Hi, I need help with the following. Please use schema for assistance. So at the moment, im learning relational algebra and need help with this HW question...... 8.18. Consider the LIBRARY relational database schema shown in Figure 8.14, which is used to keep track of books, borrowers, and book loans. Referential integrity constraints are shown as directed arcs in Figure 8.14, as in the notation of Figure 5.7. Write down relational expressions for the following queries:a. How many copies of the book titled *The Lost Tribe* are owned by the library branch whose name is ‘Sharpstown’?b. How many copies of the book titled *The Lost Tribe* are owned by each library branch?c. Retrieve the names of all borrowers who do not have any books checked out.d. For each book that is loaned out from the Sharpstown branch and whose Due_date is today, retrieve the book title, the borrower’s name, and the borrower’s address.e. For each library branch, retrieve the branch name and the total number of books loaned out from that branch.f. Retrieve the names, addresses, and number of books checked out for all borrowers who have more than five books checked out.g. For each book authored (or coauthored) by Stephen King, retrieve the title and the number of copies owned by the library branch whose name is Central. **Relational Database Schema for a Library Database**This diagram illustrates the structure of a relational database designed for a library. It defines relationships between various entities such as books, authors, publishers, borrowers, and library branches.### Entities and Attributes:1. **BOOK** - **Attributes:** - Book_id - Title - Publisher_name - A book can be linked to authors, copies, and loans.2. **BOOK_AUTHORS** - **Attributes:** - Book_id - Author_name - This table maps books to their authors.3. **PUBLISHER** - **Attributes:** - Name - Address - Phone - Defines the information about publishers.4. **BOOK_COPIES** - **Attributes:** - Book_id - Branch_id - No_of_copies - Details the number of copies of a book available at each branch.5. **BOOK_LOANS** - **Attributes:** - Book_id - Branch_id - Card_no - Date_out - Due_date - Tracks the loans of books, specifying borrower and loan details.6. **LIBRARY_BRANCH** - **Attributes:** - Branch_id - Branch_name - Address - Provides information about each library branch.7. **BORROWER** - **Attributes:** - Card_no - Name - Address - Phone - Contains information about the individuals who borrow books.### Relationships:- The **BOOK** table connects with both **BOOK_AUTHORS** and **BOOK_COPIES** through the `Book_id` attribute, indicating which authors wrote which books and how many copies are available at each branch.- The **PUBLISHER** table links to the **BOOK** table via `Publisher_name`, specifying the publisher of each book.- The **BOOK_LOANS** table relates to **BOOK_COPIES** and **BORROWER** through `Book_id`, `Branch_id`, and `Card_no`, managing the records of borrowed books and their borrowers.- The **LIBRARY_BRANCH** table connects to both **BOOK_COPIES** and **BOOK_LOANS** through `Branch_id`, elucidating the location of copies and loan activity within each branch.- The **BORROWER** table

Introduction: A procedural query language known as "relational algebra" accepts relation objects as input and output. With the aid of operators, it executes queries. Can also use a binary or unary operator. They generate relations as an output from relations that they get as input. The solution is given below:Answer: (Note: We will use S for SELECT, P for PROJECT, * for NATURAL JOIN, - for SET DIFFERENCE, F for AGGREGATE FUNCTION) (a) A <-- BOOKCOPIES * LIBRARY-BRANCH * BOOK RESULT <-- P No_Of_Copies ( S BranchName='Sharpstown' and Title='The Lost Tribe' (A) ) Note: A better query would be to do the SELECTs before the JOIN as follows: A <-- P No_Of_Copies ( ( S BranchName='Sharpstown' (LIBRARY-BRANCH) ) * (BOOKCOPIES * ( S Title='The Lost Tribe' (BOOK) ) ) ) (b) P BranchID,No_Of_Copies ( ( S Title='The Lost Tribe' (BOOK)) * BOOKCOPIES ) (c) NO_CHECKOUT_B <-- P CardNo (BORROWER) - P CardNo (BOOK_LOANS) RESULT <-- P Name (BORROWER * NO_CHECKOUT_B) (d) S <-- P BranchId ( S BranchName='Sharpstown' (LIBRARY-BRANCH) ) B_FROM_S <-- P BookId,CardNo ( ( S DueDate='today' (BOOKLOANS) ) * S ) RESULT <-- P Title,Name,Address ( BOOK * BORROWER * B_FROM_S ) (e) R(BranchId,Total) <-- BranchId FCOUNT(BookId,CardNo) (BOOK_LOANS) RESULT <-- P BranchName,Total (R * LIBRARY_BRANCH) (f) B(CardNo,TotalCheckout) <-- CardNo F COUNT(BookId) (BOOK_LOANS) B5 <-- S TotalCheckout > 5 (B) RESULT <-- P Name,Address,TotalCheckout ( B5 * BORROWER) (g) SK(BookId,Title) <-- ( sAuthorName='Stephen King' ( BOOK_AUTHORS)) * BOOK CENTRAL(BranchId) <-- sBranchName='Central' ( LIBRARY_BRANCH ) RESULT <-- P Title,NoOfCopies ( SK * BOOKCOPIES * CENTRAL )