Relaxations for two-level multi-item lot-sizing problems

Abstract

We consider several variants of the two-level lot-sizing problem with one item at the upper level facing dependent demand, and multiple items or clients at the lower level, facing independent demands. We first show that under a natural cost assumption, it is sufficient to optimize over a stock-dominant relaxation. We further study the polyhedral structure of a strong relaxation of this problem involving only initial inventory variables and setup variables. We consider several variants: uncapacitated at both levels with or without start-up costs, uncapacitated at the upper level and constant capacity at the lower level, constant capacity at both levels. We finally demonstrate how the strong formulations described improve our ability to solve instances with up to several dozens of periods and a few hundred products.

Appendix: Proof of Theorem 3

Similar to Theorem 1, the proof is in three steps. In the first step, we show that the polyhedron $SCC$ defined by (49)–(55) is integral for $m=1$. Then we extend the result to $m>1$ and finally prove that adding constraints (48) does not destroy integrality.

Let $SSC_I$ denote the set of integral solutions in $SSC$ and consider the case $m=1$. As we did in the proof of Theorem 2, for a given non-zero objective function $\min \sum _{u=1}^n g^0_u \zeta _u +\sum _{u=1}^n g^1_u \delta _u + \sum _{i=0}^1 \sum _{u=0}^n q^i_u y^i_u+ \sum _{i=0}^1 \sum _{u=1}^n \bar{q}^i_u z^i_u$ with bounded optimal value, we determine one inequality among (49)–(55) that is satisfied at equality by all optimal solutions.

We use the following observation. Let $\alpha _1 \in [0,n+1]$, $\alpha _2\in [\alpha _1, n+1]$, $\beta _2 \in [\alpha _1, n+1]$, $\beta _1 \in [0,\beta _2]$ with $\beta _1=n+1$ if $\beta _2=n+1$, $\gamma _2 \in [1,\alpha _1-1] \cup \{\beta _2\}$, $\gamma _1 \in [0,\gamma _2]$ if $\gamma _2\le \alpha _1-1$ and $\gamma _1=\beta _1$ if $\gamma _2=\beta _2$. The $y$ and $z$ vectors in the extreme points of $conv(SSC_I)$ are of the following form: $y^0_u=1$ for $u\in [\alpha _1, \alpha _2]$, $z^0_{\alpha _1}=1$, $y^1_u=1$ for $u\in [\gamma _1, \gamma _2]\cup [\beta _1,\beta _2]$, $z^1_{\gamma _1}=z^1_{\beta _1}=1$, the other entries of $y$ and $z$ vectors are zero. In the sequel, we use the values $\alpha _1, \alpha _2,\gamma _1,\gamma _2,\beta _1,\beta _2$ to represent the corresponding extreme points.

• a. Let $\bar{q}^0_0=\bar{q}^1_0=0$. We need $g^0, g^1\ge 0$, $\bar{q}^0_t + \sum _{u=t}^{t+k} q^0_u \ge 0$ and $\bar{q}^1_t + \sum _{u=t}^{t+k} q^1_u \ge 0$ for all $t\in [0,n]$, $k\in [0, n-t]$ for the problem to be bounded.

• b. For $i=0,1$, if there exists $u\in [1,n]$ with $q^i_u<0$, then $y^i_u=z^i_u + y^i_{u-1}$.

• c. For $i=0,1$, if there exists $u\in [1,n]$ with $\bar{q}^i_u<0$, then $z^i_u=y^i_u$.

• d. For $i=0,1$, if $q^i_0<\bar{q}^i_1$, then $z^i_1=0$. If $q^i_0 > \bar{q}^i_1$, then $y^i_0=0$. So $q^i_0=\bar{q}^i_1$. In the remaining, we study the case where $g^0, g^1,q^0, q^1, \bar{q}^0, \bar{q}^1\ge 0$.

• e. If $g^0=q^0=\bar{q}^0=0$, then the problem is single-level and the result is known to hold [21].

• f. Suppose that $g^0=0$. If $q^0_0>0$, then $y^0_0=0$. If there exists $u\in [1,n]$ with $q^0_u+\bar{q}^0_u>0$, then $z^0_u=0$. In the remaining, we assume that there exists $l$ such that $g^0_l>0$. Let $l$ be the highest such index.

• g. If $q^0_u+\bar{q}^0_u < q^0_{u+1}+\bar{q}^0_{u+1}$ for some $u\in [1,n-1]$, then $z^0_{u+1}=0$. Let $t\in [1,n]$ be the largest index with $q^0_t+\bar{q}^0_t > 0$. If no such $t$ exists, then let $t=0$. If $t>l$, then $z^0_t=0$. So we assume that $t\le l$.

• h. If there exist $k \in [1,l]$, $m_1\in [0,k]$ and $m_2\in [k, n]$ such that $\bar{q}^0_k+q^0_k+\bar{q}^1_{m_1} +\sum _{u=m_1}^{m_2} q^1_u < g^0_l$, then $\zeta _l=0$. Therefore, as $\bar{q}^0_k+q^0_k=0$ for $k>t$, we assume that $\bar{q}^1_{m_1} +\sum _{u=m_1}^{m_2} q^1_u \ge g^0_l>0$ for all $m_2\in [t+1,n]$ and $m_1\in [0,m_2]$.

• i. If $q^1_{t+1}>0$, we show that with $t,l$ chosen in this way, the inequality (49) if $t=0$, (50) if $t\in [1,l-1]$ or (51) if $t=l$ is satisfied at equality by all optimal solutions. Note that the cost assumptions imply that all rays with non-zero contribution in this inequality have positive cost. Let $( \zeta , \delta , y^0,y^1,z^0,z^1)$ be an extreme point optimal solution. Suppose that the inequality (49), (50), or (51) corresponding to the above choice of $t$ and $l$ is not tight.

  • (a) Case $\zeta _l=1$. If $y^1_{t+1}+\sum _{u=t+2}^{l}z^1_u \ge 1$, then $t+1\le \beta _2$ and $\sum _{u=1}^t y^0_u=0$. In this case, setting $\alpha _1, \alpha _2 \leftarrow t+1$ and $\zeta _l=0$ decreases the cost by $g^0_l>0$. If $y^1_{t+1}+\sum _{u=t+2}^{l}z^1_u =0$, then $y^0_{1}+\sum _{u=2}^{t}z^0_u =1$ and $\beta _1>l$. Now, setting $\alpha _1, \alpha _2 \leftarrow t+1$ yields a better solution since $q^0_u+\bar{q}^0_u>0$ for all $u\in [1,t]$ and $q^0_0=\bar{q}^0_1$.

  • (b) Case $\zeta _l=0$. If $y^0_{1}+\sum _{u=2}^{t}z^0_u =0$, then $y^1_{t+1} +\sum _{u=t+2}^l z^1_u \ge 2$. Hence $t+1\le \gamma _2$ and $\beta _1\ge t+2$. Now setting $\alpha _1,\alpha _2 \leftarrow \gamma _2$, $\beta _1 \leftarrow \gamma _1$ and $\beta _2 \leftarrow \gamma _2$ decreases the cost by $\bar{q}_{\beta _1}+\sum _{u=\beta _1}^{\beta _2}q^1_u$, which is positive since $\beta _1\in [t+2,l]$. If $y^0_{1}+\sum _{u=2}^{t}z^0_u =1$, then $y^1_{t+1} +\sum _{u=t+2}^l z^1_u \ge 1$. If $\beta _1\ge t+1$, then setting $\alpha _1, \alpha _2 \leftarrow t+1$ gives a better solution. If $\beta _1\le t$, then $y^1_{t+1}=1$ and $\sum _{u=t+1}^{\beta _2} q^1_u >0$, so it is better to set $\beta _2 \leftarrow t$.

• j. If $g^0_t>0$ (and $q^1_{t+1}=0$, but this is not necessary here), then inequality $\zeta _t\ge 1 - y^0_1-\sum _{u=2}^t z^0_u$ (i.e., of type (51)) is satisfied at equality. Indeed, if not, then $\zeta _t=1$, $y^0_1+\sum _{u=2}^t z^0_u =1$, and $\beta _1\ge t+1$. Then setting $\alpha _1, \alpha _2 \leftarrow t+1$ gives a better solution.

• k. If $q^1_{t+1}=0$ and $g^0_t=0$, then $z^0_t=0$, or equivalently $\alpha _1 \ne t$, in any optimal solution. Indeed, if $\alpha _1=t$ and $\beta _2>t$, then setting $\alpha _1, \alpha _2 \leftarrow t+1$ gives a better solution. If $\alpha _1=t$ and $\beta _2=t$, then setting $y^1_{t+1}=1$ at zero cost (and therefore $\beta _2=t+1$) and $\alpha _1, \alpha _2 \leftarrow t+1$ gives a better solution.

Thus we can conclude that all optimal solutions lie on a face defined by one of the inequalities (49)–(55). This proves that $SSC$ is integral when $m=1$.

To prove that the result is true for $m>1$, we need a variant of Proposition 3. First, we observe that in an extreme point of $conv(SSC_I)$, we have $y^0_0+\sum _{j=1}^n z^0_j \le 1$ and that given any $( \zeta , \delta , y^0,y^1,z^0,z^1)\in SSC$, the solution $( \zeta , \delta , \bar{y}^0,y^1,\bar{z}^0,z^1)$ is also in $SSC$ where $\bar{y}^0_0 =\min \{y^0_0, 1\}$, $\bar{y}^0_1=\min \{z^0_1+y^0_0,1\}$, $\bar{z}^0_j =\min \{(1-y^0_0-z^0_{1j-1})^+, z^0_j\}$ for $j=1, \ldots ,n$, and $\bar{y}^0_j=\bar{z}^0_j$ for $j=2, \ldots ,n$. Now, we can use similar arguments to those of Proposition 3 to obtain the result.

Finally we need to show that adding constraints (48) does not destroy integrality. As in the proof of Theorem 1, the key argument is that when such an inequality is tight, constraint (52) is dominated.