Using Relational Algebra, I need help with the database management question belo

Question

Using Relational Algebra, I need help with the database management question below:

Please explain each step because I am still trying to really get the hang of Relational Algebra queries.

Print the packing list for each model. This consists of the Model#, Model Name, Part#, Length, and Number Used.

Here are the tables:

Furniture (FID, Manufacturer Name, Model#, Model Name, Description, I#)

Parts (Part#)

Wood (Part#, Wood Piece Length, Wood Piece Width, Wood Piece Height, Wood Piece Type [Of Wood])

Nuts and Bolts (Part#, Length, Diameter, Thread Count)

Instructions (I#)

Instruction Step (I#, Step#, Text)

Warehouse Location (Aisle, Shelf, Bin#, Part#)

Used In (I#, Step#, Part#¸Number Used)

Package (FID, Part#, Number Used)

Thank you!

Explanation / Answer

Furniture (FID, Manufacturer Name, Model#, Model Name, Description, I#)

Parts (Part#)

Wood (Part#, Wood Piece Length, Wood Piece Width, Wood Piece Height, Wood Piece Type [Of Wood])

Nuts and Bolts (Part#, Length, Diameter, Thread Count)

Instructions (I#)

Instruction Step (I#, Step#, Text)

Warehouse Location (Aisle, Shelf, Bin#, Part#)

Used In (I#, Step#, Part#¸Number Used)

Package (FID, Part#, Number Used)

for Relational Algebra following notaion is used:

for showing all list of table following Relational Algebra query use :

=> find all Part# in Wood (values of Part# present).
   Part# = values of Part# present(Wood)
  
=> Conditions can be combined together using ^ (AND) and v (OR). For example, all Manufacturer Name in Furniture id(FID) 1 called `Smith':
       FID = 1 ^ Manufacturer Name = `Smith'(Furniture)
      
=> The use of the symbolic notation can lend itself to brevity. Even better, when the JOIN is a natural join, the JOIN condition may be omitted from |x|.
The earlier example resulted in:

mysql query is:
PROJECT Manufacturer_Name (Furniture JOIN FID = FID (
PROJECT Number_Used (Package JOIN Part# = Part# (
PROJECT Wood_Piece_Length,Wood_Piece_Type (Wood JOIN Part# = Part# (
PROJECT Diameter (SELECT Thread_Count = 2 Nuts and Bolts)))))))

becomes

Manufacturer_Name (Furniture |×| (
Number_Used (Package |×| (
Wood_Piece_Length,Wood_Piece_Type (Wood |×| (
Diameter (Thread_Count = 2 Nuts and Bolts) ))))))
      

i hope it will help you to understand:

thanks

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