The figure below shows a network of one way streets. The direction the traffic i
ID: 3864190 • Letter: T
Question
Explanation / Answer
a. & b.
x2 + 550 = x3 + 350
x3 + 400 = x4 + 300
x4 + 200 = x1 + 600
x1 + 250 = x2 + 150
The above system of linear equations can be simplified as follows
x2 – x3 = -200
x3 – x4 = -100
x1 – x4 = -400
x1 – x2 = -100
The above system of linear equations can be represented in the following matrix form
0 1 -1 0 -200
0 0 1 -1 -100
1 0 0 -1 -400
1 -1 0 0 -100
The reduched echleon form is
1 0 0 -1 -400
0 1 0 -1 -300
0 0 1 -1 -100
0 0 0 0 0
And clearly, x4 is a free variable.
This system has infinitely many solutions, which are (with parameters ri):
x1 = r1 – 400
x2 = r1 – 300
x3 = r1 – 100
x4 = r1
If we want the traffic to flow in the forward direction, then r1>=0
c.
When we restrict the flow from A to B, we substitute a value of x4 in above equations such that remaining all flows should be positive, so when we substittue (x4 >= 400) we get
x1 >= 0
x2 >= 100
x3 >= 200
The minimum flow between A and B must be 400 to keep the traffic flowing in the forward direction on other roads
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