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PROGRAM TITLE: HIGHER NATIONAL DIPLOMA IN COMPUTING
UNIT TITLE: UNIT 18 - DISCRETE MATHS
ASSIGNMENT NUMBER: 03
ASSIGNMENT NAME: BOOLEAN ALGEBRA
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PART 1
Discuss two real-world binary problems in two different fields using applications of
Boolean algebra
Example 1: Boolean logic used in the control of traffic
In the control of traffic, people can use a truth table to represent boolean logic circuits. After
that, the table will be well programmed into a real-life traffic light system. And that system
can take in different input conditions (i.e: roads), and estimate the output(i.e: traffic lights) to
control the traffic.
For example, we will create a logic circuit for the expression AB + BC + AC
Figure 1 : Logic circuit of AB + BC + AC
Example 2: Boolean logic used for elevator
Said, we have a two-floor elevator, it needs to decide whether it should move to the other
floor. There are two buttons: one outside the elevator on the first floor and one inside the
elevator. Assuming these two buttons are connected to one line in the system.
We’ll have three inputs:
- “f” for the 1st-floor button
- “s” for the 2nd-floor button (“1” is when the button is clicked and “0” is the opposite)
- “p” represents the current floor indicator where “1” is the 1st floor and “0” is the 2nd
floor.
As for the output, “m” represents the move function of the elevator
f s p m
The orange highlighted row indicates that the elevator is on the 2
nd
floor and is moving down.
The green highlighted row indicates that the elevator is on the 1
st
floor and is moving up.
From the truth table above, we can conclude the following Boolean expression: 𝑚 = 𝑓𝑠̅ 𝑝̅ +
sp
This is the logical circuit:
b) “If it rains and you don't open your umbrella then you will get wet.”
Let:
- Rain = A
- Opening umbrella = B
- Getting wet = C
C = A × B
A B
C = A × B
2. Produce truth tables for given Boolean expressions.
A B C 𝐀
) = D
A B C
D
PART 3
Find the simplest form of given Boolean expressions using algebraic methods
1. A (A + B) + B (B + C) + C (C + A)
= AA + AB + BB+ BC +CC + CA
= A +AB + B + BC + C + CA
= A(1 + B) + B(1 + C) + C(1 + A)
= A + B + C
2. (A + 𝐁
)(B + C) + (A + B)(C + 𝐀
= AB + AC + B
B + B
C + AC + AA
+ BC + BA
= AB + AC + B
C + AC + BC + BA
= AB + AC + B
C + BC + BA
= AB +AC + C (B
+ B) + BA
= AB +AC + C + BA
= B (A + A
) + AC + C
= B + C (A +1)
= B + C
