Question

我一直在试验Simpy中的总线故障示例手册，我真的很难理解为什么我创建多个Bus实例，最后一个实例似乎得到了“ 序列“在第一次修复后。我修改了示例代码手册非常略低于initialize（）语句下面的创建两个总线实例（Bus1和Bus2）。这是我的代码：

from SimPy.Simulation import *

class Bus(Process):

  def operate(self,repairduration,triplength):    # PEM
     tripleft = triplength
        # "tripleft" is the driving time to finish trip
        # if there are no further breakdowns
     while tripleft > 0:
        yield hold,self,tripleft      # try to finish the trip
            # if a breakdown intervenes
        if self.interrupted():
              print self.interruptCause.name, 'at %s' %now()
              tripleft=self.interruptLeft
                # update driving time to finish
                # the trip if no more breakdowns
              self.interruptReset()        # end self-interrupted state
                # update next breakdown time
              reactivate(br,delay=repairduration)
                # impose delay for repairs on self
              yield hold,self,repairduration
              print '%s repaired at %s' %(self.name, now())
        else:   # no breakdowns intervened, so bus finished trip
              break
     print 'Bus has arrived at %s' %now()

class Breakdown(Process):
   def __init__(self,myBus):
       Process.__init__(self,name='Breakdown '+myBus.name)
       self.bus=myBus

   def breakBus(self,interval):      # Process Execution Method
       while True:
          yield hold,self,interval   # driving time between breakdowns
          if self.bus.terminated(): break
            # signal "self.bus" to break itself down
          self.interrupt(self.bus)

initialize()
for i in range(1,5):
  b=Bus('Bus%s' %i)                   # create a Bus object "b" called "Bus"
  activate(b,b.operate(repairduration=20,triplength=1000))
      # create a Breakdown object "br" for bus "b", and
  br=Breakdown(b)
      # activate it with driving time between
      # breakdowns equal to 300
  activate(br,br.breakBus(300))

simulate(until=4000)
print 'SimPy: No more events at time %s' %now()

以上给出了以下输出：

Breakdown Bus1 at 300
Breakdown Bus2 at 300
Bus1 repaired at 320
Bus2 repaired at 320
Breakdown Bus1 at 600
Bus1 repaired at 620
Breakdown Bus2 at 620
Bus2 repaired at 640
Breakdown Bus1 at 900
Bus1 repaired at 920
Breakdown Bus2 at 920
Bus2 repaired at 940
Bus has arrived at 1060
Bus has arrived at 1060
SimPy: No more events at time 1240

现在，问题是：在t = 600点，为什么总线1会被修复在2号巴士发生故障之前？我原以为两辆公共汽车都会在“锁定步骤”中分解并修复。此外，如果我创建四个总线，前三个失败并得到修复在“锁步”中如下图所示;然而，4号巴士下车顺序为20 第一次修复后。我无法弄清楚为什么会发生这种情况欣赏任何人都可以提供的任何见解。它永远发生在最后一个例子。

Breakdown Bus1 at 300
Breakdown Bus2 at 300
Breakdown Bus3 at 300
Breakdown Bus4 at 300
Bus1 repaired at 320
Bus2 repaired at 320
Bus3 repaired at 320
Bus4 repaired at 320
Breakdown Bus1 at 600
Breakdown Bus2 at 600
Breakdown Bus3 at 600
Bus1 repaired at 620
Bus2 repaired at 620
Bus3 repaired at 620
Breakdown Bus4 at 620
Bus4 repaired at 640
Breakdown Bus1 at 900
Breakdown Bus2 at 900
Breakdown Bus3 at 900
Bus1 repaired at 920
Bus2 repaired at 920
Bus3 repaired at 920
Breakdown Bus4 at 920
Bus4 repaired at 940
Bus has arrived at 1060
Bus has arrived at 1060
Bus has arrived at 1060
Bus has arrived at 1060
SimPy: No more events at time 1240

谢谢，西摩

Answer 1

这似乎按预期工作。必须在总线类中初始化和激活故障过程。

from SimPy.Simulation import *

class Bus(Process):

  def __init__(self,name):
    Process.__init__(self,name)
    self.name = name
    self.br=Breakdown(self)
    activate(self.br,self.br.breakBus(300))


  def operate(self,repairduration,triplength):    # PEM
     tripleft = triplength
     while tripleft > 0:
        yield hold,self,tripleft      # try to finish the trip
        if self.interrupted():
              print self.interruptCause.name, 'at %s' %now()
              tripleft=self.interruptLeft
              self.interruptReset()        # end self-interrupted state
                # update next breakdown time
              reactivate(self.br,delay=repairduration)
              yield hold,self,repairduration
              print '%s repaired at %s' %(self.name, now())
        else:   # no breakdowns intervened, so bus finished trip
              break
     print 'Bus has arrived at %s' %now()

class Breakdown(Process):
   def __init__(self,myBus):
       Process.__init__(self,name='Breakdown '+myBus.name)
       self.bus=myBus

   def breakBus(self,interval):      # Process Execution Method
       while True:
          yield hold,self,interval   # driving time between breakdowns
          if self.bus.terminated(): break
            # signal "self.bus" to break itself down
          self.interrupt(self.bus)

initialize()

for i in range(1,5):
  b=Bus('Bus%s' %i)
  activate(b,b.operate(repairduration=20,triplength=1000))



simulate(until=4000)
print 'SimPy: No more events at time %s' %now()

这导致四个Bus同时断开并同时修复，如下面的输出所示：

Breakdown Bus1 at 300
Breakdown Bus2 at 300
Breakdown Bus3 at 300
Breakdown Bus4 at 300
Bus1 repaired at 320
Bus2 repaired at 320 
Bus3 repaired at 320
Bus4 repaired at 320
Breakdown Bus1 at 600
Breakdown Bus2 at 600
Breakdown Bus3 at 600
Breakdown Bus4 at 600
Bus1 repaired at 620
Bus2 repaired at 620
Bus3 repaired at 620
Bus4 repaired at 620
Breakdown Bus1 at 900
Breakdown Bus2 at 900
Breakdown Bus3 at 900
Breakdown Bus4 at 900
Bus1 repaired at 920
Bus2 repaired at 920
Bus3 repaired at 920
Bus4 repaired at 920
Bus has arrived at 1060
Bus has arrived at 1060
Bus has arrived at 1060
Bus has arrived at 1060
SimPy: No more events at time 1200

多个实例的故障/修复的简单优先级

1 个答案: