5) A small neighborhood of Old Town is serviced by only a few one-way streets. T
ID: 3138493 • Letter: 5
Question
5) A small neighborhood of Old Town is serviced by only a few one-way streets. There is only a single entrance to this neighborhood (call this intersection A) and three exits (at intersections B, C, and D). Assume that in a given hour, the number of cars entering an intersection is equal to the number of cars leaving it. Use the hourly data given below to determine how heavy the trafiic is within the neighborhood along each street (i.e. solve for the variables given) Every hour, 100 cars arrive at intersection A: T1 cars take Elm St. to intersection Every hour at intersection B, r3 cars take Maple St. to intersection C, 4 cars Every hour at intersection C, r5 cars take Willow St. to intersection D, and 50 B and r2 cars take Oak St. to intersection C take Birch St. to intersection D, and 20 cars exit the neighborhood cars exit the neighborhood Every hour at intersection D, 30 cars exit the neighborhoodExplanation / Answer
(5). The incoming traffic at each intersection must equal the outgoing traffic. Thus, at
A: x1 +x2 = 100
B: x1 = x3+x4+20 or, x1- x3-x4 =20
C: x2+x3 = x5 +50 or, x2+x3 - x5 = 50
D: x4+x5= 30
The augmented matrix of this linear system is M (say) = 1
1
1
0
0
0
100
1
0
-1
-1
0
20
0
1
1
0
-1
50
0
0
0
1
1
30
To solve the linear system, we will reduce M to its RREF as under:
Add -1 times the 1st row to the 2nd row
Multiply the 2nd row by -1
Add -1 times the 2nd row to the 3rd row
Multiply the 3rd row by -1
Add -1 times the 3rd row to the 4th row
Add -1 times the 3rd row to the 2nd row
Add -1 times the 2nd row to the 1st row
Then the RREF of M is
1
0
-1
0
1
50
0
1
1
0
-1
50
0
0
0
1
1
30
0
0
0
0
0
0
Thus, the above linear system is equivalent to x1-x3+x5 = 50 or, x1 = x3-x5+50…(1), x2+x3-x5 = 50 or, x2 = -x3+x5+50…(2) and x4+x5 = 30 or, x4 = -x5+30…(3). The 4 initial equations in 5 variables can be narrowed down to 3 equations in 5 variables so that we can solve for 3 variables x1,x2,x4 in terms of 2 variables x3 and x5. Then, we have X = (x1,x2,x3,x4,x5) =(x3-x5+50, -x3+x5+50, x3 ,-x5+30, x5).
1
1
0
0
0
100
1
0
-1
-1
0
20
0
1
1
0
-1
50
0
0
0
1
1
30
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