A simple proof technique for scheduling models with learning effects

Abstract

The majority of the previous works concerning scheduling problems with learning effects employed several complicated lemmas to prove the polynomial solvable properties of their models. To simplify these proofs, this paper proposes a simpler technique based on the famous Lagrange mean value theorem and the common adding-and-subtracting-terms operation in mathematics. The re-argumentation of known properties for a general model and the exploitation of a new property verify the effectiveness of this proposed technique and exhibit its simplicity. Moreover, the proof technique is applicable to other similar problems because the technique requires only the prerequisites of the Lagrange mean value theorem.

Appendices

Appendix A: The lemmas used by the majority of papers in scheduling with learning effects

The following lemmas are derived from [19]. However, the lemmas used by other related papers are identical or similar.

Lemma 4.1

(Lemma 2.2 in [19]) If $f:[0, +\infty ) \mapsto (0, 1]$ is a differentiable non-increasing function and its derivative $f'$ is non-decreasing over $[0, +\infty )$, then $(1-\alpha )f(a) + \alpha \lambda f(a+t) - \lambda f(a+ \alpha t) \leqslant 0$ for $\alpha \geqslant 1, 0 < \lambda \leqslant 1, a \geqslant 0$ and $t \geqslant 0$.

Proof

Define a function $F(t)=(1-\alpha )f(a) + \alpha \lambda f(a+t) - \lambda f(a+ \alpha t)$, then take its first derivative.

$$\begin{aligned} F'(t)=\alpha \lambda f'(a+t)-\alpha \lambda f'(a+\alpha t)=\alpha \lambda (f'(a+t)-f'(a+\alpha t)) \end{aligned}$$

Since $\alpha \geqslant 1, t \geqslant 0$ and $f'$ is non-decreasing, $f'(a+t)-f'(a+\alpha t) \leqslant 0$, i.e., $F'(t) \leqslant 0$. This implies that $F(t)$ is non-increasing over $[0, +\infty )$. Thus, for $\alpha \geqslant 1, 0 < \lambda \leqslant 1$ and $a \geqslant 0$,

$$\begin{aligned} F(t) \leqslant F(0)=(1-\alpha )f(a) + \alpha \lambda f(a)-\lambda f(a)=(1-\alpha )(1-\lambda )f(a) \leqslant 0. \end{aligned}$$

This completes the proof. $\square$

Lemma 4.2

(Lemma 2.3 in [19]) If $f:[0, + \infty ) \mapsto (0, 1]$ is a differentiable non-increasing function and its derivative $f'$ is non-decreasing over $[0, +\infty )$, then $f(a) + \lambda {\lambda _2} tf'(a + t) - \lambda {\lambda _1}f(a + t) \geqslant 0$ for $0 < \lambda \leqslant 1, 0 \leqslant {\lambda _1} \leqslant {\lambda _2} \leqslant 1, a \geqslant 0$ and $t \geqslant 0$.

Proof

Define a function $F(\lambda _1,\lambda _2, t)=f(a) + \lambda {\lambda _2} tf'(a + t) - \lambda {\lambda _1}f(a + t)$, then $\frac{\partial F}{\partial \lambda _1}=-\lambda f(a+t) \leqslant 0$, $\frac{\partial F}{\partial \lambda _2}=\lambda tf'(a+t) \leqslant 0$. Thus, for $0 < \lambda \leqslant 1, 0 \leqslant {\lambda _1} \leqslant {\lambda _2} \leqslant 1$ and $a \geqslant 0$,

$$\begin{aligned} F(\lambda _1,\lambda _2, t) \geqslant F(1,1,t)=f(a)+\lambda tf'(a+t)-\lambda f(a+t). \end{aligned}$$

By Lagrange mean value theorem, $\exists \xi \in (a, a+t)$, s.t., $f(a+t)-f(t)=f'(\xi )t$. Because of $0 < \lambda \leqslant 1, t \geqslant 0$ and $f'$ is non-decreasing,

$$\begin{aligned} F(\lambda _1,\lambda _2, t)\geqslant & {} f(a)+\lambda tf'(a+t)-\lambda f(a+t)\\\geqslant & {} \lambda f(a)+\lambda tf'(a+t)-\lambda f(a+t)\\= & {} \lambda t \left( f'(a+t)-f'(\xi ) \right) \geqslant 0. \end{aligned}$$

This completes the proof. $\square$

Lemma 4.3

(Lemma 2.4 in [19]) If $f:[0, + \infty ) \mapsto (0, 1]$ is a differentiable non-increasing function and its derivative $f'$ is non-decreasing over $[0, + \infty )$, then $(\alpha - 1)f(a) + \lambda {\lambda _2}f(a + \alpha t) - \alpha \lambda {\lambda _1}f(a + t) \geqslant 0$ for $\alpha \geqslant 1, 0 < \lambda \leqslant 1, 0 \leqslant {\lambda _1} \leqslant {\lambda _2} \leqslant 1,a \geqslant 0$ and $t \geqslant 0$.

Proof

Define a function $F(\alpha )=(\alpha - 1)f(a) + \lambda {\lambda _2}f(a + \alpha t) - \alpha \lambda {\lambda _1}f(a + t)$, then

$$\begin{aligned} F'(\alpha )=f(a)+\lambda {\lambda _2}f'(a + \alpha t)-\lambda {\lambda _1}f(a + t). \end{aligned}$$

By the second lemma above, $F'(\alpha ) \geqslant F'(1)=f(a)+\lambda {\lambda _2}f'(a + t)-\lambda {\lambda _1}f(a + t) \geqslant 0$. This implies that $F(\alpha )$ is non-decreasing. Thus, for $0 < \lambda \leqslant 1, 0 \leqslant {\lambda _1} \leqslant {\lambda _2} \leqslant 1$ and $a \geqslant 0$, we have

$$\begin{aligned} F(\alpha )\geqslant F(1)=\lambda {\lambda _2}f(a + t) - \lambda {\lambda _1}f(a + t)=\lambda f(a+t)(\lambda _2 - \lambda _1) \geqslant 0. \end{aligned}$$

This completes the proof. $\square$

Appendix B: Two examples together used the Lagrange mean value theorem and the adding-and-subtracting-terms operation

Example 1

If a continuous function is differentiable on an open interval but unbounded, then its first derivative is also unbounded.

Proof

Let $f(x)$ be a differentiable and unbounded function on $(a,\,b)$. Fix a point $t \in (a,\,b)$. By the Lagrange mean value theorem, $\forall x \in (a,\,b)$, $x \ne t$, there exists a point $\xi$ between $x$ and $t$ such that $f(x)-f(t)=f'(\xi )(x-t)$.

Assume that $f'(x)$ is bounded, i.e., $\exists N>0$ such that $|f'(x)|<N$.

$$\begin{aligned} |f(x)|= & {} |f(x)-\underline{f(t)}+\underline{f(t)}| \leqslant |f(x)-f(t)|+|f(t)|\nonumber \\ \quad= & {} |f'(\xi )(x-t)|+|f(t)| \leqslant N(b-a)+|f(t)| \end{aligned}$$

(4)

The inequality Eq. (4) indicates that function $f$ is bounded at point $x$. Since $x$ is an arbitrary point on the open interval $(a,b)$, $f(x)$ is bounded on $(a,\,b)$. This leads to a contradiction with the known condition: its unboundedness. $\square$

Example 2

The second derivative of function $f(x)$ is negative on $[b,\, \infty )$. If $f(b)>0$ and $f'(b)<0$, then equation $f(x)=0$ has a single root on $(b,\, \infty )$.

Proof

$f'(x)<f'(b)<0$ because of $f''(x)<0$. By the Lagrange mean value theorem, we have

$$\begin{aligned} f(x)=f(x){-} \underline{f(b)} + \underline{f(b)} = f'(\xi )(x{-}b)+f(b)<f'(b)(x{-}b)+f(b). \end{aligned}$$

(5)

$\mathop {\lim }\limits _{x \rightarrow \infty }{f(x)} \leqslant \mathop {\lim }\limits _{x \rightarrow \infty }{f'(b)(x-b)+f(b)} = - \infty$, thus $\mathop {\lim }\limits _{x \rightarrow \infty }{f(x)}=- \infty$. Note $f(b)>0$. By the intermediate value theorem, $\exists x_0 \in (b, \, \infty )$ such that $f(x_0)=0$.

Assume that there is another point $x_1 \in (b, \, \infty )$, $x_1 \ne x_0$, such that $f(x_1)=0$. By the Lagrange mean value theorem, there is a point $\zeta$ between $x_0$ and $x_1$ such that $f'(\zeta )=0$. This contradicts with $f'(x)<0$ at $[b,\, \infty )$. Hence, equation $f(x)=0$ has a single root on $(b,\, \infty )$. $\square$