Applications of Linear System
NETWORK ANALYSIS
Network Concepts and Flow Analysis
Network:
A network is a set of branches through which
something flows, such as electricity, water, oil,
traffic, or money.
Nodes/Junctions:
Points where branches meet and flow divides.
Examples: Electrical wires join at nodes, streets
intersect in traffic networks, and banks act as
junctions in financial networks.
Flow Rate Measurement
Flow rate varies by network type:
Electricity: Measured in amperes.
Water/Oil: Measured in gallons per minute.
Traffic: Measured in vehicles per hour.
Currency: Measured in units (e.g., Euros) per day.
Key Idea
Flow Conservation:
Flow into each node equals flow out, preventing
build-up and ensuring free movement through the
network.
Network Analysis:
Key goal: Use known flow rates in certain branches
to determine unknown flow rates in other branches.
ALOGARITHM TO SOLVE PROBLEM
Step 1:
Assigned arbitrary directions to the figure unknown
flow rates like x1, x2, and x3.
We need not be concerned if some of the directions
are incorrect, since an incorrect direction will be
signaled by a negative value for the flow rate when
we solve for the unknowns.
ALOGARITHM TO SOLVE PROBLEM
Step 2:
Following the conservation of flow at each node, we will
write linear equations for each node.
If there is 4 nodes then there will be a system of 4
equations.
Step 3:
Solve the system of linear equations for the unknown flow
rates.
Step 4:
The negative value tells us that the direction assigned to
that flow is incorrect; that is, the flow in that branch is in
opposite direction.
Exercise
A network with four nodes in which the flow rate and
direction of Figure flow in certain branches are
known. Find the flow rates and directions of flow in
the remaining branches.
Solution
Step 1: Assign the arbitrary values and directions to unknown flows.
Figure.2
Solution
Step 2: Using law of conversation(flow in= flow
out), we will write equation for each node
Nodes Flow in Flow out Equations
A 50 𝑥1 + 𝑥2 𝑥1 + 𝑥2 =50
B 𝑥2 + 𝑥3 30 𝑥2 + 𝑥3 =30
C 40 𝑥3 +50 𝑥3 +50=40
D 𝑥1 +50 60 𝑥1 +50=60
Solution
Step 3: These four conditions produce the linear system
𝑥1 + 𝑥2 =50 (1)
𝑥2 + 𝑥3 =30 (2)
𝑥3 =-10 (3)
𝑥1 =10 (4)
Put eq. (4) in (1), we obtain
𝑥2 =40
Solution
Step 4: The solution of linear system is
𝑥1 =10, 𝑥2 =40, 𝑥3 =-10 or
(𝑥1 , 𝑥2 , 𝑥3 )=(10, 40, -10)
The fact that 𝑥3 is negative tells us that the direction
assigned to that flow in Figure.2 is incorrect; that is,
the flow in that branch is into node C .
Solution
Exercise
A network with four nodes in which the flow rate and
direction of Figure flow in certain branches are
known. Find the flow rates and directions of flow in
the remaining branches.
Design of Traffic Patterns
Exercise
The accompanying figure shows a network of one-way streets
with traffic flowing in the directions indicated. The flow rates
along the streets are measured as the average number of
vehicles per hour.
(a) Set up a linear system whose solution provides the unknown
flow rates.
(b) Solve the system for the unknown flow rates.
(c) If the flow along the road from A to B must be reduced for
construction, what is the minimum flow that is required to
keep traffic flowing on all roads?
Exercise
(a)
Step 1:
Solution
Step 2:
The flow in must equal the flow out at each. For this to
happen, the following conditions must be satisfied:
Nodes Flow in Flow out Equations
A 𝑥3 + 750 250+𝑥4 𝑥3 − 𝑥4 =-500
B 𝑥4 + 200 𝑥1 + 300 𝑥4 − 𝑥1 = 100
C 𝑥1 + 100 𝑥2 +400 𝑥1 − 𝑥2 = 300
D 𝑥2 + 300 𝑥3 +400 𝑥2 − 𝑥3 = 100
Solution
Step 3: The linear system, we obtain is
𝑥3 − 𝑥4 =-500
𝑥4 − 𝑥1 = 100
𝑥1 − 𝑥2 = 300
𝑥2 − 𝑥3 = 100
(b) To find the solution of system, we will solve it
through Gaussian elimination method,(we can also
solve through Gauss Jordan Method, it’s our choice)
Solution
Step 3: The augmented matrix is
0 0 1 1 500
[A|b] 1 0 0 1 100
1 1 0 0 300
0 1 1 0 100
1 1 0 0 300
1 0 0 1 100
byR13
0 0 1 1 500
0 1 1 0 100
1 1 0 0 300
0 1 0 1 400
byR2 R1
0 0 1 1 500
0 1 1 0 100
Solution
1 1 0 0 300 1 1 0 0 3 0 0
0 1 0 1 400
by R2 0 1 0 1 4 0 0 b y R
0 0 1 1 500 R
0 0 1 1 5 0 0 4 3
0 1 1 0 100 0 0 0 0 0
[rank(A)=3, rank(A|b)=3,
1 1 0 0 300
0 1 0 1 400 As rank(A)=rank(A|b),
byR4 R2 (consistent)
0 0 1 1 500
No. of unknowns =n= 4
0 0 1 1 500
rank(A|b) <n(infinite solution)]
Solution
The corresponding system Backward Substitution,
of equations is Put x2 value in x1,we
x1 x2 300 (1) obtain
x2 x4 400 (2) x1 x4 400 300
x3 x4 500 (3) x1 x4 100
Solving for the leading Assign arbitrary values to
variables, we obtain the free variables( 𝑥4 = 𝑡).
x1 x2 300
The general solution of a
system parametrically, is
x2 x4 400 x1 t 100
x3 x4 500 x2 t 400
x3 t 500
x4 t
Solution
(c) If the flow along the road from A to B must be
reduced for construction, what is the minimum flow
that is required to keep traffic flowing on all roads?
As t=500 is the minimum value for which the value of
equations(flow) is non-negative.
To keep the traffic flowing on all roads, the flow from A
to B must exceed 500 vehicles per hour.
Exercise
The network in Figure shows a proposed plan for the traffic flow
around a new park that will house the Liberty Bell in Philadelphia,
Pennsylvania. The plan calls for a computerized traffic light at the
north exit on Fifth Street, and the diagram indicates the average
number of vehicles per hour that are expected to flow in and out of the
streets that border the complex. All streets are one-way.
(a) How many vehicles per hour should the traffic light let through to
ensure that the average number of vehicles per hour flowing into the
complex is the same as the average number of vehicles flowing out?
(b) Assuming that the traffic light has been set to balance the total
flow in and out of the complex, what can you say about the average
number of vehicles per hour that will flow along the streets that border
the complex?.
Exercise
Solution
(a)
we let x denote the number of vehicles per hour that
the traffic light must let through, then the total
number of vehicles per hour that flow in and out of the
complex will be
Flowing in: 500 + 400 + 600 + 200 = 1700
Flowing out: x + 700 + 400
Equating the flows in and out shows that the traffic
light should let x = 600 vehicles per
hour pass through.
Balancing Chemical
Equations
Chemical Formulas: Represent the atomic
composition of compounds (e.g., water is H₂O,
indicating two hydrogen atoms and one oxygen
atom).
Chemical Reactions: Compounds can combine
and rearrange their atoms to form new compounds
under certain conditions.
Example Reaction: When methane (CH₄) burns
with oxygen (O₂), it forms carbon dioxide (CO₂) and
water (H₂O):
2CH4+4O2→2CO2+4H2O
Balancing Chemical
Equations
Balancing Equations: Each type of atom must have
the same count on both sides of the reaction to follow
conservation laws.
Reactants and Products: Compounds on the left side
of the arrow are reactants, while those on the right are
products.
Balanced Equations: A balanced chemical equation
ensures complete reaction with no leftover reactants by
maintaining equal atom counts for each element on both
sides.
Systematic Balancing: For complex equations, we use
systems of linear equations to determine the smallest
integers that balance the atoms.
Exercise
Write a balanced equation for the given chemical
reaction.
C𝟑 H𝟖 + O𝟐 → CO𝟐 + H𝟐 O (propane combustion)
Solution
Let x𝟏 , x𝟐 , x𝟑 , and x𝟒 be positive integers that balance
the equation
x𝟏 (C𝟑 H𝟖 ) + x𝟐 (O𝟐 ) −→ x𝟑 (CO𝟐 ) + x𝟒 (H𝟐 O) (1)
Equating the number of atoms of each type on the two
sides yields
Solution
Carbon (C) 3x𝟏 = 1x𝟑
Hydrogen (H) 8x𝟏 = 2x𝟒
Oxygen (O) 2x𝟐 = 2x𝟑 + 1x𝟒
from which we obtain the homogeneous linear system
3 x1 x3 0
8 x1 2 x4 0
2 x2 2 x3 x4 0
Solution
The augmented matrix of
linear system is 1 0 1 1 0
3 0 1 0 0 byR R
3 0 1 0 0 1 2
8
0 0 2 0
0 2 2 1 0
0 2 2 1 0
3 0 1 0 0 1 0 1 1 0
4 0 3 0 byR2 3R1
0 by R2
1 4
0 0 1 0
2 0 2 1 0
0 2 2 1 0 2
4 0 0 1 0 1 0 1 1 0
3 0 0 byR12 0
0 1
2 2 1 0 byR23
0 2 2 1 0 0 0 4 3 0
Solution
The augmented matrix of
linear system is 1 0 1 1 0
3 0 1 0 0 byR R
3 0 1 0 0 1 2
8
0 0 2 0
0 2 2 1 0
0 2 2 1 0
3 0 1 0 0 1 0 1 1 0
4 0 3 0 byR2 3R1
0 by R2
1 4
0 0 1 0
2 0 2 1 0
0 2 2 1 0 2
4 0 0 1 0 1 0 1 1 0
3 0 0 byR12 0
0 1
2 2 1 0 byR23
0 2 2 1 0 0 0 4 3 0
Solution
The augmented matrix of
linear system is 1 0 1 1 0
3 0 1 0 0 byR R
3 0 1 0 0 1 2
8
0 0 2 0
0 2 2 1 0
0 2 2 1 0
3 0 1 0 0 1 0 1 1 0
4 0 3 0 byR2 3R1
0 by R2
1 4
0 0 1 0
2 0 2 1 0
0 2 2 1 0 2
4 0 0 1 0 1 0 1 1 0
3 0 0 byR12 0
0 1
2 2 1 0 byR23
0 2 2 1 0 0 0 4 3 0
Solution
This is the reduced row echelon form of
1 0 1 1 0
the augmented matrix of his system.
The corresponding system of equations
0 1 1
1
0 by R2
1
2 2 is
x1 x3 x4 0
0 0 4 3 0 x2 x3
1
x4 0
2
3
x3 x4 0
1 0 1 1 0 4
Solving for the leading variables, we
0 1 1
1
0 by R3
1
2 4 obtain
3
0 0 1 0 x1 x3 x4
4 1
x2 x3 x4
2
3
x3 x4
4
Solution
This is the reduced row echelon form of
1 0 1 1 0
the augmented matrix of his system.
The corresponding system of equations
0 1 1
1
0 by R2
1
2 2 is
x1 x3 x4 0
0 0 4 3 0 x2 x3
1
x4 0
2
3
x3 x4 0
1 0 1 1 0 4
Solving for the leading variables, we
0 1 1
1
0 by R3
1
2 4 obtain
3
0 0 1 0 x1 x3 x4
4 1
x2 x3 x4
2
3
x3 x4
4
Solution
Backward Substitution, 1
x1 t
Put x3 value in x1 and 4
x2,we obtain 5
x2 t
4
1
x1 x4 3
4 x3 t
5 4
x2 x4 To obtain the smallest
4
x3
3
x4
positive integers that
Assign arbitrary
4 values to balance the equation,
the free variables( 𝑥4 = 𝑡).
The general solution of a
system parametrically, is
Solution
we let t = 4, in which case we obtain x𝟏 = 𝟏, x𝟐 = 𝟓,
x𝟑 = 𝟑, and x𝟒 = 𝟒.
Substituting these values in (1) produces the balanced
equation
1(C𝟑 H𝟖 ) + 5(O𝟐 ) −→ 𝟑(CO𝟐 ) + 𝟒(H𝟐 O)
Exercise
Write a balanced equation for the given chemical
reaction.
1. CH3COF + H2O → CH3COOH + HF
2.CO2 + H2O → C6H12O6 + O2 (photosynthesis)
Exercise
