# Which is Faster — Bus or Walking?

Consider the daily trek I make from my apartment to places on campus. The eternal question beckons: which is faster — busing or walking? Here I will make a mathematical argument for why it could be the latter or former.

The time $t_b$ it takes to bus is given as follows:

$t_b = W_0 + \sum_i \left(\frac{L_i}{v_i} + S_i \right )$,

where $W_0$ is the time you wait for the bus, and the quantity in the summation is the time it takes to travel on the bus — moving time $L_i / v_i$ plus waiting time at stops and traffic lights $S_i$. The summation over index $i$ is over stop/go cycles of the bus.

However, let’s say $v_i = \text{constant} \equiv v_b$, the speed of the bus, so it can be pulled out of the summation. It also goes out without saying that

$\sum_i L_i = L$,

the total length of the trip. Hence, our original equation for $t_b$ becomes

$t_b = W_0 + \frac{L}{v_b} + \sum_i S_i$.

Now, the time it takes to walk is given as follows, assuming few stops and jaywalking at traffic lights:

$t_w = \frac{L}{v_w}$,

where $v_w$ is the walking speed. Taking the ratio of the two speeds:

$\frac{t_b}{t_w} = \frac{v_w}{L}W_0 + \frac{v_w}{v_b} + \frac{v_w}{L} \sum_i S_i = \frac{v_w}{v_b} + \frac{v_w}{L} \left(W_0 + \sum_i S_i \right)$.

Now it’s time to account for statistical uncertainties in stop times and wait times. Let $\left \langle x \right \rangle$ be the expectation value from a random variable $x$. In our formula, we assume speeds to be known to a relatively high precision (a very tenuous statement, but it simplifies our analysis). Taking all this to account, we take the expectation value of both sides,

$\left \langle \frac{t_b}{t_w} \right \rangle = \frac{v_w}{v_b} + \frac{v_w}{L} (\left \langle W_0 \right \rangle + \sum_i \left \langle S_i \right \rangle)$.

It is fairly easy to show that, if we do not know bus times, such that we leave our wait time completely to chance,

$\left \langle W_0 \right \rangle = \frac{T}{2}$,

where $T$ is the average time interval between successive buses. (We assume only one bus takes the route; a multi-bus analysis would require a more complex analysis.)

And since the expectation value is a constant, we have

$\sum_i \left \langle S_i \right \rangle = N\left\langle S_i \right \rangle$,

where $N$ is the number of stop/go cycles = number of stops.

Putting all these results together, we have

$\left \langle \frac{t_b}{t_w} \right \rangle = \frac{v_w}{v_b} + \frac{v_w}{L} (\frac{T}{2} + N\left \langle S_i \right \rangle)$.

Now we look at when our expected bus time to walk time ratio is greater than 1 — that is, when the bus takes longer than walking. In other words, we have

$\frac{v_w}{v_b} + \frac{v_w}{L} (\frac{T}{2} + N\left \langle S_i \right \rangle) \geq 1$.

Solving this inequality for $L$, we get values where walking is faster than busing when

$L \leq \frac{T/2 + N\left \langle S_i \right \rangle}{1/v_w - 1/v_b}$.

And this result is what we were looking for all along. For small $L$, walking is indeed faster than busing. Of course, for $L \rightarrow \infty$, busing is faster, and walking becomes very tedious. Now we find numerical values for $L$. We use the following values:

$v_w \approx 3 \text{ mph}$
$v_b \approx 20 \text{ mph}$ (pretty accurate in Berkeley).

For the popular 51B in Berkeley, the following values were recorded yesterday:

$T \approx 10 \text{ min}$
$\left \langle S_i \right \rangle \approx 0.33 \text{ min}$

Now, we assume that the number of stops $N$ is a function of the length $L$. In Berkeley, we can make this simpler by changing units in the equation to the number of blocks, which increase linearly with length because Berkeley’s streets are aligned in a grid. Let

$L = b \ell_b$,

where $b$ is the number of blocks and $\ell_b$ is the distance per block, and

$N \approx \frac{b}{2}$.

Regarding the latter, we can think of this two ways. Thinking of stop lights, on average half the stop lights will be red. Thinking of bus stops, the 51B stops are spaced about every two blocks. Plugging these into our equation for $L$,

$\left( \frac{1}{v_w} - \frac{1}{v_b} \right) b \ell_b \leq \frac{T}{2} + \frac{b}{2}S_i$,

where we have dropped the expectation value brackets. Solving this equation for $b$, we have walking being faster than busing when

$b \leq T \left[ 2 \ell_b \left(\frac{1}{v_w} - \frac{1}{v_b}\right) - S_i \right]^{-1}$.

In Berkeley,

$\ell_b \approx 0.125 \text{ mi}$ (between N-S roads);
$\ell_b \approx 0.065 \text{ mi}$ (between E-W roads).

Now if we plug all our values into the between N-S (i.e. traveling W-E) case, we get 2.5 blocks as our threshold distance, which is the same as from Telegraph to Tang Center. But that’s for a bus interval of 10 min, which only applies to the 51B at rush hour. Suppose we increase it to 17 min (many AC buses come with 15-20 min intervals). Then we get 4.3 blocks as our threshold, which means a W-E trip from Shattuck to Telegraph would be, on average, faster walking than busing. As would a trip from Asian Ghetto to Northside Ghetto — IF we had a bus route that went directly through campus. But of course we don’t, which means the bus would take even longer. Another example: taking the bus from South Berkeley (south of Southside) to Campus. In that case, given the 20-min bus intervals of the 1 and 18 on Sunday, walking would be on average as fast as the bus for a distance between Oregon/Stewart St. and Bancroft Way — a total of 10-11 blocks. That is the almost distance between Berkeley Bowl and the Berkeley Public Library. Or, between Willard Middle School and Campus. Translating it north, we see it’s the same distance as that between Dwight and Hearst.

In many cases, this is a conservative estimate. We often take shortcuts and routes on foot that cannot be done via bus. Additionally, there is walking time before/after getting off the bus, if it does not stop next to your departure location/destination. And this model assumes instantaneous acceleration between stop and go… finite accelerations will add time to the bus calculation. So if only one bus serves the route, this ratio is a near best-case scenario for bus users, for locations around campus at least.

Yes, this analysis is hand-wavy, and only accounts for one bus, but it serves as a rough estimate for when walking might be faster than busing. Perhaps in future posts, I can go into “higher-order corrections” to this formula, or minimize the maximum loss by characterizing the distribution of possible busing times.

Anywho, the verdict is: if you take the bus for distances on the order of campus length, either time it well, or take a bus that comes at a high frequency. If neither can be attainable, it might serve you better to walk.