Chapter 6 - Balanced Incomplete Block Design (BIBD)

Chapter 6
Balanced Incomplete Block Design (BIBD)


The designs like CRD and RBD are the complete block designs. We now discuss the balanced
incomplete block design (BIBD) and the partially balanced incomplete block design (PBIBD) which
are the incomplete block designs.


A balanced incomplete block design (BIBD) is an incomplete block design in which
- b blocks have the same number k of plots each and
- every treatment is replicated r times in the design.
- Each treatment occurs at most once in a block, i.e., nij  0 or 1 where nij is the number of

times the jth treatment occurs in ith block, i  1, 2,..., b; j  1, 2,..., v.
- Every pair of treatments occurs together is  of the b blocks.


Five parameters denote such design as D(b, k , v, r;  ) .
The parameters b, k , v, r and  are not chosen arbitrarily.
They satisfy the following relations:
( I ) bk  vr
( II )  (v  1)  r (k  1)
( III ) b  v (and hence r  k ).


Hence  nij  k for all i
j

nj
ij r for all j

nij
and n1 j nij '  n2 j nij '  ...  nb j nb j '   for all j  j '  1, 2,..., v. Obviously cannot be a constant for all
r
j. So the design is not orthogonal.




Analysis of Variance | Chapter 6 | Balanced Incomplete Block Design | Shalabh, IIT Kanpur
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,Example of BIBD
In the design D(b, k; v, r;  ) : consider b  10 (say, B1 ,..., B10 ), v  6 (say, T1 ,..., T6 ), k  3, r  5,   2
Blocks Treatments
B1 T1 T2 T3
B2 T1 T2 T4
B3 T1 T3 T4
B4 T1 T4 T6
B5 T1 T5 T6
B6 T2 T3 T6
B7 T2 T4 T5
B8 T2 T5 T6
B9 T3 T4 T5
B10 T3 T4 T6


Now we see how the conditions of BIBD are satisfied.
(i ) bk  10  3  30 and vr  6  5  30
 bk  vr
(ii )  (v  1)  2  5  10 and r ( k  1)  5  2  10
  (v  1)  r ( k  1)
(iii ) b  10  6


Even if the parameters satisfy the relations, it is not always possible to arrange the treatments in blocks
to get the corresponding design.


The necessary and sufficient conditions to be satisfied by the parameters for the existence of a BIBD
are not known.


The conditions (I)-(III) are some necessary condition only. The construction of such design depends
on the actual arrangement of the treatments into blocks and this problem is handled in combinatorial
mathematics. Tables are available, giving all the designs involving at most 20 replication and their
method of construction.


Theorem:
( I )bk  vr
( II )  (v  1)  r (k  1)
( III ) b  v.

Analysis of Variance | Chapter 6 | Balanced Incomplete Block Design | Shalabh, IIT Kanpur
2

, Proof: (I)
Let N  ( nij ) : b  v the incidence matrix

Observing that the quantities E1b NEv1 and E1v N ' Eb1 are the scalars and the transpose of each other, we

find their values.
Consider
 n11 n21  nb1  1
 
n n22  nb 2  1
E1b NEv1  (1,1,...,1)  12
      
  
 n1v n2 v  nbv  1
  n1 j 
 j 
 
  n2 j 
 (1,1,...,1)  j 
 
 
  nbj 
 j 
k 
 
k
 (1,1,...,1)1b   = bk .
 
 
k 
Similarly,
 n11 n21  nb1  1
 
n n22  nb 2   
E1v N ' Eb1  (1,...,1)  12
      
  
 n1v n2v  nbv  1
  ni 
 i  r
=(1,1,...,1)    (1,1,...,1)    vr
1v  
 
  niv  r
 
 i 


But
E1b NEv1  E1v N ' Eb1 as both are scalars.
Thus bk  vr.




Analysis of Variance | Chapter 6 | Balanced Incomplete Block Design | Shalabh, IIT Kanpur
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