DBMA Indexing

Question

Records Size in bytes:- 30,75,40,65,50,20

Block Size:- 100 bytes

If unspanning  organization is used,then what is % of wastage in disk space?

Comments

@rahul @nitish

I did in this way

Considering order : (30)(75) (40) (65) (50 20)

% waste = (220/500)*100 = 44%

Not considering order:

(50 20 30) (75) (40) (65)

% waste = (120/400) *100 = 30%

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yes, correct.

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@Ashwani Kumar 2

yes correct.

Answer 1

Considering order:
(30)(75)(40)(65)(50 20)
(100-30)+(100-75)+(100-40)+(100-65)+(100-50-20)
% waste = (220/500)*100 = 44%

Not considering order:
(75,20) (65,30)(50,40) 
(100-75-20)+(100-65-30)+(100-50-40)
% waste = (20/300) *100 = 6.6667%