Question

For a given traffic stream, the speed-density relationship is given as:

\[ v = v_o \ln\left(\frac{k_j}{k}\right) \]

where \(v\) is the mean speed (in km/h), and \(k\) is the density (in veh/km).

Considering \(v_o\) as 45 km/h, and \(k_j\) as 200 veh/km, the maximum flow (in veh/h) for the given stream is (rounded off to the nearest integer).

Hint:

Maximize \(q = kv_o\ln(k_j/k)\) by differentiating with respect to \(k\); the peak occurs at \(k = k_j/e\), where \(v = v_o\).

Answer 1

Correct Answer: 3311

For Greenberg's model $v = v_o \ln(k_j/k)$, there is a shortcut: the density that maximizes flow always works out to $k_j/e$, so instead of differentiating from scratch, we can use that known result and verify it.

Flow is $q = kv = k v_o \ln(k_j/k)$. Writing $x = k_j/k$ so that $k = k_j/x$, flow becomes:

\[ q = \frac{k_j}{x} v_o \ln x \]

This is maximized where $\dfrac{d}{dx}\left(\dfrac{\ln x}{x}\right) = 0$, and a standard calculus result is that $\ln x / x$ peaks at $x = e$. So the maximum flow condition is $k_j/k = e$, giving $k = k_j/e$, the same result reached by direct differentiation.

Substituting $x = e$ back:

\[ q_{max} = \frac{k_j}{e} v_o \ln(e) = \frac{k_j v_o}{e} \times 1 = \frac{k_j v_o}{e} \]

So the maximum flow has a clean closed form, $q_{max} = k_j v_o / e$, without needing to separately compute the speed and density at the peak.

Plugging in $k_j = 200$ veh/km and $v_o = 45$ km/h:

\[ q_{max} = \frac{200 \times 45}{2.71828} = \frac{9000}{2.71828} = 3310.9 \text{ veh/h} \]

Let's summarize:

• For Greenberg's model, maximum flow always occurs at $k = k_j/e$, where the speed equals $v_o$ exactly.
• The maximum flow has the compact form $q_{max} = k_j v_o / e$.
• Numerically this gives about 3311 veh/h after rounding.

So the maximum flow for this stream is $q_{max} \approx 3311$ veh/h.