Question: A network of one-way roads connects the intersections below. Each value/variable in the diagram represents a flow rate in cars per hour. Assume at each intersection the flow rate in must equal the flow rate out.
59 \to A \leftarrow y \to C \to 55
x \downarrow \quad z \searrow \downarrow v
B \leftarrow 32 \quad u \leftarrow D \leftarrow 28
a) Find the missing flow rates so that the traffic flow is balanced.
\u2022 Construct your coefficient matrix with columns ordered xyzuv.
\u2022 Enter free variables as themselves and all basic variables in terms of the free variables.x = 4-z-v
x = 4 - z - v cars per hour, top-right
y = 55+z+v
y = 55 + z + v cars per hour, top-right
z = z
z = z cars per hour, top-right
u = 28+v
u = 28 + v cars per hour, top-right
v = v
v = v cars per hour, top-right
b) If the value of v is zero, find the minimum value of z.
min z
z = 0
cars per
hour
z = 0 cars per hour
Incorrect.
1. **State the problem:** We have a network of one-way roads with flow rates represented by variables $x,y,z,u,v$ and constants. The flow in equals flow out at each intersection.
2. **Given relations:**
- $x = 4 - z - v$
- $y = 55 + z + v$
- $z = z$ (free variable)
- $u = 28 + v$
- $v = v$ (free variable)
3. **Goal:** Find missing flow rates so traffic flow is balanced.
4. **Part a) Express variables:**
- $x$ and $y$ are expressed in terms of free variables $z$ and $v$.
- $u$ also depends on $v$.
5. **Part b) If $v=0$, find minimum $z$:**
Since $v=0$, then:
- $x = 4 - z$
- $y = 55 + z$
- $u = 28$
6. **Constraints:** Flow rates must be non-negative.
So $x = 4 - z \geq 0 \implies z \leq 4$.
Also, $z \geq 0$ since flow rates cannot be negative.
7. **Minimum $z$ with $v=0$:**
Since $z \geq 0$ and $z \leq 4$, minimum $z$ is $0$.
8. **Check if $z=0$ is valid:**
- $x = 4 - 0 = 4 \geq 0$
- $y = 55 + 0 = 55 \geq 0$
- $u = 28 \geq 0$
All valid.
**Final answer:**
- $x = 4 - z - v$
- $y = 55 + z + v$
- $z = z$
- $u = 28 + v$
- $v = v$
For $v=0$, minimum $z$ is $0$ cars per hour.