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A new series of main effects plus one plan for 2m factorial experiments with m=4λ±1 and 2m runs
Journal of Statistical Planning and Inference 141 (2011) 1567–1574
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Journal of Statistical Planning and Inference journal homepage: www.elsevier.com/locate/jspi
A new series of main effects plus one plan for 2m factorial experiments with m = 4l 71 and 2m runs Nabaz Esmailzadeh a,, Hooshang Talebi b, Koji Momihara c, Masakazu Jimbo c a b c
Department of Statistics, University of Kurdistan, Sanandaj, Iran Department of Statistics, University of Isfahan, Isfahan, Iran Graduate School of Information Science, Nagoya University, Furo-Cho, Chikusa-ku, Nagoya 464-8601, Japan
a r t i c l e in f o
abstract
Article history: Received 17 September 2009 Received in revised form 30 October 2010 Accepted 6 November 2010 Available online 13 November 2010
An MEP.1 plan for a 2m factorial experiment is able to search and estimate one possible nonzero interaction effect along with estimating the general mean and main effects. Construction of such a plan has been tackled by several authors. Some of these works are developed for all m, in which are focused on m =2h 1 or m ¼ 4l1 for positive integers h 4 2 and l Z 2. This leads to a gap for having MEP.1 plans with well-structured runs set and probably with a reasonable number of treatments for ma2h 1 or 4l1. In this paper, we characterize and present a series of highly structured MEP.1 plans for all odd m ¼ 4l 7 1, l Z 2, with N= 2m runs. From these designs one can obtain the MEP.1 plans for even m with N =2m + 1 treatments. The exemptions are m = 8 and 9 for which we have 1 or 2 extra runs. & 2010 Elsevier B.V. All rights reserved.
Keywords: Non-negligible effect Search designs Search linear model PBIBD
1. Introduction Consider, the commonly used, main effect plans in a 2m factorial experiment and take b1 to be an ðm þ 1Þ 1 vector of the general mean and main effects, which the experimenter is interested in estimating them. By a search design, which has been introduced by Srivastava (1975), one means a design which can search and estimate k possible non-negligible factorial effects in a n2 1 vector b2 along with estimating the unknown parameters in b1 . Construction of search designs has been investigated by many authors, though a few of them, developed a method of constructing these designs for all m in general. Characterizing and constructing of such plans is presented in Ghosh (1980), Shirakura (1991) and Ghosh and Talebi (1993). More details on the history of these works, which have been developed in this area, are given in Ghosh et al. (2007). Ghosh (1980) considered the characterization problem of search design for the 2m factorial experiments when m= 2h 1 with hZ 2. He gave conditions on a follow-up design, that are required to be added to an orthogonal main effect plan (MEP) in order to provide additional information that can be used to de-alias any pair of interaction effects from the main effects. A search design with 24 runs has been constructed and presented for m= 7. Such an augmented search design carries some optimum estimation properties. However, it is poor in terms of the economical number of runs. Shirakura (1991) reduced the total number of runs to 15, for m=7, in an augmented search design by adding balanced arrays, as the follow-up designs, to the MEP given in Ghosh (1980). Ghosh and Talebi (1993) followed an analogous approach by adding balanced arrays to a saturated MEP plan and obtained search designs for m ¼ 4l1 with l Z 2. The latter result partially solves the lack of having wellstructured search designs for ma2h 1,hZ 2. It is also superior to that of previous ones in terms of the number of runs. So far, Corresponding author.
E-mail addresses: [email protected] (N. Esmailzadeh), [email protected] (H. Talebi), [email protected] (K. Momihara), [email protected] (M. Jimbo). 0378-3758/$ - see front matter & 2010 Elsevier B.V. All rights reserved. doi:10.1016/j.jspi.2010.11.008
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the lack of being well-structured search designs, with reasonable number of runs, for ma4l1 with l Z2 is open. In this paper, we investigate the problem of design construction to fill this gap. We address this problem and obtain a wellstructured factor setting, which leads to series of new search designs with smaller number of runs. This improves the result of Ghosh and Talebi (1993) by reducing the number of runs in their original designs for m ¼ 4l1 with l Z 2. We also develop and construct new search designs for all prime powers m ¼ 4l þ 1 with l 4 2. These results lead to designs with having a wellknown structural factor setting and a reasonable number of runs, for all odd m’s. 2. Preliminaries and notations In this paper, we restrict b2 to 2- and 3-factor interactions and assume that k, the number of nonzero effects in b2 , is 1. The four-factor and higher-order interactions are assumed to be zero. A search design that solves the problem of searching for a nonzero effect of b2 and estimating it along with estimation of the general mean and main effects is called a main effect plus 1 (MEP.1) plan. To deal with this problem, for an observation vector yðN 1Þ of a 2m factorial experiment, N o2m , we consider the following linear model: y ¼ X1 b1 þ X2 b2 þ e,
VarðeÞ ¼ s2 IN ,
ð1Þ
where Xi ðN ni Þ are known design matrices with elements 7 1, i= 1,2, b1 is an ðm þ 1Þ 1 vector consisting of the general m mean and main effects, b2 is a n2 1 vector consisting of 2- and 3-factor interactions, n2 ¼ ðm 2 Þ þð 3 Þ, e is an error random vector and IN is the identity matrix of order N. For noiseless case, s2 ¼ 0, Srivastava (1975) gave a necessary and sufficient condition for a design to be an MEP.1 plan. That is, for every submatrix X22 ðN 2Þ of X2, rank½X1 ; X22 ¼ mþ 3:
ð2Þ
Now, consider (0,1)-matrix TðN mÞ to be a design with m factors each at two levels and N runs, with ti as i-th column corresponding to the i-th factor, i= 1,2,y,m. Under the search linear model (1), the N ðm þ 1 þ n2 Þ model matrix for T is written as X ¼ ½1N ; U1 ; U2 ; U3 ,
ð3Þ
where 1N is an N 1 vector of 1’s corresponding to the general mean and Ui, i =1,2,3, are sets of columns, with elements +1 and 1, corresponding to the main effects, 2-factor and 3-factor interactions, respectively. Clearly, [1N;U1]= X1 and hence the i-th column in U1 is corresponding to ti. Following Chatterjee et al. (2001), to define the columns of the following submatrices T 2 and T 3 by tij = tintj and tijk = 2tintjntk + tintj + tintk +tjntk; io j o kði,j,k ¼ 1,2, . . . ,mÞ, respectively, in which stands for Hadamard product. Define the (0,1)-transformed matrix T corresponding to X as T ¼ ½1N ; T 1 ; T 2 ; T 3 ,
ð4Þ
where T 1 is the given design T, T 2 and T 3 are corresponding to U2 and U3, respectively. From the above column definition one can easily find out that an element in column tij of T 2 is 1 when the corresponding elements in both columns ti and tj are 1’s. It is also clear that an element in column tijk of T 3 is 1 when at least two of the corresponding elements in ti, tj and tk are 1’s. Otherwise, they are 0. Note that this transformation preserves the column rank condition of the model matrix X. This allows us to use matrix T directly to verify condition (2) in (3). Throughout this paper, for a set of h factors in {1,2,3,y,m}, let (a1,a2,y,ah) be a treatment combination corresponding to these factors where the treatment points, ai, are either 0 or 1 for i= 1, 2,y, h. We denote the number of runs in T consisting of a 2 ,...,ah Þ for any selected subset of h columns fi1 ,i2 , . . . ,ih g D f1,2, . . . ,mg. particular treatment combination (a1,a2,y,ah ) by Nðiða11,i,a2 ,...,i hÞ
Now, let {i1,i2,y,ic} be a set of cð r mÞ columns in T. For any subset of hð r cÞ columns fij1 ,ij2 , . . . ,ijh g Dfi1 ,i2 , . . . ,ic g and treatment combination faj1 ,aj2 , . . . ,ajh g corresponding to these columns, it is true that X ða ,a ,...,a Þ ðaj ,aj ,...,aj Þ Nij i1j ...i2 j h ¼ Ni1 i12 ...i2c c , ð5Þ 1 2
h
S
where S is the set of all permutations of treatment combinations ðaj1 ,aj2 , . . . ,ajðchÞ Þ such that jl 2 fi1 ,i2 , . . . ,ic gfij1 ,ij2 , . . . ,ijh g, for l = 1,2,y, (c h). An m 1 vector and an m n matrix with all elements 0 are denoted by 0m and OðmnÞ , respectively. 3. Main results In this section, we present main results through Theorems 1 and 2. The proofs of theorems along with some lemmas and corollaries are given in Appendix. These lemmas and corollaries are useful in proving the theorems. Now, consider the matrix Tu ¼ ½1m ; Im ; T1 , where m ¼ 4l1 in which T1 is an m m incidence matrix of a symmetric BIBD with parameters v ¼ b ¼ 4l1,r ¼ k ¼ 2l and l with l Z2. Ghosh and Talebi (1993) proved that T is an MEP.1 plan with N( =2m + 1) runs. The following theorem proves that the number of runs can still be reduced to get an MEP.1 plan. Define the matrix Tu ¼ ½Im ; T1 . Theorem 1. T is an MEP.1 plan with N =2m runs for m ¼ 4l1, l Z 2. Proof. See Appendix.
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Now, we construct a search design similar to T in Theorem 1 when m ¼ 4l þ 1 is a prime power. That is, define the design matrix " # Im T¼ , ð6Þ T2 where T2 is the m m incidence matrix of a partially balanced incomplete block design (PBIBD). We will show that the design T is an MEP.1 plan by utilizing the properties of a PBIBD and applying a result on Galois fields. This needs a more understanding of T2 in details. For an odd prime p, let m= ps and consider the ðm mÞ-matrix T2 =(di,j) whose rows and columns are labeled by the elements of GF(ps), the Galois field of order ps, defined by ( 0 if ði þjÞ 2 GFðps Þ is quadratic residue, di,j ¼ 1 if ði þjÞ 2 GFðps Þ is non-quadratic or zero: When m ¼ 4l þ 1, the matrix T2 forms an incidence matrix of a PBIBD with two associate classes and with the design parameters ðv,b,r,k, l1 , l2 Þ ¼ ð4l þ 1,4l þ 1,2l þ 1,2l þ 1, l þ1, lÞ, and the association scheme parameters ðn1 ,n2 ,p11,1 ,p21,1 ,p11,2 ,p21,2 ,p12,1 ,p22,1 ,p12,2 ,p22,2 Þ ¼ ð2l,2l, l1, l, l, l, l, l, l, l1Þ: Note that for this PBIBD, the pair of distinct elements a, b 2 GFðps Þ are 1st associate if ab is non-quadratic and 2nd associate otherwise. Then, each element of the PBIBD has ni i-th associates, i= 1, 2. Furthermore, for any two i-th associates elements a and b, the number of elements that are j-th associates of a and k-th associates of b is a constant pij,k, independent of the pair a and b. It is also known that two i-th associates elements occur together in exactly li blocks. We call this PBIBD a QR-type with parameters n ¼ b ¼ 4l þ1,k ¼ r ¼ 2l þ1, l1 ¼ l þ 1 and l2 ¼ l. Example 1. For prime odd m= 13, the quadratic residue elements in GF (13) are 1, 3, 4, 9, 10 and 12, so the following matrix T2 is the incidence matrix of a PBIBD with parameters (13, 13, 7, 7, 4, 3), 0 1 2 3 4 5 6 7 8 9 10 11 12 1 1 0 1 0 0 1 1 1 1 0 0 1 0 B 1 B0 1 0 0 1 1 1 1 0 0 1 0 1C C B C C 2 B 1 0 0 1 1 1 1 0 0 1 0 1 0 B C B 3 B0 0 1 1 1 1 0 0 1 0 1 0 1C C B C C 4 B 0 1 1 1 1 0 0 1 0 1 0 1 0 B C B 5 B1 1 1 1 0 0 1 0 1 0 1 0 0C C B C C: 6 B 1 1 1 0 0 1 0 1 0 1 0 0 1 B C B C 7 B1 1 0 0 1 0 1 0 1 0 0 1 1C B C C 8 B B1 0 0 1 0 1 0 1 0 0 1 1 1C B C 9 B0 0 1 0 1 0 1 0 0 1 1 1 1C B C B 10 B 0 1 0 1 0 1 0 0 1 1 1 1 0 C C B C 11 @ 1 0 1 0 1 0 0 1 1 1 1 0 0 A 0
12
0
0
1
0
1
0
0
1
1
1
1
0
0
1
To be more precise, from the properties of the PBIBD ð4l þ 1,2l þ 1, l þ 1, lÞ, it is true that for any two columns, say i1 and i2, of T2 8 > < ðl þ 1, l, l, lÞ ð1,1Þ ð0,1Þ ð1,0Þ ð0,0Þ ð7Þ ðNi1 i2 ,Ni1 i2 ,Ni1 i2 ,Ni1 i2 Þ ¼ or > : ðl, l þ 1, l þ 1, l1Þ: For any three columns of such a PBIBD, Proposition 2 is a direct consequence of the following proposition derived from Theorem 2.2. of Buratti and Pasotti (2009) with e= 2 and t =3. Proposition 1. Let p 1 ðmod 2Þ be a prime power with p 4 45. Then for any given triple ða1 ,a2 ,a3 Þ 2 f0,1g3 and any given triple (c1,c2,c3) of distinct elements of GF(p), there exists an element x 2 GFðpÞ such that x þci 2 Ca2i for each i, where Ca2i ¼ fa2‘ þ ai j0 r ‘ o ðp1Þ=2g and a is a primitive element of GF(p). Proposition 2. Suppose i1, i2 and i3 are three distinct columns of the incidence matrix of a symmetric QR-type PBIBD ð4l þ 1,2l þ 1, l þ 1, lÞ. The number of each triple (a1,a2,a3), where ai’s are either0 or 1, as rows of selected columns (i1, i2, i3) are at least1 for l 4 10. Now, we establish the following theorem, which gives a new series of MEP.1 plan.
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Theorem 2. The model design matrix for T in (6) satisfies condition (2), i.e. T is an MEP.1 plan with N = 2m runs, where m ¼ 4l þ 1, l 42. Proof. See Appendix. By virtue of Theorem 2, the challenge is to reduce the number of runs in contrast to the designs obtained by removing columns from those in Ghosh and Talebi (1993). Example 2. Consider T2 given in Example 1 for m= 13. The design (6) for this T2 is an MEP.1 with 26 runs. For such an m, Ghosh and Talebi (1993) gave an MEP.1 plan with at least 28 runs. Remark 1. For l ¼ 1, Theorem 2 is true by adding run (0,0,0,0,0) to design T in (6). Remark 2. An MEP.1 for an even m ða4,8Þ can be obtained by removing any one column from the MEP.1 plan of m+1 factors with N = 2m+ 1 runs. 4. Discussion A design with smaller number of runs and hence a lower cost experiment is a main purpose in design construction. It is also known that a search design is superior to numbered-resolution, e.g. IV and V, design in order to do estimation job with smaller number of observations. Many researchers have been improved MEP.1 plans by reducing the number of runs. Ghosh and Talebi (1993) gave a series of MEP.1 plans with N =2m +1, which do the search job for all m ¼ 4l1, l Z2. For m ¼ 4l1d, d ¼ 1,2,3, the MEP.1 plans are obtained by deletion of any d columns from one with m ¼ 4l1. However, one may not be satisfied with resulting plans due to their number of runs. In this paper, we develop a new series of MEP.1 plan with reasonable number of runs for 2m factorial experiments when m Z6. That is, let partition positive odd numbers to two sets, say P and Q, such that P ¼ fm : m ¼ 4l1, l Z2g and Q ¼ fm : m ¼ 4l þ1, l 4 2g. We characterize a series of (0,1)-matrix of S size ðN mÞ as MEP.1 plans with a special structure of Tu ¼ ½Im ; Ti , i=1,2, with N = 2m treatments when m 2 P Q . The matrices Ti, i =1,2 were defined in Section 3 for m 2 P and m 2 Q , respectively. Clearly, the MEP.1 plan for even m with N =2m +1 treatments can be obtained by deleting any one column from T with m+ 1 factors except for m= 4 and 8. Appendix Here, we provide some lemmas and corollaries required in proving Theorems 1 and 2. First, using these corollaries, we characterize T1 in T of Theorem 1, by the properties of symmetric BIBDð4l1,2l, lÞ. From the properties of the symmetric BIBDð4l1,2l, lÞ, for every two columns i1 and i2 of T1 it is true that ( l1 if ða1 ,a2 Þ ¼ ð0,0Þ, Niða1 i12,a2 Þ ¼ ð8Þ l if ða1 ,a2 Þ ¼ ð1,1Þ,ð1,0Þ,ð0,1Þ: The following Lemmas 1 and 2 on the properties of BIBD are immediate consequence of Lemmas 1–3 of Ghosh and Talebi (1993). Lemma 1. Suppose i1, i2 and i3 are three distinct columns of the incidence matrix of a symmetric BIBDð4l1,2l, lÞ. Then Nið1,1,0Þ Z 1,Nið1,0,1Þ Z 1 and Nið0,1,1Þ Z 1. 1 i2 i3 1 i2 i3 1 i2 i3 Lemma 2. Consider every four columns, say, i1, i2, i3 and i4, of the incidence matrix of a symmetric BIBDð4l1,2l, lÞ. Then þ Nið0,0,1,1Þ Z 1. Nið1,1,0,0Þ 1 i2 i3 i4 1 i2 i3 i4 Z1 and Nið0,0,1,1Þ Z1 if Nið1,1,1,1Þ þ Nið0,0,0,0Þ Z 1. Remark 3. It is true that Nið1,1,0,0Þ 1 i2 i3 i4 1 i2 i3 i4 1 i2 i3 i4 1 i2 i3 i4 Now, we write the matrix in (4)for design T in Theorem 1. It leads to the following corresponding column-wise partition matrix form: " # 1m Im Oðmðm2 ÞÞ Oðmðm3 ÞÞ , ð9Þ 1m T1 T12 T13 m where T12 and T13 are m ðm 2 Þ and m ð 3 Þ submatrices of T 2 and T 3 , corresponding to T1, respectively.
Corollary 1. Every m 2 submatrix of T12 has rows (1,0), (0,0) and (0,1). Proof. Suppose ti1 i2 and ti3 i4 are two columns of T12, which they are correspond to 2-factor interactions. These two columns either have a common or no common factor. If they have one common factor, say i2 =i3, then by Lemma 1 columns (i1,i2(i3),i4) have rows (1,1,0), (0,1,1) and (1,0,1). Thus, in this case the proof is complete. If they have no common factor, by Lemma 1 columns (i1,i2,i3) and (i2,i3,i4) have rows (1,1,0) and (0,1,1). Thus, by considering the rows (0,0) for columns (i2,i3) the proof is complete. &
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Table 1 Arranged T1 in terms of columns i3 and i4. i1
i2
i3
i4
1l 1l 0l1 0l
1l 0l 0l1 1l
i5
Corollary 2. Every m 2 submatrix with one column from T12 and another from T13 has rows (1,a), (0,1) and (0,0) where a is either 0 or 1. Proof. Suppose the columns in T12 and T13 are denoted by ti1 i2 and ti3 i4 i5 , where the sets {i1, i2} and {i3, i4, i5} correspond to factors. There are three possibilities for two sets: (i) no common factor, (ii) one common factor, say i2 = i3 and (iii) two common factors, say i1 =i3 and i2 = i4. For (i), we can always arrange the columns i1,i2, i3,i4 and i5 in T1 as Table 1 from (8). There must always be a run (0,a) or (a,0) for (i1,i2) in the third block row for l 43. Otherwise, in order to have l1 possible (0,0) for (i1,i3) and (i2,i3) all must occur in the fourth block row. Having l rows in this block restricts a to be 0, such that the total number of (0,0) for (i1,i2) must be at least l2. By similar argument all l1 possible (0,0) of columns (i1,i4) and (i2,i4) must occur in the second block row. Then the total number of (0,0) for columns (i1,i2) in this block must be at least l2. It leads to that have 2ðl2Þ r l1. That is, l r3 which contradicts the assumption l 4 3. Therefore, the third row of Table 1 must have some runs 0 for columns i1 and i2. So, by having a (0,a) or (a,0) run in the third row of Table 1 for (i1,i2) and from Lemma 1 and this fact that column ti1 i2 has l elements 1, the proof is clear. For l ¼ 2 and 3 the proof is simple by completing Table 1. For (ii), by considering columns i1,i2 and i4 in T1 and Lemma 1, we obtain the row (0,1) for ðti1 i2 ,ti3 i4 i5 Þ. From rows (0,0) of columns i2 and i4, row (0,0) is also obtained. Note that the number of 1’s in column ti1 i2 is l by (8). This complete the proof for this case. Finally, for (iii), by considering the rows (1,1) and (0,0) for columns i1 and i2 in T1 and by Lemma 1 the proof is clear. & Corollary 3. Every m 2 submatrix of T13 has rows in {(1,a), (b,1), (0,0)} or {(1,1), (1,0), (0,1)}, where a and b are either0 or 1, not both1 simultaneously. Proof. Let two columns in T13 are denoted by ti1 i2 i3 and ti4 i5 i6 , where the sets {i1,i2,i3} and {i4,i5,i6} correspond to factors. There are three possibilities for two sets: (i) no common factor, (ii) one common factor, say i3 =i4 and (iii) two common factors, say i2 = i4 and i3 =i5. For (i), consider four columns i2,i3,i4 and i5 and two possible cases whether at least one of the rows (1,1,1,1) or (0,0,0,0) occurs in these columns or not. If it does then by Lemma 2 and its remark we have the set of rows {(0,0), (1,0), (0,1)} or {(1,1), (1,0), (0,1)}. Otherwise, for these four columns, there exist two possible nonisomorphic row sets given in Table 1 of Lemma 3 of Ghosh and Talebi (1993). These possible nonisomorphic sets of runs give the rows in {(0,0), (b,1), (1,a)} and the proof is complete. For (ii), we arrange the matrix T1 in terms of columns i2 and i3 similar to Table 1 and consider the columns i2, i3(i4), i5 and i6. Set of rows (1,a) and (b,1) for ti1 i2 i3 and ti4 i5 i6 are obtained by applying Lemma 1 to the three columns i2, i3 (i4) and i5 and Lemma 2 to the four columns. By the argument similar to Corollary 2(i) for four columns i2,i3,i5 and i6 the row (0,0) is obtained for two columns in T13. Finally, for (iii), by considering all possible combinations for columns i2 and i3 the proof is clear from Lemma 2. & The following lemma is useful in proving Theorems 1 and 2. Lemma 3. Let W11 be an m m matrix, W12 an m q matrix, W21 an n m matrix, and W22 an n q matrix. If rank(W11)= m, then " # W11 W12 1 ¼ mþ rankðW22 W21 W11 W12 Þ: rank W21 W22 See page 98 of Harville (1997) for a proof. Proof of Theorem 1. In order to prove T is a search design we need to show that condition (2) is valid for (9), from equivalence rank property of (9) with model matrix in (3). That is, for any two distinct columns ti1 and ti2 of [T12; T13] we need to show that " # 1m Im 0m 0m rank ¼ m þ 3: 1m T1 ti1 ti2 Applying Lemma 3 by taking W11 =Im, W21 = T1, W12 = [1m; 0m; 0m] and W22 ¼ ½1m ; ti1 ; ti2 , the proof is reduced to show that rank 1 (W22 W21 W11 W12)= rank(W22)= 3. There are three possible choices in choosing two columns from [T12 ; T13]: Case (i) two columns from T12, Case (ii) one column from T12 and another from T13,
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Case (iii) two columns from T13. For the first case Corollary 1 guarantees that W22 has rank 3, as we always can identify the following submatrix of W22: 2 3 1 0 0 6 7 4 1 1 0 5, 1 a 1 whose rank is 3. The proof for cases (ii) and (iii) are obtained from Corollaries 2 and 3, respectively. This completes the proof. & Now, we give some lemmas and corollaries, which are useful in proving Theorem 2. These corollaries characterize T2 in T given in (6) using the properties of a symmetric PBIBDð4l þ 1,2l þ 1, l þ 1, lÞ. Lemma 4. Consider every four columns, say, i1, i2, i3 and i4, of the incidence matrix of a symmetric PBIBDð4l þ 1,2l þ 1, l þ 1, lÞ. Let þNið0,0,1,1Þ A ¼ Nið1,1,0,0Þ 1 i2 i3 i4 1 i2 i3 i4 8 > B þ1 if > > > > > > > B þ2 if > > > > > > B þ1 if > > > > > > B þ2 if > > > < A ¼ B þ1 if > > > > B þ3 if > > > > > > > B if > > > > > > B if > > > > > > : B1 if
and B ¼ Nið1,1,1,1Þ þ Nið0,0,0,0Þ . The following holds 1 i2 i3 i4 1 i2 i3 i4 ¼ Nið0,1Þ Nið1,1Þ 1 i3 2 i3 Nið1,1Þ 1 i3 ð1,1Þ Ni 1 i 3 Nið1,1Þ 1 i3 Nið1,1Þ 1 i3 Nið1,1Þ 1 i3 ð1,1Þ Ni 1 i 3 Nið1,1Þ 1 i3 Nið1,1Þ 1 i3
¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼
Nið0,1Þ 2 i3 ð1,1Þ Ni2 i3 Nið1,1Þ 2 i3 Nið1,1Þ 2 i3 Nið1,1Þ 2 i3 ð0,1Þ Ni2 i3 Nið1,1Þ 2 i3 Nið1,1Þ 2 i3
and
Nið1,1Þ ¼ Nið0,1Þ 1 i4 2 i4
and
Nið1,1Þ 1 i4 ð1,1Þ Ni1 i4 Nið1,1Þ 1 i4 Nið1,1Þ 1 i4 Nið1,1Þ 1 i4 ð1,1Þ Ni1 i4 Nið1,1Þ 1 i4 Nið1,1Þ 1 i4
¼ l þ1
and
¼l
and
¼l
and
¼l
and and
¼ l þ1
and
¼ l þ1
and
¼ Nið1,1Þ 2 i4 ð1,1Þ ¼ Ni 2 i 4 ¼ Nið0,1Þ 2 i4 ¼ Nið1,1Þ 2 i4 ¼ Nið1,1Þ 2 i4 ð1,1Þ ¼ Ni 2 i 4 ¼ Nið0,1Þ 2 i4 ¼ Nið1,1Þ 2 i4
ðL4:1Þ ¼l
ðL4:2Þ
¼l
ðL4:3Þ ðL4:4Þ
¼ lþ1
ðL4:5Þ
¼l
ðL4:6Þ
¼ lþ1
ðL4:7Þ ðL4:8Þ
¼ l þ 1:
ðL4:9Þ
Proof. It follows from (7) and (5) that Nið1,1Þ ¼ Nið1,1,1Þ þ Nið1,0,1Þ , 1 i3 1 i2 i3 1 i2 i3 ¼ Nið0,0,1Þ þ Nið1,0,1Þ , Nið0,1Þ 2 i3 1 i2 i3 1 i2 i3 þ Nið1,1,0Þ ¼ Nið0,0,1Þ þ Nið0,0,0Þ þ 1: Nið1,1,1Þ 1 i2 i3 1 i2 i3 1 i2 i3 1 i2 i3
ð10Þ
8 ð0,0,0Þ þ1 N > > > i1 i2 i3 < ð0,0,0Þ ð1,1,0Þ Ni1 i2 i3 ¼ Ni1 i2 i3 > > > ð0,0,0Þ : Ni1 i2 i3 þ 2
ð11Þ
So, if Nið1,1Þ ¼ Nið0,1Þ , 1 i3 2 i3 if Nið1,1Þ ¼ Nið1,1Þ ¼ l þ 1, 1 i3 2 i3 if
Nið1,1Þ 1 i3
¼
Nið1,1Þ 2 i3
¼ l:
Following (5) we have Nið1,1,0Þ ¼ Nið1,1,0,1Þ þNið1,1,0,0Þ , 1 i2 i3 1 i2 i3 i4 1 i2 i3 i4 ¼ Nið0,0,0,0Þ þ Nið0,0,0,1Þ : Nið0,0,0Þ 1 i2 i3 1 i2 i3 i4 1 i2 i3 i4 Thus, from (11) and (12) it follows that 8 ð0,0,0,0Þ þ1 Ni i i i þ Nið0,0,0,1Þ > > 1 i2 i3 i4 > < 1234 ð0,0,0,0Þ ð0,0,0,1Þ ð1,1,0,1Þ ð1,1,0,0Þ Ni1 i2 i3 i4 þNi1 i2 i3 i4 ¼ Ni1 i2 i3 i4 þ Ni1 i2 i3 i4 > > > ð0,0,0,0Þ : þ2 Ni1 i2 i3 i4 þ Nið0,0,0,1Þ 1 i2 i3 i4
ð12Þ
if Nið1,1Þ ¼ Nið0,1Þ , 1 i3 2 i3 if Nið1,1Þ ¼ Nið1,1Þ ¼ l þ1, 1 i3 2 i3 if
Nið1,1Þ 1 i3
¼
Nið1,1Þ 2 i3
¼ l:
Applying (7) and (5) we can get similar result as (10), (11) and (12) for columns i1,i2 and i4. Thus we have 8 ð0,0,0,1Þ þ Nið0,0,1,1Þ if Nið1,1Þ ¼ Nið0,1Þ , N > > 1 i2 i3 i4 1 i4 2 i4 > i1 i2 i3 i4 < ð0,0,0,1Þ ð0,0,1,1Þ ð1,1Þ ð1,1Þ ð1,1,1,1Þ ð1,1,0,1Þ þ N þ 1 if N ¼ N ¼ l þ 1, N Ni1 i2 i3 i4 þNi1 i2 i3 i4 ¼ i1 i2 i3 i4 i1 i2 i3 i4 i1 i4 i2 i4 > > > ð0,0,0,1Þ ð0,0,1,1Þ ð1,1Þ ð1,1Þ :N if Ni1 i4 ¼ Ni2 i4 ¼ l: i1 i2 i3 i4 þ Ni1 i2 i3 i4 1 Therefore, the proof is complete from (13) and (14).
&
ð13Þ
ð14Þ
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Remark 4. For the cases (L4.1)–(L4.6), Nð1,1,0,0Þ þ Nð0,0,1,1Þ Z1. Lemma 5. Consider four columns i1, i2, i3 and i4 of a symmetric PBIBDð4l þ1,2l þ 1, l þ 1, lÞ incidence matrix. Assume that N(0,0,0,0) =N(1,1,0,0) = N(0,0,1,1) =N(1,1,1,1) =0, then for both conditions (L4.7) and (L4.8), it is true that Nð1,0,1,0Þ þ Nð0,1,0,1Þ Z1:
ð15Þ
Also, suppose that (i) N(0,0,0,0) = N(1,1,0,0) =N(0,0,1,1) = 0,N(1,1,1,1) = 1, or (ii) N(1,1,1,1) =N(1,1,0,0) = N(0,0,1,1) =0,N(0,0,0,0) = 1, then for condition (L4.9), (15) holds. Proof. It follows from condition (L4.7) that Nð1,1,1,0Þ ¼ N ð0,0,1,0Þ
and
Nð1,1,0,1Þ ¼ N ð0,0,0,1Þ þ1:
ð16Þ
From (7) and (5), we have ¼ Nð1,1,1,0Þ þN ð0,0,1,0Þ þN ð1,0,1,0Þ ¼ l or l þ 1 Nið1,0Þ 3 i4
ð17Þ
¼ Nð1,1,0,1Þ þN ð0,0,0,1Þ þN ð0,1,0,1Þ ¼ l or l þ 1: Nið0,1Þ 3 i4
ð18Þ
and
So, from (16)–(18) we get 2Nð1,1,1,0Þ þ2N ð1,1,0,1Þ þ Nð1,0,1,0Þ þ Nð0,1,0,1Þ ¼ 2l þ1 or 2l þ 3: Additionally, from (7) and (5) ¼ Nð1,1,1,0Þ þN ð1,1,0,1Þ ¼ l or l þ 1: Nið1,1Þ 1 i2 Thus, N
ð1,0,1,0Þ
þN
ð0,1,0,1Þ
¼
8 <1
if Nið1,1Þ ¼ Nið1,0Þ ¼ Nið0,1Þ ¼ l or l þ 1, 1 i2 3 i4 3 i4
:3
¼ l, Nið1,0Þ ¼ Nið0,1Þ ¼ l þ 1: if Nið1,1Þ 1 i2 3 i4 3 i4
This completes the proof. The proof for conditions (L4.8) and (L4.9) are similar.
&
Lemma 6. Consider six columns i1,i2,i3,i4,i5 and i6 of T2 for l 410.
(i) If Nð0,0,0,0Þ ¼ N ð1,1,0,0Þ ¼ Nð0,0,1,1Þ ¼ N ð1,1,1,1Þ ¼ 0, then Nð1,1,0,0,0Þ þ Nð1,0,1,0,0Þ Z 1, N ð0,1,0,1,1Þ þN ð0,0,1,1,1Þ Z1 and i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i1 i2 i3 i5 i6 i1 i2 i3 i5 i6 i1 i2 i3 i5 i6 i1 i2 i3 i5 i6 Nið0,0,1,0,0Þ þNið0,0,0,1,0Þ Z1. 2 i3 i4 i5 i6 2 i3 i4 i5 i6 (ii) If Nð1,1,1,1Þ Z1 and Nð1,1,0,0Þ ¼ Nð0,0,1,1Þ ¼ Nð0,0,0,0Þ ¼ 0 then N ð0,0,1,0,0Þ þN ð0,1,0,0,0Þ Z1. i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i1 i2 i3 i4 i5 i1 i2 i3 i4 i5 (iii) If Nð0,0,0,0Þ Z 1 and Nð1,1,0,0Þ ¼ N ð0,0,1,1Þ ¼ Nð1,1,1,1Þ ¼ 0 then N ð1,1,0,1Þ þ Nð1,1,1,0Þ Z 1 and Nð0,0,1,1,1Þ þ Nð0,1,0,1,1Þ Z 1. i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i1 i2 i3 i5 i6 i1 i2 i3 i5 i6 (iv) If Nð1,1,0,0Þ Z 1 and Nð0,0,0,0Þ ¼ N ð0,0,1,1Þ ¼ Nð1,1,1,1Þ ¼ 0,then Nð0,0,1,0,0Þ þ Nð0,0,0,1,0Þ Z 1 and Nð0,1,0,1,1Þ þ Nð0,0,1,1,1Þ Z1. i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i6 i2 i3 i4 i5 i6 i1 i2 i3 i4 i5 i1 i2 i3 i4 i5 (v) If Nð0,0,1,1Þ Z 1 and Nð0,0,0,0Þ ¼ N ð1,1,0,0Þ ¼ Nð1,1,1,1Þ ¼ 0, then Nð1,1,0,1,0Þ þ Nð1,1,1,0,0Þ Z 1 and Nð0,1,0,0,0Þ þ Nð0,0,1,0,0Þ Z1. i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i6 i2 i3 i4 i5 i6 i1 i2 i3 i4 i5 i1 i2 i3 i4 i5 Proof. (i) By considering all remaining 12 among all 16 possible rows other than (0,0,0,0), (1,1,1,1), (1,1,0,0) and (0,0,1,1) for four columns (i2,i3,i4,i5) and noting that by Proposition 2 columns (i1,i4,i5) must have rows (1,0,0) and (0,1,1) and by considering row (0,0,0) for columns (i2,i3,i6) the proof is clear. (ii) By arguing similar to (i) for columns (i1,i4,i5) the proof is clear. (iii) Note that Nið1,1Þ Z l from (7) and by Proposition 2 columns (i1,i5,i6) have row (0,1,1). 2 i3 (iv) By Proposition 2 columns (i2,i3,i6) have row (0,0,0) and columns (i1,i4,i5) have row (0,1,1). (v) By Proposition 2 columns (i2,i3,i6) have row (1,1,0) and columns (i1,i4,i5) have row (0,0,0). This completes the proof. &
Note that the matrix in (4) for design (6) is the same as (9) except T12 and T13 are replaced by T22 and T23, which they are submatrix of T 2 and T 3 corresponding to T2, respectively.
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Corollary 4. Every m 2 submatrix of T22 has rows (1,0), (0,0) and (a,1), where a is either0 or 1. Proof. By Proposition 2 the proof is similar to Corollary 1.
&
Corollary 5. Every m 2 submatrix with one column from T22 and another from T23 has rows (1,a), (0,1) and (0,0) where a is either0 or 1. Proof. Consider cases (i)–(iii) in the proof of Corollary 2 for columns ti1 i2 and ti3 i4 i5 of T22 and T23, respectively. For (i), by Proposition 2, columns i2, i3 and i4 have rows (0,1,1) and (0,0,0). So, by noting that two columns i1 and i2 in T2 contain at least l copies of (1,1) the proof is clear. For (ii), columns i1, i2(i3) and i4 have rows (0,1,1) and (1,1,0) from Proposition 2. Thus, by considering the number of (0,0) for columns i2(i3) and i4 in T2 the proof is complete. At last, by considering the number of (0,0) for columns i2(i4) and i5 the proof is complete from Proposition 2. & Corollary 6. Every m 2 submatrix of T23 has rows in (1,a), (b,1), (0,0)} or {(1,0), (0,1), (1,1)} , where a and b are either0 or 1, not both1 simultaneously. Proof. Consider cases (i)–(iii) of Corollary 3 for columns ti1 i2 i3 and ti4 i5 i6 of T23. For (i), take four columns i2,i3,i4 and i5. Consider two possibilities whether at least one of the rows (1,1,0,0) or (0,0,1,1) occurs in these columns or not. If it does and at least one of rows (0,0,0,0) or (1,1,1,1) occurs then the proof is clear. If these two rows do not occur the proof is complete from Lemma 6(iv) and (v). Otherwise, by Lemma 4 and remark 4 there exist three possible cases in Lemma 6(i)–(iii) for these four columns. For the first case, by Lemma 6(i), we can always identify set of rows {(1,0), (0,1), (0,0)}. For the second case, we have set of rows {(1,0), (1,1), (0,0)} from Lemma 6(ii). For the third case, by Lemma 6(iii) we can identify the set of rows {(1,a), (0,1), (0,0)}. These complete the proof for (i). For (ii), consider four columns i2,i3(i4),i5 and i6. If N ð1,1,0,0Þ þN ð0,0,1,1Þ Z 1, by Proposition 2, we can always find set of rows {(1,1), (0,1), (0,0)} or {(1,1), (1,0), (0,0)}. Otherwise, by considering three cases in (i) of the proof and similar argument we can always find set of rows {(1,1), (0,1), (0,0)} or {(1,1), (1,0), (0,0)}. Finally, for (iii) we have four distinct columns. By Proposition 2, Remark 4 and Lemma 5, we can always identify two set of rows {(1,1), (0,1), (0,0)} or {(1,1), (1,0), (0,0)}. Thus the proof is complete. & Proof of Theorem 2. For l 4 10, the proof is similar to Theorem 1 by Corollaries 4–6. For cases 3r l r10, we have checked the condition (2) using computer programming due to complicated and intractable proof. & References Buratti, M., Pasotti, A., 2009. Combinatorial designs and the theorem of Weil on multiplicative character sums. Finite Fields Appl. 15, 332–344. Chatterjee, K., Deng, L.Y., Lin, D.k.J., 2001. Two level search design for main-effect plan plus two plan. Metrika 54, 233–245. Ghosh, S., 1980. On main effect plus one plans for 2m factorials. Ann. Statist. 8, 922–930. Ghosh, S., Talebi, H., 1993. Main effect plans with additional search property for factorial experiments. J. Statist. Plann. Inference 36, 367–384. Ghosh, S., Shirakura, T., Srivastava, J.N., 2007. Model identification using search linear models and search designs. Entropy, Search, Complexity. Bolyai Society, Mathematical Studies, vol. 16. Springer, Berlin, pp. 85–112. Harville, A.D., 1997. Matrix Algebra from a Statistician’s Perspective. Springer-Verlag, New York. Shirakura, T., 1991. Main effect plus one or two plans for 2m factorials. J. Statist. Plann. Inference 27, 65–74. Srivastava, J.N., 1975. Designs for searching non-negligible effects. In: Srivastava, J.N. (Ed.), A Survey of Statistical Design and Linear Model. North-Holland Publishing Company, Amsterdam, pp. 507–519.
Contents lists available at ScienceDirect
Journal of Statistical Planning and Inference journal homepage: www.elsevier.com/locate/jspi
A new series of main effects plus one plan for 2m factorial experiments with m = 4l 71 and 2m runs Nabaz Esmailzadeh a,, Hooshang Talebi b, Koji Momihara c, Masakazu Jimbo c a b c
Department of Statistics, University of Kurdistan, Sanandaj, Iran Department of Statistics, University of Isfahan, Isfahan, Iran Graduate School of Information Science, Nagoya University, Furo-Cho, Chikusa-ku, Nagoya 464-8601, Japan
a r t i c l e in f o
abstract
Article history: Received 17 September 2009 Received in revised form 30 October 2010 Accepted 6 November 2010 Available online 13 November 2010
An MEP.1 plan for a 2m factorial experiment is able to search and estimate one possible nonzero interaction effect along with estimating the general mean and main effects. Construction of such a plan has been tackled by several authors. Some of these works are developed for all m, in which are focused on m =2h 1 or m ¼ 4l1 for positive integers h 4 2 and l Z 2. This leads to a gap for having MEP.1 plans with well-structured runs set and probably with a reasonable number of treatments for ma2h 1 or 4l1. In this paper, we characterize and present a series of highly structured MEP.1 plans for all odd m ¼ 4l 7 1, l Z 2, with N= 2m runs. From these designs one can obtain the MEP.1 plans for even m with N =2m + 1 treatments. The exemptions are m = 8 and 9 for which we have 1 or 2 extra runs. & 2010 Elsevier B.V. All rights reserved.
Keywords: Non-negligible effect Search designs Search linear model PBIBD
1. Introduction Consider, the commonly used, main effect plans in a 2m factorial experiment and take b1 to be an ðm þ 1Þ 1 vector of the general mean and main effects, which the experimenter is interested in estimating them. By a search design, which has been introduced by Srivastava (1975), one means a design which can search and estimate k possible non-negligible factorial effects in a n2 1 vector b2 along with estimating the unknown parameters in b1 . Construction of search designs has been investigated by many authors, though a few of them, developed a method of constructing these designs for all m in general. Characterizing and constructing of such plans is presented in Ghosh (1980), Shirakura (1991) and Ghosh and Talebi (1993). More details on the history of these works, which have been developed in this area, are given in Ghosh et al. (2007). Ghosh (1980) considered the characterization problem of search design for the 2m factorial experiments when m= 2h 1 with hZ 2. He gave conditions on a follow-up design, that are required to be added to an orthogonal main effect plan (MEP) in order to provide additional information that can be used to de-alias any pair of interaction effects from the main effects. A search design with 24 runs has been constructed and presented for m= 7. Such an augmented search design carries some optimum estimation properties. However, it is poor in terms of the economical number of runs. Shirakura (1991) reduced the total number of runs to 15, for m=7, in an augmented search design by adding balanced arrays, as the follow-up designs, to the MEP given in Ghosh (1980). Ghosh and Talebi (1993) followed an analogous approach by adding balanced arrays to a saturated MEP plan and obtained search designs for m ¼ 4l1 with l Z 2. The latter result partially solves the lack of having wellstructured search designs for ma2h 1,hZ 2. It is also superior to that of previous ones in terms of the number of runs. So far, Corresponding author.
E-mail addresses: [email protected] (N. Esmailzadeh), [email protected] (H. Talebi), [email protected] (K. Momihara), [email protected] (M. Jimbo). 0378-3758/$ - see front matter & 2010 Elsevier B.V. All rights reserved. doi:10.1016/j.jspi.2010.11.008
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the lack of being well-structured search designs, with reasonable number of runs, for ma4l1 with l Z2 is open. In this paper, we investigate the problem of design construction to fill this gap. We address this problem and obtain a wellstructured factor setting, which leads to series of new search designs with smaller number of runs. This improves the result of Ghosh and Talebi (1993) by reducing the number of runs in their original designs for m ¼ 4l1 with l Z 2. We also develop and construct new search designs for all prime powers m ¼ 4l þ 1 with l 4 2. These results lead to designs with having a wellknown structural factor setting and a reasonable number of runs, for all odd m’s. 2. Preliminaries and notations In this paper, we restrict b2 to 2- and 3-factor interactions and assume that k, the number of nonzero effects in b2 , is 1. The four-factor and higher-order interactions are assumed to be zero. A search design that solves the problem of searching for a nonzero effect of b2 and estimating it along with estimation of the general mean and main effects is called a main effect plus 1 (MEP.1) plan. To deal with this problem, for an observation vector yðN 1Þ of a 2m factorial experiment, N o2m , we consider the following linear model: y ¼ X1 b1 þ X2 b2 þ e,
VarðeÞ ¼ s2 IN ,
ð1Þ
where Xi ðN ni Þ are known design matrices with elements 7 1, i= 1,2, b1 is an ðm þ 1Þ 1 vector consisting of the general m mean and main effects, b2 is a n2 1 vector consisting of 2- and 3-factor interactions, n2 ¼ ðm 2 Þ þð 3 Þ, e is an error random vector and IN is the identity matrix of order N. For noiseless case, s2 ¼ 0, Srivastava (1975) gave a necessary and sufficient condition for a design to be an MEP.1 plan. That is, for every submatrix X22 ðN 2Þ of X2, rank½X1 ; X22 ¼ mþ 3:
ð2Þ
Now, consider (0,1)-matrix TðN mÞ to be a design with m factors each at two levels and N runs, with ti as i-th column corresponding to the i-th factor, i= 1,2,y,m. Under the search linear model (1), the N ðm þ 1 þ n2 Þ model matrix for T is written as X ¼ ½1N ; U1 ; U2 ; U3 ,
ð3Þ
where 1N is an N 1 vector of 1’s corresponding to the general mean and Ui, i =1,2,3, are sets of columns, with elements +1 and 1, corresponding to the main effects, 2-factor and 3-factor interactions, respectively. Clearly, [1N;U1]= X1 and hence the i-th column in U1 is corresponding to ti. Following Chatterjee et al. (2001), to define the columns of the following submatrices T 2 and T 3 by tij = tintj and tijk = 2tintjntk + tintj + tintk +tjntk; io j o kði,j,k ¼ 1,2, . . . ,mÞ, respectively, in which stands for Hadamard product. Define the (0,1)-transformed matrix T corresponding to X as T ¼ ½1N ; T 1 ; T 2 ; T 3 ,
ð4Þ
where T 1 is the given design T, T 2 and T 3 are corresponding to U2 and U3, respectively. From the above column definition one can easily find out that an element in column tij of T 2 is 1 when the corresponding elements in both columns ti and tj are 1’s. It is also clear that an element in column tijk of T 3 is 1 when at least two of the corresponding elements in ti, tj and tk are 1’s. Otherwise, they are 0. Note that this transformation preserves the column rank condition of the model matrix X. This allows us to use matrix T directly to verify condition (2) in (3). Throughout this paper, for a set of h factors in {1,2,3,y,m}, let (a1,a2,y,ah) be a treatment combination corresponding to these factors where the treatment points, ai, are either 0 or 1 for i= 1, 2,y, h. We denote the number of runs in T consisting of a 2 ,...,ah Þ for any selected subset of h columns fi1 ,i2 , . . . ,ih g D f1,2, . . . ,mg. particular treatment combination (a1,a2,y,ah ) by Nðiða11,i,a2 ,...,i hÞ
Now, let {i1,i2,y,ic} be a set of cð r mÞ columns in T. For any subset of hð r cÞ columns fij1 ,ij2 , . . . ,ijh g Dfi1 ,i2 , . . . ,ic g and treatment combination faj1 ,aj2 , . . . ,ajh g corresponding to these columns, it is true that X ða ,a ,...,a Þ ðaj ,aj ,...,aj Þ Nij i1j ...i2 j h ¼ Ni1 i12 ...i2c c , ð5Þ 1 2
h
S
where S is the set of all permutations of treatment combinations ðaj1 ,aj2 , . . . ,ajðchÞ Þ such that jl 2 fi1 ,i2 , . . . ,ic gfij1 ,ij2 , . . . ,ijh g, for l = 1,2,y, (c h). An m 1 vector and an m n matrix with all elements 0 are denoted by 0m and OðmnÞ , respectively. 3. Main results In this section, we present main results through Theorems 1 and 2. The proofs of theorems along with some lemmas and corollaries are given in Appendix. These lemmas and corollaries are useful in proving the theorems. Now, consider the matrix Tu ¼ ½1m ; Im ; T1 , where m ¼ 4l1 in which T1 is an m m incidence matrix of a symmetric BIBD with parameters v ¼ b ¼ 4l1,r ¼ k ¼ 2l and l with l Z2. Ghosh and Talebi (1993) proved that T is an MEP.1 plan with N( =2m + 1) runs. The following theorem proves that the number of runs can still be reduced to get an MEP.1 plan. Define the matrix Tu ¼ ½Im ; T1 . Theorem 1. T is an MEP.1 plan with N =2m runs for m ¼ 4l1, l Z 2. Proof. See Appendix.
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Now, we construct a search design similar to T in Theorem 1 when m ¼ 4l þ 1 is a prime power. That is, define the design matrix " # Im T¼ , ð6Þ T2 where T2 is the m m incidence matrix of a partially balanced incomplete block design (PBIBD). We will show that the design T is an MEP.1 plan by utilizing the properties of a PBIBD and applying a result on Galois fields. This needs a more understanding of T2 in details. For an odd prime p, let m= ps and consider the ðm mÞ-matrix T2 =(di,j) whose rows and columns are labeled by the elements of GF(ps), the Galois field of order ps, defined by ( 0 if ði þjÞ 2 GFðps Þ is quadratic residue, di,j ¼ 1 if ði þjÞ 2 GFðps Þ is non-quadratic or zero: When m ¼ 4l þ 1, the matrix T2 forms an incidence matrix of a PBIBD with two associate classes and with the design parameters ðv,b,r,k, l1 , l2 Þ ¼ ð4l þ 1,4l þ 1,2l þ 1,2l þ 1, l þ1, lÞ, and the association scheme parameters ðn1 ,n2 ,p11,1 ,p21,1 ,p11,2 ,p21,2 ,p12,1 ,p22,1 ,p12,2 ,p22,2 Þ ¼ ð2l,2l, l1, l, l, l, l, l, l, l1Þ: Note that for this PBIBD, the pair of distinct elements a, b 2 GFðps Þ are 1st associate if ab is non-quadratic and 2nd associate otherwise. Then, each element of the PBIBD has ni i-th associates, i= 1, 2. Furthermore, for any two i-th associates elements a and b, the number of elements that are j-th associates of a and k-th associates of b is a constant pij,k, independent of the pair a and b. It is also known that two i-th associates elements occur together in exactly li blocks. We call this PBIBD a QR-type with parameters n ¼ b ¼ 4l þ1,k ¼ r ¼ 2l þ1, l1 ¼ l þ 1 and l2 ¼ l. Example 1. For prime odd m= 13, the quadratic residue elements in GF (13) are 1, 3, 4, 9, 10 and 12, so the following matrix T2 is the incidence matrix of a PBIBD with parameters (13, 13, 7, 7, 4, 3), 0 1 2 3 4 5 6 7 8 9 10 11 12 1 1 0 1 0 0 1 1 1 1 0 0 1 0 B 1 B0 1 0 0 1 1 1 1 0 0 1 0 1C C B C C 2 B 1 0 0 1 1 1 1 0 0 1 0 1 0 B C B 3 B0 0 1 1 1 1 0 0 1 0 1 0 1C C B C C 4 B 0 1 1 1 1 0 0 1 0 1 0 1 0 B C B 5 B1 1 1 1 0 0 1 0 1 0 1 0 0C C B C C: 6 B 1 1 1 0 0 1 0 1 0 1 0 0 1 B C B C 7 B1 1 0 0 1 0 1 0 1 0 0 1 1C B C C 8 B B1 0 0 1 0 1 0 1 0 0 1 1 1C B C 9 B0 0 1 0 1 0 1 0 0 1 1 1 1C B C B 10 B 0 1 0 1 0 1 0 0 1 1 1 1 0 C C B C 11 @ 1 0 1 0 1 0 0 1 1 1 1 0 0 A 0
12
0
0
1
0
1
0
0
1
1
1
1
0
0
1
To be more precise, from the properties of the PBIBD ð4l þ 1,2l þ 1, l þ 1, lÞ, it is true that for any two columns, say i1 and i2, of T2 8 > < ðl þ 1, l, l, lÞ ð1,1Þ ð0,1Þ ð1,0Þ ð0,0Þ ð7Þ ðNi1 i2 ,Ni1 i2 ,Ni1 i2 ,Ni1 i2 Þ ¼ or > : ðl, l þ 1, l þ 1, l1Þ: For any three columns of such a PBIBD, Proposition 2 is a direct consequence of the following proposition derived from Theorem 2.2. of Buratti and Pasotti (2009) with e= 2 and t =3. Proposition 1. Let p 1 ðmod 2Þ be a prime power with p 4 45. Then for any given triple ða1 ,a2 ,a3 Þ 2 f0,1g3 and any given triple (c1,c2,c3) of distinct elements of GF(p), there exists an element x 2 GFðpÞ such that x þci 2 Ca2i for each i, where Ca2i ¼ fa2‘ þ ai j0 r ‘ o ðp1Þ=2g and a is a primitive element of GF(p). Proposition 2. Suppose i1, i2 and i3 are three distinct columns of the incidence matrix of a symmetric QR-type PBIBD ð4l þ 1,2l þ 1, l þ 1, lÞ. The number of each triple (a1,a2,a3), where ai’s are either0 or 1, as rows of selected columns (i1, i2, i3) are at least1 for l 4 10. Now, we establish the following theorem, which gives a new series of MEP.1 plan.
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Theorem 2. The model design matrix for T in (6) satisfies condition (2), i.e. T is an MEP.1 plan with N = 2m runs, where m ¼ 4l þ 1, l 42. Proof. See Appendix. By virtue of Theorem 2, the challenge is to reduce the number of runs in contrast to the designs obtained by removing columns from those in Ghosh and Talebi (1993). Example 2. Consider T2 given in Example 1 for m= 13. The design (6) for this T2 is an MEP.1 with 26 runs. For such an m, Ghosh and Talebi (1993) gave an MEP.1 plan with at least 28 runs. Remark 1. For l ¼ 1, Theorem 2 is true by adding run (0,0,0,0,0) to design T in (6). Remark 2. An MEP.1 for an even m ða4,8Þ can be obtained by removing any one column from the MEP.1 plan of m+1 factors with N = 2m+ 1 runs. 4. Discussion A design with smaller number of runs and hence a lower cost experiment is a main purpose in design construction. It is also known that a search design is superior to numbered-resolution, e.g. IV and V, design in order to do estimation job with smaller number of observations. Many researchers have been improved MEP.1 plans by reducing the number of runs. Ghosh and Talebi (1993) gave a series of MEP.1 plans with N =2m +1, which do the search job for all m ¼ 4l1, l Z2. For m ¼ 4l1d, d ¼ 1,2,3, the MEP.1 plans are obtained by deletion of any d columns from one with m ¼ 4l1. However, one may not be satisfied with resulting plans due to their number of runs. In this paper, we develop a new series of MEP.1 plan with reasonable number of runs for 2m factorial experiments when m Z6. That is, let partition positive odd numbers to two sets, say P and Q, such that P ¼ fm : m ¼ 4l1, l Z2g and Q ¼ fm : m ¼ 4l þ1, l 4 2g. We characterize a series of (0,1)-matrix of S size ðN mÞ as MEP.1 plans with a special structure of Tu ¼ ½Im ; Ti , i=1,2, with N = 2m treatments when m 2 P Q . The matrices Ti, i =1,2 were defined in Section 3 for m 2 P and m 2 Q , respectively. Clearly, the MEP.1 plan for even m with N =2m +1 treatments can be obtained by deleting any one column from T with m+ 1 factors except for m= 4 and 8. Appendix Here, we provide some lemmas and corollaries required in proving Theorems 1 and 2. First, using these corollaries, we characterize T1 in T of Theorem 1, by the properties of symmetric BIBDð4l1,2l, lÞ. From the properties of the symmetric BIBDð4l1,2l, lÞ, for every two columns i1 and i2 of T1 it is true that ( l1 if ða1 ,a2 Þ ¼ ð0,0Þ, Niða1 i12,a2 Þ ¼ ð8Þ l if ða1 ,a2 Þ ¼ ð1,1Þ,ð1,0Þ,ð0,1Þ: The following Lemmas 1 and 2 on the properties of BIBD are immediate consequence of Lemmas 1–3 of Ghosh and Talebi (1993). Lemma 1. Suppose i1, i2 and i3 are three distinct columns of the incidence matrix of a symmetric BIBDð4l1,2l, lÞ. Then Nið1,1,0Þ Z 1,Nið1,0,1Þ Z 1 and Nið0,1,1Þ Z 1. 1 i2 i3 1 i2 i3 1 i2 i3 Lemma 2. Consider every four columns, say, i1, i2, i3 and i4, of the incidence matrix of a symmetric BIBDð4l1,2l, lÞ. Then þ Nið0,0,1,1Þ Z 1. Nið1,1,0,0Þ 1 i2 i3 i4 1 i2 i3 i4 Z1 and Nið0,0,1,1Þ Z1 if Nið1,1,1,1Þ þ Nið0,0,0,0Þ Z 1. Remark 3. It is true that Nið1,1,0,0Þ 1 i2 i3 i4 1 i2 i3 i4 1 i2 i3 i4 1 i2 i3 i4 Now, we write the matrix in (4)for design T in Theorem 1. It leads to the following corresponding column-wise partition matrix form: " # 1m Im Oðmðm2 ÞÞ Oðmðm3 ÞÞ , ð9Þ 1m T1 T12 T13 m where T12 and T13 are m ðm 2 Þ and m ð 3 Þ submatrices of T 2 and T 3 , corresponding to T1, respectively.
Corollary 1. Every m 2 submatrix of T12 has rows (1,0), (0,0) and (0,1). Proof. Suppose ti1 i2 and ti3 i4 are two columns of T12, which they are correspond to 2-factor interactions. These two columns either have a common or no common factor. If they have one common factor, say i2 =i3, then by Lemma 1 columns (i1,i2(i3),i4) have rows (1,1,0), (0,1,1) and (1,0,1). Thus, in this case the proof is complete. If they have no common factor, by Lemma 1 columns (i1,i2,i3) and (i2,i3,i4) have rows (1,1,0) and (0,1,1). Thus, by considering the rows (0,0) for columns (i2,i3) the proof is complete. &
N. Esmailzadeh et al. / Journal of Statistical Planning and Inference 141 (2011) 1567–1574
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Table 1 Arranged T1 in terms of columns i3 and i4. i1
i2
i3
i4
1l 1l 0l1 0l
1l 0l 0l1 1l
i5
Corollary 2. Every m 2 submatrix with one column from T12 and another from T13 has rows (1,a), (0,1) and (0,0) where a is either 0 or 1. Proof. Suppose the columns in T12 and T13 are denoted by ti1 i2 and ti3 i4 i5 , where the sets {i1, i2} and {i3, i4, i5} correspond to factors. There are three possibilities for two sets: (i) no common factor, (ii) one common factor, say i2 = i3 and (iii) two common factors, say i1 =i3 and i2 = i4. For (i), we can always arrange the columns i1,i2, i3,i4 and i5 in T1 as Table 1 from (8). There must always be a run (0,a) or (a,0) for (i1,i2) in the third block row for l 43. Otherwise, in order to have l1 possible (0,0) for (i1,i3) and (i2,i3) all must occur in the fourth block row. Having l rows in this block restricts a to be 0, such that the total number of (0,0) for (i1,i2) must be at least l2. By similar argument all l1 possible (0,0) of columns (i1,i4) and (i2,i4) must occur in the second block row. Then the total number of (0,0) for columns (i1,i2) in this block must be at least l2. It leads to that have 2ðl2Þ r l1. That is, l r3 which contradicts the assumption l 4 3. Therefore, the third row of Table 1 must have some runs 0 for columns i1 and i2. So, by having a (0,a) or (a,0) run in the third row of Table 1 for (i1,i2) and from Lemma 1 and this fact that column ti1 i2 has l elements 1, the proof is clear. For l ¼ 2 and 3 the proof is simple by completing Table 1. For (ii), by considering columns i1,i2 and i4 in T1 and Lemma 1, we obtain the row (0,1) for ðti1 i2 ,ti3 i4 i5 Þ. From rows (0,0) of columns i2 and i4, row (0,0) is also obtained. Note that the number of 1’s in column ti1 i2 is l by (8). This complete the proof for this case. Finally, for (iii), by considering the rows (1,1) and (0,0) for columns i1 and i2 in T1 and by Lemma 1 the proof is clear. & Corollary 3. Every m 2 submatrix of T13 has rows in {(1,a), (b,1), (0,0)} or {(1,1), (1,0), (0,1)}, where a and b are either0 or 1, not both1 simultaneously. Proof. Let two columns in T13 are denoted by ti1 i2 i3 and ti4 i5 i6 , where the sets {i1,i2,i3} and {i4,i5,i6} correspond to factors. There are three possibilities for two sets: (i) no common factor, (ii) one common factor, say i3 =i4 and (iii) two common factors, say i2 = i4 and i3 =i5. For (i), consider four columns i2,i3,i4 and i5 and two possible cases whether at least one of the rows (1,1,1,1) or (0,0,0,0) occurs in these columns or not. If it does then by Lemma 2 and its remark we have the set of rows {(0,0), (1,0), (0,1)} or {(1,1), (1,0), (0,1)}. Otherwise, for these four columns, there exist two possible nonisomorphic row sets given in Table 1 of Lemma 3 of Ghosh and Talebi (1993). These possible nonisomorphic sets of runs give the rows in {(0,0), (b,1), (1,a)} and the proof is complete. For (ii), we arrange the matrix T1 in terms of columns i2 and i3 similar to Table 1 and consider the columns i2, i3(i4), i5 and i6. Set of rows (1,a) and (b,1) for ti1 i2 i3 and ti4 i5 i6 are obtained by applying Lemma 1 to the three columns i2, i3 (i4) and i5 and Lemma 2 to the four columns. By the argument similar to Corollary 2(i) for four columns i2,i3,i5 and i6 the row (0,0) is obtained for two columns in T13. Finally, for (iii), by considering all possible combinations for columns i2 and i3 the proof is clear from Lemma 2. & The following lemma is useful in proving Theorems 1 and 2. Lemma 3. Let W11 be an m m matrix, W12 an m q matrix, W21 an n m matrix, and W22 an n q matrix. If rank(W11)= m, then " # W11 W12 1 ¼ mþ rankðW22 W21 W11 W12 Þ: rank W21 W22 See page 98 of Harville (1997) for a proof. Proof of Theorem 1. In order to prove T is a search design we need to show that condition (2) is valid for (9), from equivalence rank property of (9) with model matrix in (3). That is, for any two distinct columns ti1 and ti2 of [T12; T13] we need to show that " # 1m Im 0m 0m rank ¼ m þ 3: 1m T1 ti1 ti2 Applying Lemma 3 by taking W11 =Im, W21 = T1, W12 = [1m; 0m; 0m] and W22 ¼ ½1m ; ti1 ; ti2 , the proof is reduced to show that rank 1 (W22 W21 W11 W12)= rank(W22)= 3. There are three possible choices in choosing two columns from [T12 ; T13]: Case (i) two columns from T12, Case (ii) one column from T12 and another from T13,
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Case (iii) two columns from T13. For the first case Corollary 1 guarantees that W22 has rank 3, as we always can identify the following submatrix of W22: 2 3 1 0 0 6 7 4 1 1 0 5, 1 a 1 whose rank is 3. The proof for cases (ii) and (iii) are obtained from Corollaries 2 and 3, respectively. This completes the proof. & Now, we give some lemmas and corollaries, which are useful in proving Theorem 2. These corollaries characterize T2 in T given in (6) using the properties of a symmetric PBIBDð4l þ 1,2l þ 1, l þ 1, lÞ. Lemma 4. Consider every four columns, say, i1, i2, i3 and i4, of the incidence matrix of a symmetric PBIBDð4l þ 1,2l þ 1, l þ 1, lÞ. Let þNið0,0,1,1Þ A ¼ Nið1,1,0,0Þ 1 i2 i3 i4 1 i2 i3 i4 8 > B þ1 if > > > > > > > B þ2 if > > > > > > B þ1 if > > > > > > B þ2 if > > > < A ¼ B þ1 if > > > > B þ3 if > > > > > > > B if > > > > > > B if > > > > > > : B1 if
and B ¼ Nið1,1,1,1Þ þ Nið0,0,0,0Þ . The following holds 1 i2 i3 i4 1 i2 i3 i4 ¼ Nið0,1Þ Nið1,1Þ 1 i3 2 i3 Nið1,1Þ 1 i3 ð1,1Þ Ni 1 i 3 Nið1,1Þ 1 i3 Nið1,1Þ 1 i3 Nið1,1Þ 1 i3 ð1,1Þ Ni 1 i 3 Nið1,1Þ 1 i3 Nið1,1Þ 1 i3
¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼
Nið0,1Þ 2 i3 ð1,1Þ Ni2 i3 Nið1,1Þ 2 i3 Nið1,1Þ 2 i3 Nið1,1Þ 2 i3 ð0,1Þ Ni2 i3 Nið1,1Þ 2 i3 Nið1,1Þ 2 i3
and
Nið1,1Þ ¼ Nið0,1Þ 1 i4 2 i4
and
Nið1,1Þ 1 i4 ð1,1Þ Ni1 i4 Nið1,1Þ 1 i4 Nið1,1Þ 1 i4 Nið1,1Þ 1 i4 ð1,1Þ Ni1 i4 Nið1,1Þ 1 i4 Nið1,1Þ 1 i4
¼ l þ1
and
¼l
and
¼l
and
¼l
and and
¼ l þ1
and
¼ l þ1
and
¼ Nið1,1Þ 2 i4 ð1,1Þ ¼ Ni 2 i 4 ¼ Nið0,1Þ 2 i4 ¼ Nið1,1Þ 2 i4 ¼ Nið1,1Þ 2 i4 ð1,1Þ ¼ Ni 2 i 4 ¼ Nið0,1Þ 2 i4 ¼ Nið1,1Þ 2 i4
ðL4:1Þ ¼l
ðL4:2Þ
¼l
ðL4:3Þ ðL4:4Þ
¼ lþ1
ðL4:5Þ
¼l
ðL4:6Þ
¼ lþ1
ðL4:7Þ ðL4:8Þ
¼ l þ 1:
ðL4:9Þ
Proof. It follows from (7) and (5) that Nið1,1Þ ¼ Nið1,1,1Þ þ Nið1,0,1Þ , 1 i3 1 i2 i3 1 i2 i3 ¼ Nið0,0,1Þ þ Nið1,0,1Þ , Nið0,1Þ 2 i3 1 i2 i3 1 i2 i3 þ Nið1,1,0Þ ¼ Nið0,0,1Þ þ Nið0,0,0Þ þ 1: Nið1,1,1Þ 1 i2 i3 1 i2 i3 1 i2 i3 1 i2 i3
ð10Þ
8 ð0,0,0Þ þ1 N > > > i1 i2 i3 < ð0,0,0Þ ð1,1,0Þ Ni1 i2 i3 ¼ Ni1 i2 i3 > > > ð0,0,0Þ : Ni1 i2 i3 þ 2
ð11Þ
So, if Nið1,1Þ ¼ Nið0,1Þ , 1 i3 2 i3 if Nið1,1Þ ¼ Nið1,1Þ ¼ l þ 1, 1 i3 2 i3 if
Nið1,1Þ 1 i3
¼
Nið1,1Þ 2 i3
¼ l:
Following (5) we have Nið1,1,0Þ ¼ Nið1,1,0,1Þ þNið1,1,0,0Þ , 1 i2 i3 1 i2 i3 i4 1 i2 i3 i4 ¼ Nið0,0,0,0Þ þ Nið0,0,0,1Þ : Nið0,0,0Þ 1 i2 i3 1 i2 i3 i4 1 i2 i3 i4 Thus, from (11) and (12) it follows that 8 ð0,0,0,0Þ þ1 Ni i i i þ Nið0,0,0,1Þ > > 1 i2 i3 i4 > < 1234 ð0,0,0,0Þ ð0,0,0,1Þ ð1,1,0,1Þ ð1,1,0,0Þ Ni1 i2 i3 i4 þNi1 i2 i3 i4 ¼ Ni1 i2 i3 i4 þ Ni1 i2 i3 i4 > > > ð0,0,0,0Þ : þ2 Ni1 i2 i3 i4 þ Nið0,0,0,1Þ 1 i2 i3 i4
ð12Þ
if Nið1,1Þ ¼ Nið0,1Þ , 1 i3 2 i3 if Nið1,1Þ ¼ Nið1,1Þ ¼ l þ1, 1 i3 2 i3 if
Nið1,1Þ 1 i3
¼
Nið1,1Þ 2 i3
¼ l:
Applying (7) and (5) we can get similar result as (10), (11) and (12) for columns i1,i2 and i4. Thus we have 8 ð0,0,0,1Þ þ Nið0,0,1,1Þ if Nið1,1Þ ¼ Nið0,1Þ , N > > 1 i2 i3 i4 1 i4 2 i4 > i1 i2 i3 i4 < ð0,0,0,1Þ ð0,0,1,1Þ ð1,1Þ ð1,1Þ ð1,1,1,1Þ ð1,1,0,1Þ þ N þ 1 if N ¼ N ¼ l þ 1, N Ni1 i2 i3 i4 þNi1 i2 i3 i4 ¼ i1 i2 i3 i4 i1 i2 i3 i4 i1 i4 i2 i4 > > > ð0,0,0,1Þ ð0,0,1,1Þ ð1,1Þ ð1,1Þ :N if Ni1 i4 ¼ Ni2 i4 ¼ l: i1 i2 i3 i4 þ Ni1 i2 i3 i4 1 Therefore, the proof is complete from (13) and (14).
&
ð13Þ
ð14Þ
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Remark 4. For the cases (L4.1)–(L4.6), Nð1,1,0,0Þ þ Nð0,0,1,1Þ Z1. Lemma 5. Consider four columns i1, i2, i3 and i4 of a symmetric PBIBDð4l þ1,2l þ 1, l þ 1, lÞ incidence matrix. Assume that N(0,0,0,0) =N(1,1,0,0) = N(0,0,1,1) =N(1,1,1,1) =0, then for both conditions (L4.7) and (L4.8), it is true that Nð1,0,1,0Þ þ Nð0,1,0,1Þ Z1:
ð15Þ
Also, suppose that (i) N(0,0,0,0) = N(1,1,0,0) =N(0,0,1,1) = 0,N(1,1,1,1) = 1, or (ii) N(1,1,1,1) =N(1,1,0,0) = N(0,0,1,1) =0,N(0,0,0,0) = 1, then for condition (L4.9), (15) holds. Proof. It follows from condition (L4.7) that Nð1,1,1,0Þ ¼ N ð0,0,1,0Þ
and
Nð1,1,0,1Þ ¼ N ð0,0,0,1Þ þ1:
ð16Þ
From (7) and (5), we have ¼ Nð1,1,1,0Þ þN ð0,0,1,0Þ þN ð1,0,1,0Þ ¼ l or l þ 1 Nið1,0Þ 3 i4
ð17Þ
¼ Nð1,1,0,1Þ þN ð0,0,0,1Þ þN ð0,1,0,1Þ ¼ l or l þ 1: Nið0,1Þ 3 i4
ð18Þ
and
So, from (16)–(18) we get 2Nð1,1,1,0Þ þ2N ð1,1,0,1Þ þ Nð1,0,1,0Þ þ Nð0,1,0,1Þ ¼ 2l þ1 or 2l þ 3: Additionally, from (7) and (5) ¼ Nð1,1,1,0Þ þN ð1,1,0,1Þ ¼ l or l þ 1: Nið1,1Þ 1 i2 Thus, N
ð1,0,1,0Þ
þN
ð0,1,0,1Þ
¼
8 <1
if Nið1,1Þ ¼ Nið1,0Þ ¼ Nið0,1Þ ¼ l or l þ 1, 1 i2 3 i4 3 i4
:3
¼ l, Nið1,0Þ ¼ Nið0,1Þ ¼ l þ 1: if Nið1,1Þ 1 i2 3 i4 3 i4
This completes the proof. The proof for conditions (L4.8) and (L4.9) are similar.
&
Lemma 6. Consider six columns i1,i2,i3,i4,i5 and i6 of T2 for l 410.
(i) If Nð0,0,0,0Þ ¼ N ð1,1,0,0Þ ¼ Nð0,0,1,1Þ ¼ N ð1,1,1,1Þ ¼ 0, then Nð1,1,0,0,0Þ þ Nð1,0,1,0,0Þ Z 1, N ð0,1,0,1,1Þ þN ð0,0,1,1,1Þ Z1 and i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i1 i2 i3 i5 i6 i1 i2 i3 i5 i6 i1 i2 i3 i5 i6 i1 i2 i3 i5 i6 Nið0,0,1,0,0Þ þNið0,0,0,1,0Þ Z1. 2 i3 i4 i5 i6 2 i3 i4 i5 i6 (ii) If Nð1,1,1,1Þ Z1 and Nð1,1,0,0Þ ¼ Nð0,0,1,1Þ ¼ Nð0,0,0,0Þ ¼ 0 then N ð0,0,1,0,0Þ þN ð0,1,0,0,0Þ Z1. i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i1 i2 i3 i4 i5 i1 i2 i3 i4 i5 (iii) If Nð0,0,0,0Þ Z 1 and Nð1,1,0,0Þ ¼ N ð0,0,1,1Þ ¼ Nð1,1,1,1Þ ¼ 0 then N ð1,1,0,1Þ þ Nð1,1,1,0Þ Z 1 and Nð0,0,1,1,1Þ þ Nð0,1,0,1,1Þ Z 1. i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i1 i2 i3 i5 i6 i1 i2 i3 i5 i6 (iv) If Nð1,1,0,0Þ Z 1 and Nð0,0,0,0Þ ¼ N ð0,0,1,1Þ ¼ Nð1,1,1,1Þ ¼ 0,then Nð0,0,1,0,0Þ þ Nð0,0,0,1,0Þ Z 1 and Nð0,1,0,1,1Þ þ Nð0,0,1,1,1Þ Z1. i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i6 i2 i3 i4 i5 i6 i1 i2 i3 i4 i5 i1 i2 i3 i4 i5 (v) If Nð0,0,1,1Þ Z 1 and Nð0,0,0,0Þ ¼ N ð1,1,0,0Þ ¼ Nð1,1,1,1Þ ¼ 0, then Nð1,1,0,1,0Þ þ Nð1,1,1,0,0Þ Z 1 and Nð0,1,0,0,0Þ þ Nð0,0,1,0,0Þ Z1. i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i2 i3 i4 i5 i6 i2 i3 i4 i5 i6 i1 i2 i3 i4 i5 i1 i2 i3 i4 i5 Proof. (i) By considering all remaining 12 among all 16 possible rows other than (0,0,0,0), (1,1,1,1), (1,1,0,0) and (0,0,1,1) for four columns (i2,i3,i4,i5) and noting that by Proposition 2 columns (i1,i4,i5) must have rows (1,0,0) and (0,1,1) and by considering row (0,0,0) for columns (i2,i3,i6) the proof is clear. (ii) By arguing similar to (i) for columns (i1,i4,i5) the proof is clear. (iii) Note that Nið1,1Þ Z l from (7) and by Proposition 2 columns (i1,i5,i6) have row (0,1,1). 2 i3 (iv) By Proposition 2 columns (i2,i3,i6) have row (0,0,0) and columns (i1,i4,i5) have row (0,1,1). (v) By Proposition 2 columns (i2,i3,i6) have row (1,1,0) and columns (i1,i4,i5) have row (0,0,0). This completes the proof. &
Note that the matrix in (4) for design (6) is the same as (9) except T12 and T13 are replaced by T22 and T23, which they are submatrix of T 2 and T 3 corresponding to T2, respectively.
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Corollary 4. Every m 2 submatrix of T22 has rows (1,0), (0,0) and (a,1), where a is either0 or 1. Proof. By Proposition 2 the proof is similar to Corollary 1.
&
Corollary 5. Every m 2 submatrix with one column from T22 and another from T23 has rows (1,a), (0,1) and (0,0) where a is either0 or 1. Proof. Consider cases (i)–(iii) in the proof of Corollary 2 for columns ti1 i2 and ti3 i4 i5 of T22 and T23, respectively. For (i), by Proposition 2, columns i2, i3 and i4 have rows (0,1,1) and (0,0,0). So, by noting that two columns i1 and i2 in T2 contain at least l copies of (1,1) the proof is clear. For (ii), columns i1, i2(i3) and i4 have rows (0,1,1) and (1,1,0) from Proposition 2. Thus, by considering the number of (0,0) for columns i2(i3) and i4 in T2 the proof is complete. At last, by considering the number of (0,0) for columns i2(i4) and i5 the proof is complete from Proposition 2. & Corollary 6. Every m 2 submatrix of T23 has rows in (1,a), (b,1), (0,0)} or {(1,0), (0,1), (1,1)} , where a and b are either0 or 1, not both1 simultaneously. Proof. Consider cases (i)–(iii) of Corollary 3 for columns ti1 i2 i3 and ti4 i5 i6 of T23. For (i), take four columns i2,i3,i4 and i5. Consider two possibilities whether at least one of the rows (1,1,0,0) or (0,0,1,1) occurs in these columns or not. If it does and at least one of rows (0,0,0,0) or (1,1,1,1) occurs then the proof is clear. If these two rows do not occur the proof is complete from Lemma 6(iv) and (v). Otherwise, by Lemma 4 and remark 4 there exist three possible cases in Lemma 6(i)–(iii) for these four columns. For the first case, by Lemma 6(i), we can always identify set of rows {(1,0), (0,1), (0,0)}. For the second case, we have set of rows {(1,0), (1,1), (0,0)} from Lemma 6(ii). For the third case, by Lemma 6(iii) we can identify the set of rows {(1,a), (0,1), (0,0)}. These complete the proof for (i). For (ii), consider four columns i2,i3(i4),i5 and i6. If N ð1,1,0,0Þ þN ð0,0,1,1Þ Z 1, by Proposition 2, we can always find set of rows {(1,1), (0,1), (0,0)} or {(1,1), (1,0), (0,0)}. Otherwise, by considering three cases in (i) of the proof and similar argument we can always find set of rows {(1,1), (0,1), (0,0)} or {(1,1), (1,0), (0,0)}. Finally, for (iii) we have four distinct columns. By Proposition 2, Remark 4 and Lemma 5, we can always identify two set of rows {(1,1), (0,1), (0,0)} or {(1,1), (1,0), (0,0)}. Thus the proof is complete. & Proof of Theorem 2. For l 4 10, the proof is similar to Theorem 1 by Corollaries 4–6. For cases 3r l r10, we have checked the condition (2) using computer programming due to complicated and intractable proof. & References Buratti, M., Pasotti, A., 2009. Combinatorial designs and the theorem of Weil on multiplicative character sums. Finite Fields Appl. 15, 332–344. Chatterjee, K., Deng, L.Y., Lin, D.k.J., 2001. Two level search design for main-effect plan plus two plan. Metrika 54, 233–245. Ghosh, S., 1980. On main effect plus one plans for 2m factorials. Ann. Statist. 8, 922–930. Ghosh, S., Talebi, H., 1993. Main effect plans with additional search property for factorial experiments. J. Statist. Plann. Inference 36, 367–384. Ghosh, S., Shirakura, T., Srivastava, J.N., 2007. Model identification using search linear models and search designs. Entropy, Search, Complexity. Bolyai Society, Mathematical Studies, vol. 16. Springer, Berlin, pp. 85–112. Harville, A.D., 1997. Matrix Algebra from a Statistician’s Perspective. Springer-Verlag, New York. Shirakura, T., 1991. Main effect plus one or two plans for 2m factorials. J. Statist. Plann. Inference 27, 65–74. Srivastava, J.N., 1975. Designs for searching non-negligible effects. In: Srivastava, J.N. (Ed.), A Survey of Statistical Design and Linear Model. North-Holland Publishing Company, Amsterdam, pp. 507–519.