




The higher the load factor, the more likely we would need to search for an entry or an empty slot (for insertion)
# entries n Load factor a =  =  array size N 
# occupied entries Ҏ[ an array slot is occupied by an entry] =  Total # entries in array n =  N 
Conclusion:

Important fact from Theory of Probability:

Example:





Avg. # probes = (1  a) × 1 + a(1  a) × 2 + a^{2}(1  a) × 3 + ... = (1  a) × ( 1 + a×2 + a^{2}×3 + ... ) = (1  a) × ( 1×a^{0} + 2×a^{1} + 3×a^{2} + 4×a^{3} + ... ) 
>> sum( (k+1)*a^k, k = 0..infinity); 1  2 (1  a) 
1 Avg. # slots probed (to find empty slot) = (1  a) *  (1  a)^{2} 1 =  (1  a) 
Graphically:


Note:

O( 1/(1  a) ) ~= O(1) as long as a is small, (say < 0.5) 