# plugin-hw02 - Assignment 2 Math 417 Winter 2009 Due January...

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Assignment 2 Math 417 — Winter 2009 Due January 23 § 1.2#42. First, let’s assign variables for the traffic volume along each one of the four street segments marked with question marks. Let x 1 be the traffic on JFK St. between Winthrop and Mt. Auburn. Let x 2 be the traffic on Mt. Auburn St. between JFK and Dunster. Let x 3 be the traffic on Dunster St. between Winthrop and Mt. Auburn. Let x 4 be the traffic on Winthrop St. between Dunster and JFK. We use the convention that positive values for these variables indicates traffic flowing in the direction of the arrows, and negative values would indicate traffic flowing in the opposite direction (which is technically impossible since these are one-way streets). The cardinal rule of traffic is that all traffic flowing into an intersection must also come out of that intersection. This rule gives us one equation for each of the four intersections. JFK/Mt. Auburn: The inbound traffic volume is 300+ x 1 and the outbound traffic volume is 400 + x 2 , so we have the equation 300 + x 1 = 400 + x 2 . Mt. Auburn/Dunster: The inbound traffic volume is 100 + x 2 + x 3 and the outbound traffic volume is 250, so we have the equation 100 + x 2 + x 3 = 250 . Dunster/Winthrop: The inbound traffic volume is 120 + 150 and the outbound traffic volume is x 3 + x 4 , so we have the equation 120 + 150 = x 3 + x 4 . Winthrop/JFK: The inbound traffic volume is 300 + x 4 and the outbound traffic volume is 320 + x 1 , so we have the equation 300 + x 4 = 320 + x 1 . 1

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Gathering these equations and rearranging in standard form, we find the system x 1 - x 2 = 100 x 2 + x 3 = 150 x 3 + x 4 = 270 x 1 - x 4 = - 20 The reduced echelon form of this system is x 1 - x 4 = - 20 x 2 - x 4 = - 120 x 3 + x 4 = 270 = 0 So the system is consistent and has infinitely many solutions, which are all of the form x 1 = - 20 + t, x 2 = - 120 + t, x 3 = 270 - t, x 4 = t, where t is an arbitrary parameter. However, not all of these solutions make sense since x 1 , x 2 , x 3 , x 4 must be nonnegative.
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