1. **Problem statement:** We have a traffic network with 4 nodes (A, B, C, D) and 5 flow variables $x_1, x_2, x_3, x_4, x_5$ representing vehicle flow rates on roads. We need to set up a system of equations based on flow conservation at each node and solve for these variables. 2. **Key principle:** At each node, the total inflow equals the total outflow (conservation of vehicles). 3. **Set up equations:** - Node A: Inflows: 200 (external) + $x_2$ (from B to A vertically) Outflows: $x_1$ (to D horizontally) + 600 (external outflow from A) Equation: $200 + x_2 = x_1 + 600$ - Node B: Inflows: 400 (external) + $x_4$ (from C to B horizontally) Outflows: $x_2$ (to A vertically) Equation: $400 + x_4 = x_2$ - Node C: Inflows: $x_5$ (from D vertically) + $x_3$ (from D horizontally) Outflows: 150 (external upward) + 450 (external rightward) + $x_4$ (to B horizontally) Equation: $x_5 + x_3 = 150 + 450 + x_4$ - Node D: Inflows: $x_1$ (from A horizontally) Outflows: $x_3$ (to C horizontally) + $x_5$ (to C vertically) Equation: $x_1 = x_3 + x_5$ 4. **Rewrite equations:** (1) $200 + x_2 = x_1 + 600 \Rightarrow x_2 - x_1 = 400$ (2) $400 + x_4 = x_2 \Rightarrow x_2 - x_4 = 400$ (3) $x_5 + x_3 = 600 + x_4 \Rightarrow x_3 + x_5 - x_4 = 600$ (4) $x_1 = x_3 + x_5 \Rightarrow x_1 - x_3 - x_5 = 0$ 5. **Solve the system:** From (4): $x_1 = x_3 + x_5$ Substitute into (1): $x_2 - (x_3 + x_5) = 400 \Rightarrow x_2 - x_3 - x_5 = 400$ From (2): $x_2 - x_4 = 400 \Rightarrow x_2 = x_4 + 400$ Substitute $x_2$ into previous: $x_4 + 400 - x_3 - x_5 = 400 \Rightarrow x_4 - x_3 - x_5 = 0$ From (3): $x_3 + x_5 - x_4 = 600$ We have two equations: $a) x_4 - x_3 - x_5 = 0$ $b) x_3 + x_5 - x_4 = 600$ Add (a) and (b): $(x_4 - x_3 - x_5) + (x_3 + x_5 - x_4) = 0 + 600 \Rightarrow 0 = 600$ This is a contradiction, so re-examine the problem setup. 6. **Re-examining node C:** The problem states outflows of 150 upward and 450 rightward from C, so these are external outflows, not flows to other nodes. The flow $x_4$ is from B to C, so inflow to C includes $x_4$. So inflows at C: $x_4 + x_5 + x_3$ (since $x_3$ is from D to right, presumably to C) Outflows at C: 150 + 450 Equation: $x_4 + x_5 + x_3 = 150 + 450 = 600$ 7. **Update equations:** - Node C: $x_4 + x_5 + x_3 = 600$ - Node D: $x_1 = x_3 + x_5$ - Node B: $400 + x_4 = x_2$ - Node A: $200 + x_2 = x_1 + 600$ 8. **Rewrite:** (1) $x_2 - x_1 = 400$ (2) $x_2 - x_4 = 400$ (3) $x_4 + x_5 + x_3 = 600$ (4) $x_1 - x_3 - x_5 = 0$ 9. **From (4):** $x_1 = x_3 + x_5$ Substitute into (1): $x_2 - (x_3 + x_5) = 400 \Rightarrow x_2 - x_3 - x_5 = 400$ From (2): $x_2 = x_4 + 400$ Substitute into above: $x_4 + 400 - x_3 - x_5 = 400 \Rightarrow x_4 - x_3 - x_5 = 0$ From (3): $x_4 + x_5 + x_3 = 600$ Add the last two equations: $(x_4 - x_3 - x_5) + (x_4 + x_5 + x_3) = 0 + 600 \Rightarrow 2x_4 = 600 \Rightarrow x_4 = 300$ 10. **Find $x_3$ and $x_5$:** From $x_4 - x_3 - x_5 = 0$: $300 - x_3 - x_5 = 0 \Rightarrow x_3 + x_5 = 300$ From (4): $x_1 = x_3 + x_5 = 300$ From (2): $x_2 = x_4 + 400 = 300 + 400 = 700$ 11. **Final values:** $x_1 = 300$ $x_2 = 700$ $x_4 = 300$ $x_3 + x_5 = 300$ (cannot determine individually without more info) 12. **Interpretation:** The system is underdetermined for $x_3$ and $x_5$ individually but their sum is 300. --- **Answer:** $$x_1 = 300, \quad x_2 = 700, \quad x_4 = 300, \quad x_3 + x_5 = 300$$