Review : Systems of Equations — Question 8

Question 8 – Solution

Strategy. Conservation fixes net transfers but can leave an arbitrary circulating flow around a closed loop.

Step 1: Balance every node. Inflow equals outflow, giving $$A:\ r+3=p,\qquad B:\ p=q+1,\qquad C:\ q=r+2.$$ Equivalently $p-r=3$, $q-p=-1$, $r-q=-2$. The three left sides sum to zero, as do their right sides. Any two balances imply the third because total external inflow equals total external outflow.

Step 2: Recover the free circulation. Set $r=t$. Then $$\boxed{(p,q,r)=(t+3,t+2,t),\qquad t\ge 0.}$$ The condition $t\ge 0$ is necessary from $r\ge 0$ and sufficient for all three flows to be nonnegative. Adding the same amount to all three internal flows preserves every node balance: it is an additional circulation around the loop.

Step 3: Apply all capacities. The bounds give $t\le 2$, $t\le 2$ and $t\le 3$, respectively. Thus the entire feasible family has $\boxed{0\le t\le 2}$. The capacities on $p$ and $q$ both limit the upper endpoint; at $t=2$ they are saturated, while $r=2<3$ leaves unused capacity on the third link.

Step 4: Minimize the cost on the feasible family. Substituting the flow formulas yields $$2p+q+3r=2(t+3)+(t+2)+3t=8+6t.$$ This increases strictly on $[0,2]$. Therefore the unique minimizer is $\boxed{(p,q,r)=(3,2,0)}$, with cost rate $\boxed{8}$. It sends the required external transfers without an added loop circulation. The diagram shows directions, not arrows scaled in proportion to flow.