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我正在研究 python 2 中的排队模拟模型，该模型有工作进入系统并请求多个资源。每个到达的工作都需要不同数量的资源（而不是特定的资源！），并在不同的时间内使用这些资源。

我发现的任何此类示例都请求特定资源，例如 res[1] 和 res[2]。我只需要请求 2 个资源。

此外，我的作业仅在第一个作业完成后才运行。我知道我的 for 循环存在问题，但我不确定如何正确修复它。在这种情况下，因为有 2 个资源 a 和 b 应该能够在时间 1 运行。但是 b 等到 a 完成。奇怪的。

我将不胜感激有关请求多种资源并在适当的时间运行作业的帮助。

到目前为止，这是我的代码：

import simpy


#resource
class SuperComputer:
    def __init__(self, env):
        self.nodes = simpy.Resource(env, capacity = 2)


#users of resource
class Job:
    #enter: time the job enters the system
    #timeout is how long the job occupies a resource for
    #resources is how many resources a job needs in order to run
    def __init__(self, env, name, enter,timeout, resources):
        self.env = env
        self.name = name 
        self.enter = enter
        self.timeout = timeout
        self.resources = resources


#system
def system(env, jobs, super_computer):
    with super_computer.nodes.request() as req:

        for job in jobs:
            print('%s arrives at %s' % (job.name, job.enter)) 
            yield req
            yield env.timeout(job.enter)
            print('%s starts running with %s resources at %s' % (job.name, job.resources, env.now)) 
            yield env.timeout(job.timeout)
            print('%s completed job at %s' % (job.name, env.now)) 




env = simpy.Environment()
super_computer = SuperComputer(env) 

jobs = [
        Job(env, 'a', 1, 4, 1),
        Job(env, 'b', 1, 4, 1),
        Job(env, 'c', 1, 4, 1),
        Job(env, 'd', 1, 4, 1),
    ]


env.process(system(env, jobs, super_computer))    

env.run(50)

输出：

a arrives at 1
a starts running with 1 resources at 1
a completed job at 5
b arrives at 1
b starts running with 1 resources at 6
b completed job at 10
c arrives at 1
c starts running with 1 resources at 11
c completed job at 15
d arrives at 1
d starts running with 1 resources at 16
d completed job at 20
4

1 回答 1

1

问题是您的代码需要完成一轮 for 循环，然后才能进入下一轮。

如果将 for 循环放在进程之外，您的代码就可以工作，而是为每个作业创建一个允许它们并行操作的新进程。

import simpy


#resource
class SuperComputer:
        def __init__(self, env):
        self.nodes = simpy.Resource(env, capacity = 2)


#users of resource
class Job:
    #enter: time the job enters the system
    #timeout is how long the job occupies a resource for
    #resources is how many resources a job needs in order to run
    def __init__(self, env, name, enter,timeout, resources):
        self.env = env
        self.name = name 
        self.enter = enter
        self.timeout = timeout
        self.resources = resources


#system
def system(env, job, super_computer):
    with super_computer.nodes.request() as req:
        print('%s arrives at %s' % (job.name, job.enter)) 
        yield req
        yield env.timeout(job.enter)
        print('%s starts running with %s resources at %s' % (job.name, job.resources, env.now)) 
        yield env.timeout(job.timeout)
        print('%s completed job at %s' % (job.name, env.now)) 


env = simpy.Environment()
super_computer = SuperComputer(env) 

jobs = [
        Job(env, 'a', 1, 4, 1),
        Job(env, 'b', 1, 4, 1),
        Job(env, 'c', 1, 4, 1),
        Job(env, 'd', 1, 4, 1),
    ]


for job in jobs:
    env.process(system(env, job, super_computer))    

env.run(50)
于 2018-02-13T14:31:30.427 回答