Non-Linear Ski Rental

Abstract

We consider the following generalization of the classic ski rental problem. A task of unknown duration must be carried out using one of two alternatives called “buy” and “rent”, each with a one-time startup cost and an ongoing cost which is a function of the duration. Switching from rent to buy also incurs a one-time cost. The goal is to minimize the competitive ratio, i.e., the worst-case ratio between the cost paid and the optimal cost, over all possible durations. For linear or exponential cost functions, the best deterministic and randomized on-line trategies are well known. In this work we analyze a much more general case, assuming only that the cost functions are continuous and satisfy certain mild monotonicity conditions. For this general case we provide an algorithm that computes the deterministic strategy with the best competitive ratio, and an algorithm that, given $\epsilon >0$, computes a randomized strategy whose competitive ratio is within $(1+\epsilon )$ from the best possible, in time polynomial in $\epsilon ^{-1}$. Our algorithm assumes access to a black box that can compute the functions and their inverses, as well as find their extreme points.

Appendix: Proofs for Section 3

Proof

[Proof of Lemma 3.2] Lemma 3.2 states that, if F is a finite $\epsilon$-normal strategy, then, $\rho (F) \le (1+\epsilon )\rho _{f}(F)$. Suppose that the statement of Lemma 3.2 is false. Then, by the meaning of competitive ratio according to Definition 1.4 , there is $\hat{y}$ which satisfies: $\rho (F,\hat{y})/\rho _{f}(F) > 1+\epsilon$ .In this case, $\forall t_a \in T: \rho (F,\hat{y})/\rho _{f}(F,t_a) > 1+\epsilon$, where T is the set of the time points in $I'$. We show that this inequality contradicts the assumption in the lemma. We consider the cases $\hat{y}\in [t_0,t_1)$, $\hat{y}\in [t_1,t_{N-1})$ and $\hat{y}\ge t_{N-1}$.Let q(z) denote the probability density function of F and, for simplicity, denote $q_0=q(0)\text {and} q_{\infty }=q(\infty )$.

Case 1: $t_0\le \hat{y}<t_1$

We assign $t_a=0$, and we proceed by case analysis.

• If $d_r(0)> 0$, then

  $$\begin{aligned} \frac{\rho (F,\hat{y})}{\rho _f(F,0)}= & {} \frac{\left( q_0f_b(\hat{y})+(1-q_0)f_r(\hat{y})\right) /{f_r(\hat{y})} }{\rho _f(F,0)}\\ = & {} \frac{q_0\frac{f_b(\hat{y})}{f_r(\hat{y})}+1-q_0}{q_0\frac{d_b(0)}{d_r(0)}+1-q_0}\\ &\overset{(1)}{\le} 1 < 1+\epsilon \end{aligned}$$

  Contradiction to the assumption that $\rho (F,\hat{y})/\rho _f(F,0)>1+\epsilon$.(1) We can observe that

  $$\begin{aligned} \frac{f_b(\hat{y})}{f_r(\hat{y})}= \frac{f_b(0)+(f_b(\hat{y})-f_b(0))}{f_r(0)+(f_r(\hat{y})-f_r(0))}\le \frac{f_b(0)}{f_r(0)}=\frac{d_b(0)}{d_r(0)} , \end{aligned}$$

  where the inequality follows from the assumption that $f_{rb}$ is non-decreasing monotone.

• If $d_r(0)=0$, then $q_0=0$ (because F has a finite competitive ratio), and hence:

  $$\begin{aligned} \rho (F,\hat{y})/\rho _f(F,0) = \frac{q_0f_b(\hat{y})+(1-q_0)f_r(\hat{y})}{f_r(\hat{y})}\big /\rho_f(F,0) = 1/\rho _f(F,0)<1+\epsilon , \end{aligned}$$

  contradiction.

Case 2: $t_{1}\le \hat{y}\le t_{N-1}$In this case we assign $t_a = \max \{t_i \in T: t_{i} \le \hat{y}\}.$

$$\begin{aligned} \rho (F,\hat{y})/\rho _{f}(F,t_{a}) = & \frac{\textrm{cost}(F,\hat{y})}{\textrm{cost}(F,t_{a})}\cdot \frac{\textsc{opt}(t_a)}{\textsc {opt}(\hat{y})}\\ x \overset{(1)}{\le}&\frac{\textrm{cost}(F,\hat{y})}{\textrm{cost}(F,t_{a})}\\=& \frac{q_0f_b(\hat{y})\!+\!\sum _{1\le i\le a}q_i(f_{rb}(t_{i})\!+\!c\!+\!f_b(\hat{y})) \!+\!\sum _{i> a}q_i f_r(\hat{y})}{q_0f_b(t_a)\!+\!\sum _{1\le i\le a}q_i(f_{rb}(t_{i})\!+\!c\!+\!f_b(t_a)) \!+\!\sum_{i> a}q_i f_r(t_a)}\\\overset{(2)}{\le}&\frac{\sum _{0 \le i\le a}q_i f_b(\hat{y}) +\sum _{i> a}q_i f_r(\hat{y})}{\sum _{0\le i\le a}q_i f_b(t_a) +\sum _{i> a}q_i f_r(t_a)}\\ & \overset{(3)}{\le}1+\epsilon , \end{aligned}$$

We have contradiction.

(1) $t_a \le \hat{y} \Rightarrow \textsc {opt}(t_a)\le \textsc {opt}(\hat{y})$ by monotonicity.

(2) $f_{rb}(t_i)+c\ge f_{rb}(0)+c\ge 0$

(3) T is $\epsilon$-continuous (Condition 2)

Case 3: $\hat{y}> t_{N-1}$

• If $q_{\infty }=0$, let $t_a = t_{N-1}$. Then

  $$\begin{aligned}\begin{array}{ll} \rho (F,\hat{y})/\rho _{f}(F,t_{N-1})&=\frac{q_0f_b(\hat{y})+\sum _{i=1}^N q_i(f_{rb}(t_i)+c+ f_b(\hat{y}))}{q_0f_b(t_{N-1})+\sum _{i=1}^N q_i(f_{rb}(t_i)+c+ f_b(t_{N-1}))} \cdot \frac{f_b(t_{N-1})}{f_b(\hat{y})}\\ &\overset{(1)}{\le}\frac{f_b(\hat{y})}{f_b(t_{N-1})} \cdot \frac{f_b(t_{N-1})}{f_b(\hat{y})}=1<1+\epsilon\end{array} , \end{aligned}$$

  We have contradiction. (1) $f_{rb}(t_i)+c\ge f_{rb}(0)+c\ge 0$

• Otherwise, $q_{\infty }>0$ and we have $\sup _{y\ge t_{N-1}} \frac{f_r(y)}{f_b(y)} \le (1+\epsilon )\frac{f_r(t_{N-1})}{f_b(t_{N-1})}$ because F is finite $\epsilon$-normal strategy.We assign $t_a=t_{N-1}$:

  $$\begin{aligned}\begin{array}{ll} \rho (F,\hat{y})/\rho _{f}(F,t_{N-1})&= \frac{q_0f_b(\hat{y})+ \sum _{i=1}^{N-1} q_i(f_{rb}(t_i)+c+ f_b(\hat{y}))+q_{\infty }f_r(\hat{y})}{q_0f_b(t_{N-1})+\sum _{i=1}^{N-1} q_i(f_{rb}(t_i)+c+ f_b(t_{N-1}))+q_{\infty }f_r(t_{N-1})} \cdot \frac{f_b(t_{N-1})}{f_b(\hat{y})}\\& \overset{(1)}{\le}\frac{q_0f_b(\hat{y})+\sum _{i=1}^{N-1} q_i(f_b(\hat{y}))+q_{\infty }f_r(\hat{y})}{q_0f_b(t_{N-1}) +\sum _{i=1}^{N-1} q_i(f_b(t_{N-1}))+q_{\infty }f_r(t_{N-1})} \cdot \frac{f_b(t_{N-1})}{f_b(\hat{y})}\\&= \frac{1-q_{\infty }+q_{\infty }\frac{f_r(\hat{y})}{f_b(\hat{y})}}{1-q_{\infty }+q_{\infty }\frac{f_r(t_{N-1})}{f_b(t_{N_1})}}\\ &\le \frac{f_r(\hat{y})f_b(t_{N-1})}{f_r(t_{N-1})f_b(\hat{y})} \overset{(2)}{\le}1+\epsilon,\end{array} \end{aligned}$$

  (1) $f_{rb}(t_i)+c\ge f_{rb}(0)+c\ge 0$(2) $\sup _{y\ge t_{N-1}} \frac{f_r(y)}{f_b(y)} \le (1+\epsilon )\frac{f_r(t_{N-1})}{f_b(t_{N-1})}$

We have contradiction.$\square$

Proof

[Proof of Lemma 3.3] We start by establishing some notation:

$$\begin{aligned} A_{z,y} \overset{\textrm{def}}{=} f_{rb}(z)+c+f_b(y)=f_r(z)+c+(f_b(y)-f_b(z)) \end{aligned}$$

and prove a useful lemma with this notation:

Lemma 6.1

Let $\epsilon >0$, let I be a instance, and let $I'$ be a finite instance which is $\epsilon$-compatible with I. Then for all $i\ge 2$ and $t_i\le t$ we have $\frac{A_{t_i,t}}{A_{t_{i-1},t}}\le 1+\epsilon$.

Proof

$$\begin{aligned} \frac{A_{t_i,t}}{A_{t_{i-1},t}}= & {} \frac{f_{rb}(t_{i})+c+f_b(t)}{f_{rb}(t_{i-1})+c+f_b(t)}\\\le & {} \frac{f_{rb}(t_{i})+c+f_b(t_i)}{f_{rb}(t_{i-1})+c+f_b(t_i)}\\= & {} \frac{f_{r}(t_{i})+c}{f_{rb}(t_{i-1})+c+f_b(t_i)}\\\le & {} \frac{f_{r}(t_{i})+c}{f_{r}(t_{i-1})+c}\\ \le & {} 1+\epsilon , \end{aligned}$$

by the monotonicity of $f_r$ and $f_b$, and by the fact that T is $\epsilon$-continuous w.r.t to I.$\square$

Proof

[Proof of Lemma 3.3] Recall that Lemma 3.3 says that if Q is the $\epsilon$-quantized strategy of S, and T is $\epsilon$-continuous, then $\rho _{f}(Q) \le (1+\epsilon ) \rho (S).$ Suppose the lemma is false. Then there exists $t_a\in T$ which satisfies $\rho _{f}(Q,t_a)/\rho (S) > 1+\epsilon ,$ implying that

$$\begin{aligned} \forall \hat{y} \in \mathbb {R}^+_0:\quad \rho _{f}(Q,t_a) > (1+\epsilon )\rho (S,\hat{y}). \end{aligned}$$

We prove that this leads to contradiction. Let $(q_0,q(z),q_{\infty })$ denote the probability density function of S, and $q'(z)$ denote the probability function of Q, as in Definition 3.8. We by case analysis based on the value of $t_a \in T$: $t_a=t_0=0$, $t_a \in [t_{1},t_{N-1}]$, $t_a= t_{N-1}$, and $t_a=t_{N}=\infty$.

Case 1: $t_a=t_0=0$. If $d_r(0)>0$, then let $\hat{y}=0$. In this case we have:

$$\begin{aligned} \rho _{f}(Q,0)/\rho (S,0)= &\frac{q'(0)\frac{d_b(0)}{d_r(0)}+1-q'(0)}{\frac{q_0f_b(0)+(1-q_0)f_r(0)}{f_r(0)}}\\ \overset{(1)}{=} &1 < 1+\epsilon, \text {contradiction.} \end{aligned}$$

(1) Because $q'(0) =q_0$, and $d_b(0)/d_r(0)=f_b(0)/f_r(0)$.

When $d_r(0)=0$, we deduce that $q_0=q'(0)=0$, because we can assume S has finite competitive ratio (otherwise the lemma is trivial) based on lemma 3.1. It follows that

$$\begin{aligned} \rho _{f}(Q,0)/\rho (S) =1 /\rho (S) \le 1 \le 1+\epsilon . \end{aligned}$$

Case 2: $t_{1}\le t_a< t_{N-1}$. We assign $\hat{y}=t_a$:

$$\begin{aligned} \frac{\rho _{f}(Q,t_a)}{\rho (S,t_a)} =& \frac{q'(0) f_b(t_a)+\sum _{i=1}^aq'(t_i)(f_{rb}(t_i)+c+f_b(t_a))+\sum _{i=a+1}^Nq'(t_i)f_r(t_a) }{q_0f_b(t_a)+\sum _{i=1}^a\int _{t_{i-1}}^{t_i}[f_{rb}(z)+c+f_b(t_a)]q(z)\textrm{d}z+ \left( \sum _{i=a+1}^N\int _{t_{i-1}}^{t_i}q(z)\textrm{d}z+q_{\infty }\right) f_r(t_a) }\\ =& \frac{q_0 f_b(t_a)+\sum _{i=1}^a\int _{t_{i-1}}^{t_i}[f_{rb}(t_i)+c+f_b(t_a)]q(z)\textrm{d}z+ \left( \sum _{i=a+1}^N\int _{t_{i-1}}^{t_i}q(z)\textrm{d}z+q_{\infty }\right) f_r(t_a) }{q_0f_b(t_a)+\sum _{i=1}^a\int _{t_{i-1}}^{t_i}[f_{rb}(z)+c+f_b(t_a)]q(z)\textrm{d}z+ \left( \sum _{i=a+1}^N\int _{t_{i-1}}^{t_i}q(z)\textrm{d}z+q_{\infty }\right) f_r(t_a)}\\ \le& \frac{\int _{t_{0}}^{t_1}[f_{rb}(t_1)+c+f_b(t_a)]q(z)\textrm{d}z+ \sum _{i=2}^a\int _{t_{i-1}}^{t_i}[f_{rb}(t_i)+c+f_b(t_a)]q(z)\textrm{d}z}{\int _{t_{0}}^{t_1}[f_{rb}(z)+c+f_b(t_a)]q(z)\textrm{d}z+ \sum _{i=2}^a\int _{t_{i-1}}^{t_i}[f_{rb}(z)+c+f_b(t_a)]q(z)\textrm{d}z}\\ \overset{(1)}{\le} & \frac{\int _{t_{0}}^{t_1}[f_{rb}(t_1)+c+f_b(t_a)]q(z)\textrm{d}z +\sum _{i=2}^a\int _{t_{i-1}}^{t_i}[f_{rb}(t_i)+c+f_b(t_a)]q(z)\textrm{d}z}{\int _{t_{0}}^{t_1}[f_{rb}(t_0)+c+f_b(t_a)]q(z)\textrm{d}z+ \sum _{i=2}^a\int _{t_{i-1}}^{t_i}[f_{rb}(t_{i-1})+c+f_b(t_a)]q(z)\textrm{d}z}\\ \overset{(2,3)}{\le} & 1+\epsilon , \text {contradiction.} \end{aligned}$$

(1) $f_{rb}$ is monotone non decreasing function(2) T is $\epsilon$-continuous (Condition 1)(3) Lemma 6.1Case 3: $t_a= t_{N-1}$. When Condition 3a holds (i.e we quantized right) the proof is similar to the above case. Otherwise $q'(\infty )=0$ because Q is an $\epsilon$-quantized strategy and therefore it is $\epsilon$-normal. In this case, for $\hat{y} = t_{N-1}$ we obtain:

$$\begin{aligned} \frac{\rho _{f}(Q,t_{N-1})}{\rho (S,t_{N-1})} =& \frac{q'(0) f_b(t_{N-1})+\sum _{i=1}^{N-1}q'(t_i)(f_{rb}(t_i)+c+f_b(t_{N-1})) }{q_0f_b(t_{N-1})+\sum _{i=1}^{N-1}\int _{t_{i-1}}^{t_i} \left( f_{rb}(z)+c+f_b(t_{N-1})\right) q(z)\textrm{d}z +(\int _{t_{N-1}}^{\infty }\!\!\!\!\!\!q(z)\textrm{d}z +q_{\infty })f_r(t_{N-1}) }\\ \le& \frac{\sum _{i=1}^{N-1}q'(t_i)(f_{rb}(t_i)+c+f_b(t_{N-1})) }{\sum _{i=1}^{N-1}\int _{t_{i-1}}^{t_i}[f_{rb}(z)+c+f_b(t_{N-1})]q(z)\textrm{d}z +(\int _{t_{N-1}}^{\infty }q(z)\textrm{d}z+q_{\infty })f_r(t_{N-1}) }\\ =& \frac{\sum _{i=1}^{N-1}\int _{t_{i-1}}^{t_i}[f_{rb}(t_i)+c+f_b(t_{N-1})]q(z)\textrm{d}z +(\int _{t_{N-1}}^{\infty }q(z)\textrm{d}z +q_{\infty })(f_r(t_{N-1})+c) }{\sum _{i=1}^a\int _{t_{i-1}}^{t_i}[f_{rb}(z)+c+f_b(t_{N-1})]q(z)\textrm{d}z +(\int _{t_{N-1}}^{\infty }q(z)\textrm{d}z+q_{\infty })f_r(t_{N-1}) } \end{aligned}$$

We split this expression in two and bound each part separately. First:

$$\begin{aligned} \frac{\sum _{i=1}^{N-1} \int _{t_{i-1}}^{t_i} \left( f_{rb}(t_i)+c+f_b(t_{N-1})\right) q(z)\textrm{d}z }{\sum _{i=1}^{N-1} \int _{t_{i-1}}^{t_i}[f_{rb}(z)+c+f_b(t_{N-1})]q(z)\textrm{d}z } \le 1+\epsilon \end{aligned}$$

by Lemma 6.1. And second,

$$\begin{aligned} \frac{\left( \int _{t_{N-1}}^{\infty }q(z)\textrm{d}z+q_{\infty }\right) (f_r(t_{N-1})+c) }{\left( \int _{t_{N-1}}^{\infty }q(z)\textrm{d}z+q_{\infty }\right) f_r(t_{N-1}) } = \frac{f_r(t_{N-1})+c}{f_r(t_{N-1}) } \le 1+\epsilon \end{aligned}$$

by condition 3b. We conclude that in this case, $\rho _{f}(Q,t_{N-1})/\rho (S,t_{N-1}) \le 1+\epsilon$, contradiction.Case 4: $t_a=t_{N}=\infty$. First we deal with the special cases of Definition 3.3. If $d_b(\infty )=\infty$ and $q_{\infty }>0$, consider $\hat{y}=t_{N-1}$. Note that because $q[\infty ]>0$, Condition 3b cannot satisfied.

$$\begin{aligned} \frac{\rho _{f}(Q,\infty )}{\rho (S,t_{N-1})} =& \frac{1-q'_{\infty } +\lambda _{\infty }q'_{\infty } }{q_0 f_b(t_N-1) +\sum _{i=1}^{N-1} \int _{t_{i-1}}^{t_i}(f_{rb}(z)+c+f_b(t_{N-1}))q(z)\textrm{d}z +\left( \int _{t_{N-1}}^{\infty }\!\!q(z)\textrm{d}z + q_{\infty }\right) f_r(t_{N-1}) }\\ & \cdot \frac{1}{f_b(t_{N-1})} \\ \overset{(1)}{\le} & \frac{1-q'_{\infty } + \lambda _{\infty }q'_{\infty } }{q_0 f_b(t_N-1) +\sum _{i=1}^{N-1}\int _{t_{i-1}}^{t_i}f_b(t_{N-1})q(z)\textrm{d}z +(\int _{t_{N-1}}^{\infty }\!\!q(z)\textrm{d}z +q_{\infty })f_r(t_{N-1}) } \cdot \frac{1}{f_b(t_{N-1})} \\ \le & \frac{1-q'_{\infty } +\lambda _{\infty }q'_{\infty } }{q_0+\sum _{i=1}^{N-1}\int _{t_{i-1}}^{t_i}q(z)\textrm{d}z +(\int _{t_{N-1}}^{\infty }q(z)\textrm{d}z +q_{\infty })f_r(t_{N-1})/f_b(t_{N-1}) } \\ \overset{(2)}{=} & \frac{1-q'_{\infty }+\lambda _{\infty }q'_{\infty } }{(1-q'_{\infty })+q'_{\infty }f_r(t_{N-1})/f_b(t_{N-1})} \\ \overset{(3)}{\le }& 1+\epsilon , \text{contradiction.} \end{aligned}$$

(1) $f_{rb}(z)+c \ge 0$ (2) Q is $\epsilon$-quantized strategy and $q'_{\infty } = \int _{t_{N-1}}^{\infty }q(z)\textrm{d}z+q_{\infty }$ due to condition 3a (3) $\lambda _{\infty } = \lim _{y\rightarrow \infty } \frac{f_r(y)}{f_b(y)}$ and using condition  3a we have $\lambda _{\infty }<1+\epsilon$

If $d_b(\infty )=\infty$,$\lambda _{\infty }=0$ and $q[\infty ]=0$ then $\rho _{f}(Q,\infty )=1$ and the contradiction is trivial.Next we deal with the main case, assuming that Condition 3a holds. In that case we assign $\hat{y}=t_{N-1}$.

$$\begin{aligned} \frac{\rho _{f}(Q,\infty )}{\rho (S,t_{N-1})}&= \frac{\sum _{i=0}^{N} q'(t_i) \textrm{cost}(t_i,\infty ) }{\int _0^{\infty }\textrm{cost}(z,t_{N-1})q(z)\textrm{d}z \cdot {f_b(t_{N-1})}/{d_b(\infty )}}\\ &=\frac{\sum _{i=0}^{N}q'(t_i)\textrm{cost}(t_i,\infty )}{\left( \sum _{i=0}^{N} \int _{t_{i-1}}^{t_{i}}\textrm{cost}(z,t_{N-1})q(z)\textrm{d}z +q_{\infty }f_r(t_{N-1})\right) {f_b(t_{N-1})}/{d_b(\infty )}}\\ x&\le \frac{f_b(t_{N-1})}{d_b(\infty )}\cdot \max \left( \left\{ \frac{q'(t_i)\textrm{cost}(t_i,\infty )}{\int _{t_{i-1}}^{t_{i}}\!\!\!\!\textrm{cost}(z,t_{N-1})q(z)\textrm{d}z} \mid i<N \right\} , \frac{q'(\infty )\textrm{cost}(\infty ,\infty )}{\int _{t_{N-1}}^{\infty }\!\!\!\!\textrm{cost}(z,t_{N-1})q(z)\textrm{d}z +q_{\infty }f_r(t_{N-1}) }\right) \end{aligned}$$

Note that $\forall i<N:\,q'(t_i) = \int _{t_{i-1}}^{t_i}q(z)\,\textrm{d}z$, and that $q'(\infty )=\int _{t_{N-1}}^{\infty }q(z)\,\textrm{d}z+q_{\infty }$. We can observe that:

$$\begin{aligned} \frac{q'(\infty )\textrm{cost}(\infty ,\infty )}{\int _{t_{N-1}}^{\infty }\textrm{cost}(z,t_{N-1})q(z)\,\textrm{d}z+q_{\infty }f_r(t_{N-1}))} \cdot \frac{f_b(t_{N-1})}{d_b(\infty )}= & {} \frac{q'(\infty )d_r(\infty )}{\int _{t_{N-1}}^{\infty }f_r(t_{N-1})q(z)\,\textrm{d}z+q_{\infty }f_r(t_{N-1}))}\cdot \frac{f_b(t_{N-1})}{d_b(\infty )}\\= & {} \frac{d_r(\infty )f_b(t_{N-1})}{d_b(\infty )f_r(t_{N-1})}\\\le & {} 1+\epsilon , \end{aligned}$$

where the last equality is due to Condition 3a. Moreover, for all $i<N$, it holds that

$$\begin{aligned} \frac{q'(t_i)\textrm{cost}(t_i,\infty )}{\int _{t_{i-1}}^{t_{i}}\textrm{cost}(z,t_{N-1})q(z)\,\textrm{d}z}\cdot \frac{f_b(t_{N-1})}{d_b(\infty )}= & {} \frac{q'(t_i)(f_{rb}(t_i)+c+d_b(\infty ))}{\int _{t_{i-1}}^{t_{i}}(f_{rb}(z)+c+f_b(t_{N-1}))q(z)\,\textrm{d}z}\cdot \frac{f_b(t_{N-1})}{d_b(\infty )}\\\le & {} \frac{q'(t_i)(f_{rb}(t_i)+c+d_b(\infty ))}{\int _{t_{i-1}}^{t_{i}}(f_{rb}(t_{i-1})+c+f_b(t_{N-1}))q(z)\,\textrm{d}z}\cdot \frac{f_b(t_{N-1})}{d_b(\infty )}\\= & {} \frac{f_{rb}(t_i)+c+d_b(\infty )}{f_{rb}(t_{i-1})+c+f_b(t_{N-1})}\cdot \frac{f_b(t_{N-1})}{d_b(\infty )}\\ &\overset{(1)}{\le}\frac{f_{rb}(t_{i-1})+c+d_b(\infty )}{f_{rb}(t_{i-1})+c+f_b(t_{N-1})}\cdot \frac{f_b(t_{N-1})}{d_b(\infty )}\\ &\overset{(2)}{\le }\frac{d_b(\infty )}{f_b(t_{N-1})}\cdot \frac{f_b(t_{N-1})}{d_b(\infty )}=1\\ \end{aligned}$$

(1) $f_{rb}$ is non-decreasing monotone function.(2) $f_{rb}(t_{i-1})+c\ge 0$.Note that if $d_b(\infty )$ is infinity then necessarily $q'(\infty )=0$, assuming the competitive ratio is finite. We can conclude that $\rho _{f}(Q,\infty )/\rho (S,t_{N-1})\le 1+\epsilon$.

If Condition 3b holds, consider $\hat{y}$ which tends to infinity.

$$\begin{aligned} \lim _{y\rightarrow \infty }&\frac{\rho _{f}(Q,\infty )}{\rho (S,y)} = \lim _{y\rightarrow \infty } \sum _{i=0}^{N-1}q'(t_i)\textrm{cost}(t_i,\infty ) \Big /\int _0^{\infty }\textrm{cost}(z,y)q(z)\textrm{d}z \cdot \frac{f_b(y)}{d_b(\infty )} \\ =&\lim _{y\rightarrow \infty } \sum _{i=0}^{N-1}q'(t_i)\textrm{cost}(t_i,\infty ) \Big /\left( \sum _{i=0}^{N} \int _{t_{i-1}}^{t_{i}}\textrm{cost}(z,y)q(z)\textrm{d}z +q_{\infty }f_r(y)\right) \frac{f_b(y)}{d_b(\infty )} \\ \le&\lim _{y\rightarrow \infty } \max \left( \left\{ \frac{q'(t_i)\textrm{cost}(t_i,\infty ) }{\int _{t_{i-1}}^{t_{i}}\!\!\!\!\textrm{cost}(z,y)q(z)\textrm{d}z }\mid i<N-1 \right\} \right. \cup \left. \left\{ \frac{q'(t_{N-1})\textrm{cost}(t_{N-1},\infty ) }{\int _{t_{N-2}}^{\infty }\!\!\!\!\!\!\textrm{cost}(z,y)q(z)\textrm{d}z +q_{\infty }f_r(y) } \right\} \right) \frac{f_b(y)}{d_b(\infty )} \end{aligned}$$

Note that $\forall i<N-1$, we have $q'(t_i) = \int _{t_{i-1}}^{t_i}q(z)\textrm{d}z$, $q'(t_{N-1})=\int _{t_{N-2}}^{\infty }q(z)\,\textrm{d}z+q_{\infty }$ and $q'(\infty )=0$. We can observe that:

$$\begin{aligned} \lim _{y\rightarrow \infty }&\frac{q'(t_{N-1})\textrm{cost}(t_{N-1},\infty ) }{\int _{t_{N-2}}^{\infty }\textrm{cost}(z,y)q(z)\,\textrm{d}z +q_{\infty }f_r(y) } \cdot \frac{f_b(y)}{d_b(\infty )} \\ &= \lim _{y\rightarrow \infty } \frac{q'(t_{N-1})(f_{rb}(t_{N-1})+c+d_b(\infty ))}{\int _{t_{N-2}}^{t_{N-1}}(f_{rb}(z)+c+f_b(y))q(z)\,\textrm{d}z+\int _{t_{N-1}}^{\infty }f_{r}(y)q(z)\,\textrm{d}z+q_{\infty }f_r(y)}\cdot \frac{f_b(y)}{d_b(\infty )}\\ &\le \lim _{y\rightarrow \infty } \max \left( \frac{f_{rb}(t_{N-1})+c+f_b(y) }{f_{rb}(t_{N-2})+c+f_b(y)} ,\frac{f_{r}(t_{N-1})+c }{f_{r}(t_{N-1}) }\right) \cdot \frac{f_b(y)}{d_b(\infty )}\\ &\le 1+\epsilon \end{aligned}$$

And we can moreover observe that for all $i<N-1$ we have

$$\begin{aligned} \lim _{y\rightarrow \infty } \frac{q'(t_i)\textrm{cost}(t_i,\infty )}{\int _{t_{i-1}}^{t_{i}}\textrm{cost}(z,y)q(z)\,\textrm{d}z} \cdot \frac{f_b(y)}{d_b(\infty )} =&\lim _{y\rightarrow \infty } \frac{q'(t_i)(f_{rb}(t_i)+c+d_b(\infty ))}{\int _{t_{i-1}}^{t_{i}}(f_{rb}(z)+c+f_b(y))q(z)\,\textrm{d}z}\cdot \frac{f_b(y)}{d_b(\infty )}\\ \le&\lim _{y\rightarrow \infty } \frac{q'(t_i)(f_{rb}(t_i)+c+d_b(\infty ))}{\int _{t_{i-1}}^{t_{i}}(f_{rb}(t_{i-1})+c+f_b(y))q(z)\,\textrm{d}z}\cdot \frac{f_b(y)}{d_b(\infty )}\\ =&\lim _{y\rightarrow \infty } \frac{f_{rb}(t_i)+c+d_b(\infty )}{f_{rb}(t_{i-1})+c+f_b(y)}\cdot \frac{f_b(y)}{d_b(\infty )}\\ \overset{(1)}{\le} &\lim _{y\rightarrow \infty } \frac{f_{rb}(t_{i-1})+c+d_b(\infty )}{f_{rb}(t_{i-1})+c+f_b(y)}\cdot \frac{f_b(y)}{d_b(\infty )}\\ \overset{(2)}{\le} &\lim_{y\rightarrow \infty } \frac{d_b(\infty )}{f_b(y)}\cdot \frac{f_b(y)}{d_b(\infty )} = 1 .\\ \end{aligned}$$

(1) $f_{rb}$ is non-decreasing monotone function.(2) $f_{rb}(t_{i-1})+c\ge 0$.We can therefore conclude that $\lim _{y\rightarrow \infty }\rho _{f}(Q,\infty )/\rho (S,y)\le 1+\epsilon$ .$\square$

Proof

[Proof of Lemma 3.4] First we prove that T is $\epsilon$-continuous w.r.t I as claimed.

Condition 1: $f_r(t_1)+c \le (1+\epsilon )(f_{rb}(t_0) +c+f_b(t_1))$. We can observe that if we assign 0 instead of $t_1$, the inequality is strict for any $\epsilon >0$ , because $0<f_r(0)+c = f_{rb}(0)+f_b(0) +c$. Therefore, there exists $z\ge 0$ which satisfies this condition because the functions are continuous. If $z>1$, we assign $t_1=1$. Line 3 of Algorithm 1 corresponds to this analysis.

Condition 2: $f_r(t_{i+1})\le (1+\epsilon )f_r(t_i)$ and $f_b(t_{i+1})\le (1+\epsilon )f_b(t_i)$.Since $f_r$ and $f_b$ are monotone non-decreasing, the inverse functions well-defined, with range including $\infty$.The assignments at Lines 6, 13 of Algorithm 1 ensures that this condition is satisfied. Because $t_{N-1}$ is finite (see below) and because $f_r$ and $f_b$ are continuous, the loop terminates in finite time.

Condition 3: Clearly, line 3 of Algorithm 1 corresponds to conditions 3a or 3b or both. We need to show that $t_{N-1}$ is finite.If $f_r$ is unbounded then $f_{r}^{-1}(c/ \epsilon )$ is finite, implying that $t_{N-1}$ is finite.If $f_r$ is bounded then $f_r(t)/f_b(t)$ is also bounded (note that by assumption 5, $\forall t\ge 0:\,f_b(t)>0$). Because $f_b$ is monotone non-decreasing we conclude that $f_b(x)\ge 1$ for any $x>1$. Therefore we can always find finite x that satisfies $1\le x< \infty$ such that $\forall y\ge x:\,\frac{f_r(y)}{f_b(y)}\le (1+\epsilon )\frac{f_r(x)}{f_b(x)}$.

In order to construct $I'$ we define also the parameters $(d_r,d_b,c',\lambda _{\infty })$ as follows:

$$\begin{aligned} c' =& c&\lambda _{\infty } =&\lim _{y\rightarrow \infty } \frac{f_r(y)}{f_b(y)} \\ d_r(\tau ) =&{\left\{ \begin{array}{ll} \lim _{y \rightarrow \infty } f_r(y) &{}\tau =\infty \\ f_r(\tau ) &{} \tau<\infty \end{array}\right. }&d_b(\tau ) =&{\left\{ \begin{array}{ll} \lim _{y \rightarrow \infty } f_b(y) &{}\tau =\infty \\ f_b(\tau ) &{} \tau <\infty \end{array}\right. } \end{aligned}$$

Note that by our assumptions we can find such parameters.

We now turn to analyze the time complexity. We show that the running time of Algorithm 1 is $O(\epsilon ^{-1}\log (\epsilon ^{-1}))$. The complexity of our approach is O(|T|), since the algorithm consists of O(|T|) iterations, and each iteration requires O(1) running time. We upper bound the points $t_1$ and $t_{N-1}$ of this set.From by Condition 1 we have that:

$$\begin{aligned} f_r(t_1)+c= & {} (1+\epsilon )(f_{rb}(0) +c+f_b(t_1) ) \\\ge & {} (1+\epsilon ) (f_r(0)+c)\\\ge & {} c+\epsilon c , \end{aligned}$$

implying that $f_r(t_1) \ge \epsilon c$. From Line 6 of algorithm 1, we have:

$$\begin{aligned} |\{t:t\in T,t\le 1\}|\le & {} \left\lceil \log _{1+\epsilon }({f_r(1)}/{f_r(t_1)}) \right\rceil + \left\lceil \log _{1+\epsilon }({f_b(1)}/{f_b(t_1)}) \right\rceil \\ \le & {} 2\left\lceil \log _{1+\epsilon }\left( {1}/{f_r(t_1)}\right) \right\rceil\\= & {} O(\epsilon ^{-1}\log (1/f_r(t_1)) \end{aligned}$$

where the last inequality follows from $f_r(1)=f_b(1)=1$ and $f_r(t_1)\le f_b(t_1)$.Similarly, from Line 10 we deduce that $f_{r}(t_{N-1})\le \frac{c}{\epsilon }$. Then we can observe, that

$$\begin{aligned} |\{t:t\in T,t\ge 1\}|\le & {} \left\lceil \log _{1+\epsilon }(\frac{f_r(t_{N-1})}{f_r(1)}) \right\rceil + \left\lceil \log _{(1+\epsilon )}(\frac{f_b(t_{N-1})}{f_b(1)}) \right\rceil \\\le & {} 2 \left\lceil \log _{1+\epsilon }f_r(t_{N-1}) \right\rceil . \end{aligned}$$

Given the above bounds, we can bound |T|.

$$\begin{aligned} |T|\le & {} 2\left\lceil \log _{1+\epsilon }({1}/{f_r(t_1)}) \right\rceil +2\left\lceil \log _{1+\epsilon }(f_r(T_{N-1}) \right\rceil \\\le & {} 2\,\log _{1+\epsilon }\frac{f_r(T_{N-1})}{f_r(t_1)}+O(1)\\ \le & {} \log _{1+\epsilon }\epsilon ^{-2}+O(1)\\= & {} O(\epsilon ^{-1}\log (1/\epsilon ))\\ \end{aligned}$$

for $0<\epsilon \le 1$.

According to definition 15, $I'$ is $\epsilon$-compatible w.r.t I, and the proof is complete.$\square$

Proof

[Proof of Lemma 3.5] Denote the solution to (LP1) problem by q[z] and $\rho$. Let O denote the finite strategy with q[z] as its probability function. We claim that O has competitive ratio $\rho$. We also prove that O gets the best competitive ratio achievable by finite strategies for any T.$\square$

We prove two statements that yield the lemma.

Claim 1

If q[z] is $\epsilon$-normal strategy and a feasible solution to (LP1), then for all $y\in T$ we have $\rho (S_{q[z]},y) = \sum _{z\in T'} \rho _{zy}q[z]$.

We prove the claim for each case of Definition 3.4.

• 1. $y=0, q[0]>0$. In this case, $d_r(0)>0$ (because otherwise $q[0]=0$). Therefore:

  $$\begin{aligned} \rho _f(S,0)= & {} q[0]\frac{d_b(0)}{d_r(0)}+1-q[0]\\= & {} q[0]\cdot \rho (S_0,0)+1-q[0]\\= & {} q[0]\rho _{00}+\sum _{z\in T'\setminus \{0\}}\rho _{z0}(1-q[0]) = \sum _{z\in T'}\rho _{z0}q[z] \end{aligned}$$

• 2. $y=0, d_r(0)=0, q[0]=0$ . In this case $0\notin T'$, and therefore, $\rho _f(S,0) =1 = \sum _{z\in T'}\rho _{z0}q[z]$.

• 3. $y=\infty , d_b(\infty )=\infty , q[\infty ] >0$ .Since $q[\infty ]>0$, condition 3a must be satisfied, i.e $\lambda _{\infty }<\infty$, so

  $$\begin{aligned} \rho _f(S,\infty )= & {} q[\infty ]\lambda _{\infty }+1-q[\infty ]\\= & {} q[\infty ]\rho (S_{\infty },\infty )+\sum _{z\in T'\setminus \{\infty \}}q[z]\\= & {} q[\infty ] \rho _{\infty \,\infty }+\sum _{z\in T'\setminus \{\infty \}}\rho _{z\,\infty }\,q[z]\\= & {} \sum _{z\in T'}\rho _{z\,\infty }\,q[z] \end{aligned}$$

• 4. $y=\infty , d_b(\infty )=\infty , \lambda _{\infty }=\infty , q[\infty ]=0$. Since $\lambda _{\infty }=\infty$ condition 3b must be satisfied, and $\infty \notin T'$. Therefore, $\rho _f(S,\infty ) =\sum _{z\in T'}q[z]=\sum _{z\in T'}\rho _{z\infty }q[z]$ .

• 5. Otherwise, we have

  $$\begin{aligned} \rho _f(S,y) =\frac{\textrm{cost}(S,y)}{\textsc {opt}(y)}=\frac{\sum _{z\in T'}\textrm{cost}(S_{z},y)q[z]}{\textsc {opt}(y)} = \sum _{z\in T'}\rho _{zy}q[z]. \end{aligned}$$

  Note that in this case we have no infinite summands.

Claim 2

There is no $\epsilon$-normal strategy with competitive ratio better than $\rho$.

Assume to the contrary that there is an $\epsilon$-normal strategy $O'$ with parameter p[z] and competitive ratio $\eta <\rho$. We prove that $O'$ is an optimal solution to (LP1), in contradiction to the assumption that O is the optimal solution. We will show that $O'$ satisfies all constraints of LP1.

• 1. $d_r(0)=0 \Rightarrow q_0=0$ .If $q_0>0$ then the competitive ratio of $O'$ at time $y=0$ is:

  $$\rho _f(O',0) = q[0]\frac{d_b(0)}{d_r(0)}+1-q[0]=\infty ,$$

  in contradiction to the assumption that $\eta <\rho$.

• 2. $\lambda _{\infty }=\infty$ (In this case we must have $q[\infty ]=0$).If $O'$ is $\epsilon$-normal then a stronger constraint must be satisfied: $\frac{f_r(t_{N-1})}{f_b(t_{N-1})} \ge (1+\epsilon )\max _{y\ge t_{N-1}} \frac{f_r(y)}{f_b(y)}$, and hence $q_{\infty }=0$ by Definition 3.5.

• 3. $\forall y\in T: \,\sum _{z\in T'} \rho _{zy}p[z]-\eta \le 0$ .From the first statement we conclude that

  $$\forall y\in T:\qquad \rho _f(S_{q},y) = \sum _{z\in T'} \rho _{zy}q[z]$$

  When $S_{q}$ is the probabilistic strategy which decides to switch by the discrete probability density function q[z] and by our assumption we get:

  $$\begin{aligned} \rho _f(S_{q},y) \le \max _{y\in T}\rho _f(S_{q},y)=\max _{y \in T}\sum _{z\in T'} \rho _{zy}q[z] = \eta \end{aligned}$$

  as required.

• 4. $\sum _{z\in T'} p[z]=1, p[z]\ge 0\text { and } \eta \ge 0$Since $\forall z\in T\setminus T', p[z]=0$, as we observed, we can deduce that $\sum _{z\in T'} q[z]=\sum _{z\in T} q[z]=1 .$

• 5. The constraints $p[z]\ge 0 , \rho \ge 1$ are necessarily true for every strategy.From both statements we can deduce that the best $\epsilon$-normal strategy is the solution to (LP1) and its competitive ratio is $\rho$.$\square$

Lemma 6.2

The running time of Algorithm 2 is bounded by $O(\epsilon ^{-2}\cdot \delta ^{-3}\cdot \log ^4 (1/\epsilon )) .$

Proof

For the analysis of the time complexity of (LP1), we use the result of Lemma 3.5, which states that $|T| \le O(\epsilon ^{-1}\log (\epsilon ^{-1}))$. To solve the linear program, we replace it with the following equivalent packing version.

$$\begin{aligned} \text {maximize } \sum _{z\in T'} q[z],\text { subject to:}\\ \sum _{z\in T'} \rho _{zy}q[z]\le 1 &\text { for all }y\in T\nonumber \\ q[z]\ge 0&\text { for all } z\in T'\nonumber \\ q[z]=0&\text { for all }z\in T\setminus T' \nonumber \end{aligned}$$

(LP2)

In this version, we do not normalize the probability vector q, i.e., its components do not necessarily sum up to 1. After normalization, the competitive ratio computed by LP2 is $\rho =1/\sum _{z\in T'} q[z]$. We note that LP2 is a positive linear program, as studied by [21]. We apply the methods of [3]. Using approximation factor $(1+\delta )$, the time complexity is of the solution is $O(\delta ^{-3}\cdot N\log ^2N)$, where N is the number of non zero coefficients. In our case $N=O(|T|^2)$, and and since by lemma 3.4 we know that $|T| \le O(\epsilon ^{-1}\log (\epsilon ^{-1}))$, we can conclude that our algorithm finds a solution which approximates the optimal strategy with factor $(1+\epsilon )^2(1+\delta )$ and time complexity of $O(\epsilon ^{-2}\cdot \delta ^{-3}\cdot \log ^4 (1/\epsilon ))$. This completes the proof of Lemma 3.3.$\square$