Single-Level Uncapacitated Lotsizing Problem

This problem is also known as the Wagner-Whitin problem.


a single item is considered
demands are dynamic (period-specific, days, weeks, etc.)
the capacity of the resource is neglected

There are several mathematical formulations for this problem. The most common model reads as follows:

$\mathrm{Minimize Z}=\displaystyle{\sum_{t=1}^T \big( {h\cdot y_t}+{ s\cdot \gamma _t} \big)}

subject to:

$y_{t-1}+q_t-y_t=d_t \qquad {t=1,2,\ldots,T}$

$q_t-M\cdot \gamma _t\le 0 \qquad {t=1,2,\ldots,T}$

$q_t\geq 0 \qquad {t=1,2,\ldots,T}$

$y_t\geq 0 \qquad {t=1,2,\ldots,T}$

$\gamma _t\in \left\{ {0,\,1} \right\} \qquad {t=1,2,\ldots,T}$


$d_t$ demand in period $t$
$h$ holding costs
$M$ big number (M must be greater than the maximum lot size)
$s$ setup costs
$T$ length of the planning horizon
$q_t$ lot size in period $t$
$y_t$ inventory at the end of period $t$
$\gamma_t$ binäry setup variable

Every optimum solution to this model satifies the so-called ZIO (zero-inventory-ordering) condition. This means that in any period $t$ either the beginning inventory $y_{t-1}$ or the lot size $q_t$ are zero, i.e. $y_{t-1}\cdot q_t=0$. This property allows to reformulate the above model as a shortest-route model.

Example: Consider a product with the following demands: $d_1=20$, $d_2=80$, $d_3=160$, $d_4=85$, $d_5=120$ and $d_6=100$. The setup costs are $s=500$ and the holding costs are $h=1$.

The corresponding shortest-route network is:


For each period there is a node. An arrow starting at node $i$ and ending at node $j$ means: the lot produced in period $i$ covers the demand from period $i$ to period $j-1$. The numbers associated to the arrows are the setup and holding costs that result from the corresponding lot size.

Using any appropriate shortest-route algorithm (e.g. Dijkstra'a algorithm), we get the following solution:

SR Solution

Going backwards from node "Dummy" to node "1" we find, that the optimum predecessor of "Dummy" is "3" (production quantity in period 3: $q_3=465$). Next, the optimum predecessor of "3" is "1" (production quantity in period 1: $q_1=100$). The total costs are 1705.

In industrial practice, heuristic solution procedures are used, such as

  • Silver-Meal heuristic
  • Least unit-cost heuristic
  • Part-period heuristic
  • Groff heuristic

and many others.

» See also: MLCLSP - Multi-level capacitated dynamic lotsizing problem