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管理科学代写|决策论代写Management Science Models for Decision Making代考|Convexity of PL Functions of a Single Variable

We discuss convexity of PL functions next. As these functions are not differentiable at points where there slopes change, the arguments used in the previous section based on differentiability do not apply.

Result 2.2. Let $\theta(\lambda)$ be a PL function of a single variable $\lambda \in R^1$. Let $\lambda_1, \ldots, \lambda_r$ be the various breakpoints in increasing order where its slope changes. $\theta(\lambda)$ is convex iff at each breakpoint $\lambda_t$ its slope to the right of $\lambda_t$ is strictly greater than its slope to the left of $\lambda_t$; that is, iff its slopes are monotonic increasing with the variable.

Proof. Suppose at a breakpoint $\lambda_t, c_t=$ the slope of $\theta(\lambda)$ to the right of $\lambda_t$ is $\lambda_l$, where its slope is $c_t$. Then the graph of $\theta(\lambda)$ in the neighborhood of $\lambda_t$ will be as shown by the solid line in Fig. 2.5. The chord of the function in the interval $\bar{\lambda} \leq \lambda \leq \tilde{\lambda}$ shown by the dashed line segment is below the function, violating Jensen’s inequality for convex functions. So, $\theta(\lambda)$ cannot be convex.

If the slopes of the function satisfy the condition mentioned in the Result, then it can be verified that every chord lies above the function, establishing its convexity.
The corresponding result for concave functions is: a PL function of one variable is concave iff its slope to the right of every breakpoint is less than its slope to the left of that breakpoint, that is, its slopes are monotonic decreasing with the variable. These results provide a convenient way to check whether a PL function of one variable is convex, or concave, or neither. For example, the PL function in Example $2.1$ has monotonically increaasing slopess, so it is convex. For thé oné in Example 2.2, thé slope is not monotone, so it is neither convex nor concave.

管理科学代写|决策论代写Management Science Models for Decision Making代考|PL Convex and Concave Functions in Several Variables

Let $f(x)$ be a PL function of variables $x=\left(x_1, \ldots, x_n\right)^T$ defined over $R^n$. So, there exists a partition $R^n=\cup_{t=1}^r K_t$, where $K_t$ is a convex polyhedral set for all $t$, the interiors of $K_1, \ldots, K_r$ are mutually disjoint, and $f(x)$ is affine in each $K_t$; that is, we have vectors $c^t$ and constants $c_0^t$ such that
$$f(x)=c_0^T+c^t x \text { for all } x \in K_t, t=1 \text { to } r .$$

Checking the convexity of $f(x)$ on $R^n$ is not as simple as in the one-dimensional case (when $n=1$ ), but the following theorem explains how it can be done.

Theorem 2.5. Let $K_1 \cup \ldots \cup K_r$ be a partition of $R^n$ into convex polyhedral regions, and $f(x)$ the PL function defined by the above equation (2.1). Then $f(x)$ is convex iff for each $t=1$ to $r$, and for all $x \in K_t$
$$c_0^t+c^t x=\operatorname{Maximum}\left{c_0^p+c^p x: \quad p=1, \ldots, r .\right}$$
In effect, this says that $f(x)$ is convex iff for each $x \in R^n$
$$f(x)=\text { Maximum }\left{c_0^p+c^p x: p=1, \ldots, r .\right}$$
Proof. Suppose $f(x)$ satisfies the condition (2.2) stated in the theorem. Let $x^1, x^2 \in$ $R^n$ and $0 \leq \alpha \leq 1$. Suppose
$f\left(x^1\right)=\operatorname{Miximum}\left{c_0^p+c^p x^1: p=1, \ldots, r.\right}=c_0^1+c^1 x^1$,
$f\left(x^2\right)=\operatorname{Maximum}\left{c_0^p+c^p x^2: p=1, \ldots, r.\right}=c_0^2+c^2 x^2$,
and $f\left(\alpha x^1+(1-\alpha) x^2\right)=\max \left{c_0^p+c^p\left(\alpha x^1+(1-\alpha) x^2\right): p=1, \ldots, r\right}$
$f\left(\alpha x^1+(1-\alpha) x^2\right)=\alpha\left(c_0^a+c^a x^1\right)+(1-\alpha)\left(c_0^a+c^a x^2\right)$,
$\leq \alpha\left(c_0^1+c^1 x^1\right)+(1-\alpha)\left(c_0^2+c^2 x^2\right)$
from (2.3), (2.4),
$=\alpha f\left(x^1\right)+(1-\alpha) f\left(x^2\right)$.
As this holds for all $x^1, x^2 \in R^n$ and $0 \leq \alpha \leq 1, f(x)$ is convex by definition. Now suppose that $K_1 \cup \ldots \cup K_r$ is a partition of $R^n$ into convex polyhedral regions, and $f(x)$ the PL function defined by $f(x)=c_0^t+c^t x$ for all $x \in K_t$, $t=1$ to $r$, is convex. Let $\bar{x}$ be any point in $R^n$, suppose $\bar{x} \in K_{p_b}$. Let $x^1 \in$ $K_1, x^2 \in K_2$ be any two points such that $\bar{x}$ is on the line segment $L$ joining them, that is, $\bar{x}=\bar{\lambda} x^1+\left(1-\bar{\lambda} x^2\right)$ for some $0<\bar{\lambda}<1$. For $0 \leq \lambda \leq 1$ let $f\left(\lambda x^1+(1-\lambda) x^2\right)=\theta(\lambda)$.

决策论代考

管理科学代写|决策论代写管理科学决策模型代考|PL多变量中的凸和凹函数

. .

$$f(x)=c_0^T+c^t x \text { for all } x \in K_t, t=1 \text { to } r .$$

$$c_0^t+c^t x=\operatorname{Maximum}\left{c_0^p+c^p x: \quad p=1, \ldots, r .\right}$$

$$f(x)=\text { Maximum }\left{c_0^p+c^p x: p=1, \ldots, r .\right}$$

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