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Gnoma [55]
3 years ago
12

Create a class named BaseballGame that contains data fields for two team names and scores for each team in each of nine innings.

names should be an array of two strings and scores should be a two-dimensional array of type int; the first dimension indexes the team (0 or 1) and the second dimension indexes the inning. Create get and set methods for each field. The get and set methods for the scores should require a parameter that indicates which inning’s score is being assigned or retrieved. Do not allow an inning score to be set if all the previous innings have not already been set. If a user attempts to set an inning that is not yet available, issue an error message. Also include a method named display in DemoBaseballGame.java that determines the winner of the game after scores for the last inning have been entered. (For this exercise, assume that a game might end in a tie.) Create two subclasses from BaseballGame: HighSchoolBaseballGame and LittleLeagueBaseballGame. High school baseball games have seven innings, and Little League games have six innings. Ensure that scores for later innings cannot be accessed for objects of these subtypes.
Computers and Technology
1 answer:
m_a_m_a [10]3 years ago
6 0

Answer:

Check the explanation

Explanation:

BaseballGame:

public class BaseballGame {

  protected String[] names = new String[2];

  protected int[][] scores;

  protected int innings;

  public BaseballGame() {

      innings = 9;

      scores = new int[2][9];

      for(int i = 0; i < 9; i++)

          scores[1][i] = scores[0][i] = -1;

  }

 

  public String getName(int team) {

      return names[team];

  }

  public void setNames(int team, String name) {

      names[team] = name;

  }

 

  public int getScore(int team, int inning) throws Exception {

      if(team < 0 || team >= 2)

          throw new Exception("Team is ut of bounds.");

      if(inning < 0 || inning >= innings)

          throw new Exception("Inning is ut of bounds.");

     

      return scores[team][inning];

  }

  public void setScores(int team, int inning, int score) throws Exception {

      if(team < 0 || team >= 2)

          throw new Exception("Team is ut of bounds.");

      if(inning < 0 || inning >= innings)

          throw new Exception("Inning is ut of bounds.");

      if(score < 0)

          throw new Exception("Score is ut of bounds.");

      for(int i = 0; i < inning; i++)

          if(scores[team][i] == -1)

              throw new Exception("Previous scores are not set.");

     

      scores[team][inning] = score;

  }

 

}

HighSchoolBaseballGame:

public class HighSchoolBaseballGame extends BaseballGame {

  public HighSchoolBaseballGame() {

      innings = 7;

      scores = new int[2][7];

      for(int i = 0; i < 7; i++)

          scores[1][i] = scores[0][i] = -1;

  }

}

LittleLeagueBaseballGame:

public class LittleLeagueBaseballGame extends BaseballGame {

  public LittleLeagueBaseballGame() {

      innings = 6;

      scores = new int[2][6];

      for(int i = 0; i < 6; i++)

          scores[1][i] = scores[0][i] = -1;

  }

}

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Computer Networks - Queues
lyudmila [28]

Answer:

the average arrival rate \lambda in units of packets/second is 15.24 kbps

the average number of packets w waiting to be serviced in the buffer is 762 bits

Explanation:

Given that:

A single channel with a capacity of 64 kbps.

Average packet waiting time T_w in the buffer = 0.05 second

Average number of packets in residence = 1 packet

Average packet length r = 1000 bits

What are the average arrival rate \lambda in units of packets/second and the average number of packets w waiting to be serviced in the buffer?

The Conservation of Time and Messages ;

E(R) = E(W) + ρ

r = w + ρ

Using Little law ;

r = λ × T_r

w =  λ × T_w

r /  λ = w / λ  +  ρ / λ

T_r =T_w + 1 / μ

T_r = T_w +T_s

where ;

ρ = utilisation fraction of time facility

r = mean number of item in the system waiting to be served

w = mean number of packet waiting to be served

λ = mean number of arrival per second

T_r =mean time an item spent in the system

T_w = mean waiting time

μ = traffic intensity

T_s = mean service time for each arrival

the average arrival rate \lambda in units of packets/second; we have the following.

First let's determine the serving time T_s

the serving time T_s  = \dfrac{1000}{64*1000}

= 0.015625

now; the mean time an item spent in the system T_r = T_w +T_s

where;

T_w = 0.05    (i.e the average packet waiting time)

T_s = 0.015625

T_r =  0.05 + 0.015625

T_r =  0.065625

However; the  mean number of arrival per second λ is;

r = λ × T_r

λ = r /  T_r

λ = 1000 / 0.065625

λ = 15238.09524 bps

λ ≅ 15.24 kbps

Thus;  the average arrival rate \lambda in units of packets/second is 15.24 kbps

b) Determine the average number of packets w waiting to be serviced in the buffer.

mean number of packets  w waiting to be served is calculated using the formula

w =  λ × T_w

where;

T_w = 0.05

w = 15238.09524 × 0.05

w = 761.904762

w ≅ 762 bits

Thus; the average number of packets w waiting to be serviced in the buffer is 762 bits

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