Question

我正在研究python 2中的排队模拟模型，该模型具有进入系统并请求多个资源的作业。到达的每个作业都会请求不同数量的资源（而不是特定资源！）并将资源用于不同的时间。

我找到的任何这样的例子都会请求特定的资源，比如res [1]和res [2]。我只需要2个资源。

此外，我的作业仅在第一个作业完成后才会运行。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等待直到完成。奇。

我很感激请求多个资源并在适当的时间运行作业。

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

Answer 1

问题是你的代码需要完成一轮for循环才能继续下一个。

如果for循环放在流程之外，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运行的新流程。

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)

Simpy：请求多个（非特定）资源和请求顺序

1 个答案: