When should you leave for work and allow for enough time that you won't be late for work? Economists have something to say about that. Whenever travel time is uncertain, there is a non-zero probability that you are late. You cannot completely avoid that risk. However, you can reduce this risk economically. Here is some math that shows when you should leave for work.

Let's introduce a bit of notation first. You want to arrive no later than at time \(T^A\), and thus you decide to depart at time \(T^D\). Travel time is a random variable with mean \(T^M\) and standard deviation \(T^S\). What sort of distribution we may choose for the trip duration matters. The log-normal distribution with a positive support range is a good candidate, as travel time cannot be negative, and there is a "long tail" of long delays. The log-normal distribution with mean \(T^M\) and standard deviation \(T^S\) has a mean and variance of the travel time's natural logarithm that is equal to \(\mu=\ln(T^M)-\sigma^2/2\) and \(\sigma^2=\ln(1+(T^s/T^M)^2)\), respectively.

Economists put a value on the cost of time. Travel and wait time incurs an hourly cost of \(w\). There is also a fixed cost \(v\), a penalty, for being late. One can think of it as the cost of missing a flight connection, or your employer for penalizing you for being late to work. This penalty has to be large enough to matter, so \(v\gg w\). Total travel cost is then \[C=w\cdot(T^A-T^D)+v\cdot \mathcal{P}\left\{T+T^D>T^A\right\}\] where \(\mathcal{P}(\cdot)\) indicates probability. When our travel time \(T\) is less than the time we have allowed between departure time and ideal arrival time, \(T^\ast\equiv T^A-T^D\), we are too early and we have to wait until work starts. Ideally, we want to minimize the amount of time we are too early and spend idling. On the other hand, we are trying to avoid incurring the penalty of being late. Another way to model this would be to incur a per-minute penalty of being late, so it would matter how many minutes we are late. The math for this is a bit more challenging, so I will stick to the simple case of a fixed penalty. All numbers in the table are in physical units of "minutes" and correspond to a mean travel time of 30 minutes.

Using our assumption of a log-normal distribution, we find that \[ \mathcal{P}\left\{T>T^\ast\right\}= 1-\Phi\left(\frac{\ln(T^\ast)-\mu}{\sigma}\right)\] where \(\Phi(\cdot)\) is the standard normal cumulative distribution function. To determine the optimal advance time \(T^\ast\equiv T^A-T^D\) when to leave, the first-order condition for a cost minimum is \[\frac{\mathrm{d}C}{\mathrm{d}T^\ast}=w-v\frac{\mathrm{d}\Phi}{\mathrm{d}T^\ast}=0\quad\Rightarrow\quad \frac{\mathrm{d}\Phi}{\mathrm{d}T^\ast}=\phi(T^\ast)=\frac{w}{v}\] Thus, using the definition of the standard normal density function \(\phi(\cdot)\), we can solve directly for the optimal planning time \(T^\ast\): \[\ln(T^\ast)=\ln(T^M)+\sigma^2\left[ \sqrt{2+\frac{2}{\sigma^2}\ln\left[\frac{v/w}{\sqrt{2\pi}\sigma T^M}\right]} -\frac{3}{2}\right]\]

Consider a penalty ratio of \(v/w=240\) and assume it takes \(T^M=30\) minutes on average to drive to work, but with a standard deviation of \(T^S=6\) minutes. Thus \(\sigma=0.198\). This results in \(T^\ast=45.3\) minutes of leaving before the desired arrival time, and on average about 15 minutes of waiting after driving. The table below shows how a 30-minute drive can easily take you over an hour. If the penalty ratio is 300 and the standard deviation of travel time 15 minutes, you should depart a full hour before the targeted arrival time.

Travel Time Std. Deviation |
Penalty Ratio | ||||||
---|---|---|---|---|---|---|---|

120 | 180 | 240 | 300 | 360 | 420 | 480 | |

3 | 37.4 | 38.0 | 38.4 | 38.8 | 39.0 | 39.2 | 39.4 |

6 | 42.6 | 44.2 | 45.3 | 46.1 | 46.8 | 47.3 | 47.8 |

9 | 45.9 | 48.7 | 50.6 | 52.1 | 53.3 | 54.3 | 55.2 |

12 | 47.5 | 51.6 | 54.5 | 56.7 | 58.5 | 60.0 | 61.3 |

15 | 47.9 | 53.2 | 57.0 | 59.9 | 62.2 | 64.3 | 66.0 |

18 | 47.3 | 53.7 | 58.3 | 61.8 | 64.8 | 67.3 | 69.5 |

21 | 46.1 | 53.4 | 58.7 | 62.8 | 66.3 | 69.2 | 71.8 |

24 | 44.7 | 52.6 | 58.5 | 63.1 | 67.0 | 70.3 | 73.2 |

The penalty ratio is measured in minutes because penalty [measured in dollars] is divided by time cost [measured in dollars per minute]. Thus a penalty ratio of 300 means 300 minutes or 5 hours; being late would cost you five hours of your cost of time. The penalty ratio can be thought of as a measure of a motorist's schedule inflexibility. The higher the ratio, the less flexible the time schedule.

The intuition behind leaving early is easy to grasp: if there is a cost of being late, leave sufficiently early to reduce that risk up to the point where the extra cost of waiting is more of a bother than the penalty of being late. What we often don't know explicitly is the distribution of travel times. It is easy to see how people can be late quite often if they underestimate the standard deviation of travel time.

There is a more important implication of the above discussion.
When it comes to finding an optimal road toll, we usually assume that
drivers choose the least-cost route in terms of pure travel time.
That is wrong. Modern work on road tolling takes into account that
one also needs to take the **reliability** of routes into account.
Two roads that have equal average travel time are not the same if
one has much higher variance than the other. A motorist will choose
the route with the lower variance. At the margin, motorists will trade
off a slightly higher average travel time for a lower variance of travel
time. This means that there is a price for reliability. An optimal road
toll should be higher if it helps reduce the average travel time *and*
the variance of travel time. Road pricing therefore faces two types of
heterogeneity of motorists: their **time cost** may vary with income,
and their **schedule flexibility** may vary with occupation. The
two are not necessarily correlated. My own time cost tends to be quite high,
but I have a flexible work schedule. An sales clerk on the way to the store
may have a low time cost but faces an inflexible work schedule.

Agent heterogeneity is an important element in the road tolling literature. My economist colleague Jonathan Hall, from the University of Toronto, recently published an article on the effects of separate lane pricing, where he finds that congestion pricing requires variation across time of day as well as pricing lanes differently. Understanding rich and poor drivers, and flexible and inflexible drivers, makes a big difference in designing optimal tolls. Jonathan Hall's work makes intuitive sense also from a political perspective: if we want to implement congestion pricing, we have to account for the distributional effects among rich and poor commuters. With a tolled lane and a free lane, motorists have a choice. They can either pay with money or with time. Poor motorists may prefer to pay with time. Jonathan Hall calls the tolled lanes the "Lexus Lanes". That moniker may well catch on in the literature. The key to "good" (politically viable) tolling is price differentation. The explicit cost (toll) is borne only by the Lexus drivers, while the invisible cost (time) is borne by all.

When we measure traffic along a particular road, we use that information to estimate volume-speed relationships. Reduce volume and thus reduce congestion and travel time. What we overlook is that the volume-speed relationship is quite noisy. Even for a given volume, there is considerable variation in travel time. Some roads or bridges have more variation than others due to idiosyncractic features or conditions. Because motorists take reliability into account, we need to impose higher tolls on unreliable roads than on reliable roads in order to get the same net reduction in travel costs. Empirically, we need to observe measures of reliability (variance) and put these into our frameworks for optimal tolling. Without such metrics, our work is incomplete.

Returning to the question about when you should leave for work, here is one suggestion on changing the economics of it slightly. If you arrive early, make your wait time more valuable. Have something that allows you to turn this wait time from "wasted time" into "useful time"—whatever that may be. Even if it is just time to read a good newspaper or a book.

Further readings:

- Jonathan D. Hall: Pareto improvements from Lexus Lanes: The effects of pricing a portion of the lanes on congested highways,
*Journal of Public Economics*158, February 2018, pp. 113-125. - Kenneth A. Small, Clifford Winston, and Jia Yan: Uncovering the Distribution of Motorists' Preferences for Travel Time and Reliability,
*Econometrica*73(4), July 2005, pp. 1367-1382. - David Brownstone and Kenneth A. Small: Valuing time and reliability: assessing the evidence from road pricing demonstrations,
*Transportation Research Part A: Policy and Practice*39(4), May 2005, pp. 279-293. - Daniel A. Brent and Austin Gross: Dynamic road pricing and the value of time and reliability,
*Journal of Regional Science*58(2), March 2018, pp. 330-349.

*Technical Postscript*

The log-normal distribution is not the only appropriate distribution function. I had chosen the log-normal type because it leads to an algebraically tractable solution. Whether this distribution fits the data well is a matter for empirical tests. It is straight-forward to use the Gamma distribution instead with shape parameter \(k=(\mu/\sigma)^2\) and scale parameter \(\theta=\sigma^2/\mu\). Then the optimal advance time for departure is \[\frac{(T^\ast)^{k-1}\exp(-T^\ast/\theta)}{\Gamma(k)\theta^k} =\frac{v}{w}\] This expression can be solved numerically quite easily, but the algebraic form is a bit unwieldy. The table below replicates the table from above, but with the gamma distribution replacing the log-normal distribution. The results are similar, but as the standard deviation and penalty ratio rises, the advance time tends to be slightly higher than with the log-normal distribution.

Advance Time (using Gamma Distribution) | |||||||
---|---|---|---|---|---|---|---|

Travel Time Std. Deviation |
Penalty Ratio | ||||||

120 | 180 | 240 | 300 | 360 | 420 | 480 | |

3 | 37.3 | 37.9 | 38.3 | 38.6 | 38.8 | 39.0 | 39.2 |

6 | 42.6 | 44.0 | 45.0 | 45.7 | 46.3 | 46.8 | 47.2 |

9 | 46.4 | 49.0 | 50.7 | 52.0 | 53.0 | 53.8 | 54.6 |

12 | 48.9 | 52.7 | 55.3 | 57.3 | 58.8 | 60.1 | 61.2 |

15 | 50.1 | 55.4 | 58.9 | 61.6 | 63.7 | 65.5 | 67.0 |

18 | 50.1 | 56.9 | 61.5 | 65.0 | 67.7 | 70.1 | 72.0 |

21 | 49.0 | 57.4 | 63.1 | 67.4 | 70.8 | 73.7 | 76.1 |

24 | 47.1 | 57.0 | 63.7 | 68.8 | 72.9 | 76.4 | 79.4 |

Another aspect is whether a commuter worries about the probability of being late (i.e., incurring a fixed penalty) or whether the the commuter worries about how many minutes arrival is delayed. If \(v\) represents the per-minute cost of being late, then the commuter minimizes \[ C= w\cdot T^\ast+v\cdot\left(\mu+\sigma\frac{\phi(T^\ast)}{1-\Phi(T^\ast)}-T^\ast\right)\] where the expression in parentheses is the expected value of the minutes arriving late. The first-order condition for minimizng \(C\) can then be solved for \(T^\ast\). The key difference is that there is less urgency to arrive on time unless the penalty \(v\) is very high. For this economic problem, the distributional assumptions matter more because a "fatter" tail of the distribution will increase the cost of being late and will thus encourage earlier departure.