1报童模型的定义和阐述
每天早上，报童以批发价 c c c元/份采购当天的报纸，然后以零售价 p p p元/份售卖。如果当天报纸没有卖完，则以 s s s元/份的价格卖给废品回收站。不失一般性，假设 p > c > s p > c > s p>c>s。用随机变量 D D 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 ) F(d)和f(d) F(d)和f(d)。求使得期望收益最大的采购量 x x x。
拓展：未满足需要将支付惩罚成本 r r r。
2求解过程
推导可得：
m
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max(x,D)+min(x,D)=x+D; min(x,0）=-max(-x,0)
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：
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π(x,D)
π(x,D)
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=p⋅min(x,D)+s⋅max(x−D,0)−c⋅x
=p⋅min(x,D)+s⋅max(x−D,0)−c⋅x
=
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=p⋅min(x,D)+s⋅[max(x,D)−D]−c⋅x
=p⋅min(x,D)+s⋅[max(x,D)−D]−c⋅x
=
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=p⋅min(x,D)+s⋅[x+D−min(x,D)−D]−c⋅x
=p⋅min(x,D)+s⋅[x+D−min(x,D)−D]−c⋅x
=
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=(p−s)⋅min(x,D)−(c−s)⋅x
=(p−s)⋅min(x,D)−(c−s)⋅x
拓展的利润函数求解：
π
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π(x,D)
π(x,D)
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=p⋅min(x,D)+s⋅max(x−D,0)-r⋅max(D-x,0)-c⋅x
=p⋅min(x,D)+s⋅max(x−D,0)−r⋅max(D−x,0)−c⋅x
=
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=p⋅min(x,D)+s⋅[max(x,D)−D]−r⋅[-min(x-D,0)]-c⋅x
=p⋅min(x,D)+s⋅[max(x,D)−D]−r⋅[−min(x−D,0)]−c⋅x
=
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=p⋅min(x,D)+s⋅[x+D−min(x,D)−D]-r⋅[-min(x,D)+D]−c⋅x
=p⋅min(x,D)+s⋅[x+D−min(x,D)−D]−r⋅[−min(x,D)+D]−c⋅x
=
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=p⋅min(x,D)+s⋅[x+D−min(x,D)−D]-r⋅[max(x,D)-x-D+D]−c⋅x
=p⋅min(x,D)+s⋅[x+D−min(x,D)−D]−r⋅[max(x,D)−x−D+D]−c⋅x
=
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=(p−s)⋅min(x,D)-r⋅max(x,D)−(c−s-r)⋅x
=(p−s)⋅min(x,D)−r⋅max(x,D)−(c−s−r)⋅x
2.2期望利润函数
原模型的期望利润函数：
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E[π(x,D)]
E[π(x,D)]
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=(p−s)⋅E[min(x,D)]−(c−s)⋅x
=(p−s)⋅E[min(x,D)]−(c−s)⋅x
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∫
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= (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：
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E[π(x,D)]
E[π(x,D)]
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=(p−s)⋅min(x,D)-r⋅max(x,D)−(c−s-r)⋅x
=(p−s)⋅min(x,D)−r⋅max(x,D)−(c−s−r)⋅x
=
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∫
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= (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 0 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 ∞ m i n ( x , d ) f ( d ) d d ) − ( 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 ( ∫ 0 x d ⋅ f ( d ) d d + ∫ x ∞ x ⋅ f ( d ) d d ) − ( 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∂(∫0xd⋅f(d)dd+∫x∞x⋅f(d)dd)−(c−s)
= ( p − s ) ⋅ [ ∫ 0 x ∂ ( d ⋅ f ( d ) ) ∂ x d d + ∫ x ∞ ∂ ( x ⋅ f ( d ) ) ∂ x d d ] − ( 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 ) d d − ( 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 =0 =0时， F ( x ) = p − c p − s = γ F(x)=\frac{p−c}{p−s}=γ F(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：文章来源：https://www.toymoban.com/news/detail-597790.html
∂ E [ π ( x , D ) ] ∂ x = 0 \frac{∂ E [ π ( x , D ) ] }{∂ x}=0 ∂x∂E[π(x,D)]=0
= ( p − s ) ⋅ ∂ ∂ x ( ∫ 0 ∞ m i n ( x , d ) f ( d ) d d ) − r ⋅ ∂ ∂ x ( ∫ 0 ∞ m a x ( x , d ) f ( d ) d d ) − ( 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 ( ∫ 0 x d ⋅ f ( d ) d d + ∫ x ∞ x ⋅ f ( d ) d d ) − r ⋅ ∂ ∂ x ( ∫ 0 x x ⋅ f ( d ) d d + ∫ x ∞ d ⋅ f ( d ) d d ) − ( 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∂(∫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 ) ⋅ [ ∫ 0 x ∂ ( d ⋅ f ( d ) ) ∂ x d d + ∫ x ∞ ∂ ( x ⋅ f ( d ) ) ∂ x d d ] − r ⋅ [ ∫ 0 x ∂ ( x ⋅ f ( d ) ) ∂ x d d + ∫ x ∞ ∂ ( d ⋅ f ( d ) ) ∂ x d d ] − ( 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 ) d d − r ⋅ ∫ 0 x f ( d ) d d − ( 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⋅∫0xf(d)dd−(c−s)=(p−s)⋅[1−F(x)]−r⋅F(x)−(c−s−r)
显然，上式 = 0 =0 =0时， F ( x ) = p − c + r p − s + r = γ F(x)=\frac{p−c+r}{p−s+r}=γ F(x)=p−s+rp−c+r=γ，则 x = F − 1 ( γ ) x=F^{-1}(γ) x=F−1(γ)
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)文章来源地址https://www.toymoban.com/news/detail-597790.html
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