Copula-based Randomized Mechanisms for Truthful Scheduling on Two Unrelated Machines

Abstract

We design a Copula-based generic randomized truthful mechanism for scheduling on two unrelated machines with approximation ratio within [1.5852,1.58606], offering an improved upper bound for the two-machine case. Moreover, we provide an upper bound 1.5067711 for the two-machine two-task case, which is almost tight in view of the known lower bound of 1.506 for the scale-free truthful mechanisms. Of independent interest is the explicit incorporation of the concept of Copula in the design and analysis of the proposed approximation algorithm. We hope that techniques like this one will also prove useful in solving other related problems in the future.

Appendices

Appendix A: Proof of Theorem 2

In this section, we present the main part of our theoretical proof of Theorem 2, which follows as a result of the next lemma.

Lemma 4

Let a = 1.715 and b = 0.76. Suppose F(⋅) is defined as in ( 9 ), and φ(⋅, ⋅) is defined by

$$ \begin{array}{c} \varphi(x,y)=1+y- F(x)-yF(y)+\left(1+\frac1x\right)F(x)F(y)\,. \end{array} $$

(21)

Then for any $x,y\in \mathbb R_{+}$ with xy ≥ 1, it holds that φ(x, y) < ρ ^∗ = 1.58606.

Proof

Recall from (9) that F(⋅) is a continuous piecewise algebraic function defined by six different expressions in six intervals I ₁, I ₂,…,I ₆, respectively. Note that φ(x, y) is continuous in $\mathbb {R}_{+} \times \mathbb {R}_{+}$. To prove the lemma, we exhaust all possible cases of x ∈ I _i , y ∈ I _j (1 ≤ i, j ≤ 6) with x y ≥ 1, and upper bound the function value φ(x, y) at the critical points (i.e., when the derivatives of φ(x, y) are equal to zero) and at the demarcation points yielding the six intervals. Clearly, we may skip any point (x, y) at which φ does not attain its maximum.

CASE 1. x ≥ a. It follows from (9) that F(x) = 1 and from (21) that $\varphi (x,y) = y+(1+\frac 1x-y)F(y)$, where $F(y)=1-\frac {2(1-b)(a-y)}{a-1}$ for $y\in \left .\left [\frac {a+1}2,a\right .\right )$, and $F(y)=\frac 12+\frac {(2b-1)(y-1)}{a-1}$ for $y\in \left (1,\frac {a+1}2\right )$.

If $y\le 1+\frac 1x$ then $\varphi (x,y)\le y+(1+\frac 1x-y) =1+\frac 1x\le 1+\frac 1a<1.584<\rho ^{\ast }$, where the third to last inequality follows from x ≥ a. If y ≥ a then $\varphi (x,y)=1+y-F(x)-y+\left (1+\frac 1x\right )F(x)\le 1+\frac 1x<\rho ^{\ast }$. Now notice that $y>1+\frac 1x$ and x ≥ a implies that y∈(1,a).

In case of $y\in \left .\left [\frac {a+1}2,a\right .\right )$, we have $\frac {\partial \varphi }{\partial x}(x,y)=-\frac {2400y-541}{3575x^{2}}$, which is negative. Thus from the first-order necessary optimality conditions (i.e., the KKT conditions [2]), we deduce that φ(x, y) < ρ ^∗ for any (x, y) in this case, unless maybe for x = a. When x = a, note that $\varphi \left (a,\frac {a+1}2\right )<1.53$, and that φ(a, y) has a unique critical point $y=\frac {a^{2}+a+1}{2a}$ in $\left (\frac {a+1}2,a\right )$ with corresponding critical value less than 1.58602 < ρ ^∗.

In case of $y\in \left (1,\frac {a+1}2\right )$, it suffices to consider the case where x = a as $\frac {\partial \varphi }{\partial x}(x,y)=-\frac {16y-5}{22x^{2}}<0$. Note that $\frac {\partial \varphi }{\partial y}(a,y)=\frac {17949}{7546}-\frac {16}{11}y>2.3-\frac {16}{11}\left (\frac {a+1}2\right ) >0$, which implies that the maximum value of φ(x, y) is not attained in this case.

CASE 2. y ≥ a > x ≥ 0. Note that $\varphi (x,y)=1+y- F(x)-y +(1+\frac 1x)F(x)=1+\frac {F(x)}x$ is a function of single variable x. It is easy to check that function $1+\frac {F(x)}x$ has derivative $\frac {541}{3575x^{2}}$ in $\left (\frac {a+1}2,a\right )$, $\frac {5}{22x^{2}}$ in $\left (1,\frac {a+1}2\right )$, $\frac {32-27x}{22x^{3}}$ in $\left (\frac 2{a+1},1\right )$, & $\frac {12(400-343x)}{3575x^{3}}$ in $x\in \left (\frac 1a,\frac 2{a+1}\right )$. Note that the derivative of $1+\frac {F(x)}x$ is positive for all $x\in \left (\frac 1a,a\right )\setminus \left \{\frac 2{a+1},1,\frac {a+1}2\right \}$. The continuity of φ implies $\varphi (x,y)\le \varphi (a,y)=1+\frac 1a<1.584<\rho ^{\ast }$ for all $x\in \left (\frac 1a,a\right )$. When $x\in \left .\left (0,\frac {1}a\right .\right ]$, it is clear that φ(x, y) = 1.

Cases 1 and 2 above show that φ(x, y) < ρ ^∗ when x or y belongs to I ₁. For the remaining cases, we have x, y < a. As x y ≥ 1, we have $x,y>\frac 1a$ both contained in $(I_{2}\cup I_{3}\cup I_{4})\cup \left (I_{5}\setminus \left \{\frac 1a\right \}\right )$. We distinguish among Cases 3 – 6, where Case i + 1 deals with for x ∈ I _i , i = 2, 3 ,4 and Case 6 deals with $x\in I_{5}\setminus \left \{\frac 1a\right \}$.

CASE 3. $x\in I_{2}=\left .\left [\frac {a+1}2,a\right .\right )$. We distinguish among four subcases depending on which interval y belongs to: $y\in \left .\left [\frac {a+1}2,a\right .\right ), y\in \left .\left [1,\frac {a+1}2\right .\right ), y\in \left .\left [\frac 2{a+1},1\right .\right )$, and $y\in \left (\frac {1}{a},\frac {2}{a+1}\right )$, respectively.

CASE 3.1. $y\in [\frac {a+1}2,a)$. (9) and (21) imply $\varphi (x,y) = 1 + y - \left (1 - \frac {2(1 - b)(a - x)}{a-1}\right ) -y\left (1-\frac {2(1-b)(a-y)}{a-1} \right ) +\left (1+\frac 1x\right ) \left (1-\frac {2(1-b)(a-x)}{a-1} \right )\left (1-\frac {2(1-b)(a-y)}{a-1} \right )$ and

$$\begin{array}{@{}rcl@{}} 2500(a-1)^{2}x^{2}\cdot\frac{\partial\varphi}{\partial x}(x,y)&=&(576x^{2}-129.84)y-987.84x^{2}-29.2681\,,\\ \frac{2500(a-1)^{2}x}{ b-1 }\cdot\frac{\partial\varphi}{\partial y}(x,y)&=&7150xy-2400x^{2}-7990.125x+541\,. \end{array} $$

In case of $x,y\in (\frac {a+1}2,a)$, solving $\frac {\partial \varphi }{\partial x}(x,y)=0=\frac {\partial \varphi }{\partial y}(x,y)$ gives

$$\frac{987.84x^{2}+29.2681}{576x^{2}-129.84}=y=\frac{2400x^{2}+7990.125x-541}{7150x},$$

which implies 2764800x ⁴−4921488x ³+1656341.83x−140486.88=0. Among the four real roots of the biquadratic equation, only one root x ₀≈1.5419 belongs to $I_{2}=\left .\left [\frac {a+1}2,a\right .\right )=\left .\left [1.3572,1.715\right .\right )$, and the other three 0.6970..., 0.0866..., −0.5455... are much less than $\frac {a+1}2$. So function φ(x, y) has a unique critical point (x ₀, y ₀), where $y_{0}=\frac {987.84x_{0}^{2}+29.2681}{576x_{0}^{2}-129.84}\approx 1.586$, giving critical value φ(x ₀, y ₀)<1.585 < ρ ^∗.

In case of $x=\frac {a+2}2$, function $\varphi \left (\frac {a+1}2,y\right )$ has a unique critical point $y_{0}=\frac {a^{2}+a+ab+3b}{2a+2}\approx 1.5174$ in $\left (\frac {a+1}2,a\right )$, giving critical value smaller than 1.586059 < ρ ^∗. Note that $\varphi \left (\frac {a+1}2,\frac {a+1}2\right )<1.57<\rho ^{\ast }$.

In case of $y=\frac {a+1}2$ and $x\in \left (\frac {a+1}2,a\right )$, the derivative of $\varphi \left (x,\frac {a+1}2\right )$ is $\frac {10279}{89375x^{2}}-\frac {576}{3575}<\frac {10279}{89375 }-\frac {576}{3575}<0$, implying that the maximum value of φ(x, y) is not attained in this case.

CASE 3.2. $y\in [1,\frac {a+1}2)$. (9) and (21) give $\varphi (x,y) \!= \!1 + y - \left (1 - \frac {2(1 - b)(a - x)}{a-1} \right ) - y\left (\frac 12+\frac {(2b-1)(y-1)}{a-1} \right ) +\left (1+\frac 1x\right ) \left (1-\frac {2(1-b)(a-x)}{a-1} \right )$, and

$$\begin{array}{@{}rcl@{}} 2500(a-1)^{2}x^{2}\cdot\frac{\partial\varphi}{\partial x}(x,y)&=&(624x^{2}+140.66)y-1053x^{2}-43.95625\,,\\ 2\times 10^{4}(a-1)^{2}x\cdot\frac{\partial\varphi}{\partial y}(x,y)&=&-14872xy+4992x^{2}+16414.97x-1125.28\,. \end{array} $$

In case of $x\in \left (\frac {a+1}2,a\right )$ and $y\in \left (1,\frac {a+1}2\right )$, solving $\frac {\partial \varphi }{\partial x}(x,y)=0=\frac {\partial \varphi }{\partial y}(x,y)$ gives $\frac {1053x^{2}+43.95625}{624x^{2}+140.66}=y= \frac {4992x^{2}+16414.97x-1125.28}{14872x}$, which implies

$$92160000x^{4}-160274400x^{3}+48970779x-4682896=0.$$

Among the four roots of the biquadratic equation, only one root x ₀≈1.5249 belongs to $I_{2}=\left (\frac {a+1}2,a\right )$, the other three 0.642..., 0.098..., −0.526... are much smaller than $\frac {a+1}2=1.3572$. Thus φ(x, y) has a unique critical point (x ₀, y ₀), where $y_{0}= \frac {1053{x_{0}^{2}}+43.95625}{624{x_{0}^{2}}+140.66}\approx 1.566$, giving critical value φ(x ₀, y ₀)<1.583 < ρ ^∗.

In case of $x=\frac {a+1}2$, function φ attains its critical value $\varphi \left (\frac {a+1}2,y_{0}\right )<1.585<\rho ^{\ast }$ at $y_{0}=\frac {a^{2}-3+2ab-2b-2a+4ab^{2}+12b^{2}}{4(a+1)(2b-1)} \approx 1.5037$; at the boundary, $\varphi \left (\frac {a+1}2,1\right )<1.4$.

In case of y = 1 and $x\in \left (\frac {a+1}2,a\right )$, the derivative of φ(x,1) is $\frac {541}{7150x^{2}}-\frac {48}{143}$, which is less than $\frac {541 }{7150 }-\frac {48}{143}<0$, implying that the maximum value of φ(x, y) is not attained in this case.

CASE 3.3. $y\in \left .\left [\frac 2{a+1},1\right .\right )$. (9) and (21) give $\varphi (x,y)=1+y- \left (1-\frac {2(1-b)(a-x)}{a-1} \right ) -y\left (\frac 12-\frac {(2b-1)(1/y-1)}{a-1} \right ) +\left (1+\frac 1x\right ) \left (1-\frac {2(1-b)(a-x)}{a-1} \right )\left (\frac 12-\frac {(2b-1)(1/y-1)}{a-1} \right )$, and

$$\begin{array}{@{}rcl@{}} 2\times10^{4}(a-1)^{2}x^{2}y\cdot\frac{\partial\varphi}{\partial x}(x,y)&=&(1560y-4992)x^{2}+1898.91y-1125.28 \,. \end{array} $$

If $\frac {\partial \varphi }{\partial x}(x,y)=0$, then (1560y−4992)x ²+1898.91y−1125.28=0, which implies $x^{2} =\frac {5.41}{24}\left (\frac {70.4}{16-5y}-5.4\right )<\frac {5.41}{24}\left (\frac {70.4}{16-5}-5.4\right )<0.3$, contradicting the hypothesis $x\in \left .\left [\frac {a+1}2,a\right .\right )$ of Case 3. Thus $\frac {\partial \varphi }{\partial x}(x,y)\ne 0$, and it suffices to consider the case where $x=\frac {a+1}2$. Note that the derivative of $\varphi \left (\frac {a+1}2,y\right )$ is $\frac {143336}{149325y^{2}}-\frac {5}{22}>\frac {143336}{149325}-\frac {5}{22}>0$. We deduce that the maximum value of φ(x, y) is not attained in Case 3.3.

CASE 3.4. $y\in \left (\frac 1a,\frac 2{a+1}\right )$. (9) and (21) imply $\varphi (x,y)=1+y- \left (1-\frac {2(1-b)(a-x)}{a-1} \right ) -y\left (\frac {2(1-b)(a-1/y)}{a-1} \right ) +\left (1+\frac 1x\right )$ $\left (1-\frac {2(1-b)(a-x)}{a-1} \right )$ $\left (\frac {2(1-b)(a-1/y)}{a-1} \right )$, and

$$\begin{array}{@{}rcl@{}} \frac{2500(a-1)^{2}x^{2}y}{b-1}\cdot\frac{\partial\varphi}{\partial x}(x,y)=(2400-541y)x^{2}-927.815y+541\,. \end{array} $$

If $\frac {\partial \varphi }{\partial x}(x,y)=0$, then (2400−541y)x ²−927.815y + 541=0 implies $x^{2} =\frac {3575}{2400-541y}-1.715 <\frac {3575}{2400-541}-1.715<0.21$, contradicting the hypothesis $x\in \left .\left [\frac {a+1}2,a\right .\right )$ of Case 3. Thus $\frac {\partial \varphi }{\partial x}(x,y)\ne 0$, and it suffices to consider the case where $x=\frac {a+1}2$. Note that the derivative of $\varphi \left (\frac {a+1}2,y\right )$ is $\frac {573344}{647075y^{2}}-\frac {541}{3575}>\frac {573344}{647075}-\frac {541}{3575}>0$. We deduce that the maximum value of φ(x, y) is not attained in Case 3.4.

CASE 4. $x\in I_{3}=[1,\frac {a+1}2)$. It follows from x y ≥ 1 that $y>\frac 2{a+1}$ for which we distinguish among three subcases depending on which interval y belongs to: $y\in \left .\left [\frac {a+1}2,a\right .\right ), y\in \left .\left [1,\frac {a+1}2\right .\right )$ and $y\in \left .\left [\frac 2{a+1},1\right .\right )$, respectively.

CASE 4.1. $y\in [\frac {a+1}2,a)$. (9) and (21) give $\varphi (x,y)=1+y-\left (\frac 12+\frac {(2b-1)(x-1)}{a-1} \right ) -y\left (1-\frac {2(1-b)(a-y)}{a-1} \right ) +\left (1+\frac 1x\right ) \left (\frac {1}{2}+\frac {(2b-1)(x-1)}{a-1} \right ) \left (1-\frac {2(1-b)(a-y)}{a-1} \right )$, and

$$\begin{array}{@{}rcl@{}} 62500(a-1)^{2}x^{2}\cdot\frac{\partial\varphi}{\partial x}(x,y)&=&(15600x^{2}+4875)y-26754x^{2}- \frac{35165}{32}, \vspace{0.5mm} \\ \frac{200(a-1)^{2}x}{b-1}\cdot\frac{\partial\varphi}{\partial y}(x,y)&=& 572xy-208x^{2}-633.49x+65. \end{array} $$

For $x\in \left (1,\frac {a+1}2\right )$ and $y\in \left (\frac {a+1}2,a\right )$, solving $\frac {\partial \varphi }{\partial x}(x,y)=0=\frac {\partial \varphi }{\partial y}(x,y)$ gives

$$\frac{26754x^{2} +{35165}/{32}}{15600x^{2}+4875}=y=\frac{208x^{2}+633.49x-65}{572x},$$

which implies 153600x ⁴−256608x ³+116435x−15000=0. Among the four real roots of this biquadratic equation, only one root x ₀≈1.2027 belongs to $I_{3}=\left .\left [1,\frac {a+1}2\right .\right )$, the other three 0.964..., 0.133..., −0.629... are less than 1. Thus function φ(x, y) has a unique critical point of (x ₀, y ₀), where $y_{0}=\frac {26754{x_{0}^{2}} +{35165}/{32}}{15600{x_{0}^{2}}+4875}\approx 1.4504$, yielding critical value φ(x ₀, y ₀)<1.5854 < ρ ^∗.

Note that $\varphi \left (1,\frac {a+1}2\right )=1.5858<\rho ^{\ast }$. For x = 1 and $y\in \left (\frac {a+1}2,a\right )$, the derivative of φ(1,y) is $\frac {\partial \varphi }{\partial y}(1,y)<\frac {b-1}{200(a-1)^{2}}\left (572\frac {a+1}2-208-633.49+65\right )=0$. For $y=\frac {a+1}2$ and $x\in \left (1,\frac {a+1}2\right )$, the derivative of $\varphi \left (x,\frac {a+1}2\right )$ is $\frac {\partial \varphi }{\partial x}\left (x,\frac {a+1}2\right )= \frac {19}{110x^{2}}-\frac {48}{275}<\frac {19}{110 }-\frac {48}{275}<0$. The negative derivatives in both cases imply that the maximum value of φ(x, y) is attained in neither case.

CASE 4.2. $y\in \left .\left [1,\frac {a+1}2\right .\right )$. (9) and (21) give $\varphi (x,y)=1+y- \left (\frac 12+\frac {(2b-1)(x-1)}{a-1} \right )$ $-y\left (\frac 12+\frac {(2b-1)(y-1)}{a-1} \right ) +\left (1+\frac 1x\right ) \left (\frac 12+\frac {(2b-1)(x-1)}{a-1} \right ) \left (\frac 12+\frac {(2b-1)(y-1)}{a-1} \right )$, and

$$10^{4}(a-1)^{2}x^{2}\cdot\frac{\partial\varphi}{\partial x}(x,y)=(2704y-4563)x^{2}+845y-\frac{4225}{16}\,. $$

If $\frac {\partial \varphi }{\partial x}(x,y)=0$, then the above equation implies $x^{2}=\frac {6.875}{27-16y}-\frac 5{16}<\frac {6.875}{27-8(a+1)}-\frac 5{16} <1$, contradicting the hypothesis x ≥ 1 of Case 4. So it suffices to consider x = 1. When $y\in \left (1,\frac {a+1}2\right )$, function φ(1,y) attains its unique critical value $\varphi (1,\frac {43}{32})<1.586<\rho ^{\ast }$ at $y=\frac {43}{32}$. At the boundary, we have φ(1, 1) = 1.5 < ρ ^∗.

CASE 4.3. $y\in (\frac 2{a+1},1)$. (9) and (21) give $\varphi (x,y)=1+y- \left (\frac 12+\frac {(2b-1)(x-1)}{a-1} \right )$ $-y\left (\frac 12-\frac {(2b-1)(1/y-1)}{a-1} \right ) +\left (1+\frac 1x\right ) \left (\frac 12+\frac {(2b-1)(x-1)}{a-1} \right ) \left (\frac 12-\frac {(2b-1)(1/y-1)}{a-1} \right )$, and

$$\begin{array}{@{}rcl@{}} 32000(a-1)^{2}x^{2}y\cdot\frac{\partial\varphi}{\partial x}(x,y)=(2704y-8652.8)x^{2}+4563y-2704\,. \end{array} $$

If $\frac {\partial \varphi }{\partial x}(x,y)=0$, then the above equation implies $x^{2}=\frac {80-135y}{80y- 256}$, which along with x ≥ 1 enforces $y\ge \frac {336}{215}$, a contradiction to y<1. Therefore we may assume x = 1. Since the derivative of φ(1,y) is $\frac {16-5y^{2}}{22y^{2}}>0$, we deduce that the maximum value of φ(x, y) is not attained in this case.

CASE 5. $x\in I_{4}=\left .\left [\frac {2}{a+1},1\right .\right )$. It follows from x y ≥ 1 that y > 1. We distinguish between two subcases depending on whether y is at least $\frac {a+1}2$ or not.

CASE 5.1. $y\in [\frac {a+1}2,a)$. By (9) and (21) we have $\varphi (x,y)=1 + y - \left (\frac 12-\frac {(2b-1)(1/x-1)}{a-1} \right ) \!- y\left (1-\frac {2(1-b)(a-y)}{a-1} \right ) +\left (1+\frac 1x\right ) \left (\frac 12-\frac {(2b-1)(1/x-1)}{a-1} \right ) \left (1-\frac {2(1-b)(a-y)}{a-1} \right )$, and

$$\begin{array}{@{}rcl@{}} 2500(a-1)^{2}x^{2}\cdot\frac{\partial\varphi}{\partial y}(x,y)=-1716x^{2}y+2524.47 x^{2}+429x-624\,. \end{array} $$

If $\frac {\partial \varphi }{\partial y}(x,y)=0$, then the above equation implies $y=\frac {2524.47 x^{2}+429x-624}{1716x^{2}}$, which along with $y\ge \frac {a+1}2$ enforces 5x ²+11x−16≥0 implying x ≤ −3.2 or x ≥ 1, a contradiction to the hypothesis x ∈ I ₄ of Case 5. So we may assume $y=\frac {a+1}2$. When $x\in (\frac 2{a+1},1)$, the unique critical point of $\varphi \left (x,\frac {a+1}2\right )$ is $x=\frac {608}{609}$, giving critical value less than 1.586 < ρ ^∗. At the boundary, we have $\varphi (\frac {2}{a+1},\frac {a+1}2)<1.5159<\rho ^{\ast }$.

CASE 5.2. $y\in (1,\frac {a+1}2)$. (9) and (21) give $\varphi (x,y)=1+y- \left (\frac 12-\frac {(2b-1)(1/x-1)}{a-1} \right ) -y\left (\frac 12+\frac {(2b-1)(y-1)}{a-1} \right ) +\left (1+\frac 1x\right ) \left (\frac 12-\frac {(2b-1)(1/x-1)}{a-1} \right ) \left (\frac 12+\frac {(2b-1)(y-1)}{a-1} \right )$, and

$$\begin{array}{@{}rcl@{}} 10^{4} (a-1)^{2}x^{3}\cdot\frac{\partial\varphi}{\partial x}(x,y)&=&(5408-1859x)y-\frac{50193}{16}x -1690\,,\vspace{1mm}\\ 10^{4} (a-1)^{2}x^{2}\cdot\frac{\partial\varphi}{\partial y}(x,y)&=&-7436x^{2}y+ \frac{86697}8x^{2} +1859x-2704\,. \end{array} $$

If $\frac {\partial \varphi }{\partial x}(x,y)=0=\frac {\partial \varphi }{\partial y}(x,y)$, we have $\frac {50193x/16+1690}{5408-1859x}=y=\frac {86697x^{2}/8+1859x-2704}{7436x^{2}}$, which implies 12177x ³−11928x ²−4224x + 4096=0. Among the three real roots of the cubic equation, only one root x ₀≈0.9850 belongs to $I_{4}=\left .\left [\frac 2{a+1},1\right .\right )$, the other two −0.587... and 0.581... are less than $0.59<0.73<\frac 2{a+1}$. Hence, when $x\in \left (\frac 2{a+1},1\right )$, function φ attains its unique critical value φ(x ₀, y ₀)<1.58603 < ρ ^∗ at $x_{0}\approx 0.985 , y_{0}=\frac {50193x_{0}/16+1690}{5408-1859x_{0}}\approx 1.3364$. When $x=\frac 2{a+1}$, function $\varphi \left (\frac 2{a+1},y\right )$ has a unique critical value $\varphi \left (\frac 2{a+1},1.12665\right )<1.5546<\rho ^{\ast }$.

CASE 6. $x\in I_{5}\setminus \{\frac 1a\}=\left (\frac {1}{a},\frac {2}{a+1}\right )$. It follows from x y ≥ 1 that $y\in \left (\frac {a+1}{2},a\right )$. From (9) and (21) we obtain

$$\begin{array}{@{}rcl@{}} \varphi(x,y) &=& 1+y- \left( \frac{2(1-b)(a-1/x)}{a-1} \right) -y\left( 1-\frac{2(1-b)(a-y)}{a-1} \right)\\ &&+\left(1+\frac1x\right) \left( \frac{2(1-b)(a-1/x)}{a-1} \right) \left (1-\frac{2(1-b)(a-y)}{a-1} \right). \end{array} $$

Note that

$$\begin{array}{@{}rcl@{}} \frac{1250 (a-1)^{2}x^{3}}{b-1}\cdot\frac{\partial\varphi}{\partial x}(x,y)=\left(\frac{637637}{400}+858y\right)x-2400y+541\,, \end{array} $$

saying that $\frac {\partial \varphi }{\partial x}(x,y)$ is continuous. If $\frac {\partial \varphi }{\partial x}(x,y)=0$, then the above equation implies $x=\frac {2400y-541}{637637/400+858y}$, which along with $x\le \frac 2{a+1}$ enforces $y\le \frac {4657}{4800}<1$, a contradiction. Since $\frac {\partial \varphi }{\partial x}(x,y)$ is a continuous function, we deduce that $\frac {\partial \varphi }{\partial x}(x,y)$ is always positive or always negative, implying that the maximum value of φ(x, y) is not attained in Case 6.

Among all cases analyzed above (see Table 2 for a partial summary), the bottleneck 1.586058... ( < ρ ^∗) is attained by Case 3.1 with $\varphi \left (\frac {a+1}2,\frac {a^{2}+a+ab+3b}{2a+2}\right )$ =φ(1.3575,1.51742..), completing the proof. □

Appendix B: Details omitted in the proof of Theorem 3

We point out that the computational accuracy of the value λ ₄ is crucial to obtaining the upper bound 0.1143 on $F(\frac 1\beta )$ in (19), which is the key for us to derive the desired contradiction. The computational accuracy of λ ₄ depends on that of λ ₃, λ ₂, λ ₁ recursively. It is a routine matter to check the following exact expressions:

$$\begin{array}{l} \lambda_{1}=\frac{383}{500}+\frac{\sqrt{154595269}}{105500},\\ [6pt] \lambda_{2}=\frac{169(3830\lambda_{1}-2367)}{2500(294\lambda_{1}-169)} =\frac{18069396176}{32281745375}+\frac{2054533\sqrt{154595269}}{129126981500},\\ \lambda_{3}=\frac{169}{250}-\frac{227/2500}{2\lambda_{2}-1}=\frac{17274798609857}{22964052192750}-\frac{466378991\sqrt{154595269}}{22964052192750},\text{~and}\\ \lambda_{4}=\frac12-\frac{0.0908}{2\lambda_{3}-0.648}= \frac{7635461853+9926731\sqrt{154595269}}{137555892260+32555020\sqrt{154595269}}. \end{array} $$

Appendix C: Proof of Theorem 4

By the setting of a and b, we see that a−1, 1−b, 2b−1, 3b−2, 2a b−a−1, a + 1−4b, 2a−3a b + b, 3a + 1−4a b are all positive. We will use this fact implicitly in our analysis. In view of Lemma 2, it suffices to consider x y ≥ 1, F(x)≥F(1/y) = 1−F(y), and

$$\begin{array}{@{}rcl@{}} \varphi(x,y)&=&1+y- F(x)-yF(y)+ \left(1+\frac1x\right)(F(x)+F(y)-1) \\ &=&y-\frac1x+\frac1xF(x)+\left(1+\frac1x-y\right)F(y) \end{array} $$

(22)

We prove φ(x, y) < ρ ^∗ = 1.5067711 for any $x,y\in \mathbb R_{+}$ with x y ≥ 1. In the following, we will consider x ≥ a in Case 1, y ≥ a in Case 2, and max{x, y} < a in Cases 3 – 6. We upper bound by ρ ^∗ the function value φ(x, y) at the critical points and at the demarcation points. We may skip any point (x, y) as soon as we show φ(x, y) is not the maximum.

• CASE 1. x ≥ a. It follows from F(x) = 1 that $\varphi (x,y)=y+\left (1+\frac 1x-y\right )F(y)$. In case of $y\le 1+\frac 1x$ or y ≥ a, we have $\varphi (x,y)\le y+\left (1+\frac 1x-y\right ) =1+\frac 1x\le 1+\frac 1a<1.5$. In case of $y>1+\frac 1x$ and y < a, we have y∈(1,a).

  When $y\in \left .\left [\frac {a+1}2,a\right .\right )$, it follows from (9) and (22) that $\frac {\partial \varphi }{\partial x}(x,y)=-\frac {2ab-a-1+2(1-b)y}{(a-1)x^{2}} <0$.

  By the first-order necessary optimality conditions (i.e., the KKT conditions [2]), we deduce that φ(x, y) < ρ ^∗ for any (x, y) in this case, unless maybe for x = a. In case of x = a, function φ(a, y) has a unique critical point $y=\frac {a^{2}+a+1}{2a}$ in $\left (\frac {a+1}2,a\right )$, at which the critical value is less than 1.50677 < ρ ^∗. At the boundary point , we have $\varphi \left (a,\frac {a+1}2\right )<1.5<\rho ^{\ast }$.

  When $y\in \left (1,\frac {a+1}2\right )$, since $\frac {\partial \varphi }{\partial x}(x,y)=-\frac {a+1-4b+(4b-2)y}{2(a-1)x^{2}} <0$, it suffices to consider the case where x = a. Note that $\frac {\partial \varphi }{\partial y}(a,y)=\frac {17770909}{11672126}- \frac {869}{1039}y>\frac {3}{2}-\frac {869}{1039}\left (\frac {a+1}2\right ) >0.1$, which implies that the maximum of φ(x, y) is not attained in this case.

• CASE 2. y ≥ a > x ≥ 0. It follows from F(y) = 1 that $\varphi (x,y)=1+\frac {F(x)}x$ is a function of single variable x. When $x\in \left [0,\frac {1}a\right ]$, it is clear that φ(x, y) = 1. The derivative of φ(x, y) is $\frac {2(1-b)(2-ax)}{(a-1)x^{3}}>0$ for $x\in \left (\frac 1a,\frac 2{a+1}\right ),-\frac {a+1-4b}{(a-1)x^{2}}<0$ for $x\in \left (1,\frac {a+1}2\right )$, and $-\frac {2ab-a-1}{(a-1)x^{2}}<0$ for $x\in \left (\frac {a+1}2,a\right )$. So we may assume $x\in \left [\frac 2{a+1},1\right ]$ Within $x\in \left (\frac 2{a+1},1\right )$, the derivative of φ(x, y) has a unique root $x_{0}=\frac {4(2b-1)}{a-3+4b}$. So we only need to consider φ(x ₀, y), φ(1,y), and $\varphi \left (\frac 2{a+1},y\right )=\frac {3+a-b-ab}2$. All the three values are smaller than 1.505 < ρ ^∗.

  In the following case analysis, we consider only x, y < a. As x y ≥ 1, we have $x,y>\frac {1}{a}$. We distinguish among four cases for $x\in \left .\left [\frac {a+1}2,a\right .\right ), x\in \left .\left [1,\frac {a+1}2\right .\right ),x\in \left .\left [\frac 2{a+1},1\right .\right )$, and $x\in \left (\frac {1}{a},\frac {2}{a+1}\right )$, respectively.

• CASE 3. $x\in [\frac {a+1}2,a)$. We distinguish among four subcases depending on which interval y belongs to: $y\in \left .\left [\frac {a+1}2,a\right .\right ),y\in \left .\left [1,\frac {a+1}2\right .\right ), y\in \left .\left [\frac 2{a+1},1\right .\right )$, and $y\in \left (\frac {1}{a},\frac {2}{a+1}\right )$, respectively.

• CASE 3.1. $y\in \left .\left [\frac {a+1}2,a\right .\right )$. It follows from (9) and (22) that

  $$\frac{\partial\varphi}{\partial x}(x,y)=\frac{2(b-1)y-4ab+3a+1}{(a-1)x^{2}}$$

  and $\frac {\partial \varphi }{\partial y}(x,y)=\frac {2(1-b)(ax+x-2yx+1)}{(a-1)x}$.

  In case of $x,y\in \left .\left [\frac {a+1}2,a\right .\right )$, the unique solution of $\frac {\partial \varphi }{\partial x}(x,y)=0=\frac {\partial \varphi }{\partial y}(x,y),x_{0}= \frac {1-b}{2a-3ab+b}$ and $y_{0}=\frac {3a-4ab+1}{2(1-b)}$, gives the critical value φ(x ₀, y ₀) < 1.5061 < ρ ^∗.

  In case of $x=\frac {a+1}2$ and $y\in \left (\frac {a+1}2,a\right )$, function $\varphi \left (\frac {a+1}2,y\right )$ has a unique critical point $y_{0}=\frac { a^{2}+2a+3}{2(a+1)}$, giving critical value $\varphi \left (\frac {a+1}2,y_{0}\right )<1.5067711=\rho ^{\ast }$. ¹.

  In case of $y=\frac {a+1}2$, function $\varphi \left (x,\frac {a+1}2\right )$ has positive derivative $\frac {2a-3ab+b}{(a-1)x^{2}} \!>\!0$ for all $x\in \left (\frac {a+1}2,a\right )$, saying that the maximum of φ(x, y) is not attained in this case.

• CASE 3.2. $y\in \left .\left [1,\frac {a+1}2\right .\right )$. In case of $1<y<\frac {a+1}2<x<a$, solving $\frac {\partial \varphi }{\partial x}(x,y)=0=\frac {\partial \varphi }{\partial y}(x,y)$, we obtain a unique critical point $x_{0}=\frac {2(2b-1)}{5a+3-8ab}$, $y_{0}=\frac {3a-4ab+4b-1}{2(2b-1)}$ of φ, giving critical value φ(x ₀, y ₀)<1.504.

  In case of $1<y<\frac {a+1}2=x$, function $\varphi \left (\frac {a+1}2,y\right )$ attains its critical value $\varphi \left (\frac {a+1}2,y_{0}\right )<1.504<\rho ^{\ast }$ at its unique critical point $y_{0}=\frac {a^{2}+8ab-4a+16b-9}{ 4(2b-1)(a+1)}$.

  In case of y = 1 and $x\in \left (\frac {a+1}2,a\right )$, the derivative of φ(x,1) is $\frac {3a+1-4ab}{2(a-1)x^{2}} >0$, implying that the maximum of φ(x, y) is not attained in this case.

• CASE 3.3. $y\in [\frac 2{a+1},1)$. It follows from (9) and (22) that

  $\frac {\partial \varphi }{\partial x}(x,y)=\frac {2(1-b)yx^{2}+(a+1-4b)y +4b-2}{2(a-1)yx^{2}}$.

  If $\frac {\partial \varphi }{\partial x}(x,y)=0$, then $x^{2}=-\frac {(a+1-4b)y +4b-2}{2(1-b)y} <0$ shows a contradiction. Thus $\frac {\partial \varphi }{\partial x}(x,y)\ne 0$, and it suffices to consider the case where $x=\frac {a+1}2$. Note that the derivative of $\varphi \left (\frac {a+1}2,y\right )$ is $\frac {(a+1)(a+1-4b)}{2(a^{2}-1)}+\frac {(a+3)(2b-1)}{(a^{2}-1)y^{2}}>0$. We deduce that the maximum of φ(x, y) is not attained in Case 3.3.

• CASE 3.4. $y\in \left (\frac 1a,\frac 2{a+1}\right )$. Since $\frac {\partial \varphi }{\partial x}(x,y)=\frac {2(1-b)}{(a-1)x^{2}y}>0$, we deduce that the maximum of φ(x, y) is not attained in this case.

• CASE 4. $x\in \left .\left [1,\frac {a+1}2\right .\right )$. It follows from x y ≥ 1 that $y>\frac 2{a+1}$, for which we distinguish between two subcases for $y\in \left .\left [\frac {a+1}2,a\right .\right )$ and $y\in \left .\left [\frac 2{a+1},\frac {a+1}2\right .\right )$, respectively.

  Consider the subcase of $y\in \left .\left [\frac {a+1}2,a\right .\right )$. When $1<x<\frac {a+1}2<y<a$, solving $\frac {\partial \varphi }{\partial x}(x,y)=0=\frac {\partial \varphi }{\partial y}(x,y)$ gives the unique critical point $x_{0}=\frac {2(1-b)}{a-2ab+6b-3}$, $y_{0}=\frac {3a-4ab+4b-1}{4(1-b)}$, and the corresponding critical value φ(x ₀, y ₀) < 1.503 < ρ ^∗. When x = 1, the unique critical point of φ(1,y) is $y =\frac {a+2}2$, giving critical value $\varphi \left (1,\frac {a+2}2\right )<1.506<\rho ^{\ast }$. When $y=\frac {a+1}2$, the derivative of $\varphi \left (x,\frac {a+1}2\right )$ is $\frac {(2b-1)(3-a)}{2(a-1)x^{2}}>0$ for all $x\in \left (1,\frac {a+1}2\right )$, saying that the maximum of φ(x, y) is not attained in this case.

  Consider the subcase of $y\in \left .\left [\frac 2{a+1},\frac {a+1}2\right .\right )$. Note that $\frac {\partial \varphi }{\partial x}(x,y)=\frac {(2b-1)(2-y)}{(a-1)x^{2}}>0$ for all $y\in \left (1,\frac {a+1}2\right )$, and $\frac {\partial \varphi }{\partial x}(x,y)=\frac {2b-1}{(a-1)yx^{2}}>0$ for all $y\in \left (\frac 2{a+1},1\right )$. We deduce that the maximum of φ(x, y) is not attained in this case.

• CASE 5. $x\in \left .\left [\frac {2}{a+1},1\right .\right )$. It follows from x y ≥ 1 that y > 1. We distinguish between two subcases depending on whether y is at least $\frac {a+1}2$ or not.

  Consider the subcase of $y\in \left .\left [\frac {a+1}2,a\right .\right )$. When $\frac {2}{a+1}<x<\frac {a+1}2<y$, solving $\frac {\partial \varphi }{\partial x}(x,y)=0=\frac {\partial \varphi }{\partial y}(x,y)$, we obtain a unique critical point $x_{0}=\frac {2(5b-3)}{2ab+2b-a-1}$, $y_{0}=\frac { (a+1) (12b-7)}{4(5b-3)}$ corresponding critical value φ(x ₀, y ₀) < 1.50677 < ρ ^∗. When $y=\frac {a+1}2$. the derivative of $\varphi \left (x,\frac {a+1}2\right )$ is $\frac {8b-4-(a-1)(2b-1)x}{2(a-1)x^{3}}> \frac {8b-4-(a-1)(2b-1)}{2(a-1)x^{3}} =\frac {(3-a)(2b-1)}{2(a-1)x^{3}} >0$ for all $x\in \left (\frac {2}{a+1},1\right )$, saying that the maximum of φ(x, y) is not attained in this case. When $x=\frac 2{a+1}$, the derivative of $\varphi \left (\frac 2{a+1},y\right )=\frac {3a-4y+3}{(1-b)(a-1)}> \frac {3a-4a+3}{(1-b)(a-1)}>0$ for all $y\in \left (\frac {a+1}2,a\right )$, saying that the maximum of φ(x, y) is not attained in this case.

  Consider the subcase of $y\in (1,\frac {a+1}2)$. When $x<\frac {2}{a+1}$, solving $\frac {\partial \varphi }{\partial x}(x,y)=0=\frac {\partial \varphi }{\partial y}(x,y)$, we obtain a unique critical point $x_{0}=\frac {6(2b-1)}{a+8b-5},y_{0}=\frac {a+8b-5}{3 (2 b-1)}$, corresponding critical value φ(x ₀, y ₀)<1.506 < ρ ^∗. When $x=\frac 2{a+1}$, the derivative of function $\varphi (\frac 2{a+1},y)$ is $\frac {ab+5b-3-2(2b-1)y}{a-1}>\frac {ab+5b-3-(2b-1)(a+1)}{a-1}\,=\,\frac {a(1-b)+3b-2}{a-1}\!>\!0$, saying that the maximum of φ(x, y) is not attained in this case.

• CASE 6. $x\in \left (\frac 1a,\frac {2}{a+1}\right )$. It follows from x y ≥ 1 that $y>\frac {a+1}2$. If $\frac {\partial \varphi }{\partial x}(x,y)=0$, then it can be deduced that $x=\frac 2y$, which along with $x\le \frac 2{a+1}$ enforces y ≥ a + 1, a contradiction to y < a. Thus $\frac {\partial \varphi }{\partial x}(x,y)$ is always positive or always negative, saying that that the maximum of φ(x, y) is not attained in Case 6.

  Among all cases analyzed above, the bottleneck 1.506771... is attained in Case 3.1 by $\varphi \left (\frac {a+1}2,\frac {a^{2}+2a+3}{2(a+1)}\right )$. The proof is completed.

D More accurate data for Table 3

Table 4 Long digital expressions of data from Table 3, where the values of a, b and δ are omitted