E R model

Question

why not b) option

Answer 1

From the above question, you will clearly understand about weak Entity . 

Let us revise what is weak Entity ?

1 . A weak entity is one which is depend upon other strong entity for its existence 

2. The attributes in weak entity are not sufficient to identify tuples . so it need primary key of strong entity 

3. If you have a relation of strong and weak entity . then the weak entity always have TOTAL participation . converse may not be true => every entity which has a total participation need not be a weak entity (check in the above figure )

4. you need to have an identifying relationship (double diamond box ) which happen to be in between PARTS and PRODUCTS and NOT between PARTS AND TRAINER 

the above question check point no 3 and point no 4 

Hence now it should be pretty clear that option a is true :)

Comments

i have one doubt,as Parts and TRainer has M:1 relationship,so should'nt Trainer -id also be the part of key for Parts Entity??

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Question states "attribute set for one of the minimized RDBMS table". In the case of a many-to-one relationship, it is only a convention (and a good practice) to include the PK of the entity type with higher cardinality as a FK in the other entity's relation, and not a necessity. A new relation can also be created for the relationship, which aligns with our goal of minimizing the table 'Parts'. Hence, (a) is the correct option.

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"Minimized RDBMS" means Minimum number of attributes,what the question want to ask  and so,option a) is answer, right?

And in case of minimum number of tables answer should be B).

Answer 2

First of all weak entity should not have any primary key.Parts id attribute is the discriminator.it depends on the strong entity set Products for its existence because “belongs to” is the identifying relationship.So the schema for the weak entity set Parts will be

Parts(Part-id,P-id,Part-name) with (Part-id,Part-name) as the primary key.

For further reference follow the book Database Systems Concept