Question

我用一条跑道模拟机场。飞机等待起飞和等待起飞的飞机有2个排队。在跑道上一次只能有一架飞机，空中的所有飞机必须在任何飞机起飞前降落。这就是我到目前为止所做的事情（当我确定我不需要它们时，我将会摆脱一些无用的变量）：

public class Runway<E> {
private LinkedBlockingQueue<Plane> takeoff;
private LinkedBlockingQueue<Plane> landing;
private LinkedBlockingQueue<Plane> runway;
private int planesLanded;
private int planesTookoff;
private double averageTakeOffWait;
private double averageLandWait;
private int totalTakeoffWait;
private int totalLandingWait;
private int planesWaitingToTakeOff;
private int planesWaitingToLand;
private int maxLandingQueueLength;
private int maxTakeOffQueueLength;

public int timeToLand = 5;
public int timeToTakeoff = 4;
public double landingProbability = .1;
public double takeOffProbability = .1;
public int simulationLength = 1440;

public Runway() {
    takeoff = new LinkedBlockingQueue<>();
    landing = new LinkedBlockingQueue<>();
    runway = new LinkedBlockingQueue<>();

    planesLanded = 0;
    planesTookoff = 0;
    averageTakeOffWait = 0;
    averageLandWait = 0;
    totalTakeoffWait = 0;
    totalLandingWait = 0;
    planesWaitingToTakeOff = 0;
    planesWaitingToLand = 0;
    maxLandingQueueLength = 0;
    maxTakeOffQueueLength = 0;
    //reset();
}

public void simulate(int duration, double takeoffRate, double landingRate, int landingTime, int takeoffTime) {
    for(int count = 0; count < duration; count++) {
        if(Math.random() < takeoffRate) {
            Plane p = new Plane(landingTime, takeoffTime, count);
            takeoff.offer(p);
        }
        if(Math.random() < landingRate) {
            Plane p = new Plane(landingTime, takeoffTime, count);
            landing.offer(p);
        }

        if(runway.size() == 0) {
                if(landing.peek() != null) {
                    Plane landingPlane = landing.poll();
                    runway.offer(landingPlane);
                    planesLanded++;
                    int landTimeWaited = count - landingPlane.getArrivalTime();
                    totalLandingWait += landTimeWaited; 
                    runway.poll();
                } else if(takeoff.peek() != null) {
                    Plane takeoffPlane = takeoff.poll();
                    runway.offer(takeoffPlane);
                    planesTookoff++;
                    int takeoffTimeWaited = count - takeoffPlane.getArrivalTime();
                    totalLandingWait += takeoffTimeWaited;
                    runway.poll();
                }
            }
    }
}

public void report() {
    System.out.println(planesLanded + " planes have landed.");
    System.out.println(planesTookoff + " planes have taken off.");
    System.out.println(landing.size() + " planes still waiting to land.");
    System.out.println(takeoff.size() + " planes still waiting to takeoff.");

    System.out.println(1.0*totalTakeoffWait/planesTookoff + " average take off wait time.");
    System.out.println(1.0*totalLandingWait/planesLanded + " average landing wait time."); 

  }

}

我的问题是飞机几乎不必等待做任何事情。队列中永远不会留下任何飞机。我知道我必须以某种方式使它成为5＆＃34;分钟＆＃34;在下一架飞机做任何事之前通过，但我该怎么做？这是一个试运行：

Runway myAirport = new Runway();
    myAirport.simulate(1440, .1, .1, 5, 4);
    myAirport.report();

146 planes have landed.
155 planes have taken off.
0 planes still waiting to land.
0 planes still waiting to takeoff.
0.0 average take off wait time.
0.14383561643835616 average landing wait time.

Answer 1

在循环中，您始终清除跑道队列，这意味着您将始终清除您添加的平面。你应该做什么，是用一些时间来检查更新。

if(runway.size()==0){
    //decide to get from takeoff or landing, your current choice is always take off
    runway.offer(chosenPlane);
    inUseTime = 100;
}else{
    Plane p = runway.peek();
    if(p.getRunwayTime()>=inUseTime){
        runway.poll();
    } else{
        p.updateRunwayTime();
    }
}

在此示例中，如果飞机已完成时间，则该飞机仅从跑道中移除。这意味着在等待当前完成时，更多计划可以进入其他队列。

一个简单的机场模拟

1 个答案: