## Block Designs Properties

The necessary conditions that the parameters of a BIB design must satisfy are

• $bk = vr$, where $r=\frac{bk}{v}$ each treatment has $r$ replications
• no treatment appears more than once in any block
• all unordered pairs of treatments appear exactly in $\lambda$ blocks (equiconcurrence)
where $\lambda=\frac{r(k-1)}{v-1}=\frac{bk(k-1}{v(v-1)}$ is often referred to as the concurrence parameter of a BIB design.

A design say $d$ with parameters $(v, b, r, k, \lambda)$ can be represented as a $v \times b$ treatment block incidence matrix (having $v$ rows and $b$ columns). Let denote it by $N=n_{ij}$ whose elements $n_{ij}$ signify the number of units in block $j$ allocated to treatment $i$. The rows of incidence matrix are labeled with varieties (treatments) of the design and the columns with the blocks. We have to put 1 in the ($i$, $j$)th cell of the matrix if variety $i$ is contained in block $j$ and 0 otherwise. Each row of the incidence matrix has $r$ 1’s and each column has $k$ 1’s and each pair of distinct rows have $\lambda$ column 1’s, which lead to a useful identity matrix.
The matrix $NN’$ have $v$ rows and $v$ columns, referred to as concurrence matrix of design $d$ and its entries, the concurrence parameters are denoted by $\lambda_{dij}$. For a BIBD, $n_{ij}$ is either one or zero and $n_{ij}^2= n_{ij}$.

Theorem: If $N$ is the incidence matrix of a $(v, b, r, k, \lambda)$-design then $NN’=(r-\lambda)I+\lambda J$ where $I$ is $v\times v$ identity matrix and $J$ is the $v\times v$ matrix of all 1’s.

Example: For design {1,2,3}, {2,3,4}, {3,4,1}, {4,1,2} construct incidence matrix  Denoting the elements of $NN’$ by $q_{ih}$, we see that $q_{ii}=\sum_j n_{ij}^2$ and $q_{ih}=\sum_j n_{ij} n_{hj}, (i \ne h)$. For any block design $NN’$, the treatment concurrence with diagonal elements equal to $q_{ii}=r$ and off diagonal elements are $q_{ih}=\lambda, (i\ne h)$ equal to the number of times any pairs of treatment occur together within block. In a balanced design, the off-diagonal entries in $NN’$ are all equal to a constant $\lambda$ i.e., the common replication for a BIBD is $r$, and the common pairwise treatment concurrence is $\lambda$.

As $N$ is a matrix of $v$ rows and $b$ columns so that $r(N)\le min(b, c)$. Hence, $t\le min(b, v)$. If design is symmetric $b=v$ and $N$ is square the $|NN’|=|N|^2$, so $(r-\lambda)^{v-1}r^2$ is a perfect square.