---
title: "RA: increase loans over 10000"
subject: "Database Management System"
chapter: 3
key: "semester_6/database_management_system:ra-increase-loan-10"
source: https://www.bctengineer.com/q/semester_6/database_management_system/ra-increase-loan-10
site: BCT Engineering
---

# RA: increase loans over 10000

Database Management System · Chapter 3: Relational Languages & Relational Model

- Asked in: year not given
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/database_management_system/ra-increase-loan-10
- Whole chapter: https://www.bctengineer.com/md/semester_6/database_management_system/chapter-3

## Question

Consider the relational schema:
Acc (a_no, b_name, bal)
Branch (b_name, b_city, assets)
Customer (c_name, c_street, c_city)
Loan (l_no, b_name, amt)
Depositor (c_name, a_no)
Borrower (c_name, l_no)
Increase loan amount by 10% whose amt is greater than 10000. (Write a relational algebra expression.)

## Short answer

Remove loans with amt > 10000, apply generalized projection to multiply amt by 1.1, union with unchanged loans, and assign back to Loan.

## Answer

Relational algebra symbols used: σ select, π project, ⋈ natural join, ∪ union, ∩ intersection, − difference, ρ rename, 𝒢 aggregate.

Updating in relational algebra uses the generalized projection with an assignment: the tuples to be changed are removed, and their changed versions are added back.

#### Relational algebra expression
```text
Loan ← ( Loan − σ amt > 10000 (Loan) )
        ∪ π l_no, b_name, amt * 1.1 ( σ amt > 10000 (Loan) )
```

#### Explanation
1. σ amt > 10000 (Loan) selects the loans to change.
2. Loan − (that set) keeps the loans that stay as they are.
3. The generalized projection π l_no, b_name, amt*1.1 outputs the same loans with the amount increased by 10% (amt × 1.1 = amt + 10%).
4. The union of the two parts is assigned back to Loan, so the relation now holds all the loans, with the large ones increased.

Example: a loan of 20000 becomes 22000; a loan of 9000 is unchanged.

An equivalent form uses a selection instead of the difference:
```text
Loan ← π l_no, b_name, amt * 1.1 ( σ amt > 10000 (Loan) )  ∪  σ amt ≤ 10000 (Loan)
```

#### Equivalent SQL
```sql
UPDATE Loan SET amt = amt * 1.1 WHERE amt > 10000;
```

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How does the expression update loans?
   A. Change tuples in place with σ
   B. Remove the tuples to change and add back their modified versions
   C. Use intersection
   D. Delete all loans and insert none
   Answer: B. Algebra has no in-place update, so it uses difference and union. (Key point: Update approach)

2. What does π l_no,b_name,amt*1.1 do?
   A. Decreases by 10%
   B. Doubles
   C. Selects loans above 1.1
   D. Outputs the same loans with amount increased by 10%
   Answer: D. amt*1.1 = amt + 10%. (Key point: Increase factor)

3. What happens to a loan of 9000 and one of 20000?
   A. Both increased
   B. 9000 stays; 20000 becomes 22000
   C. 9000 becomes 9900
   D. 20000 becomes 21000
   Answer: B. Only amt > 10000 is changed, by 10%. (Key point: Example)

## Close variants set in other exams

- [RA: delete loans under 1000](https://www.bctengineer.com/question.md?key=semester_6%2Fdatabase_management_system%3Ara-delete-small-loans) — Database Management System, chapter 3

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