1报童模型的定义和阐述

每天早上，报童以批发价$c$元/份采购当天的报纸，然后以零售价$p$元/份售卖。如果当天报纸没有卖完，则以$s$元/份的价格卖给废品回收站。不失一般性，假设$p>c>s$。用随机变量$D$表示当天的需求量，并已知其概率分布函数和密度分布函数分别为$F(d)和f(d)$。求使得期望收益最大的采购量$x$。

拓展：未满足需要将支付惩罚成本$r$。

2求解过程

推导可得：
$max(x,D)+min(x,D)=x+D;min(x,0）=-max(-x,0)$

2.1利润函数

将上述公式带入利润函数求解：
$\pi (x,D)$
$=p\cdot min(x,D)+s\cdot max(x-D,0)-c\cdot x$
$=p\cdot min(x,D)+s\cdot [max(x,D)-D]-c\cdot x$
$=p\cdot min(x,D)+s\cdot [x+D-min(x,D)-D]-c\cdot x$
$=(p-s)\cdot min(x,D)-(c-s)\cdot x$

拓展的利润函数求解：
$\pi (x,D)$
$=p\cdot min(x,D)+s\cdot max(x-D,0)-r\cdot max(D-x,0)-c\cdot x$
$=p\cdot min(x,D)+s\cdot [max(x,D)-D]-r\cdot [-min(x-D,0)]-c\cdot x$
$=p\cdot min(x,D)+s\cdot [x+D-min(x,D)-D]-r\cdot [-min(x,D)+D]-c\cdot x$
$=p\cdot min(x,D)+s\cdot [x+D-min(x,D)-D]-r\cdot [max(x,D)-x-D+D]-c\cdot x$
$=(p-s)\cdot min(x,D)-r\cdot max(x,D)-(c-s-r)\cdot x$

2.2期望利润函数

原模型的期望利润函数：
$E[\pi (x,D)]$
$=(p-s)\cdot E[min(x,D)]-(c-s)\cdot x$
=(p−s)∫0∞min(x,d)f(d)dd−(c−s)⋅x.= (p-s)\int_0^\infty min(x,d)f(d)dd\,−(c−s)⋅x.=(p−s)∫0∞​min(x,d)f(d)dd−(c−s)⋅x.

拓展的期望利润函数：
$E[\pi (x,D)]$
$=(p-s)\cdot min(x,D)-r\cdot max(x,D)-(c-s-r)\cdot x$
=(p−s)∫0∞min(x,d)f(d)dd−r∫0∞max(x,d)f(d)dd−(c−s−r)⋅x.= (p-s)\int_0^\infty min(x,d)f(d)dd\,-r\int_0^\infty max(x,d)f(d)dd\,−(c−s-r)⋅x.=(p−s)∫0∞​min(x,d)f(d)dd−r∫0∞​max(x,d)f(d)dd−(c−s−r)⋅x.

2.3求解期望收益最大时的采购量

为使原函数期望收益最大，对期望收益求采购量的偏导，并令其值为$0$，即

∂E[π(x,D)]∂x=0\frac{∂ E [ π ( x , D ) ] }{∂ x} = 0∂x∂E[π(x,D)]​=0

∂E[π(x,D)]∂x\frac{∂ E [ π ( x , D ) ] }{∂ x}∂x∂E[π(x,D)]​

=(p−s)⋅∂∂x(∫0∞min⁡(x,d)f(d)dd)−(c−s)= ( p − s ) ⋅ \frac{∂} {∂x} ( \int_0^\infty min ⁡ ( x , d ) f ( d ) d d\, ) − ( c − s )=(p−s)⋅∂x∂​(∫0∞​min⁡(x,d)f(d)dd)−(c−s)

=(p−s)⋅∂∂x(∫0xd⋅f(d)dd+∫x∞x⋅f(d)dd)−(c−s)= ( p − s ) ⋅ \frac{∂} {∂x} (\int_0^x d ⋅ f ( d ) d d +\int_x^\infty x ⋅ f ( d ) d d ) − ( c − s )=(p−s)⋅∂x∂​(∫0x​d⋅f(d)dd+∫x∞​x⋅f(d)dd)−(c−s)

=(p−s)⋅[∫0x∂(d⋅f(d))∂xdd+∫x∞∂(x⋅f(d))∂xdd]−(c−s)= ( p − s ) ⋅ [ \int_0^x \frac{∂ ( d ⋅ f ( d ) ) }{∂ x} d d +\int_x^\infty \frac{∂ ( x ⋅ f ( d ) ) }{∂ x} d d ] − ( c − s )=(p−s)⋅[∫0x​∂x∂(d⋅f(d))​dd+∫x∞​∂x∂(x⋅f(d))​dd]−(c−s)

=(p−s)⋅∫x∞f(d)dd−(c−s)=(p−s)⋅[1−F(x)]−(c−s)= ( p − s ) ⋅ \int_x^\infty f ( d ) d d − ( c − s ) = ( p − s ) ⋅ [ 1 − F ( x ) ] − ( c − s )=(p−s)⋅∫x∞​f(d)dd−(c−s)=(p−s)⋅[1−F(x)]−(c−s)

显然，上式$=0$时，F(x)=p−cp−s=γF(x)=\frac{p−c}{p−s}=γF(x)=p−sp−c​=γ，则

x=F−1(γ)x=F^{-1}(γ)x=F−1(γ)

​同理，我们使拓展利润函数期望收益最大：

∂E[π(x,D)]∂x=0\frac{∂ E [ π ( x , D ) ] }{∂ x}=0∂x∂E[π(x,D)]​=0

=(p−s)⋅∂∂x(∫0∞min⁡(x,d)f(d)dd)−r⋅∂∂x(∫0∞max(x,d)f(d)dd)−(c−s−r)= ( p − s ) ⋅ \frac{∂} {∂x} ( \int_0^\infty min ⁡ ( x , d ) f ( d ) d d\, ) -r⋅\frac{∂} {∂x} ( \int_0^\infty max(x,d)f(d)dd\, ) − ( c − s -r)=(p−s)⋅∂x∂​(∫0∞​min⁡(x,d)f(d)dd)−r⋅∂x∂​(∫0∞​max(x,d)f(d)dd)−(c−s−r)

=(p−s)⋅∂∂x(∫0xd⋅f(d)dd+∫x∞x⋅f(d)dd)−r⋅∂∂x(∫0xx⋅f(d)dd+∫x∞d⋅f(d)dd)−(c−s−r)= ( p − s ) ⋅ \frac{∂} {∂x} (\int_0^x d ⋅ f ( d ) d d +\int_x^\infty x ⋅ f ( d ) d d \, )-r ⋅ \frac{∂} {∂x} (\int_0^x x⋅ f ( d ) d d +\int_x^\infty d ⋅ f ( d ) d d ) − ( c − s-r )=(p−s)⋅∂x∂​(∫0x​d⋅f(d)dd+∫x∞​x⋅f(d)dd)−r⋅∂x∂​(∫0x​x⋅f(d)dd+∫x∞​d⋅f(d)dd)−(c−s−r)

=(p−s)⋅[∫0x∂(d⋅f(d))∂xdd+∫x∞∂(x⋅f(d))∂xdd]−r⋅[∫0x∂(x⋅f(d))∂xdd+∫x∞∂(d⋅f(d))∂xdd]−(c−s−r)= ( p − s ) ⋅ [ \int_0^x \frac{∂ ( d ⋅ f ( d ) ) }{∂ x} d d +\int_x^\infty \frac{∂ ( x ⋅ f ( d ) ) }{∂ x} d d ] -r ⋅ [ \int_0^x \frac{∂ ( x ⋅ f ( d ) ) }{∂ x} d d +\int_x^\infty \frac{∂ ( d ⋅ f ( d ) ) }{∂ x} d d ] − ( c − s-r )=(p−s)⋅[∫0x​∂x∂(d⋅f(d))​dd+∫x∞​∂x∂(x⋅f(d))​dd]−r⋅[∫0x​∂x∂(x⋅f(d))​dd+∫x∞​∂x∂(d⋅f(d))​dd]−(c−s−r)

=(p−s)⋅∫x∞f(d)dd−r⋅∫0xf(d)dd−(c−s)=(p−s)⋅[1−F(x)]−r⋅F(x)−(c−s−r)= ( p − s ) ⋅ \int_x^\infty f ( d ) d d-r ⋅ \int_0^x f ( d ) d d− ( c − s ) = ( p − s ) ⋅ [ 1 − F ( x ) ] – r ⋅ F(x) − ( c − s-r )=(p−s)⋅∫x∞​f(d)dd−r⋅∫0x​f(d)dd−(c−s)=(p−s)⋅[1−F(x)]−r⋅F(x)−(c−s−r)

显然，上式$=0$时，F(x)=p−c+rp−s+r=γF(x)=\frac{p−c+r}{p−s+r}=γF(x)=p−s+rp−c+r​=γ，则x=F−1(γ)x=F^{-1}(γ)x=F−1(γ)

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参考博客：
报童问题（3）-深入分析
报童问题的简单解法
报童问题 (The Newsvendor Problem)