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Former Member

[PP]Alternative item group - BOM使用strategy 1(可能性百分数)的疑问

各位好:

我在BOM中对两个子项物料A和B定义了Alternative item group ,假设为G。

使用的策略是1-Manual maintenance/by usage probability。

父项基本数量1ea,子项A和B在BOM中的需求数和可能性百分数分别为:

A 10ea 30%可能性

B 5ea 50%可能性

(注意这里的可能性总和是80%,不是100%)

我想请问在生产订单中对该两组件的需求数是怎样分配的?提供两种分配方法(假设订单数1ea):

1. 各自需求数 = 各自在CS03的需求数 * 各自可能性百分比 [遇小数点进位]

即: A需求数 = 10ea * 30% = 3ea;

B需求数 = 5ea * 50% = 2.5ea,进位至3ea。

2. 各自需求数 = 各自在CS03的需求数 * 各自可能性百分比占总百分比的比例 [遇小数点进位]

即: A需求数 = 10ea * 30 % / 80 % = 3.75ea ,进位至4ea。

B需求数 = 5ea * 30 % / 50% = 3ea。

实际上,我在ECC6.0里测试的算法跟1的算法是一致的。但是网上有人测试的算法跟2的算法是一致的。

有朋友能给出算法的逻辑和官方一点的说明吗?

多谢了!

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1 Answer

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    Former Member
    Sep 20, 2010 at 06:32 AM

    你的是对的。

    Examples

    Withdrawal based on usage probability

    Strategy: value 1

    An alternative item group is maintained for each assembly. This

    alternative item group contains four materials with differing usage

    probabilities. Dependent requirements are passed on with 100 as

    quantity.

    Material Usage probability (%) Requirement

    M-1 70 70

    M-2 20 20

    M-3 20 20

    M-4 0 no requirement

    In this case, the materials are planned in excess since the sum of the

    usage probabilities is greater than 100%.

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