 So we're going to do a second example of calculating power flow on a three node network. This network has two generators. There's a generator at node one, a generator at node two, and then there's a load at node three. In this network, the line connecting node one to node three has resistance r, the line connecting node one to two has resistance r, and the line connecting node two to node three has resistance two r. So the resistance on line two to three is twice as high as the resistance on either of the other two lines. So the first step is to calculate the distribution factors for g1. So first we're going to do the distribution factor on line one to three from g1. And so the numerator in this is going to be the inverse of the resistance of the path that goes from node one to node three. So this is one path with resistance r. So that's going to go in the numerator over here. And then in the denominator, we're going to have the sum of the inverse resistances for both paths. So the resistance of the direct path from one to three is one over r, and the resistance of the indirect path that goes from node one to node two down to node three has total resistance three r. So the second term in the denominator of these distribution factor equations is going to be one over three r, and we're going to get three quarters. Now we're going to calculate the distribution factor on line one to from g1. And so the numerator is going to be the total resistance of the long path that goes from one to node two to node three, which is three r. So the inverse of that is one over three r. And then in the denominator, we're going to have one over r plus one over three r, and we're going to get one fourth there. And last, we're going to do the distribution factor of line two to three for g1, which, because line two to three is on that same indirect path as line one to two, we're going to have the same distribution factor on line two to three for g1 as we did on line one to for g1. So we're going to have one over three r divided by one over r plus one over three r, and we're going to get one quarter. So the next step is to calculate the distribution factors for g2. And from g2, some of the power injected at g2 is going to go directly from node two to node three. So this is the direct path, which is going to have resistance to r. And some of the power injected at g2 is going to go from node two through node one to node three. So this indirect path, which is going to have total resistance to r. Okay, so now we can calculate the distribution factors. So the distribution factor on line one to three from g2, for that in the numerator, we're going to have the inverse of the total resistance of the path that contains line one three. So this is this indirect path with resistance to r. And then in the denominator, we're going to have the sum of the inverse resistances on both paths. So we have the indirect path with resistance one over two r. And we have the direct path, which also has resistance one over two r. And so when we work this out, we're going to get a half. The distribution factor on line one to two from g2. To calculate this, the first thing we need to remember is that power going from node two to node one is in what we had defined as the negative direction. So there's going to be a negative sign in front of the distribution factor. And remember that on these long indirect paths, both of the lines that make up the path are going to have the same distribution factor in magnitude. So the distribution factor for line one two is going to have the same magnitude as the distribution factor for line one three, except there's going to be a negative sign in front of it. So to see how that works out, we can write out the whole formula. So in the numerator, we're going to have one over two r. And in the denominator, we're going to have one over two r plus one over two r. And here we're going to get negative a half. So the distribution factor on line one two is the same magnitude as line one three just has a negative sign. And then finally, we're going to calculate the distribution factor on line two three from g2. And so in the numerator here, we're going to have the inverse of the resistance of the path that goes directly from two to three. So that's one over two r. And then in the denominator, we're going to have the same thing as in the other distribution factors. And again, we're going to get one half. So the last step is we're going to use the distribution factors and the assumed outputs of the two generators to calculate power flows. So in this case, we're going to assume that generator one produces 50 megawatts and generator two also produces 50 megawatts. So we're going to use our flow equation here, which is that the flow on line jk is equal to the distribution factor on line jk from g1 times the output of g1 plus the distribution factor on line jk from g2 times the output of g2. So we're going to take our outputs for g1 and g2, our distribution factors that we've already calculated, and we're going to calculate the flows for each of the three lines. So the flow on line one to three is going to be equal to the distribution factor on line one three from generator one, which we had said was three quarters times 50, the output of g1, plus the distribution factor on line one to three from g2, which we had said was one half times 50. And so when we add this together, we're going to get 62 and one half megawatts. Now for line one to two, we're going to take the distribution factor on line one to two from g1, which we had said was a quarter times 50. Then the distribution factor on line one to two from g2, which remember was equal to minus one half, because the flow in this case is going from node two to node one, which we defined as the negative direction. So this is going to be equal to minus a half times 50. And when we multiply this out, we're going to get minus 12 and a half megawatts. And remember what that minus sign means. It doesn't mean that we have negative energy. All that minus sign means is that it's telling us that 12 and a half megawatts is flowing on this line, but the direction is going from node two to node one. So remember, don't get freaked out by the minus sign. That just tells us the direction that the power is flowing. So now we'll do the flow on line two to three. So the distribution factor on line two to three from generator one, we had said was one fourth times 50, plus the distribution factor on line two to three from generator two, which we had said was one half times 50. And so when we multiply this out, we get 37 and a half megawatts. So in summary, what we get is the total flow from node one to node three is 62 and a half megawatts. The total flow from node two to node three is 37 and a half megawatts. And the total flow on line one two is going from node two to node one, and the magnitude of that flow is 12 and a half megawatts.