Affordable Access

Some optimal block designs with nested rows and columns for research on alternative methods of limiting slug damage

Authors
Disciplines
  • Design

Abstract

nestedheader4.dvi The CRC Handbook of Combinatorial Designs Edited by Charles J. Colbourn Department of Computer Science and Engineering Arizona State University Jeffrey H. Dinitz Department of Mathematics and Statistics University of Vermont AUTHOR PREPARATION VERSION 21st April 2006 .36 Nested Designs 1 36 Nested Designs J. P. Morgan 36.1 NBIBDs: Definition and Example 36.1 Definition If the blocks of a BIBD (V,D1) with v symbols in b1 blocks of size k1 are each partitioned into sub-blocks of size k2, and the b2 = b1k1/k2 sub-blocks themselves constitute a BIBD (V,D2), then the system of blocks, sub-blocks and symbols is a nested balanced incomplete block design (nested BIBD or NBIBD) with parameters (v, b1, b2, r, k1, k2), r denoting the common replication. (V,D1) and (V,D2) are the component BIBDs of the NBIBD. 36.2 Example An NBIBD(16,24,48,15,10,5). Sub-blocks are separated by |. (0, 1, 2, 3, 4|5, 6, 7, 8, 9) (0, 1, 2, 3, 5|4, 6, 10, 11, 12) (0, 1, 2, 3, 6|4, 5, 13, 14, 15) (0, 1, 10, 11, 12|2, 3, 7, 8, 9) (0, 2, 13, 14, 15|1, 3, 7, 8, 9) (0, 3, 13, 14, 15|1, 2, 10, 11, 12) (0, 4, 5, 7, 11|1, 8, 10, 13, 14) (0, 4, 5, 9, 10|1, 7, 12, 13, 15) (0, 4, 5, 8, 12|1, 9, 11, 14, 15) (0, 6, 7, 10, 13|2, 4, 8, 11, 14) (0, 6, 9, 12, 15|2, 4, 7, 10, 13) (0, 6, 8, 11, 14|2, 4, 9, 12, 15) (0, 7, 8, 10, 15|3, 5, 6, 12, 14) (0, 7, 9, 12, 14|3, 5, 6, 11, 13) (0, 8, 9, 11, 13|3, 5, 6, 10, 15) (1, 5, 7, 12, 14|2, 6, 8, 10, 15) (1, 5, 9, 11, 13|2, 6, 7, 12, 14) (1, 5, 8, 9, 15|2, 6, 9, 11, 13) (1, 4, 6, 7, 13|3, 8, 11, 12, 15) (1, 4, 6, 9, 15|3, 7, 10, 11, 14) (1, 4, 6, 8, 14|3, 4, 8, 12, 13) (2, 5, 7, 11, 15|3, 4, 8, 12, 13) (2, 5, 9, 10, 14|3, 4, 7, 11, 15) (2, 5, 8, 12, 13|3, 4, 9, 10, 14) 36.2 NBIBDs: Existence 36.3 Remarks The necessary conditions for existence of a NBIBD are those for the two component BIBDs (V,D1) and (V,D2). Together they are: b1 ≥ v, v|b1k1, v(v − 1)|b1k1(k1 − 1), and v(v − 1)|b1k1(k2 − 1). The necessary conditions are sufficient for k1 = 4 [4]

There are no comments yet on this publication. Be the first to share your thoughts.