Flexible sliced Latin hypercube designs with slices of different sizes

Abstract

Sliced Latin hypercube designs (SLHDs) are widely used in various computer experiments. Literatures concerning the construction of SLHDs are all about constructing SLHDs with slices of the same size. However, in some cases, e.g. when an experiment with multiple computer models having different complexities or a sequential experiment with varying costs in different periods is considered, SLHDs with slices of different sizes are preferable. In this paper, we propose a new class of SLHD, named the flexible SLHD, in which the whole design is a Latin hypercube design (LHD), and each slice is also an LHD when its levels being properly collapsed, the difference lies in that its slices may have different run sizes. Several methods for constructing such designs are developed. Theoretical results on the constructed designs are derived, and discussion on the slice sizes of the constructed flexible SLHDs is provided. Furthermore, an optimization algorithm is developed to improve the space-filling property of the constructed SLHDs. The newly proposed flexible SLHD is also a special nested LHD (Qian in Biometrika 96:957–970, 2009), each of its slice can be viewed as a small LHD nested in the whole design.

Appendix: Proof of Theorem 6

• (i) It is known that $N=m_1 +m_2$, and $m_1 | N,\ m_2 | N$. So we have $m_1 | m_2$ and $m_2|m_1$, as all the numbers are positive integers, therefore, $m_1=m_2=N/2$.

• (ii) For the positive integers $m_1$, $m_2$, and $m_3$, since $m_1 \le m_2 \le m_3$, $N=m_1+m_2+m_3$ and $m_i |N$ for $i=1,2,3$, then we have

  $$\begin{aligned} m_1+m_2 \le 2 m_3\ \text{ and }\ m_3 | (m_1+m_2), \end{aligned}$$

  which imply

  $$\begin{aligned}&\displaystyle m_1+m_2 = 2 m_3, \text{ or } \end{aligned}$$

  (7)
  $$\begin{aligned}&\displaystyle m_1+m_2 = m_3. \end{aligned}$$

  (8)

  From Eq. (7), it is easy to deduce that $m_1=m_2=m_3$. From Eq. (8) and $m_2|N$, we have $m_2 | 2(m_1+m_2)$, which implies $m_2 | 2 m_1$; since $m_1 \le m_2$, $m_1$ and $m_2$ have the following relation,

  $$\begin{aligned}&\displaystyle 2 m_1=m_2, \text{ or } \end{aligned}$$

  (9)
  $$\begin{aligned}&\displaystyle 2 m_1=2 m_2. \end{aligned}$$

  (10)

  If Eq. (9) holds, it is easy to deduce that $m_2=2 m_1$, $m_3=3 m_1$, and $m_1<m_2<m_3$. Alternatively, if Eq. (10) holds, it is easy to deduce that $m_3=2 m_1$ and $m_1=m_2<m_3$. This completes the proof. $\square$