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Anylogic：如何通过概率来限制行？

[英]Anylogic: How to block a line by a probability?

原文 2019-01-11 14:12:00 4 1 simulation/ anylogic

So I'm modelling a production line (simple, with 5 processes which I modelled as Services). 因此，我正在对一条生产线进行建模（简单，有5个我建模为服务的流程）。 I'm simulating for 1 month, and during this one month, my line stops approximately 50 times (due to a machine break down). 我模拟了1个月，在这一个月中，我的生产线停止了大约50次（由于机器故障）。 This stop can last between 3 to 60 min and the avg = 12 min (depending on a triangular probability). 此停止可以持续3到60分钟，平均= 12分钟（取决于三角概率）。 How could I implement this to the model? 如何将其实施到模型中？ I'm trying to create an event but can't figure out what type of trigger I should use. 我正在尝试创建一个事件，但无法弄清楚应该使用哪种类型的触发器。

1 个解决方案

Have your services require a resource. 让您的服务需要资源。 If they are already seizing a resource like labor, that is ok, they can require more than one. 如果他们已经抢占了劳动力之类的资源，那就可以了，他们可能需要不止一个。 On the resourcePool, there is an area called "Shifts, breaks, failures, maintenance..." Check "Failures/repairs:" and enter your downtime distribution there. 在resourcePool上，有一个名为“班次，中断，故障，维护...”的区域。选中“故障/维修：”，然后在此处输入停机时间分布。

If you want to use a triangular, you need min/MODE/max, not min/AVERAGE/max. 如果要使用三角形，则需要min / MODE / max，而不是min / AVERAGE / max。 If you really wanted an average of 12 minutes with a minimum of 3 and maximum of 60; 如果您真的想要平均12分钟，最少3分钟，最多60分钟； then this is not a triangular distribution. 那么这不是三角形分布。 There is no mode that would give you an average of 12. 没有任何一种模式可以让您平均获得12分。

Average from triangular, where X is the mode: ( 3 + X + 60 ) / 3 = 12 三角的平均值，其中X为众数：（3 + X + 60）/ 3 = 12

Means X would have to be negative - not possible for there to be a negative delay time for the mode. 意味着X必须为负-模式的延迟时间不可能为负。

Look at using a different distribution. 查看使用其他发行版。 Exponential is used often for time between failures (or poisson for failures per hour). 指数通常用于表示两次失败之间的时间（或针对每小时的一次失败进行泊松）。