Selling apples for two people

Question

Let $A=\{0,1,\dots,n\}$. We have $n$ apples to be divided between two people. The two people have the same nondecreasing function $f:A\rightarrow\mathbb{R}_{\ge 0}$ that indicates the value upon receiving a certain number of apples. This function satisfies $f(0)=0$ and $f(x+y)\leq f(x)+f(y)$ for any $x,y\in A$.

Let $M$ denote the optimal total value from dividing the apples.

Define $m$ as follows. We are allowed to set a price $p$ per apple (possibly depending on $f$) and do the following to the first person, then the second person: If there are $k$ apples left, the next person will choose the number $l\in[0,k]$ of apples that maximizes the profit $f(l)-p\cdot l$. Let $m$ be the optimal total value from selling the apples in this way.

Clearly, $m\leq M$. On the other hand, by setting price $p=0$ we can see that $m\geq \frac{1}{2}M$. Is there a constant $\frac{1}{2}

Answer 1

I will argue $m\geq \tfrac23 M.$ Set the price $p=M/3n.$ Denote the number of apples chosen by the first and second person by $n_1$ and $n_2$ respectively, so $m\geq f(n_1)+f(n_2).$

If the optimal quantities in an optimal solution are $n_1^*$ and $n-n_1^*,$ the midpoint of $(n_1^*,f(n_1^*))$ and $(n-n_1^*,f(n-n_1^*))$ is $(\tfrac n2,\tfrac M2).$ One of the points must lie on or above the line $y=M\tfrac{n+x}{3n}$ because this line passes through $(\tfrac n2,\tfrac M2).$ So $(n_1,f(n_1))$ also lies above this line: $$f(n_1)\geq M\tfrac{n+n_1}{3n}.\tag{1}$$

If $n_1\leq \tfrac n2$ then $n_2=n_1$ and $m\geq 2f(n_1)\geq \tfrac {2M}{3}.$

For $n_1>\tfrac n2,$ we want to show that $f(n_2)\geq \tfrac 23 M-f(n_1).$ This will be the case as long as there is a point $(x,f(x)),$ with $x\leq n-n_1,$ lying on or above the line $y=\tfrac 23 M-f(n_1)+\tfrac M{3n}x.$ Let $k=\lfloor n_1/(n-n_1)\rfloor$ so that $$n_1=k(n-n_1)+((k+1)n_1-kn)$$ with $0\leq (k+1)n_1-kn

• $f(n-n_1)<\tfrac 23 M-f(n_1)+\tfrac M{3n}(n-n_1),$ and
• $f((k+1)n_1-kn)<\tfrac 23 M-f(n_1)+\tfrac M{3n}((k+1)n_1-kn)$

Subadditivity gives $f(n_1)\leq kf(n-n_1)+f((k+1)n_1-kn),$ so \begin{align*} f(n_1)&\leq (k+1)(\tfrac 23 M-f(n_1))+\tfrac {kM}{3n}(n-n_1)+\tfrac M{3n}((k+1)n_1-kn)\\ (k+2)f(n_1)&\leq (k+1)\tfrac 23 M+\tfrac{Mn_1}{3n}. \end{align*} Using (1) we get $$(k+2)M\tfrac{n+n_1}{3n}\leq (k+1)\tfrac 23 M+\tfrac{Mn_1}{3n}.$$ Dividing by $M/3n$ gives $(k+2)(n+n_1)\leq 2(k+1)n+n_1,$ which is $(k+1)n_1-nk<0,$ a contradiction. This proves that $f(n_2)\geq \tfrac 23M-f(n_1)$ and hence $m\geq \tfrac 23M.$

For an upper bound $c\leq \tfrac 2 3$, take a large even integer $n,$ a small $\epsilon>0,$ and consider the function defined by $f(x)=n+x(1-\epsilon)$ for $0

Note that even if the prices are allowed to vary ("non-uniform pricing"), there is a bound of $\tfrac 3 4$ given by Ezra et al, "Pricing Identical Items" Proposition 5.6.

Answer 2

I can propose to MSE community the first and not very checked improvement of a trivial upper bound for $c$. Namely, we can show $c\le\frac{2n-3}{2n-2}$ for $n\ge 4$. Indeed, pick any number $1\le r<\frac{n-1}{n-2}$ and put $f(1)=1$, $f(2)=f(3)=\dots=f(n-1)=r$, and $f(n)=r+1$. Then for each division $n=x+y$ with $x,y\in A$ we have $f(x)+f(y)=2r$ iff both $x$ and $y$ are at least $2$ and $f(x)+f(y)=r+1$, otherwise. Thus $M=2r$. If a price $p$ forces a divison $n=x+y$ with $x\ge 2$ then $r-xp\ge 1-p$ and $r-xp\ge r+1-np$. That is $\frac{r-1}{x-1}\ge p\ge\frac 1{n-x}$ and $r\ge\frac{n-1}{n-x}\ge \frac{n-1}{n-2}$, which is impossible. Thus $m=r+1$. Hence $$c\le \frac{1+\frac{n-1}{n-2}}{\frac{n-1}{n-2}}=\frac{2n-3}{2n-2}.$$

The above example was plainly generalized from a case $n=4$. I guess a different function $f$ can provide better upper bounds for $c$, but I didn’t figured (yet) such a function $f$. I guess patterns for it can be suggested by computer search for $n\le 10$.