Abstract
We study a periodic-review inventory model. In each period, the buyer can source from N suppliers that have random capacities. The optimal ordering policy for this problem has been shown to be a reorder point policy in a setting that includes price-dependent demand (Feng and Shi in Prod Oper Manag 21(3):547–563, 2012). We show that if one of the suppliers is reliable, then the optimal order quantity from each unreliable and cheaper supplier stays constant when inventory level is less than the reorder point of the reliable supplier. This implies that having one reliable supplier would result in a relatively stable ordering policy, despite the unreliability of the rest of the suppliers. For the special case of sourcing from two suppliers, we first prove that the optimal policy is monotone in the inventory level. Then, we provide insights into order allocation between an expensive but more reliable supplier and a cheaper but less reliable one. Specifically, the optimal policy tends to allocate more units to the more reliable one when inventory level is low. For the special case of all-or-nothing supply, there is a threshold such that if the inventory is below the threshold, then the expensive supplier’s order quantity is larger than that of the cheaper. Finally, we show that the marginal cost of holding inventory increases, while the optimal order quantities decrease over the planning horizon.
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Notes
Without loss of generality, we assume that if there are multiple minimizers, we always pick the optimal solution with the smallest total order quantity \(q_{1,t}^*(I),\ldots , q_{N,t}^*(I)\); further if there are multiple minimizers with the same smallest order quantity, we always prefer ordering more from the supplier with the smaller index.
In the view of the proof, we have that \(J_t\) has lower bound which guarantees the existence of the minimizer.
\(J_t(I, \mathbf {q})\) may not be jointly convex in \((I, \mathbf {q})\), which is different with \(\pi _t(I, \mathbf {q})\).
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Appendix: Proofs
Appendix: Proofs
Lemma 1
For every convex function f, a random variable K with distribution \(F_{K}\) and \({\tilde{x}}=\arg \min _{x\ge 0} f(x)\), it should be true that
Proof of Lemma 1
Considering every realization of \(\check{k}\), we have two cases.
If \(\check{k}<{\tilde{x}}\), then we have
The inequality comes from \(\check{k}<{\tilde{x}}\).
Notice that f is a convex function and \(\lambda (x_1\wedge \check{k}) + {\bar{\lambda }}(x_2\wedge \check{k})\le \check{k}<{\tilde{x}}\) where \({\tilde{x}}=\arg \min _{x\ge 0} f(x)\) . Hence, \(\check{k}\) is closer to the minimum point than \(\lambda (x_1\wedge \check{k}) + {\bar{\lambda }}(x_2\wedge \check{k})\) and that both are on the left side of a convex function. Thus,
If \(\check{k}\ge {\tilde{x}}\), then we notice that \({\tilde{x}}=\arg \min _{x\ge 0} f(x)\). Hence,
From the discussion above two scenarios, we have
for every \(x_1, x_2\ge 0\), every random variable K with distribution \(F_{K}\) and every convex function f. \(\square \)
Lemma 2
The following properties hold:
-
(i)
\(V_t(I)\) is convex in I and \(V_t(I)\ge 0\) for every I.Footnote 2
-
(ii)
\(\pi _t(I, \mathbf {q})\) is jointly convex in \((I, \mathbf {q})\).
-
(iii)
\(L_t(I)\) is convex in I.
-
(iv)
\(J_t(I, \mathbf {q})\) is convex in \(I, q_1,\ldots ,\) or \(q_N\) in each dimension.Footnote 3
Proof of Lemma 2
We use mathematical induction to proof. we have \(V_{T+1}(I)=c_1 \max \{-I,0\} + c_e \max \{I,0\}\ge 0\), which is convex in I. We assume \(V_{t+1}(I)\) is convex in I and \(V_{t+1}(I)\ge 0\). We first show that \(V_t(I)\ge 0\). Notice that
where \(\sum _{i=1}^N c_i\mathbf{E}_{K_i}(q_i\wedge K_i)\ge 0\), \(H(x)\ge 0\) and \(V_{t+1}(x)\ge 0\) for every x. Hence, \(V_t(I)\ge 0\).
Notice that
where H(.) and \(V_{t+1}(.)\) are convex and D follows a continuous distribution. Hence \(L_t(x)\) is convex in x. We also notice that
From the convexity of \(L_t(.)\) and linear combination of \(I + \sum _{i=1}^N q_i\), we have \(L_t(I + \sum _{i=1}^N q_i)\) is jointly convex in \((I,\mathbf {q})\). Clearly, \(\sum _{i=1}^N c_i q_i\) is jointly convex in \((I,\mathbf {q})\). Therefore, \(\pi _t(I,\mathbf {q})\) is jointly convex in \((I,\mathbf {q})\).
Notice that \(J_t(I,\mathbf {q})=\mathbf{E}_{K_i,1\le i\le N}[\pi _t(I,q_1\wedge K_1,\ldots ,q_N\wedge K_N)]\). we have \(J_t(I, \mathbf {q})\) is convex in \(I, q_1,\ldots ,\) or \(q_N\) in each dimension.
For every \(I^a\), \(I^b\) and \(\lambda \in [0,1]\), denote \({\bar{\lambda }}=1-\lambda \) and \({\hat{I}}=\lambda I^a+{\bar{\lambda }} I^b\). We have
From the definitions above and Lemma 1, we have
The explanation of the first inequality is as follows. For every given \(({\hat{I}}, q_{2,t}^*(I^a),q_{2,t}^*(I^b),q_{3,t}^*(I^a),\) \(q_{3,t}^*(I^b),\ldots ,q_{N,t}^*(I^a),q_{N,t}^*(I^b))\), the expression \(\mathbf{E}_{K_j, \forall j\ne 1}[\pi _t({\hat{I}}, x, \,\,\lambda (q_{2,t}^*(I^a)\wedge K_2) + {\bar{\lambda }}(q_{2,t}^*(I^b)\wedge K_2), \, \ldots , \,\lambda (q_{N,t}^*(I^a)\wedge K_N) + {\bar{\lambda }}(q_{N,t}^*(I^b)\wedge K_N))]\) can been looked as a one-dim convex function about x. Hence, from Lemma 1, we have the first inequality. That is to say, the first inequality holds for every given \(({\hat{I}}, q_{2,t}^*(I^a),q_{2,t}^*(I^b), q_{3,t}^*(I^a),\) \(q_{3,t}^*(I^b),\ldots ,q_{N,t}^*(I^a),q_{N,t}^*(I^b))\).
The explanation of the second inequality is as follows. For every given \(({\hat{I}}, q_1,q_{3,t}^*(I^a),q_{3,t}^*(I^b),\ldots ,\) \(q_{N,t}^*(I^a),q_{N,t}^*(I^b))\), the expression \(\mathbf{E}_{K_j, \forall j\ne 2} [\pi _t({\hat{I}}, q_1\wedge K_1, x, \,\lambda (q_{3,t}^*(I^a)\wedge K_3) + {\bar{\lambda }}(q_{3,t}^*(I^b)\wedge K_3), \,\ldots , \, \lambda (q_{N,t}^*(I^a)\wedge K_N) + {\bar{\lambda }}(q_{N,t}^*(I^b)\wedge K_N))]\) can been looked as a one-dim convex function about x. Hence, from Lemma 1, we have the second inequality. That is to say, the second inequality holds for every given \(({\hat{I}}, q_1,q_{3,t}^*(I^a),q_{3,t}^*(I^b),\ldots ,\) \(q_{N,t}^*(I^a),q_{N,t}^*(I^b))\).
The explanation of the third inequality is as follows. For every given \(({\hat{I}}, q_1, q_2, q_{4,t}^*(I^a), q_{4,t}^*(I^b),\ldots ,\) \(q_{N,t}^*(I^a),q_{N,t}^*(I^b))\), the expression \(\mathbf{E}_{K_j, \forall j\ne 3}[\pi _t({\hat{I}}, q_1\wedge K_1, q_2\wedge K_2, x, \, \lambda (q_{4,t}^*(I^a)\wedge K_4) + {\bar{\lambda }}(q_{4,t}^*(I^b)\wedge K_4), \, \ldots , \, \lambda (q_{N,t}^*(I^a)\wedge K_N) + {\bar{\lambda }}(q_{N,t}^*(I^b)\wedge K_N))]\) can been looked as a one-dim convex function about x. Hence, from Lemma 1, we have the third inequality. That is to say, the third inequality holds for every given \(({\hat{I}}, q_1, q_2, q_{4,t}^*(I^a),q_{4,t}^*(I^b),\ldots ,\) \(q_{N,t}^*(I^a),q_{N,t}^*(I^b))\).
The explanation of the sixth inequality is as follows. For every given \(({\hat{I}}, q_1, q_2,\ldots , q_i, q_{i+2,t}^*(I^a),q_{i+2,t}^*(I^b),\ldots ,\) \(q_{N,t}^*(I^a),q_{N,t}^*(I^b))\), the expression \(\mathbf{E}_{K_j, \forall j\ne i+1} [\pi _t({\hat{I}}, q_1\wedge K_1,q_2\wedge K_2, \ldots , q_i\wedge K_i, x, \,\lambda (q_{i+2,t}^*(I^a)\wedge K_{i+2}) + {\bar{\lambda }}(q_{i+1,t}^*(I^b)\wedge K_{i+1}), \ldots , \, \lambda (q_{N,t}^*(I^a)\wedge K_N) + {\bar{\lambda }}(q_{N,t}^*(I^b)\wedge K_N))]\) can been looked as a one-dim convex function about x. Hence, from Lemma 1, we have the sixth inequality. That is to say, the sixth inequality holds for every given \(({\hat{I}}, q_1, q_2,\ldots ,q_i, q_{i+2,t}^*(I^a),q_{i+2,t}^*(I^b),\ldots ,\) \(q_{N,t}^*(I^a),q_{N,t}^*(I^b))\).
The explanation of the ninth inequality is as follows. For every given \(({\hat{I}}, q_1, q_2,\ldots ,q_{N-1})\), the expression \(\mathbf{E}_{K_j, \forall j\ne i+1} [\pi _t({\hat{I}}, q_1\wedge K_1,q_2\wedge K_2, \ldots , q_{N-1}, x)]\) can been looked as a one-dim convex function about x. Hence, from Lemma 1, we have the ninth inequality. That is to say, the ninth inequality holds for every given \(({\hat{I}}, q_1, q_2,\ldots ,q_{N-1})\).
The tenth inequality comes from sequential minimization is bigger than the global minimization.
Combining with (4), we have
Therefore \(V_t(I)\) is convex in I. By induction, \(V_t\) is convex \(\forall t\), concluding the proof. Parts (ii) and (iii) follow from the discussion above. \(\square \)
Theorem 7
The optimal ordering policy is a reorder point policy; i.e.,
and \(V_t^{\prime }(I_{i,t})= -c_i\); \(1 \le i \le N\).
Proof of Theorem 7
From the first order conditions (FOC) with respect to \(q_i\) for all i, we have
From convexity of \(L_t\), we have \(\phi _{i,t}(I,\mathbf {q})\) increasing in \(q_l\), for all \(1\le l\le N\).
In the following discussion, we will show that the optimal ordering policy for Supply i is a reorder-point policy.
Denote
We next will show that \(q_{i,t}^*(I)=0\) if and only if \({\tilde{q}}_{i,t}^i(I)=0\).
If \({\tilde{q}}_{i,t}^i=0\), then
The first inequality comes from the optimality of \(({\tilde{q}}_{1,t}^i(I), \ldots , {\tilde{q}}_{N,t}^i(I))\). The second inequality comes from Jensen’s inequality. Recall that we always pick the smallest optimal solution. Therefore, \(q_{i,t}^*(I)=0\).
If \(q_{i,t}^*(I)=0\), then we have
That is to say \((q_{1,t}^*(I),\ldots , q_{N,t}^*(I))\) is also the minimizer of \(\mathfrak {J}_{t}^i\). Hence, \({\tilde{q}}_{i,t}^i=0\).
From the discussion above, we have \(q_{i,t}^*(I)=0\) if and only if \({\tilde{q}}_{i,t}^i(I)=0\). Hence, to show that the optimal ordering policy for Supply i is a reorder-point policy, we only need to show that the optimal ordering quantity, \({\tilde{q}}_{i,t}^i(I)\), of Supply i in minimizing \(\mathfrak {J}_{t}^i\) is a reorder-point policy. Denote
From convexity of \(L_t\), we have convexity of f. We have
Notice that the right hand side is supermodular in \((I,q_i)\). That is to say, \({\tilde{q}}_{i,t}^i(I)\) is decreasing in I. Combining with the result that \(q_{i,t}^*(I)=0\) if and only if \({\tilde{q}}_{i,t}^i(I)=0\), we have the optimal ordering policy for Supply i is a reorder-point policy; i.e., \(q_{i,t}^*(I)>0\) for \(I<I_{i,t}\) and \(q_{i,t}^*(I)=0\) for \(I\ge I_{i,t}\).
Next we will show that \(V_t^{\prime }(I_{i,t})= -c_i\). From \(q_{i,t}^*(I)>0\) for \(I<I_{i,t}\) and \(q_{i,t}^*(I)=0\) for \(I\ge I_{i,t}\), we have
Hence,
concluding the proof. \(\square \)
Proof of Theorem 1
To see Part (i), from \(K_{j^{\prime }}\equiv +\infty \), then \({\bar{G}}_{j^{\prime }}(q_{j^{\prime }})=1\) and \(q_{j^{\prime }}=q_{j^{\prime }}\wedge K_{j^{\prime }}\) for every \(q_{j^{\prime }}\). Thus, for \(I\le I_{j^{\prime },t}\),
Hence,
We also have if \(q_i>0\), then for every \(i>j^{\prime }\),
The first inequality comes from the convexity of \(L_t\) and the fact that \(q_{i,t}^*(I)\ge q_{i,t}^*(I)\wedge K_j\). The second inequality comes from \(c_i>c_{j^{\prime }}\), \(\forall i>j^{\prime }\). Therefore, \(q_{i,t}^*(I)=0\) for every I and \(i>j^{\prime }\).
To see Part (ii), we have shown that \(V_t^{\prime }(I)= -c_{j^{\prime }}\) for all \(I\le I_{j^{\prime },t}\). We also have \(q_{i,t}^*(I)=q_{i,t}^*(I_{j^{\prime },t})=0\), for every \(i>j^{\prime }\). Next we focus on \(i\le j^{\prime }\). The FOC at \(I=I_{j^{\prime },t}\) can be written as
From the result in Part (i), the objective function can be rewritten as
where
From FOC, we have
Combining with (6) and (7), we have \(I\le I_{j^{\prime }}\)
That is to say, for \(I\le I_{j^{\prime }}\), \(q_{j^{\prime },t}^*=I_{j^{\prime },t}-I\) and \(q_{i,t}^*(I)=q_{i,t}^*(I_{j^{\prime },t})\) for every \(i<j^{\prime }\). Combining results with \(q_{i,t}^*(I)=0\) for \(i>j^{\prime }\), we conclude the proof. \(\square \)
Proposition 4
\(I_{1,t}>I_{2,t}>\ldots >I_{N,t}\) for \(c_i \le c_{i+1}\).
Proof of Proposition 4
From \(c_i\le c_{i+1}\) and Theorem 7, we have \(V_t^{\prime }(I_{i,t})>V_t^{\prime }(I_{i+1,t})\). Combining with the convexity of \(V_t\), we have \(I_{i,t}>I_{i+1,t}\) for all i. \(\square \)
Proof of Proposition 1
Assume there exist I and i satisfying that \(q_{i,t}^*({\tilde{I}})< q_{i+1,t}^*({\tilde{I}})\). Combining with Proposition 4, there must exist \({\tilde{I}}\) and \({\tilde{q}}\) satisfying that \(q_{i,t}^*({\tilde{I}})=q_{i+1,t}^*({\tilde{I}})={\tilde{q}}>0\).
From \(K_i \succeq _{st} K_{i+1}\), we have \(\big ({\tilde{q}} \wedge K_i\big )\succeq _{st} \big ({\tilde{q}} \wedge K_{i+1}\big )\). Combining it with the convexity of \(L_t\), we have
From \(q_{i,t}^*({\tilde{I}})=q_{i+1,t}^*({\tilde{I}})={\tilde{q}}>0\) and
Combining the above with (8) and \(c_i<c_{i+1}\), we have
We reach a contradiction. Hence, we conclude the proof. \(\square \)
Proof of Theorem 2
To simplify the following discussion, we define \(\mathbf {q}^s=(q,\ldots ,q)\). To see part (i)& (ii), we have that
Hence, we have
Thus, \(J_t(I,\mathbf {q}^s)\) is supermodular in (I, q). Therefore \(q_{x,t}^*(I)\) decreases in I.
To see parts (iii–v), we consider two different cases. Case (a) \(q_{x,t}^*(I)=0\) and Case (b) \(q_{x,t}^*(I)>0\).
Case (a), \(q_{x,t}^*(I)=0\). Combining with (11) and \(c_N>c_i\) for \(1\le i <N\),
Thus, \(\phi _{N,t}(I,0,\ldots , 0)\ge 0\). Then, \(q_{N,t}^*(I)=0\). Therefore \(q_{N,t}^*(I)=q_{x,t}^*(I)\le q_{1,t}^*(I)\).
Case (b), \(q_{x,t}^*(I)>0\). If \(I\ge I_{1,t}\), then we have \( c_i+L_t^{\prime }(I)>c_1+L_t^{\prime }(I)\ge 0, \,\, \forall i>1 \). Hence,
We reach a contradiction. Hence, \(I<I_{1,t}\) when \(q_{x,t}^*(I)>0\).
If \(\min \{q_{i,t}^*(I), 1\le i\le N\}>q_{x,t}^*(I)\), then from (5), we have
Combining with (11), we have
Therefore, \((q_{x,t}^*(I),\ldots ,q_{x,t}^*(I))\) is also the minimizer of \(J_t(I,q_1, \ldots , q_N)\). It contradicts with the assumption that we always pick the smallest as the optimal solution.
If \(\max \{q_{i,t}^*(I), 1\le i\le N\}<q_{x,t}^*(I)\), then from (5), we have
Combining with (11), we have
Let \({\tilde{q}}=\max \{q_{i,t}^*(I), 1\le i\le N\}\). From the convexity of \(L_t\), we have
Hence,
Notice that \({\tilde{q}}<q_{x,t}^*(I)\) which contradicts with the fact that \(q_{x,t}^*(I)\) is the smallest minimizer of \(J_t(I, q, \ldots , q)\). Combining the discussions on Case a) & b), we complete the proof for Parts (iii)-(iv). \(\square \)
Proof of Theorem 3
To see part (i),(ii)& (iii), we will show \(\frac{dq_{i,t}^*(I)}{dI}\in [-1,0]\), \(i=1,2\). First, the first order condition for \(q_i\) are expressed as follows. For \(i=1,2\) and \(j\ne i\)
where \(G_i(\cdot )\) is the distribution of \(K_i\) with \({\bar{G}}_i(x)=1-G_j(x)\), \(i=1,2\), and
We can obtain the convexity of \(L_t\) from the convexity of \(V_{t+1}\) and H. Hence, \(\phi _{i,t}(I, q_1, q_2)\) (\(i=1,2\)) increases in \(q_1\) and \(q_2\). From the envelop theorem, we have
To simplify the notation, denote
From the convexity of \(L_t\), we have \(x_i, \delta _i\ge 0\) for \(i=1,2\). From (14) and (15), we have
It is easy to see, \(\frac{dq_{1,t}^*(I)}{dI}\in [-1,0]\). By a similar argument, we have \(\frac{dq_{2,t}^*(I)}{dI}\in [-1,0]\) proving part (i). Part (ii) directly follows from (i) and part (iii) directly follows from \(\frac{dq_{i,t}^*(I)}{dI}\in [-1,0]\).
To see part (iv), we consider the following two cases: Case (a), \(I\le I_{2,t}\) and Case (b), \(I\in [I_{2,t}, I_{1,t}]\).
Case (a), we will show that \(1+ \frac{dq_{1,t}^*(I)}{dI}+\frac{dq_{2,t}^*(I)}{dI}\le I\) when \(I\le I_{2,t}\). From the above discussion, we have
That is to say, \(1+ \frac{dq_{1,t}^*(I)}{dI}+\frac{dq_{2,t}^*(I)}{dI}\le I\).
Case (b) \(I\in [I_{2,t}, I_{1,t}]\). We have
Hence, \(q_{1,t}^*(I)=I_{1,t}-I\) when \(I\in [I_{2,t}, I_{1,t}]\). Therefore, \(I+q_{1,t}^*(I)+q_{2,t}^*(I)=I_{1,t}\). We conclude the proof. \(\square \)
Proof of Theorem 4
Notice that when \(q_{x,t}^*(I)=0\), \(f_t(I,q_{x,t}^*(I))=0<c_2-c_1\). From Proposition 4 and Theorem 2, we have \(q_{2,t}^*=q_{x,t}^*(I)=0\le q_{1,t}^*(I)=0\).
Next, we focus on discussing that \(q_{x,t}^*(I)>0\). The next discussions are based on the following two equations.
To see part (i), if \(f_t(I,q_{x,t}^*(I))= c_2-c_1\), then from (16) and (17) we have \(\phi _{1,t}(I,q_{x,t}^*(I),q_{x,t}^*(I))=\phi _{2,t}(I,q_{x,t}^*(I),q_{x,t}^*(I))=0\). Hence, \(q_{2,t}^*(I)=q_{1,t}^*(I)=q_{x,t}^*(I)\).
To see part (ii), if \(f_t(I,q_{x,t}^*(I))>c_2-c_1\), then from (16) and (17) we have \(\phi _{1,t}(I,q_{x,t}^*(I),q_{x,t}^*(I))>0\) and \(\phi _{2,t}(I,q_{x,t}^*(I),q_{x,t}^*(I))<0\). Combining with Proposition 4 & Theorem 2 and \(q_{x,t}^*(I)>0\), we have
Then
If \(q_{1,t}^*(I)=q_{2,t}^*(I)\), then we have \(q_{1,t}^*(I)=q_{2,t}^*(I)=q_{x,t}^*(I)\). It contradicts with (18) and (19).
If \(q_{2,t}^*(I)<q_{1,t}^*(I)\), then let \(\delta _1=q_{x,t}^*(I)-q_{2,t}^*(I)\) and \(\delta _2=q_{1,t}^*(I)-q_{x,t}^*(I)\). Further we have \(\delta _1\ge 0\) and \(\delta _2\ge 0\) from Theorem 2.
If \(\delta _1\ge \delta _2\), then we have
The first inequality comes from \((q_{x,t}^*(I)-\delta _2)\wedge {\hat{K}}_2\ge q_{x,t}^*(I)\wedge {\hat{K}}_2-\delta _2\) for every realization of \({\hat{K}}_2\). The second inequality comes from the convexity of \(L_t\) and \(\delta _1\ge \delta _2\). It contradicts with (18).
If \(\delta _1\le \delta _2\), then we have
The first inequality comes from \((q_{x,t}^*(I)+\delta _1)\wedge {\hat{K}}_1\le q_{x,t}^*(I)\wedge {\hat{K}}_1 + \delta _1\) for every realization of \({\hat{K}}_1\). The second inequality comes from the convexity of \(L_t\) and \(\delta _1\le \delta _2\). It contradicts with (19).
Combining the above discussion, we have that \(q_{2,t}^*(I)>q_{1,t}^*(I)\) if \(f_t(I,q_{x,t}^*(I))> c_2-c_1\).
To see part (iii), if \(f_t(I,q_{x,t}^*(I))<c_2-c_1\), then from (16) and (17), we have \(\phi _{1,t}(I,q_{x,t}^*(I),q_{x,t}^*(I))<0\) and \(\phi _{2,t}(I,q_{x,t}^*(I),q_{x,t}^*(I))>0\). We assume \(q_{2,t}^*(I)>q_{1,t}^*(I)\) and then reach a contradiction.
From the assumption \(q_{2,t}^*(I)>q_{1,t}^*(I)\) and Proposition 4, we have \(I<I_{2,t}\) and \(q_{2,t}^*(I)>q_{1,t}^*(I)>0\). Hence,
Then
let \(\delta _3=q_{x,t}^*(I)-q_{1,t}^*(I)\) and \(\delta _4=q_{2,t}^*(I)-q_{x,t}^*(I)\). Then we have \(\delta _4\ge 0\) and \(\delta _5\ge 0\).
If \(\delta _4\le \delta _3\), then we have
The first inequality comes from \((q_{x,t}^*(I)+\delta _4)\wedge {\hat{K}}_2\le q_{x,t}^*(I)\wedge {\hat{K}}_2 + \delta _4\) for every realization of \({\hat{K}}_2\). The second inequality comes from the convexity of \(L_t\) and \(\delta _4\le \delta _3\). It contradicts with (20).
If \(\delta _4\ge \delta _3\), then we have
The first inequality comes from \((q_{x,t}^*(I)-\delta _3)\wedge {\hat{K}}_1\ge q_{x,t}^*(I)\wedge {\hat{K}}_1-\delta _3\) for every realization of \({\hat{K}}_1\). The second inequality comes from convexity of \(L_t\) and \(\delta _4\ge \delta _3\). It contradicts with (21).
Combining the above discussion, we have that \(q_{2,t}^*(I)\le q_{1,t}^*(I)\) if \(f_t(I,q_{x,t}^*(I))<c_2-c_1\). \(\square \)
Proof of Theorem 5
If \(\theta _1\ge \theta _2\), then \(K_1\succeq _{st} K_2\). From Proposition 1, we have \(q_{1,t}^*(I)\ge q_{2,t}^*(I)\) for all I. That is to say \(I_{e,t}=+\infty \).
Next we will focus on discussing the scenario with \(\theta _1<\theta _2\).
To see (i), when both suppliers are all-or-nothing type, the objective function can be written as
where \(x=I+q\) and \(y=I+2q\). From linear combinations and the convexity of \(L_t\), we have J(I, q, q) is supermodular in \((I,I+q)\) and submodualar in \((I,I+2q)\). Hence, when both suppliers are all-or-nothing, \(I + q_{x,t}^*(I)\) increases in I and \(I + 2 q_{x,t}^*(I)\) decreases in I.
We also notice that in the all-or-nothing case
From facts that \(I + q_{x,t}^*(I)\) increases in I and \(I + 2 q_{x,t}^*(I)\) decreases in I, we have \(f_t(I,q_{x,t}^*(I))\) decreases in I. Combining with Theorem 4, we conclude the proof.
To see (ii), firstly we always pick the minimizers such that \(q_{j,t}^*\le {\bar{k}}\) for \(j=1,2,x\). From Theorem 2 and Theorem 3, let \({\hat{I}} =\max \{I: q_{1,t}^*(I)={\bar{k}} \hbox { or } q_{2,t}^*(I)={\bar{k}} \}\).
When \(I\ge {\hat{I}}\), we still have
Following a similar argument to that in Part (i), we have \(f_t(I,q_{x,t}^*(I))\) decreasing in I when \(I\ge {\hat{I}}\). Combining with Theorem 4, we either have \(q_{2,t}^*(I)\le q_{1,t}^*(I)\) for all \(I\le {\hat{I}}\) or find an \(I_{e,t} \in (I_{2,t}, {\hat{I}}_t)\) satisfying that
Next we consider when \(I\ge {\hat{k}}\). If it is under the first scenario, then combining the above argument with Theorem 2, we have \(q_{2,t}^*(I)\le {\bar{k}}=q_{1,t}^*(I)\) for all \(I\ge {\hat{I}}\). If it is under the second scenario, then combining the above argument with Theorem 2, we have \(q_{2,t}^*(I)={\bar{k}}\ge q_{1,t}^*(I)\) for all \(I\ge {\hat{I}}\). Taken together, we conclude the proof. \(\square \)
Proof of Proposition 2
To see Part (i), we assume \(q_{i,t}^{*}(I)>q_{i,t}^{\delta *}(I)\) and then reach a contradiction. From \(q_{i,t}^{*}(I)>q_{i,t}^{\delta *}(I)\), we have \({\bar{K}}_i>q_{i,t}^{\delta *}(I)\) and
Hence,
Combining with the convexity of \(L_t\), we have \(q_{j,t}^{*}(I)< q_{j,t}^{\delta *}(I)\). Therefore,
Hence,
From \(q_{i,t}^{*}(I)>q_{i,t}^{\delta *}(I)\) and \(q_{j,t}^{*}(I)<q_{j,t}^{\delta *}(I)\), we have for all the realizations of \({\hat{K}}_j\) and \({\hat{K}}_i\)
It contradicts with (26). Hence, we have \(q_{i,t}^{*}(I)\le q_{i,t}^{\delta *}(I)\).
To see Part (ii), we assume \(q_{j,t}^{*}(I)<q_{j,t}^{\delta *}(I)\) and reach a contradiction. We first have
Together with \(q_{j,t}^{*}(I)<q_{j,t}^{\delta *}(I)\), we have \(q_{i,t}^*(I)\ge q_{i,t}^{\delta *}(I)\). Together with the result in Part (i), we have \(q_{i,t}^*(I)= q_{i,t}^{\delta *}(I)\). Combining with \(q_{j,t}^{*}<q_{j,t}^{\delta *}\) and (28), we have
If \(q_{i,t}^*(I)= q_{i,t}^{\delta *}(I)=0\), then combining it with (29), we have
Combining with (31), we have \((q_{1,t}^*(I),q_{2,t}^*(I))\) is not only the optimal solution of \(j_t\) but also the optimal solution of \(J_t^{\delta }\). It contradicts with \(q_{j,t}^{*}<q_{j,t}^{\delta *}\).
If \(q_{i,t}^*(I)= q_{i,t}^{\delta *}(I)>0\), then \(q_{i,t}^*(I) + {\hat{K}}_j=q_{i,t}^{\delta *}(I) + {\hat{K}}_j\) for every realization of \({\hat{K}}_j\). Combining with (30), we have
Therefore
It contradicts with \(q_{i,t}^{\delta *}(I)>0\). Therefore, we have \(q_{j,t}^{*}\ge q_{j,t}^{\delta *}\).
To see Part (iii), we assume \(q_{1,t}^*(I)+q_{2,t}^*(I)>q_{1,t}^{\delta *}(I)+q_{2,t}^{\delta *}(I)\) and reach a contradiction in the following discussion. From the result of Part (ii), we have \(q_{j,t}^*(I)\ge q_{j,t}^{\delta *}(I)\). If \(q_{j,t}^*(I)=q_{j,t}^{\delta *}(I)\), then we obtain that \(q_{i,t}^*(I)>q_{i,t}^{\delta *}(I)\) from \(q_{1,t}^*(I)+q_{2,t}^*(I)>q_{1,t}^{\delta *}(I)+q_{2,t}^{\delta *}(I)\). It contradicts with the result of Part (i) \(q_{i,t}^*(I)\le q_{i,t}^{\delta *}(I)\).
Next, we discuss \(q_{j,t}^*(I)>q_{j,t}^{\delta *}(I)\). Hence we have \(q_{j,t}^*(I)+{\hat{K}}_i>q_{j,t}^{\delta *}(I)+{\hat{K}}_i\) for every realization \({\hat{K}}_i\). Combining with \(q_{1,t}^*(I)+q_{2,t}^*(I)>q_{1,t}^{\delta *}(I)+q_{2,t}^{\delta *}(I)\), we have for every realization \({\hat{K}}_i\),
From \(q_{j,t}^*(I)>q_{j,t}^{\delta *}(I)\), we also have
Combining with the discussion of (32), we have
In the following discussion, we consider two different cases.
Case (a), if \(c_i+\mathbf{E}_{K_j} L_t^{\prime }(I+q_{j,t}^*(I)\wedge K_j+q_{i,t}^*(I))>0\) then there exists
satisfying that
Together with (33) and (34) and the convexity of \(L_t\),
Further, we have
Therefore, \((q_{i,t}^*(I),\, q_{j,t}^*(I)-\rho _1)\) is also the optimal solution of \(J_t\). It contradicts with \((q_{i,t}^*(I),q_{j,t}^*(I))\) being the the smallest minimizer.
Case (b), if \(c_i+\mathbf{E}_{K_j} L_t^{\prime }(I+q_{j,t}^*(I)\wedge K_j+q_{i,t}^*(I))=0\), then there exist
and \(\rho _2>\rho _3>0\) satisfying that
Hence, we have
From \(q_{i,t}^*(I)\le q_{i,t}^{\delta *}(I)\) and the definition of \(\rho _2\) and \(\rho _3\), we have
Together with (33) and (34) and the convexity of \(L_t\),
Further, we have
Therefore, \((q_{i,t}^*(I)+\rho _3,\, q_{j,t}^*(I)-\rho _2)\) is also the optimal solution of \(J_t\) and \(\rho _3-\rho _2<0\). It contradicts with the assumption that \((q_{i,t}^*(I),q_{j,t}^*(I))\) is the minimizer with the smallest total order quantity. \(\square \)
Proof of Proposition 3
We only need to show that the statement holds for a small enough \(\delta \). From \(p^{\delta }>p\),
Case (i), if \(q_{i,t}^{*}(I)=0\) for \(i=1,2\), then
Hence, \(q_{i,t}^{\delta *}(I)=0\) for \(i=1,2\).
Case (ii), if \(q_{i,t}^*(I)=0\) and \(q_{j,t}^*(I)>0\), then
From \(q_{j,t}^*(I)>0\), then from \(\delta \) is a small enough number, there exists \(\rho _1\in [0,q_{i,t}^*(I))\) satisfying that
we also have that
The first inequality comes from (35). The second inequality comes from (36). Therefore \((0,q_{j,t}^*(I)-\rho _1)\) is the optimal solution for \(J_t^{\delta }\).
Case (iii), if \(q_{i,t}^*(I)>0\) for \(i=1,2\), then since \(\delta \) is a small enough number, there exist \(\rho _2\in [0,q_{1,t}^*(I))\) and \(\rho _3\in [0,q_{2,t}^*(I))\) satisfying that
Therefore, \((q_{1,t}^*(I)-\rho _2, q_{2,t}^*(I)-\rho _3)\) is the optimal solution for \(J_t^{\delta }\). We conclude the proof. \(\square \)
Proof of Theorem 6
We will use mathematical induction to obtain the result.
First, we will show that for \(V_{T}^{\prime }(I)\le V_{T+1}^{\prime }(I)\) almost everywhere and \(q_{1,T}^*(0)>0\). Recall that \(V_{T+1}(I)=c_1 I^{-} + c_e I^+\) where \(c_e >h/(1-\alpha )\).
Hence, \(I_{1,t}>0\) and \(q_{1,T}^*(0)>0\).
Therefore, when \(I\le I_{1,t}\), combining it with Theorem 7, we have
The inequality comes from that \(V_{T+1}(I)=c_1 I^{-} + c_e I^+\). Hence, above inequality holds almost everywhere. When \(I\ge I_{1,t}\), we have \(I>0\) and
The first inequality comes from the expression of H and \(V_{T+1}\). The second inequality comes from \(c_e> h/(1-\alpha )\). From the above discussion, we have \(V_{T}^{\prime }(I)\le V_{T+1}^{\prime }(I)\) all most everywhere and \(q_{1,T}^*(0)>0\).
Next we will show that if \(V_{t+1}^{\prime }(I)\le V_{t+2}^{\prime }(I)\) almost everywhere and \(q_{1,t+1}^*(0)>0\), then \(V_{t}^{\prime }(I)\le V_{t+1}^{\prime }(I)\), \(q_{1,t}^*(0)>0\) and \(q_{i,t}^*(I)\ge q_{t+1}^*(I)\).
From \(V_{t+1}^{\prime }(I)\le V_{t+2}^{\prime }(I)\) almost everywhere, we have
Comparing the first order conditions at time t and \(t+1\), we have \(i=1,2\) and \(j\ne i\),
From (37) and following a similar argument as Proposition , we can compare the two equation systems and obtain \(q_{i,t}^*(I)\ge q_{i,t+1}^*(I)\) for \(i=1,2\). Thus \(q_{1,t}^*(0)\ge q_{1,t+1}^*(0)>0\). Next we will show that \(V_{t}^{\prime }(I)\le V_{t+1}^{\prime }(I)\). From \(q_{i,t}^*(I)\ge q_{i,t+1}^*(I)\) for \(i=1,2\), we have
From (37), we also have
Thus,
Therefore,
We conclude the proof by the mathematical induction. \(\square \)
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Chen, W., Tan, B. Dynamic procurement from multiple suppliers with random capacities. Ann Oper Res 317, 509–536 (2022). https://doi.org/10.1007/s10479-016-2285-2
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DOI: https://doi.org/10.1007/s10479-016-2285-2