编写代数方程

发布于 2024-07-15 04:15:47 字数 1032 浏览 11 评论 0 原文

在另一篇文章中，MSN 为我提供了解决代数问题的良好指南（计算出价总成本）。现在，尽管我可以手动计算它，但我完全不知道如何用伪代码或代码编写它。有人可以给我一个快速提示吗？顺便说一句，我想根据最终成本计算出价。

usage cost(bid) = PIN(bid*0.10, 10, 50)
seller cost(bid) = bid*.02
added cost(bid) = PIN(ceiling(bid/500)*5, 5, 10) + PIN(ceiling((bid - 1000)/2000)*5, 0, 10)
storing cost(bid) = 100
So the final cost is something like:

final cost(bid) = PIN(bid*.1, 10, 50) + pin(ceiling(bid/500)*5, 5, 20) + PIN(ceiling((bid - 1000)/2000)*10, 0, 20) + bid*.02 + 100 + bid
Solve for a particular value and you're done.

For example, if you want the total cost to be $2000:

2000 = PIN(bid*.1, 10, 50) + pin(ceiling(bid/500)*5, 5, 10) + PIN(ceiling((bid - 1000)/2000)*5, 0, 10) + bid*.02 + 100 + bid.
Bid must be at least > 1500 and < 2000, which works out nicely since we can make those PIN sections constant:

2000 = 50 + 10 + 5 + 100 + bid*1.02
1835 = bid*1.02
bid = 1799.0196078431372549019607843137

原文

in another post, MSN gave me a good guide on solving my algebra problem (Calculating bid price from total cost). Now, even though I can calculate it by hand, I'm completely stuck on how to write this in pseudocode or code. Anyone could give me a quick hint? By the way, I want to calculate the bid given the final costs .

usage cost(bid) = PIN(bid*0.10, 10, 50)
seller cost(bid) = bid*.02
added cost(bid) = PIN(ceiling(bid/500)*5, 5, 10) + PIN(ceiling((bid - 1000)/2000)*5, 0, 10)
storing cost(bid) = 100
So the final cost is something like:

final cost(bid) = PIN(bid*.1, 10, 50) + pin(ceiling(bid/500)*5, 5, 20) + PIN(ceiling((bid - 1000)/2000)*10, 0, 20) + bid*.02 + 100 + bid
Solve for a particular value and you're done.

For example, if you want the total cost to be $2000:

2000 = PIN(bid*.1, 10, 50) + pin(ceiling(bid/500)*5, 5, 10) + PIN(ceiling((bid - 1000)/2000)*5, 0, 10) + bid*.02 + 100 + bid.
Bid must be at least > 1500 and < 2000, which works out nicely since we can make those PIN sections constant:

2000 = 50 + 10 + 5 + 100 + bid*1.02
1835 = bid*1.02
bid = 1799.0196078431372549019607843137

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‖放下 2024-07-22 04:15:47

该函数简化为：

                  / 1.02 * bid + 115   bid <   100
                  | 1.12 * bid + 105   bid <=  500
final cost(bid) = | 1.02 * bid + 160   bid <= 1000
                  | 1.02 * bid + 165   bid <= 3000
                  \ 1.02 * bid + 170   otherwise

如果您将每个部分视为一个单独的函数，则它们可以反转：

bid_a(cost) = (cost - 115) / 1.02
bid_b(cost) = (cost - 105) / 1.12
bid_c(cost) = (cost - 160) / 1.02
bid_d(cost) = (cost - 165) / 1.02
bid_e(cost) = (cost - 170) / 1.02

如果您将成本代入每个函数，您将获得该范围的估计出价。您必须检查该值是否确实在该函数的有效范围内。

示例：

cost = 2000

bid_a(2000) = (2000 - 115) / 1.02 = 1848  Too big! Need to be < 100
bid_b(2000) = (2000 - 105) / 1.12 = 1692  Too big! Need to be <= 500
bid_c(2000) = (2000 - 160) / 1.02 = 1804  Too big! Need to be <= 1000
bid_d(2000) = (2000 - 165) / 1.02 = 1799  Good. It is <= 3000
bid_e(2000) = (2000 - 170) / 1.02 = 1794  Too small! Need to be > 3000

Just to check:

final cost(1799) = 1.02 * 1799 + 165 = 2000   Good!

由于原始函数是严格递增的，因此这些函数中最多有一个会给出可接受的值。但对于某些输入，它们都不会给出良好的值。这是因为原始函数跳过了这些值。

final cost(1000) = 1.02 * 1000 + 160 = 1180
final cost(1001) = 1.02 * 1001 + 165 = 1186

因此，没有函数可以给出可接受的值，例如 cost = 1182。

The function simplifies to:

                  / 1.02 * bid + 115   bid <   100
                  | 1.12 * bid + 105   bid <=  500
final cost(bid) = | 1.02 * bid + 160   bid <= 1000
                  | 1.02 * bid + 165   bid <= 3000
                  \ 1.02 * bid + 170   otherwise

If you consider each piece as a separate function, they can be inverted:

bid_a(cost) = (cost - 115) / 1.02
bid_b(cost) = (cost - 105) / 1.12
bid_c(cost) = (cost - 160) / 1.02
bid_d(cost) = (cost - 165) / 1.02
bid_e(cost) = (cost - 170) / 1.02

If you plug your cost into each function you get an estimated bid value for that range. You must check that this value indeed is within that functions valid range.

Example:

cost = 2000

bid_a(2000) = (2000 - 115) / 1.02 = 1848  Too big! Need to be < 100
bid_b(2000) = (2000 - 105) / 1.12 = 1692  Too big! Need to be <= 500
bid_c(2000) = (2000 - 160) / 1.02 = 1804  Too big! Need to be <= 1000
bid_d(2000) = (2000 - 165) / 1.02 = 1799  Good. It is <= 3000
bid_e(2000) = (2000 - 170) / 1.02 = 1794  Too small! Need to be > 3000

Just to check:

final cost(1799) = 1.02 * 1799 + 165 = 2000   Good!

Since the original function is strictly increasing, at most one of those functions will give an acceptable value. But for some inputs none of them will give a good value. This is because the original function jumps over those values.

final cost(1000) = 1.02 * 1000 + 160 = 1180
final cost(1001) = 1.02 * 1001 + 165 = 1186

So no function will give an acceptable value for cost = 1182 for example.

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葬﹪忆之殇 2024-07-22 04:15:47

由于使用了PIN和ceiling，我没有看到一个简单的方法来反转计算。假设 bid 具有固定精度（我猜点后面有两位小数），您始终可以使用二分搜索（因为函数是单调的）。

编辑：经过更多思考，我观察到，采用x = bid*1.02 + 100，我们得到最终成本在x+15（不含）和x+70（含）之间（即x+15 <最终成本）。考虑到此范围的大小 (70-15=55) 以及 bid 的特殊值（参见下面的注释）的差距都大于此，您可以取x+15 =最终成本和x+70 =最终成本，获得正确的使用情况/值和增加的成本，并简单地求解该方程（不再有其中的 PIN 或 ceiling）。

为了说明这一点，让最终成本为 222。从x+15 = 222得出bid = 107/1.02 = 104.90。那么我们可以得到使用成本由 bid*0.1 给出，额外成本为 5。换句话说，我们得到最终成本 = bid*0.1 + bid*0.02 + 5 + 100 + bid = bid*1.12 + 105，因此 bid = (222-105)/1.12 = 104.46。由于此 bid 值意味着采用了正确的使用值和额外费用，因此我们知道这就是解决方案。

但是，如果我们首先查看 x+70 = 222，我们会得到以下结果。首先，我们得出这个假设 bid = 52/1.02 = 50.98。这意味着使用成本为 10，额外成本为 5。因此，我们得到最终成本 = 10 + bid*0.02 + 5 + 100 + bid = bid*1.02 + 115，因此 bid = (222-115)/1.02 = 104.90 。但如果 bid 为 104.90 则使用成本不是 10 而是 bid*0.1，因此这不是正确的解决方案。

我希望我解释得足够清楚。如果没有，请告诉我。

注意：对于特殊值，我指的是那些定义使用值和附加成本的函数发生变化的值。例如，对于使用成本，这些值为 100 和 500：低于 100 使用 10，高于 100 使用 10 >500 您使用 50，中间您使用 bid*0.1。