Two-stage stochastic lot-sizing problem under cost uncertainty

Abstract

For production planning problems, cost parameters can be uncertain due to marketing activities and interest rate fluctuation. In this paper, we consider a single-item two-stage stochastic lot-sizing problem under cost parameter uncertainty. Assuming cost parameters will increase or decrease after time period p each with certain probability, we minimize the total expected cost for a finite horizon problem. We develop an extended linear programming formulation in a higher dimensional space that can provide integral solutions by showing that its constraint matrix is totally unimodular. We also project this extended formulation to a lower dimensional space and obtain a corresponding extended formulation in the lower dimensional space. Final computational experiments demonstrate that the extended formulation is more efficient and performs more stable than the two-stage stochastic mixed-integer programming formulation.

Appendix:  Proof of Proposition 2

To show that the constraint matrix described by constraints (4), (5), (10)–(14), and (16) is totally unimodular, we order variables f _i , g _i , and z _i with loop i ranging from 1 to $|\mathcal{V}|$. Variable $m^{i}_{t}$ is ordered with an outer loop i ranging from 1 to $|\mathcal{V}|$, and an inner loop t ranging from t(i)+1 to T. Variable $n^{i}_{t}$ is ordered with an outer loop i ranging from 1 to $|\mathcal{V}|$, and an inner loop t ranging from 1 to t(i). Table 8 shows the constraint matrix corresponding to Fig. 2.

Table 8 The matrix of constraints (4), (5), (10) to (14), and (16) for the example in Fig. 2

As the submatrix corresponding to variable $n^{i}_{t}$ is an identity matrix for $i\in\mathcal{V}$ and t≤t(i), we do not need to consider $n^{i}_{t}$ in our construction. As the submatrix corresponding to variable $m^{i}_{t}$ is an identity matrix for t(i)≥t(p)+2, $i\in\mathcal{V}$, and t≥t(i)+1, we only need to consider $m^{i}_{t}$, 1≤t(i)≤t(p)+1, t≥t(i)+1 in our construction. In the following, we only study the constraint submatrix for the variables we need to consider, denoted as matrix A.

We show that for any column subset J of matrix A, there exist partitions J ₁ and J ₂ of J such that

$$ \biggl|\sum_{j\in J_1} a_{ij} - \sum_{j\in J_2} a_{ij}\biggr| \leq1 $$

(27)

for all i. We partition variables f, g, z, m in J starting from branching node p and then, extend it in both directions to nodes after p and before p.

First, we define M(i) as the closest ancestor of node i such that z _M(i)∈J and m(i) as the closest descendant of node i such that z _m(i)∈J.

In the following Steps 1 to 5, we allocate the decision variables m and z to J ₁ and J ₂:

Step 1.:

Allocate $m^{p}_{t}$ to J ₁, z _p to J ₁, and $z_{q_{w}}$ to J ₂, where t≥t(p)+1, $q_{w} \in\mathcal{C}(p)$.

Step 2.:

Allocate $m^{q_{w}}_{t}$, t≥t(p)+2 to the same set as $z_{q_{w}}$ (if $z_{q_{w}} \in J$), or to the same set as $m^{p}_{t}$ (if $z_{q_{w}} \notin J$ and $m^{p}_{t}\in J$), or to J ₁ (if $z_{q_{w}}\notin J$ and $m^{p}_{t}\notin J$).

Step 3.:

Allocate z _i , t(i)≥t(p)+2, to the opposite set of z _M(i), if M(i) exists and t(M(i))≥t(p). Otherwise, allocate z _i to the opposite set of $m^{p}_{t}$ if $m^{p}_{t}\in J$. If $m^{p}_{t} \notin J$, then allocate z _i to J ₂.

Step 4.:

Allocate z _i , 1≤i≤p−1, to the opposite set of z _m(i), if m(i) exists and t(m(i))≤t(p). Otherwise, allocate z _i to J ₁.

Step 5.:

Allocate $m^{i}_{t}$, 1≤i≤p−1, to the opposite set of z _m(i), if m(i) exists and t(m(i))≤t(p). Otherwise, allocate $m^{i}_{t}$ to J ₁.

In the following Steps 6 and 7, we allocate the decision variables f and g to J ₁ and J ₂:

Step 6.:

Allocate f _i to the same set of z _M(i) if M(i) exits; allocate f _i to the opposite set of z _i if M(i) does not exist and z _i ∈J; allocate f _i to the opposite set of z _m(i) if m(i) exists, M(i) does not exist, and z _i ∉J; allocate f _i to J ₁ if m(i) and M(i) do not exist and z _i ∉J.

Step 7.:

Allocate g _i to the same set of z _m(i) if m(i) exists; allocate g _i to the opposite set of z _i if m(i) does not exist and z _i ∈J; allocate g _i to the opposite set of z _M(i) if m(i) does not exist, M(i) exists, and z _i ∉J; allocate g _i to J ₂ if m(i) and M(i) do not exist and z _i ∉J.

Following the above partition steps, we observe the following two properties:

Claim 1

If z _i ∈J and M(i) exists, z _i goes to the opposite set of z _M(i) for all $i\in{\mathcal{V}}\setminus\{1\}$.

Proof of Claim 1

If 1≤t(M(i))≤t(i)≤t(p), because the closest descendant of M(i) is i and t(i)≤t(p), m(M(i))=i, z _M(i) goes to the opposite set of z _i based on Step 4.

If 1≤t(M(i))<t(p)<t(i)≤T, z _M(i) goes to J ₁ based on Step 4 and z _i goes to J ₂ based on Step 3 (if t(i)≥t(p)+2) or based on Step 1 (if t(i)=t(p)+1).

If t(p)=t(M(i))<t(i)≤T, z _M(i) goes to J ₁ based on Step 1 (i.e., M(i)=p), z _i goes to J ₂ based on Step 1 (if i=q _w ) or Step 3 (if t(i)≥t(p)+2). Thus, z _i goes to the opposite set of z _M(i).

If t(p)+1≤t(M(i))<t(i)≤T, z _i goes to the opposite set of z _M(i) based on Step 3.

Therefore, Claim 1 holds. □

Claim 2

If $j(w_{k_{1}})$ and $j(w_{k_{2}})$ are the first second-stage nodes corresponding to scenarios $w_{k_{1}}$ and $w_{k_{2}}$ such that $z_{j(w_{k_{1}})}, z_{j(w_{k_{2}})} \in J$, then $z_{j(w_{k_{1}})}$ and $z_{j(w_{k_{2}})}$ go to the same set.

Proof of Claim 2

For any first second-stage node j(w), if j(w)=q _w , then z _j(w) goes to J ₂ based on Step 1, otherwise j(w) goes to J ₂ based on Step 3 due to the fact that j(w) is the first second-stage node in J and on the branch corresponding to w and t(p)≤t(j(w))≤T. Thus, the claim holds. □

Now, we verify that (27) holds for constraints (4), (5), (10)–(14), and (16) under the above partition. First, corresponding to each row, if J contains at most one decision variable in A, then it is obvious that (27) holds. In the following, we consider the case in which J contains at least two decision variables in each row of A.

• 1. For constraint (4), we discuss the following four cases:

  • 1.1 {f _i ,g _i }⊆J and z _i ∉J

    • 1.1.1 If both M(i) and m(i) exist, f _i goes to the same set of z _M(i) based on Step 6; g _i goes to the same set of z _m(i) based on Step 7. Because M(m(i))=M(i) (due to z _i ∉J), z _m(i) goes to the opposite set of z _M(i) based on Claim 1. Thus, f _i goes to the opposite set of g _i . Then, (27) holds.

    • 1.1.2 If M(i) exists and m(i) does not exist, f _i goes to the same set of z _M(i) based on Step 6; g _i goes to the opposite set of z _M(i) based on Step 7. Thus, f _i goes to the opposite set of g _i .

    • 1.1.3 If m(i) exists and M(i) does not exist, g _i goes to the same set of z _m(i) based on Step 7; f _i goes to the opposite set of z _m(i) based on Step 6. Thus, f _i goes to the opposite set of g _i .

    • 1.1.4 If neither m(i) nor M(i) exists, f _i and g _i go to J ₁ and J ₂ respectively based on Steps 6 and 7.

  • 1.2 {g _i ,z _i }⊆J and f _i ∉J

    • 1.2.1 If m(i) exists, then g _i goes to the same set of z _m(i) based on Step 7. Because M(m(i))=i, z _m(i) goes to the opposite set of z _i based on Claim 1. Thus, g _i goes to the opposite set of z _i . Then, (27) holds.

    • 1.2.2 If m(i) does not exist, g _i goes to the opposite set of z _i based on Step 7. Thus, (27) holds.

  • 1.3 {f _i ,z _i }∈J and g _i ∉J

    • 1.3.1 If M(i) exists, f _i goes to the same set of z _M(i) based on Step 6; z _i goes to the opposite set of z _M(i) based on Claim 1. Then, f _i goes to the opposite set of z _i . Thus, (27) holds.

    • 1.3.2 If M(i) does not exist, f _i goes to the opposite set of z _i based on Step 6. Thus, (27) holds.

  • 1.4 {f _i ,g _i ,z _i }∈J. This conclusion directly follows 1.3, because f _i goes to the opposite set of z _i . Then, (27) holds no matter where g _i goes.

• 2. For constraint (5), we discuss the following four cases:

  • 2.1 If {g _i ,z _i }∈J and g _a(i)∉J this condition is the same as 1.2. Thus, (27) holds.

  • 2.2 If {g _a(i),z _i }∈J and g _i ∉J, g _a(i) goes to the same set as z _i due to z _i =z _m(a(i)) based on Step 7. Thus, (27) holds.

  • 2.3 If {g _i ,g _a(i)}∈J and z _i ∉J, we first have m(i)=m(a(i)). If (a) m(i)=m(a(i)) exists, then g _i and g _a(i) go to the same set as z _m(i) based on Step 7. If (b) m(i)=m(a(i)) does not exist and z _a(i)∈J, then g _a(i) goes to the opposite set of z _a(i) based on Step 7. Also, because z _a(i)∈J, we have z _M(i)=z _a(i). Then, g _i goes to the opposite set of z _a(i) based on Step 7. Thus, g _i and g _a(i) go to the same set. If (c) m(i)=m(a(i)) does not exist, z _a(i)∉J and M(a(i)) exists, then z _M(i)=z _M(a(i)). Because z _i ∉J, g _i and g _a(i) go to the opposite set of z _M(i)=z _M(a(i)) based on Step 7. Thus, g _i and g _a(i) go to the same set. If (d) m(i)=m(a(i)), M(a(i)) does not exist, z _a(i)∉J, then both g _i and g _a(i) go to J ₂ based on Step 7.

  • 2.4 If {g _i ,g _a(i),z _i }∈J, this conclusion directly follows 2.2, because g _a(i) and z _i go to the same set. Then, (27) holds no matter where g _i goes.

• 3. For constraint (10), we consider two cases:

  • 3.1 t(i)≥t(p)+2. For this case, we do not need to consider $m^{i}_{t}$ based on the identity matrix argument at the beginning of the proof. Then, based on Step 6, f _ξ(i,t,w) goes to the same set of z _M(ξ(i,t,w)), where z _M(ξ(i,t,w))∈J and the corresponding M(ξ(i,t,w)) is the largest-index such node in path ${\mathcal{P}}^{i}_{t}(w)$ with ${\mathcal{P}}^{i}_{t}(w)$ representing the set of nodes on the path from node i to a node at time period t and on the branch corresponding to scenario w. Note that there must exist at least one such M(ξ(i,t,w)) based on the assumption that we have at least two elements in J for each constraint. Based on Step 4, z _j alternatively goes to J ₁ and J ₂ based on Step 4, where t(p)+2≤t(j)≤t. Thus, (27) holds.

  • 3.2 t(i)=t(p)+1. For this case, i=q _w , for some 1≤w≤W. We discuss the following two cases:

    • 3.2.1 If $m^{q_{w}}_{t}\notin J$, this argument is the same as 3.1. Thus, (27) holds.

    • 3.2.2 If $m^{q_{w}}_{t}\in J$, under this condition, we discuss $f_{\xi(q_{w},t,w)}\notin J$ and $f_{\xi(q_{w},t,w)}\in J$ respectively.

      • 3.2.2.1 $f_{\xi(q_{w},t,w)}\notin J$. Under this case, if $z_{q_{w}}\in J$, $m^{q_{w}}_{t}$ and $z_{m(q_{w})}$ go to the same set and the opposite set of $z_{q_{w}}$ based on Steps 2 and 3 respectively; if $z_{q_{w}}\notin J$, $m^{q_{w}}_{t}$ goes to the same set as $m^{p}_{t}$ (i.e., J ₁) if $m^{p}_{t}\in J$ or J ₁ if $m^{p}_{t}\notin J$ based on Step 2 and similarly $z_{m(q_{w})}$ goes to J ₂ based on Step 3. Thus, $m^{q_{w}}_{t}$ and $z_{m(q_{w})}$ go to the opposite sets. Besides these, z _j ∈J alternatively goes to J ₁ and J ₂ where t(p)+2≤t(j)≤t based on Step 3. Thus, (27) holds.

      • 3.2.2.2 $f_{\xi(q_{w},t,w)}\in J$. Under this case, we discuss two cases depending on if there exists a node $j\in{\mathcal{P}}^{i}_{t}(w)\setminus\{i, \xi(i, t, w)\}$ such that z _j ∈J.

        3.2.2.2.1  If no such node j exists, then $\{f_{\xi(q_{w},t,w)}, m^{q_{w}}_{t}\}\in J$, based on our assumption that at least two elements in each constraint in matrix A. If $z_{q_{w}}\in J$, M(ξ(q _w ,t,w))=q _w due to z _j ∉J for each $j \in{\mathcal{P}}^{i}_{t}(w)\setminus\{i, \xi(i, t, w)\}$. Based on Steps 6 and 2, $f_{\xi(q_{w},t,w)}$ and $m^{q_{w}}_{t}$ go to the same set of $z_{q_{w}}$. If $z_{q_{w}}\notin J$, based on Step 2, $m^{q_{w}}_{t}$ goes to J ₁. In the following, we prove that $f_{\xi (q_{w}, t,w)}$ goes to J ₁ in this case. Based on Step 6, $f_{\xi(q_{w}, t, w)}$ goes to (a) the same set of $z_{M(\xi(q_{w}, t, w))}$ if M(ξ(q _w ,t,w)) exists, (b) the opposite set of $z_{\xi(q_{w}, t, w)}$ if M(ξ(q _w ,t,w)) does not exist and $z_{\xi(q_{w}, t,w)}\in J$, (c) the opposite set of $z_{m(\xi(q_{w}, t,w))}$ if m(ξ(q _w ,t,w)) exists, M(ξ(q _w ,t,w)) does not exist, and $z_{\xi(q_{w}, t,w)} \notin J$, or (d) J ₁ if M(ξ(q _w ,t,w)) and m(ξ(q _w ,t,w)) do not exist, and $z_{\xi (q_{w}, t,w)} \notin J$. For (a), M(ξ(q _w ,t,w)) exists and $z_{q_{w}} \notin J$. Based on Step 4, $z_{M(\xi(q_{w}, t, w))}$ goes to J ₁, because 1≤t(M(ξ(q _w ,t,w)))≤t(p). For (b), $z_{\xi (q_{w}, t, w)}\in J$, $z_{q_{w}}\notin J$, and M(ξ(q _w ,t,w)) does not exist. Then, $z_{\xi(q_{w}, t, w)}$ goes to J ₂ based on Step 3 and accordingly $f_{\xi(q_{w}, t, w)}$ goes to J ₁. For (c), $z_{m(\xi(q_{w},t,w))}\in J$, and $z_{\xi(q_{w}, t, w)}, z_{M(\xi(q_{w}, t, w))}, z_{q_{w}}\notin J$. Then, $z_{m(\xi(q_{w},t,w))}$ goes to J ₂ based on Step 3. Thus, accordingly $f_{\xi(q_{w}, t, w)}$ goes to J ₁. Then, $f_{\xi(q_{w}, t, w)}$ and $m^{q_{w}}_{t}$ go to the same set.

        3.2.2.2.2  If there exists such a node j, then $f_{\xi(q_{w},t,w)}$ goes to the same set of $z_{M(\xi(q_{w}, t, w))}$ based on Step 6. If $z_{q_{w}}\in J$, $m^{q_{w}}_{t}$ goes to the same set of $z_{q_{w}}$ based on Step 2, and $z_{m(q_{w})}$ goes to the opposite set of $z_{q_{w}}$ based on Step 3. If $z_{q_{w}}\notin J$, $m^{q_{w}}_{t}$ goes to J ₁ based on Step 2 and $z_{m(q_{w})}$ goes to J ₂ based on Step 3. Thus, for both $z_{q_{w}}\in J$ and $z_{q_{w}}\notin J$, $m^{q_{w}}_{t}$ goes to the opposite set of $z_{m(q_{w})}$. Besides these, z _j alternatively goes to J ₁ and J ₂, $j\in {\mathcal{P}}^{i}_{t}(w)\setminus\{i,\xi(i,t,w)\}$. Thus, (27) holds.

• 4. For constraint (11), we only need to consider the case in which $\{m^{i}_{t}, f_{i+1}\}\in J$.

  • 4.1 If M(i+1) exists, f _i+1 goes to the same set as z _M(i+1) based on Step 6. If m(i) exists and t(m(i))≤t(p), $m^{i}_{t}$ goes to the opposite set of z _m(i) based on Step 5. Because m(i)=m(M(i+1)), z _M(i+1) and z _m(i) go to the opposite set based on Step 4. Thus, f _i+1 and $m^{i}_{t}$ go to the same set. If m(i) does not exist, or m(i) exists and t(m(i))>t(p), then m(M(i+1)),t(m(M(i+1)))≤t(p), does not exist. Thus, z _M(i+1) and $m^{i}_{t}$ go to J ₁ based on Steps 4 and 5. Thus, $m^{i}_{t}$ and f _i+1 go to the same set. Therefore, (27) holds.

  • 4.2 If M(i+1) does not exist and z _i+1∈J, then m(i)=i+1. Thus, $m^{i}_{t}$ and f _i+1 go to the opposite set of z _i+1 based on Steps 5 and 6. Thus, (27) holds.

  • 4.3 If M(i+1) does not exist, z _i+1∉J, and m(i+1) exists, then, m(i)=m(i+1). If 1≤t(m(i+1))≤t(p), $m^{i}_{t}$ and f _i+1 go to the opposite set of z _m(i+1) based on Steps 5 and 6; otherwise, if t(m(i+1))>t(p), $m^{i}_{t}$ goes to J ₁ and f _i+1 goes to the opposite set of z _m(i+1). Since z _m(i+1) goes to J ₂ based on Step 1 (if t(m(i+1))=t(p)+1) or Step 3 (under this case, M(m(i+1)) does not exist), f _i+1 and $m^{i}_{t}$ go to the same set. Then, (27) holds.

  • 4.4 If neither M(i+1) nor m(i+1) exists and z _i+1∉J, then $m^{i}_{t}$ and f _i+1 go to J ₁ based on Steps 5 and 6. Then, (27) holds.

• 5. For constraint (12), we discuss the following four cases:

  • 5.1 $\{m^{i}_{t},z_{i+1} \}\subseteq J$, and $m^{i+1}_{t}\notin J$. Based on Step 5, $m^{i}_{t}$ goes to the opposite set of z _i+1 since z _i+1 is the closest descendant of node i. Thus, (27) holds.

  • 5.2 $\{m^{i+1}_{t},z_{i+1} \}\subseteq J$, and $m^{i}_{t} \notin J$. Under this case, $m^{i+1}_{t}$ and z _i+1 will go to the same set, because both z _i+1 and $m^{i+1}_{t}$ are in the opposite set of z _m(i+1), if m(i+1) exists and t(m(i+1))≤t(p) or in J ₁ otherwise, based on Steps 4 and 5.

  • 5.3 $\{m^{i}_{t},m^{i+1}_{t} \}\subseteq J$, and z _i+1∉J. Because z _i+1∉J, $m^{i}_{t}$ and $m^{i+1}_{t}$ are in the same set based on Step 5.

  • 5.4 $\{m^{i}_{t},m^{i+1}_{t},z_{i+1} \}\subseteq J$. The conclusion follows from Step 5. Variable $m^{i}_{t}$ goes to the opposite set of z _i+1, since z _i+1 is the closest descendant of node i. Then, (27) holds no matter where $m^{i+1}_{t}$ goes.

• 6. For constraint (13), we only need to consider the case in which $\{f_{q_{w}}, m^{p}_{t}\} \subseteq J$. Based on Step 1, $m^{p}_{t}$ goes to J ₁. In the following, we show that $f_{q_{w}}$ goes to J ₁.

  • 6.1 If M(q _w ) exists, then $z_{M(q_{w})}$ goes to J ₁ based on Step 1 if M(q _w )=p, or Step 4 if t(m(M(q _w )))≤t(p)−1. Thus, based on Step 6, $f_{q_{w}}$ goes to the same set of $z_{M(q_{w})}$, which is J ₁.

  • 6.2 If $z_{q_{w}}\in J$, M(q _w ) does not exist, then $z_{q_{w}}$ goes to J ₂ based on Step 1. Based on Step 6, $f_{q_{w}}$ goes to the opposite set of $z_{q_{w}}$, which is J ₁.

  • 6.3 If m(q _w ) exists, $z_{q_{w}} \notin J$ and M(q _w ) does not exist, then t(m(q _w ))≥t(p)+2 and $z_{m(q_{w})}$ goes to the opposite set of $m^{p}_{t}$ based on Step 3. Based on Step 6, we have $f_{q_{w}}$ goes to the opposite set of $z_{m(q_{w})}$. Thus, $f_{q_{w}}$ and $m^{p}_{t}$ go to the same set, which is J ₁.

  • 6.4 If m(q _w ) and M(q _w ) do not exist and $z_{q_{w}}\notin J$, then $f_{q_{w}}$ goes to J ₁ based on Step 6.

• 7. For constraint (14), we need to consider the following four cases:

  • 7.1 $\{m^{p}_{t},z_{q_{w}} \}\subseteq J$ and $m^{q_{w}}_{t}\notin J$. Based on Step 1, $m^{p}_{t}$ and $z_{q_{w}}$ go to J ₁ and J ₂, respectively. Thus, (27) holds.

  • 7.2 $\{m^{q_{w}}_{t}, z_{q_{w}} \}\subseteq J$ and $m^{p}_{t}\notin J$. $m^{q_{w}}_{t}$ and $z_{q_{w}}$ go to the same set based on Step 2.

  • 7.3 $\{m^{p}_{t},m^{q_{w}}_{t} \}\subseteq J$ and $z_{q_{w}}\notin J$. Under this case, $m^{p}_{t}$ and $m^{q_{w}}_{t}$ go to the same set based on Step 2.

  • 7.4 $\{m^{p}_{t},m^{q_{w}}_{t}, z_{q_{w}}\}\subseteq J$. Under this case, $m^{p}_{t}$ and $z_{q_{w}}$ go to J ₁ and J ₂, respectively based on Step 1. Then, (27) holds no matter where $m^{q_{w}}_{t}$ goes.

• 8. For constraint (16), we do not need to consider $n^{i}_{t}$ based on the identity matrix argument at the beginning of the proof. Based on Step 7, g _η(i,t) goes to the same set of z _m(η(i,t)). Note here, there must exist m(η(i,t)) based on the assumption that we have at least two elements in J. If t(m(η(i,t)))≤t(p) or t(η(i,t))≥t(p)+1, then η(i,t) and m(η(i,t)) are one-to-one corresponding. Otherwise, all z _m(η(i,t)) in different scenarios go to the same set based on Claim 2 because t(m(η(i,t)))≥t(p)+1. Besides these, z _j alternatively goes to J ₁ and J ₂ based on Claim 1. Thus, (27) holds.

Therefore, the desired property (27) holds for constraints (4), (5), (10)–(14), and (16), and the corresponding constraint matrix is totally unimodular.