Linear System - Network Flow Matrices

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SUMMARY

The discussion focuses on solving a linear system representing traffic flow in a network of streets with intersections labeled a, b, c, d, and e. The system of equations derived from observed traffic includes variables x1 through x6, with specific relationships defined. The reduced row-echelon form leads to a general solution where x2 is set to zero due to roadwork on street ED. The minimum flow along AC, represented by x6, is determined to be zero under the condition that all variables must be non-negative, specifically requiring t to be at least 20.

PREREQUISITES
  • Understanding of linear algebra concepts, particularly systems of equations.
  • Familiarity with reduced row-echelon form and its application in solving linear systems.
  • Knowledge of traffic flow modeling and constraints in network systems.
  • Ability to manipulate algebraic expressions and inequalities.
NEXT STEPS
  • Study the application of linear programming in network flow problems.
  • Learn about the simplex method for solving linear optimization problems.
  • Explore the concept of flow conservation in directed graphs.
  • Investigate the implications of variable constraints in linear systems.
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Students and professionals in mathematics, engineering, and operations research who are involved in traffic modeling, optimization, and linear algebra applications.

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Homework Statement


Consider the network of streets with intersections a,b,c,d and e below. The arrows indicate the direction of traffic flow along the oneway streets, and the numbers refer to the exact number of cars observed to enter or leave a,b,c,d and e during one minute. Each xi denotes the unknown number of cars which passed along the indicated streets during the same period.
See Attached Image
Linear System:
x1+x6-x5=55
x5-x4=35
x6+x3-x4=60
x3-x2=40
x1-x2=70

Reduced row-echelon form
1 0 0 0 -1 1 | 55
0 1 0 0 -1 1 |-15
0 0 1 0 -1 1 |25
0 0 0 1 -1 0 |-35
0 0 0 0 0 0 |0
s t
General Solution:
x1=55+s-t
x2=-15+s-t
x3=25+s-t
x4=-35-s
x5=s
x6=t

The question is: if ED were closed due to roadwork, find the minimum flow along AC, using your results in the general solution.
Note: x2 = ED therefor x2=0, and AC is x6

Homework Equations





The Attempt at a Solution



I know x2=0 and no idea what to do with S and T...
what I did is set x2=0 so -15+s-t=0 so S=t+15
And sub it into the general solution you get:
x1=70
x2=0
x3=40
x4=-20+t
x5=15+t
x6=t

so x4≥0 when t≥20
x6≥0 when t≥0
So minflow is 0?
 

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1LastTry said:

Homework Statement


Consider the network of streets with intersections a,b,c,d and e below. The arrows indicate the direction of traffic flow along the oneway streets, and the numbers refer to the exact number of cars observed to enter or leave a,b,c,d and e during one minute. Each xi denotes the unknown number of cars which passed along the indicated streets during the same period.
See Attached Image
Linear System:
x1+x6-x5=55
x5-x4=35
x6+x3-x4=60
x3-x2=40
x1-x2=70

Reduced row-echelon form
1 0 0 0 -1 1 | 55
0 1 0 0 -1 1 |-15
0 0 1 0 -1 1 |25
0 0 0 1 -1 0 |-35
0 0 0 0 0 0 |0
s t
General Solution:
x1=55+s-t
x2=-15+s-t
x3=25+s-t
x4=-35-s
x5=s
x6=t

The question is: if ED were closed due to roadwork, find the minimum flow along AC, using your results in the general solution.
Note: x2 = ED therefor x2=0, and AC is x6

Homework Equations





The Attempt at a Solution



I know x2=0 and no idea what to do with S and T...
what I did is set x2=0 so -15+s-t=0 so S=t+15
And sub it into the general solution you get:
x1=70
x2=0
x3=40
x4=-20+t
x5=15+t
x6=t

so x4≥0 when t≥20
x6≥0 when t≥0
So minflow is 0?

No, you are not thinking it through! You need ALL xi >= 0, so you need x4 >= 0 and so you need t >= 20. Just having t >= 0 is not good enough.

RGV
 
I am trying to find the MINIMUM flow so...
 

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