Constrained Light Deployment for Reducing Energy Consumption in Buildings

Abstract

Lighting systems account for a major part of the energy consumed by large commercial buildings. This paper aims at reducing this energy consumption by defining the Contrained Light Deployment Problem. This new problem is related to the classical Art Gallery Problem (AGP) in computational geometry. In contrast to AGP, which asks for the minimum number of guards to monitor a polygonal area, our problem, CLDP, poses a new challenging requirement: not only must each point p have an unobstructed line-of-sight to a light source, but also, the combined illuminance at p from all light sources must exceed some given threshold value. We provide evidence that our new problem is NP-hard, based on known results for AGP. Then we propose fast heuristics for floor plans shaped like orthogonal polygons, with and without holes. Our problem formulation allows lights to be placed internally, not only at vertices. Our algorithm, which combines ideas from computational geometry, clustering and binary search, computes a set of light placements that satisfies the illumination requirement. The algorithm seeks a light set of minimum size by an iterative binary search procedure that progressively tightens upper and lower bounds.

Appendix

1.1 Proof of Theorem 1

Proof

Proving Theorem 1 is equivalent to solving the following problem shown in Fig. 3: Assuming that the side length of the square is R and O is an arbitrary point in the square, we need to prove that

$$\begin{aligned} \frac{1}{1+|OA'|^{2}}+\frac{1}{1+|OB'|^{2}}+\frac{1}{1+|OC'|^{2}}+\frac{1}{1+|OD'|^{2}}\ge \frac{1}{1+R^{2}}. \end{aligned}$$

(8)

The equivalence is based on the fact that Fig. 3 illustrates the worst case³ scenario: each corner of the square is covered by a different light, and the locations of the lights (i.e., $A',B',C', D'$) lead to the lowest illuminance value at O. As the assumption in Theorem 1, the right-hand side of (8), $\frac{1}{1+R^{2}}$, is larger than the given threshold.

First, by the inequality of arithmetic and geometric means

$$\begin{aligned} \frac{x_{1}+x_{2}+\cdots +x_{n}}{n}\ge \root n \of {x_{1}\cdot x_{2}\cdot \cdots x_{n}}, \quad x_1,\cdots ,x_n>0, \end{aligned}$$

(9)

we have

$$\begin{aligned} \begin{aligned}&\frac{1}{1+|OA'|^{2}}+\frac{1}{1+|OB'|^{2}}+\frac{1}{1+|OC'|^{2}}+\frac{1}{1+|OD'|^{2}}\\&\ge \frac{4}{\root 4 \of {(1+|OA'|^{2})(1+|OB'|^{2})(1+|OC'|^{2})(1+|OD'|^{2})}} \\&\ge 4\times \frac{1}{\frac{(1+|OA'|^{2})+(1+|OB'|^{2})+(1+|OC'|^{2})+(1+|OD'|^{2})}{4}} \\&=\frac{16}{4+(|OA'|^{2}+|OB'|^{2}+|OC'|^{2}+|OD'|^{2})}\\ \end{aligned} \end{aligned}$$

(10)

Observing that $|OA'|^{2}=(R+|OA|)^{2}=R^{2}+|OA|^{2}+2R|OA|\le 2(R^{2}+|OA|^{2})$, the relation of which holds also for $|OB'|,|OC'|,|OD'|$, we have

$$\begin{aligned} \begin{aligned}&\frac{16}{4+(|OA'|^{2}+|OB'|^{2}+|OC'|^{2}+|OD'|^{2})}\\&\ge \frac{16}{4+8R^{2}+2[|OA|^{2}+|OB|^{2}+|OC|^{2}+|OD|^{2}]}\\ \end{aligned} \end{aligned}$$

(11)

Now, we indicate the arbitrary O by (x, y) where $x,y\in [-\frac{R}{2},\frac{R}{2}]$ in the Euclidean coordinate, and $A=(\frac{R}{2},\frac{R}{2})$, $B=(-\frac{R}{2},\frac{R}{2})$, $C=(-\frac{R}{2},-\frac{R}{2})$, $D=(\frac{R}{2},-\frac{R}{2})$. Thus we have

$$\begin{aligned} \begin{aligned}&|OA|^2=(x-\frac{R}{2})^{2}+(y-\frac{R}{2})^{2},|OB|^2=(x+\frac{R}{2})^{2}+(y-\frac{R}{2})^{2},\\&|OC|^2 =(x+\frac{R}{2})^{2}+(y+\frac{R}{2})^{2},|OD|^2 =(x-\frac{R}{2})^{2}+(y+\frac{R}{2})^{2}\\ \end{aligned} \end{aligned}$$

(12)

Substituting Eq. (12) into (11) and using the fact that $x^{2}+y^{2}\le (\frac{R}{2})^{2}+(\frac{R}{2})^{2}=\frac{R^{2}}{2}$, we obtain

$$\begin{aligned} \begin{aligned}&\frac{16}{4+8R^{2}+2[|OA|^{2}+|OB|^{2}+|OC|^{2}+|OD|^{2}]} \\&=\frac{16}{4+8R^{2}+2[4(x^{2}+y^{2})+2R^{2}]}\\&\ge \frac{16}{4+8R^{2}+2[2R^{2}+2R^{2}]} \ge \frac{1}{1+R^{2}}\quad \square \end{aligned} \end{aligned}$$

(13)

Fig. 3.

Illustration of Theorem 1

1.2 Proof of Proposition 1

Proof

There are four kinds of blocks generated by edge extensions:

• 1. The block has only one edge on an extended edge.

• 2. The block has two edges on the extended edges.

• 3. The block has three edges on the extended edges.

• 4. The block has four edges on the extended edges.

For the first case, the block has two possible shapes, which are shown by (a) and (b) in Fig. 4. The solid edges are line segments on the polygon’s boundary and the dotted edge is on the extended edge. Since the extended edge is a ray starting from a reflex vertex, at least one endpoint of the dotted edge is reflex vertex and the whole block is visible to this reflex vertex.

For the second case, the block has five kinds of shapes, shown by (c) to (g) in Fig. 4. In (c), although no vertex of the block is reflex, the block must be visible to the reflex vertices that emit the block’s dotted edges, because in a simple orthogonal polygon, if some other points obstruct the visibility between the reflex vertex and the block, there must be another reflex vertex that divides the block into smaller ones. In (d) to (k), at least one vertex of the polygon is reflex, and hence, the whole block is visible to the vertex.

For the third case, (h) and (i) show the two possible shapes. For the shape in (h), the block must be visible to three reflex vertices and the reason is similar to that for (c). In (i), the block is visible to at least two reflex vertices, which are its own vertices.

For the fourth case, similar to the shape in (c) and (h), the block is entirely visible to at least four reflex vertices.    $\square$

Fig. 4.

Block visibility

1.3 Proof of Proposition 4

Proof

We can prove this proposition by contradiction. Suppose there is a block that is not entirely visible to the light g, while all its four vertices are visible to the light g. Then there must be some other points or edges that obstruct the visibility from the light g. Since it is a simple orthogonal polygon, the obstacle must be part of the boundary of the orthogonal polygon. Hence, the boundary must obstruct the visibility from the light g to at least one vertex of the block, which contradicts to the condition.    $\square$

1.4 Proof of Proposition 5

Proof

This proposition can be proved by contradiction. Suppose block ABCD is a small block generated by the edge extensions of an orthogonal polygon P, and all four vertices of block ABCD (Fig. 5) are visible to light $G_{1}$ but the block is not entirely visible to light $G_{1}$. The convex polygon shaped by light $G_{1}$ and some vertices of block ABCD is $G_{1}ABCD$. Since block ABCD is not entirely visible to light $G_{1}$, there must be some vertices of the polygon P inside the polygon $G_{1}ABCD$ blocking the light from light $G_{1}$. This contradicts with the condition “there is no other vertex inside the convex polygon”.    $\square$

Fig. 5.

Figure for the proof of Proposition 5