please help in this, i included the notes as well to assist with solving.

Exercise 1.4.3 A traffic circle has five one-way streets, and vehicles enter and leave as shown in the accompany- ing diagram. 35 f4 25 fs E D 50 A C f1 f2 40 B 30 a. Compute the possible flows.1.4 An Application to Network Flow There are many types of problems that concern a network of conductors along which some sort of ow is observed. Examples of these include an irrigation network and a network of streets or freeways. There are often points in the system at which a net ow either enters or leaves the system. The basic principle behind the analysis of such systems is that the total ow into the system must equal the total ow out. In fact, we apply this principle at every junction in the system. Junction Rule At each of the junctions in the network, the total ow into that junction must equal the total ow out. This requirement gives a linear equation relating the ows in conductors emanating from the junction. Example 1.4.1 A network of one-way streets is shown in the accompanying diagram. The rate of ow of cars into intersection A is 500 cars per hour, and 400 and 100 cars per hour emerge from B and C, respectively. Find the possible ows along each street. Solution. Suppose the ows along the streets are f1, f2, f3, f4, 500 A f1 B 400 f5, and f5 cars per hour in the directions shown. Then, equating the ow in with the ow out at each intersection, ' Intersection A 500 = fl + f2 + f3 f3 f6 . Intersectlon B f1 + f4 + f6 = 400 Intersection C f3 + f5 = f5 + 100 C 100 IntersectionD f2: f4+ f5 These give four equations in the six variables f1, f2, ..., f5. f1+f2+f3 =500 f1 +f4 +f6=400 f3 +f5f6=100 h - = 0 The reduction of the augmented matrix is 111000500 100101400 100101400_>0101100 001011100 001011100 0101100 0000000 Hence, when we use f4, f5, and f5 as parameters, the general solution is f1=400f4-f6 f2=f4+f5 f3=100f5+f6 This gives all solutions to the system of equations and hence all the possible ows