Time & Work – Advanced University‑Level Study Notes
(MIT‑style: first principles → rigorous reasoning → edge cases → challenging problem)
1. Introduction
The time‑and‑work (T&W) problem is a classic exercise in linear algebra and optimization.
It asks: If several workers (or machines) work together on a job, how long will it take?
While elementary textbooks treat it with simple “rate = work/time” arithmetic, the full theory reveals:
| Why it matters |
|---|
| • Modeling heterogeneous productivity (e.g., machines, workers). |
| • Foundations of rate‐based scheduling. |
| • Connection to linear programming and queueing theory. |
We will:
- Derive the basic formula rigorously from infinitesimal increments.
- Generalize to arbitrary productivity profiles (time‑dependent, stochastic).
- Explore edge cases (zero productivity, infinite time, cyclic work).
- Solve a challenging problem that tests all concepts.
2. First Principles
2.1 Work as a Continuous Quantity
Let a job be a continuous resource (W \in \mathbb{R}_{\ge 0}) to be completed (e.g., miles of road to pave, data to process).
Define instantaneous productivity of worker (i) at time (t) as
[ r_i(t) \equiv \frac{\text{d}w_i}{\text{d}t}, ]
where (w_i(t)) is the cumulative work done by worker (i) by time (t).
Assumptions (first‑order model):
| Assumption | Description |
|---|---|
| A1 | Work is additive: (w_{\text{total}}(t) = \sum_i w_i(t)). |
| A2 | Each worker’s productivity is independent of others. |
| A3 | Productivity functions are integrable over the interval. |
Under these, the total work done by all workers up to time (T) is
[ W(T) = \int_0^T \sum_{i=1}^n r_i(t), \mathrm{d}t. ]
The job is completed when (W(T)=W_{\text{job}}).
2.2 Constant Productivity (Classical T&W)
If each worker’s productivity is constant, (r_i(t)=r_i), then
[ W(T) = \biggl(\sum_{i=1}^n r_i\biggr) T. ]
Setting (W_{\text{job}} = W) yields the classical formula
[ T = \frac{W}{\sum_{i=1}^n r_i}. \tag{1} ]
Equation (1) is the time‑to‑completion rule.
It can be rewritten as
[ \frac{1}{T} = \sum_{i=1}^n \frac{r_i}{W} = \sum_{i=1}^n \frac{1}{T_i}, ] where (T_i = W/r_i) is the time worker (i) would take alone.
Remark
The rule is linear in the reciprocals, reflecting the harmonic sum of individual rates.
3. Rigorous Derivation
3.1 Differential Equation Approach
Let (W(t)) denote remaining work at time (t) (so (W(0)=W_{\text{job}})).
The rate of change is
[ -\frac{\mathrm{d}W}{\mathrm{d}t} = \sum_{i=1}^n r_i(t). \tag{2} ]
Integrating from (0) to (T):
[ W(0) - W(T) = \int_0^T \sum_{i=1}^n r_i(t), \mathrm{d}t. ]
Since (W(T)=0) at completion,
[ W_{\text{job}} = \int_0^T \sum_{i=1}^n r_i(t), \mathrm{d}t. \tag{3} ]
Equation (3) is the most general form.
If (r_i(t)=r_i) constant, (3) reduces to (1).
3.2 Linear Algebra Interpretation
Let (\mathbf{r} = (r_1,\dots,r_n)^{!\top}).
If all workers work simultaneously for a common interval ([0,T]) with constant rates, the total work vector is
[ \mathbf{W} = T \mathbf{r}. ]
The scalar job size is (W_{\text{job}} = \mathbf{1}^{!\top}\mathbf{W}) where (\mathbf{1}) is the all‑ones vector. Hence
[ W_{\text{job}} = T,\mathbf{1}^{!\top}\mathbf{r} ;;\Longrightarrow;; T = \frac{W_{\text{job}}}{\mathbf{1}^{!\top}\mathbf{r}}. ]
This is simply (1) expressed in vector notation.
4. Generalizations
| Scenario | Model | Key Equation | Remarks |
|---|---|---|---|
| Time‑dependent rates | (r_i(t)) | (W_{\text{job}} = \int_0^T \sum_i r_i(t),dt) | Example: Shift changes, fatigue |
| Stochastic productivity | (r_i(t)\sim \mathcal{F}) | Expected completion time (E[T]) from stochastic integration | Requires probability distributions |
| Cyclic work (on/off) | (r_i(t)=r_i,\mathbf{1}_{\text{on}}(t)) | Piecewise integration | Models maintenance cycles |
| Variable job size | (W(t)) changes | Differential equation (-dW/dt = \sum r_i(t)) | E.g., “incremental” tasks |
4.1 Work Rate Functions
A typical time‑dependent productivity might be:
[ r_i(t) = r_i^{\max} \bigl( 1 -