Applications to Networks

Introduction A Network consists of branches and nodes. A typical example is the street network where the branches are the streets and the nodes are the intersections. Another example is an electrical network. Many network problems can be modeled by systems of linear equations. The basic laws are explained below.

_{1) Electrical Networks}

In such networks, Ohm’s law and Kirchhoff’s laws govern current flow, as follows:

Ohm’s Law: The voltage drop across a resistor is the product of the current and the resistance: V=IR

Kirchhoff’s first Law: The sum of the currents flowing into a node is equal to the sum of the current flowing out.

Kirchhoff’s second Law: The algebraic sum of the voltage drops around a closed loop is equal to the total voltage in the loop.

Example Determine the currents I₁, I_2, and I₃ for the following electrical network:

Applying Kirchhoff’s first Law to either of the nodes B or C, we find I₁=I₂+I₃ . In other words:

I₁-I₂-I₃ =0.

Applying Kirchhoff’s second Law to the loops BDCB and BCAB, we obtain the equations

-10I₁+10I₂=10.

20I₁+10I₂ =5.

This gives a linear system of three equations

The augmented matrix of the above system is

which can be reduced to

Therefore, the currents are:

Since I₃ is negative, the current flow is from C to B rather than B to C, as tentatively assigned in the above diagram.

_{2) Traffic Networks}

The diagram below represents the traffic flow through a certain block of streets. (The numbers are the average flows into and out of the network at peak traffic hours)

By Kirchhoff’s first Law, the flow into an intersection is equal to the flow out. This gives the following system

The augmented matrix of the above system is

Some calculations show that this matrix is row equivalent to

Therefore, the solution can be written as

For instance, if w=300 and t=1300 (in vehicles per hour), then

Suppose now that the streets from A to B and from B to C must be closed (for construction for instance), that is x=0 and y=0. How might the traffic be rerouted?

To answer this question, set x=y=0 in the above solution, we get , , and . Of course, the negative value for is not normal. In order to avoid negative flow, we must reverse directions on the streets connecting C and D; this change makes instead of .