Simpy如何让项在调用get方法之前等待任意时间？

Question

我想模拟一种通信过程，其中源不断生成消息并将其发送到接收器，而消息在到达目的地之前必须覆盖一定距离。但是，距离可能会随着时间的推移而改变，因此每条消息在电缆中的持续时间并不相同。我的第一种方法是使用env.timeout(t_i)，其中t_i是电缆中第条消息的时间。我的代码是这样的

import simpy
import random

class network:

    def __init__(self) -> None:
        self.env = simpy.Environment()
        self.cable = simpy.Store(self.env)

    def generateMessage(self):
        
        i = 0
        while True:

            message = (f'message: {i}', f'travel time {0.1 + round(random.random(), 3)}', (round(random.random(), 3)))
            i += 1
            print(f'gen at: {self.env.now} {message}')
            yield self.env.timeout(0.1)
            self.cable.put(message)

    def receieveMessage(self):

        while True:

            # wait for message to travel in the cable for 'sometime' before arrive
            
            message = yield self.cable.get()
            yield self.env.timeout(message[2]) #<- This is not the right approach

            print(f'receieve at {self.env.now}, message: {message[:2]}')


    def run(self):

        self.env.process(self.generateMessage())
        self.env.process(self.receieveMessage())

        self.env.run(until=1)

q = network()
q.run()

输出在哪里

gen at: 0 ('message: 0', 'travel time 0.22', 0.22)
gen at: 0.1 ('message: 1', 'travel time 0.71', 0.71)
gen at: 0.2 ('message: 2', 'travel time 0.377', 0.377)
gen at: 0.30000000000000004 ('message: 3', 'travel time 1.056', 1.056)
receieve at 0.32, message: ('message: 0', 'travel time 0.22')
gen at: 0.4 ('message: 4', 'travel time 0.138', 0.138)
gen at: 0.5 ('message: 5', 'travel time 0.45299999999999996', 0.45299999999999996)
gen at: 0.6 ('message: 6', 'travel time 1.024', 1.024)
gen at: 0.7 ('message: 7', 'travel time 0.695', 0.695)
gen at: 0.7999999999999999 ('message: 8', 'travel time 0.503', 0.503)
gen at: 0.8999999999999999 ('message: 9', 'travel time 0.702', 0.702)
gen at: 0.9999999999999999 ('message: 10', 'travel time 0.75', 0.75)

这不是我想要的，因为在这种方法中，receieveMessage方法的每个循环在下一次迭代之前等待travel_time，而不是在通过receieveMessage方法之前等待travel_time的消息延迟

期望的输出应该是这样的

gen at: 0 ('message: 0', 'travel time 0.207')
gen at: 0.1 ('message: 1', 'travel time 0.735')
receieve at 0.207, message: ('message: 0', 'travel time 0.207')
gen at: 0.2 ('message: 2', 'travel time 0.498')
gen at: 0.3 ('message: 3', 'travel time 0.492')
gen at: 0.4 ('message: 4', 'travel time 0.864')
gen at: 0.5 ('message: 5', 'travel time 0.241')
gen at: 0.6 ('message: 6', 'travel time 0.505')
receieve at 0.698, message: ('message: 2', 'travel time 0.498')
gen at: 0.7 ('message: 7', 'travel time 0.76')
receieve at 0.741, message: ('message: 5', 'travel time 0.241')
receieve at 0.792, message: ('message: 3', 'travel time 0.492')
gen at: 0.8 ('message: 8', 'travel time 0.815')
receieve at 0.835, message: ('message: 1', 'travel time 0.735')
gen at: 0.9 ('message: 9', 'travel time 0.104')
gen at: 1 ('message: 10', 'travel time 0.524')

# Message of generated message which should not print out be cause the util arg.
receieve at 1.004, message: ('message: 9', 'travel time 0.104')
receieve at 1.105, message: ('message: 6', 'travel time 0.505')
receieve at 1.264, message: ('message: 4', 'travel time 0.864')
receieve at 1.46, message: ('message: 7', 'travel time 0.76')
receieve at 1.524, message: ('message: 10', 'travel time 0.524')
receieve at 1.615, message: ('message: 8', 'travel time 0.815')

我应该如何做到这一点？我正在考虑为每条消息引入另一个流程，但这似乎只是一个想法

Answer 1

我认为simpy更有用，如果你可以一次发送多条信息。我还认为，当您建议创建另一个流程时，您的思路是正确的

我将生成器与发送进程分离，以便消息可以作为独立进程竞争公共资源（通道）。打开通道查看消息发送时间戳的重叠情况

"""
quick sim to send messages over a network chanel

programmer: Michael R. Gibbs
"""

import simpy
import random

class Network():

    def __init__(self, env, num_of_chanels):
        """
        The num_of_chanels limits how many messages
        can be "sending" at one time.
        If all the chanels are busy, then new messages queue up
        for next available chanel
        """
        self.env = env
        self.num_of_chanels = num_of_chanels

        self.chanels = simpy.Resource(env, num_of_chanels)

    def send_mess(self, mess):
        """
        sim process for sending a message
        All the send mess process queue up for a channel resource
        Then with the channel 'sends' the messaage
        The release the channel for the next senm mess process in the queue

        mess is a tuple of (id, message text, send time)
        """

        print(f'{self.env.now}: - message: {mess[0]} being queued')

        with self.chanels.request() as req:
            yield req

            print(f'{self.env.now}: - message: {mess[0]} being sent')

            yield self.env.timeout(mess[2])

            print(f'{self.env.now}: - message: {mess[0]} has been sent {mess[1]}')


def gen_messages(env, network):
    """
    Generates a series of messages with a random send time

    """

    id = 1

    while True:
        # change the randint params to stress the queue more
        yield env.timeout(random.randint(1,3))

        # (id, mess text, send time)
        mess = (id, 'mess ' + str(id), random.randint(1,4))
        id += 1

        # no yield here, just drop and go
        env.process(network.send_mess(mess))

# create and start sim
env = simpy.Environment()
network = Network(env,1) # start with just one chanel

env.process(gen_messages(env, network))
env.run(50)