converted to a double if it's valid- see below) char status String cityState (city and state and combine together but separated by a comma)
Sample input:
Ira Rudowsky 87654.32 F Brooklyn.NY
Jane Doe 987609.87 F NY NY
Mickey Mantle 345678.30 D Orlando FL
Fran Young 10456A.82 G Boston MA
Richard Clark 67890.32 D
Serena liams 1295609.87 D Denver, CO
Read from the file one record at a time and process as follows:
1. Any record whose status code is neither D or F should be written to the screen indicating the error. The next record should be read in
2. If the status code is D or F, validate the salary that it contains only digits and one decimal point 3 positions from the right (indicating cents)
a. If the salary is invalid, print the entire record to the screen, indicating the error (char at position x is not a digit or period is missing or period is in the wrong position)
b. If the salary is valid, convert it to a double c. If the status code is D, compute the bonus (double) as 12.5% of salary and 18% if the status code is F
3. Separate the cityState into two individual Strings named city and state. Use the comma to extract the city portion before the comma and the state after the comma
4. Each record that has a status of D should be written to a file named strOutputD and those record with status of F should be written to a file named strOutputRE.
a. For both files, each record printed should include firstName, lastName, status, salary, bonus, city and state
5. When all records have been read in, print to the screen the number of D, F and incorrect records processed
Engineering
1 answer:
stiks02 [169]3 years ago
6 0

Answer:

The program requires that you have the specified input files and it reads from each file at a time and processes salary in digits, states the city, state and bonus with respective first and last name as requested in the question. Note that you must have access to the mentioned output files for the program to work properly. Below is the java version of the program.

import java.io.File;

import java.io.FileNotFoundException;

import java.io.PrintWriter;

import java.util.Scanner;

class Driver

{

public static void main(String[] args) throws FileNotFoundException

{

Scanner sc = new Scanner(new File("strInput.txt"));

PrintWriter pd = new PrintWriter(new File("strOutputD"));

PrintWriter prf = new PrintWriter(new File("strOutputRF"));

String firstname = "", lastname = "", strSalary = "", status = "", cityState = "", city = "", state = "";

double salary = 0, bonus = 0;

int incorrectRecords = 0;

int dRecords = 0;

int fRecords = 0;

while(sc.hasNextLine())

{

firstname = sc.next();

lastname = sc.next();

strSalary = sc.next();

status = sc.next();

cityState = sc.next();

if(!status.equals("D") && !status.equals("F"))

{

System.out.println("Records is neither D nor F. Skipping this...");

incorrectRecords++;

continue;

}

else if(status.equals("D") || status.equals("F"))

{

char c = ' ';

int i = 0;

for(i=0; i<strSalary.length() && c != '.'; i++)

{

c = strSalary.charAt(i);

if(!Character.isDigit(c))

{

System.out.println("Char at position " + (i+1) + " in salary is not a digit");

incorrectRecords++;

continue;

}

}

if(c == '.')

{

if(i+1 == strSalary.length()-1)

{

if(!Character.isDigit(strSalary.charAt(i)))

{

System.out.println("Char at position " + (i+1) + " in salary is not a digit");

incorrectRecords++;

continue;

}

if(!Character.isDigit(strSalary.charAt(i+1)))

{

System.out.println("Char at position " + (i+1+1) + " in salary is not a digit");

incorrectRecords++;

continue;

}

}

else

{

System.out.println("Period is in the wrong position. Expected at " + (strSalary.length()-3) + " but found at " + (i+1));

continue;

}

}

city = cityState.split(",")[0];

state = cityState.split(",")[1];

salary = Double.parseDouble(strSalary);

if(status.equals("D"))

{

bonus = salary * 0.125;

dRecords++;

pd.write(firstname + " " + lastname + " " + status + " " + salary + " " + bonus + " " + city + " " + state);

}

else

{

bonus = salary * 0.18;

fRecords++;

prf.write(firstname + " " + lastname + " " + status + " " + salary + " " + bonus + " " + city + " " + state);

}

}

}

System.out.println("No of D records : " + dRecords);

System.out.println("No of F records : " + fRecords);

System.out.println("No of incorrect records : " + incorrectRecords);

}

}

You might be interested in
A Rankine steam power plant is considered. Saturated water vapor enters a turbine at 8 MPa and exits at condenser at 10 kPa. The
Ray Of Light [21]

Answer:

0.31

126.23 kg/s

Explanation:

Given:-

- Fluid: Water

- Turbine: P3 = 8MPa , P4 = 10 KPa , nt = 85%

- Pump: Isentropic

- Net cycle-work output, Wnet = 100 MW

Find:-

- The thermal efficiency of the cycle

- The mass flow rate of steam

Solution:-

- The best way to deal with questions related to power cycles is to determine the process and write down the requisite properties of the fluid at each state.

First process: Isentropic compression by pump

       P1 = P4 = 10 KPa ( condenser and pump inlet is usually equal )

      h1 = h-P1 = 191.81 KJ/kg ( saturated liquid assumption )

       s1 = s-P1 = 0.6492 KJ/kg.K

       v1 = v-P1 = 0.001010 m^3 / kg

       

       P2 = P3 = 8 MPa( Boiler pressure - Turbine inlet )

       s2 = s1 = 0.6492 KJ/kg.K   .... ( compressed liquid )

- To determine the ( h2 ) at state point 2 : Pump exit. We need to determine the wok-done by pump on the water ( Wp ). So from work-done principle we have:

   

                           w_p = v_1*( P_2 - P_1 )\\\\w_p = 0.001010*( 8000 - 10 )\\\\w_p = 8.0699 \frac{KJ}{kg}

- From the following relation we can determine ( h2 ) as follows:

                          h2 = h1 + wp

                          h2 = 191.81 + 8.0699

                          h2 = 199.88 KJ/kg

                           

Second Process: Boiler supplies heat to the fluid and vaporize

- We have already evaluated the inlet fluid properties to the boiler ( pump exit property ).

- To determine the exit property of the fluid when the fluid is vaporized to steam in boiler ( super-heated phase ).

              P3 = 8 MPa

              T3 = ?  ( assume fluid exist in the saturated vapor phase )

              h3 = hg-P3 = 2758.7 KJ/kg

              s3 = sg-P3 = 5.7450 KJ/kg.K

- The amount of heat supplied by the boiler per kg of fluid to the water stream. ( qs ) is determined using the state points 2 and 3 as follows:

                          q_s = h_3 - h_2\\\\q_s = 2758.7 -199.88\\\\q_s = 2558.82 \frac{KJ}{kg}

Third Process: The expansion ( actual case ). Turbine isentropic efficiency ( nt ).

- The saturated vapor steam is expanded by the turbine to the condenser pressure. The turbine inlet pressure conditions are similar to the boiler conditions.

- Under the isentropic conditions the steam exits the turbine at the following conditions:

             P4 = 10 KPa

             s4 = s3 = 5.7450 KJ/kg.K ... ( liquid - vapor mixture phase )

             

- Compute the quality of the mixture at condenser inlet by the following relation:

                           x = \frac{s_4 - s_f}{s_f_g} \\\\x = \frac{5.745- 0.6492}{7.4996} \\\\x = 0.67947

- Determine the isentropic ( h4s ) at this state as follows:

                          h_4_s = h_f + x*h_f_g\\\\h_4_s = 191.81 + 0.67947*2392.1\\\\h_4_s = 1817.170187 \frac{KJ}{kg}        

- Since, we know that the turbine is not 100% isentropic. We will use the working efficiency and determine the actual ( h4 ) at the condenser inlet state:

                         h4 = h_3 - n_t*(h_3 - h_4_s ) \\\\h4 = 2758.7 - 0.85*(2758.7 - 181.170187 ) \\\\h4 = 1958.39965 \frac{KJ}{kg} \\

- We can now compute the work-produced ( wt ) due to the expansion of steam in turbine.

                        w_t = h_3 - h_4\\\\w_t = 2758.7-1958.39965\\\\w_t = 800.30034 \frac{KJ}{kg}

- The net power out-put from the plant is derived from the net work produced by the compression and expansion process in pump and turbine, respectively.

                       W_n_e_t = flow(m) * ( w_t - w_p )\\\\flow ( m ) = \frac{W_n_e_t}{w_t - w_p} \\\\flow ( m ) = \frac{100000}{800.30034-8.0699} \\\\flow ( m ) = 126.23 \frac{kg}{s}

Answer: The mass flow rate of the steam would be 126.23 kg/s

- The thermal efficiency of the cycle ( nth ) is defined as the ratio of net work produced by the cycle ( Wnet ) and the heat supplied by the boiler to the water ( Qs ):

                        n_t_h = \frac{W_n_e_t}{flow(m)*q_s} \\\\n_t_h = \frac{100000}{126.23*2558.82} \\\\n_t_h = 0.31

Answer: The thermal efficiency of the cycle is 0.31

       

   

7 0
3 years ago
If you are interested only in the temperature range of 20° to 40°C and the ADC has a 0 to 3V input range, design a signal condit
mario62 [17]

Explanation:

Temperature range → 0 to 80'c

respective voltage output → 0.2v to 0.5v

required temperature range 20'c to 40'c

Where T = 20'c respective voltage

\begin{aligned}v_{20} &=0.2+\frac{0.5-0.8}{80} \times 20 \\&=0.2+\frac{0.3}{80} \times 20 \\V_{20} &=0.275 v\end{aligned}

\begin{aligned}\text { when } T=40^{\circ} C & \text { . } \\v_{40} &=0.2+\frac{0.5-0.2}{80} \times 40 \\&=0.35 V\end{aligned}

Therefore, Sensor output changes from 0.275v to 0.35volts for the ADC the required i/p should cover the dynamic range of ADC (ie - 0v to 3v)

so we have to design a circuit which transfers input voltage 0.275volts - 0.35v to 0 - 3v

Therefore, the formula for the circuit will be

\begin{array}{l}v_{0}=\left(v_{i n}-0.275\right) G \\\sigma=\ldots \frac{3-0}{0.35-0.275}=3 / 0.075=40 \\v_{0}=\left(v_{i n}-0.275\right) 40\end{array}

The simplest circuit will be a op-amp

NOTE: Refer the figure attached

Vs is sensor output

Vr is the reference volt, Vr = 0.275v

\begin{aligned}v_{0}=& v_{s}-v_{v}\left(1+\frac{R_{2}}{R_{1}}\right) \\\Rightarrow & \frac{1+\frac{R_{2}}{R_{1}}}{2}=40 \\& \frac{R_{2}}{R_{1}}=39 \quad \Rightarrow\end{aligned}

choose R2, R1 such that it will maintain required  ratio

The output Vo can be connected to voltage buffer if you required better isolation.

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Answer:

b

Explanation:

only if there signal is turned on

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