1. **Stating the problem:** We have a network of one-way streets with intersections A, B, C, and D. The inflow and outflow of cars per hour are known for some intersections, and the flow rates along six streets, $f_1$ to $f_6$, are unknown. We want to find all traffic flows such that the inflow and outflow at each intersection balances (conservation of cars). 2. **Set up balance equations for each intersection:** - Intersection A (inflow 500, outflow to B, C, D): $$500 = f_1 + f_2 + f_3$$ - Intersection B (outflow 400, inflow from A and D): $$f_1 + f_4 + f_6 = 400$$ - Intersection C (outflow 100, inflow from A and D): $$f_3 + f_5 = 100$$ - Intersection D (inflow from A via $f_2$, outflow to B, C): $$f_2 = f_4 + f_5 + f_6$$ 3. **Rewrite as system of linear equations:** \[ \begin{cases} f_1 + f_2 + f_3 = 500 \ f_1 + f_4 + f_6 = 400 \ f_3 + f_5 = 100 \ f_2 - f_4 - f_5 - f_6 = 0 \end{cases} \] 4. **Apply Gaussian elimination:** Express $f_1$ and $f_3$ from the first and third equations: - $f_1 = 500 - f_2 - f_3$ - From third equation, express $f_5 = 100 - f_3$ Substitute $f_1$ and $f_5$ in second and fourth equations: - Second: $(500 - f_2 - f_3) + f_4 + f_6 = 400 \Rightarrow 500 - f_2 - f_3 + f_4 + f_6 = 400$ Simplifies to: $$f_4 + f_6 - f_2 - f_3 = -100$$ - Fourth: $f_2 - f_4 - (100 - f_3) - f_6 = 0$ Simplifies to: $$f_2 - f_4 - 100 + f_3 - f_6 = 0 \Rightarrow f_2 - f_4 - f_6 + f_3 = 100$$ 5. **Add the two simplified equations:** \[ (f_4 + f_6 - f_2 - f_3) + (f_2 - f_4 - f_6 + f_3) = -100 + 100 = 0 \] This confirms system consistency. 6. **Rewrite the two equations as:** \[ \begin{cases} f_4 + f_6 - f_2 - f_3 = -100 \ f_2 - f_4 - f_6 + f_3 = 100 \end{cases} \] Rearranged: - $f_4 + f_6 = f_2 + f_3 - 100$ - $f_2 + f_3 = f_4 + f_6 + 100$ Both represent the same relation, confirming redundancy; we have 4 equations but only 3 independent because the last two are dependent. 7. **Choose free variables:** Let $f_2 = a$, $f_3 = b$, $f_6 = c$ for $a,b,c \\geq 0$ (since flow rates cannot be negative). Using $f_4 + c = a + b - 100$, we get: $$f_4 = a + b - 100 - c$$ Also from step 4, $$f_1 = 500 - a - b$$ From step 3, $$f_5 = 100 - b$$ 8. **Summary:** $$ \begin{cases} f_1 = 500 - a - b\\ f_4 = a + b - 100 - c\\ f_5 = 100 - b\\ f_2 = a\\ f_3 = b\\ f_6 = c \end{cases} $$ where $a,b,c \geq 0$ are free parameters. 9. **Physical meaning constraints:** Flow rates must be non-negative: - $f_1 \geq 0 \Rightarrow 500 - a - b \geq 0 \Rightarrow a + b \leq 500$ - $f_4 \geq 0 \Rightarrow a + b - 100 - c \geq 0 \Rightarrow c \leq a + b - 100$ - $f_5 \geq 0 \Rightarrow 100 - b \geq 0 \Rightarrow b \leq 100$ - $f_2 = a \geq 0$ - $f_3 = b \geq 0$ - $f_6 = c \geq 0$ Also note from $c \leq a + b - 100$, since $c \geq 0$ this implies $a + b \geq 100$. 10. **Final solution set:** All flows are determined by free non-negative $a,b,c$ such that $$100 \leq a + b \leq 500, \quad 0 \leq b \leq 100, \quad 0 \leq c \leq a + b - 100$$ The other flows are: $$f_1 = 500 - a - b, \quad f_4 = a + b - 100 - c, \quad f_5 = 100 - b$$