Managing Production-Inventory Systems with Scarce Resources

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1 Managing Producion-Invenory Sysems wih Scarce Resources Online Supplemen Proof of Lemma 1: Consider he following dynamic program: where ḡ (x, z) = max { cy + E f (y, z, D)}, (7) x y min(x+u,z) f (y, z, d) = max { L(y l) + py + αḡ 1 (y l, z l)}, (8) l min(d,z) and ḡ (x, z) =, for z x, where L(x) = ( p + αc)x h(x) + b( x) +. Le Ḡ(y, z) = cy + E f (y, z, D) and F (y, z, l) = L(y l) + py + αḡ 1 (y l, z l). I can be shown by inducion ha xḡ(x, z) + z ḡ(x, z) p c. Noe ha l F p αc + h α( x (y, z, l) = v 1(y l, z l) + z v 1(y l, z l)), if l y, p αc b α( x v 1(y l, z l) + z v 1(y l, z l)), if l > y. Clearly, when b p αc, l F (y, z, l) if l y; and l F (y, z, l) if l > y. This implies ha f (y, z, d) = F (min(y, d), z, d) for y z and ḡ (x, z) = max { cy + E F (min(y, D), z, D)}, x y min(x+u,z) = max x y min(x+u,z) {(p c)y + (p αc + h)e(y D)+ + αeḡ 1 ((y D) +, z min(y, d))}. Compare he above equaion wih Equaion (2), we have v (x, z) = ḡ (x, z) for all x, z and =,..., T, and V (x, z) = Ḡ(y, z) for all y, z and =,..., T, when b p αc. Therefore, we have x v (x, z) + z v (x, z) p c. To show ha V (y, z) saisfies properies (a), (b) and (c), we only need o show ha V (y, z) is L -convex, or equivalenly, Ḡ(y, z) is L -convex. In he following, we prove by inducion ha ḡ (x, z) and Ḡ(y, z) are boh L -convex for = 1,..., T. We have ḡ (x, z) = for z x. Clearly, ḡ (x, z) is L -convex. Suppose ḡ 1 (x, z) is L -convex. I can be easily verified ha L(x) is concave. By Lemma 1 of Zipkin (28), 1

2 ḡ 1 (y l, z l) is L -convex in (y, z, l) and L(y l) is L -convex in (y, l). Then, F (y, z, l) = L(y l) + py + αḡ 1 (y l, z l) is L -convex. By propery (b) of Lemma 2 of Huh and Janakiraman (21), we see ha f (y, z, d) = min { F (y, z, l)} is L -convex in (y, z) l min(d,z) for fixed d. Thus, Ḡ(y, z) = cy E f (y, z, D) is L -convex. Again, by propery (b) of Lemma 2 of Huh and Janakiraman (21), ḡ (x, z) = min x y min(x+u,z) { Ḡ(y, z)} is L -convex. This complees he inducion and he proof. Proof of Theorem 2: Le ȳ (z) = argmax V (y, z) and y (x, q ) = ȳ (x + q ). Since V (y, z) y z is concave in y, he opimal producion hreshold is given by y (x, q ) = argmax V (y, x + q ) = x y min(x +u,x +q ) Clearly, y (x, q ) depends only on he sum x + q. x, if x y (x, q ), y (x, q ), if x y (x, q ) x + u, x + u, if x + u y (x, q ). Since V (y, z) is L -convex and ȳ (z) = argmin{ V (y, z)}, by propery (c) of Lemma 2 of y z Huh and Janakiraman (21), we have z ȳ(z) 1. This proves properies (a) and (b). Proof of Proposiion 3: I can be shown by inducion ha z v (x, z, u) p c. Noe ha v (x, q, u) = v (x, x + q, u) + cx. Thus, q v (x, q, u) = z v (x, x + q, u) [, p c] and 2 q 2 v (x, q, u) = 2 z 2 v (x, x + q, u). This proves properies (a) and (b). (c) If we can show v (x, z, u) = v (x, z x, u) cx is supermodular in (z, u), hen we have 2 qu v (x, q, u) = 2 zu v (x, z, u) z=x+q. In he following, we prove by inducion ha v (x, z, u) is supermodular in (z, u), π (z, u) = v (z, z, u) is supermodular in (z, u), and x v (x, z, u) + z v (x, z, u) is nondecreasing in u. Clearly, his is rue for = since v (x, z, u) =. Now, suppose v 1 (x, z, u) is supermodular in (z, u), π 1 (z, u) is supermodular in (z, u), and x v 1(x, z, u) + z v 1(x, z, u) is nondecreasing in u. Then, i is easy o verify ha V (y, z, u) is supermodular in (z, u) and Π (z, u) = V (z, z, u) is supermodular in (z, u). Moreover, y V (y, z, u)+ z V (y, z, u) = L (y)+α y [ x v 1(y ξ, z ξ, u)+ z v 1(y ξ, z ξ, u)]φ(ξ)dξ, which is nonincreasing in u. Equivalenly, u V (ȳ + δ, z, u + δ) u V (ȳ, z, u) for δ. Similar o he proofs of Lemma 1 and Theorem 2, we can show ha he opimal producion hreshold y (x, q, u) is nonincreasing in u and u y (x, z, u) 1. Thus, for fixed x and z, here exis u 1 2

3 and u 2, wih u 1 u 2, such ha x + u, if u [, u 1 ], y (x, z, u) = ȳ (z, u), if u [u 1, u 2 ], x, if u [u 2, ). When u [, u 1 ], v (x, z, u) = V (x, z, u), which is supermodular in (z, u). When u [u 2, ), v (x, z, u) = V (x + u, z, u), which is supermodular in (z, u) since V (y, z, u) is supermodular in (z, u) and (y, z). When u [u 1, u 2 ], if ȳ (z, u) = z, hen v (x, z, u) = V (z, z, u), which is supermodular in (z, u). This also implies ha π (z, u) = v (z, z, u) = V (z, z, u) is supermodular in (z, u). If ȳ (z, u) < z, we have v (x, z, u) = V (y (z, u), z, u) and u v (x, z, u) = V u (ȳ (z, u), z, u). Thus, for δ, u v (x, z + δ, u) = u V (ȳ (z + δ, u), z + δ, u) u V (ȳ (z, u) + δ, z + δ, u) u V (ȳ, z, u) = u v (x, z, u). This implies ha v (x, z, u) is supermodular when u [u 1, u 2 ]. z, z v(x, z, u) in nondecreasing in u on [, u 1), (u 1, u 2 ), and (u 2, ). verified ha lim u u1 z v(x, z, u) lim u u 1 z v(x, z, u) and lim u u 2 z v(x, z, u) lim u u 2 Therefore, for fixed x and Moreover, i can be z v(x, z, u). Thus, z v(x, z, u) in nondecreasing in u for all u and v (x, u, z) in supermodular in (u, z). Similarly, we can show x v (x, z, u) + z v (x, z, u) is nondecreasing in u. This complees he inducion and he proof. Proof of Theorem 4: Le ˆV (y, z) = L(y) + Eˆv 1 (z y + (y D) + ). Similar o he proofs of Lemma 1 and Theorem 2, we can show ha ˆv (z) is concave, ˆV (y, z) is joinly concave and supermodular, and ŷ (z) 1. We can also show by inducion ha v (z) is coninuously differeniable, ˆv (z) p c, and ˆv () = p c. Nex, we show by inducion ha ŷ (z) saisfies he following properies: (a) if z y b, hen ŷ (z) = y b, (b) if z < y b, hen z ŷ (z) y b, (c) if z ˇy, hen ŷ (z) = z, and (d) ŷ (z) is nonincreasing in. For (a), we show by inducion ha when z y b, v (z) =. For =, v (z) =. Now, suppose ha when z ( 1)y b, v 1 (z) =. Then, when z yb, we have z ξ ( 1)y b for ξ y b. Thus, v 1 (z ξ) = for ξ yb and V (y, z) y = L (y b ) α ˆv 1(z y b )φ(ξ)dξ =. y=y b y b 3

4 This implies ha ŷ (z) = y b. Therefore, when z y b, y b v (z) = α ˆv 1(z ξ)φ(ξ)dξ α ˆv 1(z y b )φ(ξ)dξ =. y b This complees he inducion. Moreover, from he above analysis, we can see ha he fac v (z) = when z y b implies ha ŷ +1 (z) = y b. Therefore, (a) holds for = 1,..., T. Nex, we prove (c). Noe ha for z ˇy, we have V (y, z) y = L (z) α y=z z ˆv 1()φ(ξ)dξ = (1 α)p c ((1 α)p + h)φ(z). This implies ha ŷ (z) = z when z ˇy and V y (y, z) y=ŷ(z) = when z ˇy. For (b), firs noe ha for = 1,..., T, when z ˇy, we have V y y=ŷ(z) (y, z) =, i.e., L (ŷ (z)) α(1 Φ(ŷ (z)))ˆv 1 (z ŷ (z)) =. Therefore, for = 1,..., T, when z ˇy, ˆv (z) = α( ŷ(z) = L (ŷ (z)) + α ˆv 1(z ξ)φ(ξ)dξ + ŷ(z) ŷ (z) ˆv 1(z ξ)φ(ξ)dξ L (ŷ (z)) + αφ(ŷ (z))ˆv 1(z ŷ (z)) ˆv 1(z ŷ (z))φ(ξ)dξ) = L (ŷ (z)) + Φ(ŷ (z)) 1 Φ(ŷ (z)) L (ŷ (z)) 1 = 1 Φ(ŷ (z)) L (ŷ (z)). (9) Since v 1 (z), we have V (y, z) y = L (y b ) α y=y b y b ˆv 1(z y b )φ(ξ)dξ. This implies ha ŷ (z) y b. Nex we show ha ŷ (z) z when z < yb. I can be easily verified ha ŷ 1 (z) = min(z, y b ). Thus, ŷ 1 (z) z when z < y b. Now, suppose ha ŷ 1 (z) z < ( 1)y b. Then, in period, we have V y y= (y, z) z consider wo cases: 1 V y (y, z) y= z z ˇy and 1 z 1 when = L ( z ) α(1 Φ( z ))ˆv 1 ( 1 z). We z > ˇy. For he firs case, we have z 1 z ˇy. Thus, L ( z ) α(1 Φ(z ))(p l) = (1 α)p c ((1 α)p + h)φ(z ). This implies ha ŷ (z) z. For he second case, by (9), we have ˆv 1( 1 1 z) 1 Φ(ŷ 1 ( 1 z)) L (ŷ 1 ( 1 where θ(y) = L (y) 1 Φ(y) nonincreasing on [, y b ]. Therefore, V (y, z) y y= z z)) = θ(ŷ 1 ( 1 z)), = p c (p αc+h)φ(y) 1 Φ(y). I can be verified ha θ(y) is nonnegaive and = L ( z ) α(1 Φ(z ))ˆv 1( 1 z) 4

5 1 1 Φ( z )(θ(z ) αθ(ŷ 1( 1 1 z))) 1 Φ( z )(θ(z ) θ(ŷ 1( 1 z))), where he las inequaliy is due o he inducive assumpion ha ŷ 1 ( 1 z) z = z. This implies ha ŷ (z) z. For (d), we show by inducion ha ŷ (z) ŷ +1 (z) and ˆv (z) ˆv +1 (z). We have ˆv (z) = and ˆv (z) for all. Thus, ˆv (z) ˆv 1 (z). Now, suppose ˆv 1 (z) ˆv (z). When z ˇy, we have ŷ (z) = ŷ +1 (z) = z and ˆv (z) ˆv +1(z) = α z (ˆv 1(z ξ) ˆv (z ξ))φ(ξ)dξ. When z > ˇy, we have V (y, z) y = L (ŷ +1 (z)) α(1 Φ( z y=ŷ+1 (z) ))ˆv 1(z ŷ +1 (z)) This implies ha ŷ (z) ŷ +1 (z). We also have ˆv (z) ˆv +1(z) = L (ŷ ) L (ŷ +1 ) + α( L (ŷ +1 (z)) α(1 Φ( z ))ˆv (z ŷ +1 (z)) =. ŷ ŷ+1 v 1(z ξ)φ(ξ)dξ ŷ (p αc + h)(φ(ŷ ) Φ(ŷ +1 )) + α v 1(z ξ)φ(ξ)dξ ŷ +1 ŷ (p αc + h)(φ(ŷ ) Φ(ŷ +1 )) + α (p c)φ(ξ)dξ ŷ +1 = ((1 α)c + h)(φ(ŷ ) Φ(ŷ +1 )), v (z ξ)φ(ξ)dξ) where ŷ = ŷ (z), ŷ +1 = ŷ +1 (z), and he las inequaliy is due o he fac ha ŷ (z) ŷ +1 (z). Las we show ha he opimal producion hresholds of he sysem wih only he allowance consrain are exacly he same as hose of he relaxed sysem and herefore saisfy properies (a), (b), (c) and (d) in Theorem 4. We only need o show ha v (x, z) = ˆv (z) for x ŷ (z), and V y (y, z) for y > ŷ (z). This hen implies ha ŷ (z), if x y (z), argmax V (y, z) = x y z x, if x y (z). In oher words, y (x, q ) = ȳ (x + q ) = ŷ (x + q ). When =, we have v (x, z) = ˆv (z) =. Now, suppose ha v 1 (x, z) = ˆv 1 (z) for x ŷ 1 (z), and y V 1 (y, z) for y > ŷ 1 (z). Then, when x ŷ (z), since ŷ (z) 1 and ŷ 1 (z) ŷ (z), we have ŷ (z) ξ ŷ (z ξ) ŷ 1 (z ξ) for ξ ŷ (z). This implies ha (ŷ (z) D) + ŷ 1 (z y + (y D) + ). Thus, v 1 ((ŷ (z) D) +, z y + (y D) + ) = ˆv 1 (z y + (y D) + ) and ˆV (ŷ (z), z) = V (ŷ (z), z). As a consequence, ˆv (z) = ˆV (ŷ (z), z) = 5

6 V (ŷ (z), z) v (x, z) for x ŷ (z). Clearly, v (x, z) ˆv (z). Therefore, v (x, z) = ˆv (z) for x ŷ (z). Similarly, we can show ha ˆV (y, z) = V (y, z) for y ŷ (z). We can also prove by inducion ha V (y, z) is coninuously differeniable. Since V (y, z) is concave in y, hen for y > ŷ 1 (z), we have This complees he inducion and he proof. y V (y, z) y V (ŷ (z), z) = V (ȳ (z), z) =. Proof of Theorem 5: The proof is similar o he proofs of Lemma 8.5 and Theorem 8.4 in Poreus (22). Proof of Proposiion 6: Similar o he proof of Lemma 1, we can show ha v (x, z, u) is joinly concave in (x, z, u). Therefore, v (x, q, u) = v (x, x + q, u) is joinly concave in (q, u). This implies ha F T (c q, c u, x, q, u) = c q q c u u + v T (x, q, u) is joinly concave in (q, u). Proof of Proposiion 7: (a) Noe ha 2 F T 2 F T u 2 u T c q + 2 F T uc q =. Since 2 F T qc q 1 2 F T A, where A = 2 F T 2 F T u 2 q 2 u 2 2 F T u 2 qt c u q 2 q T c q = 1 and 2 F T uc q. Moreover, 2 F T qu = 2 v T qu. Therefore, q T c q and u T c u. (b) By Envelope Theorem, + 2 F T u T qu c q + 2 F T qc q =, we have q T c q = and 2 F T qt qu c q + = 1 2 F T A qu and q T c q = ( 2 F T qu )2. Since F T is joinly concave in (q, u), A and and u T c q c q F T (c q, c u, x) = q T (c q, c u, x) and. Similarly, we have c u F T (c q, c u, x) = u T (c q, c u, x). Moreover, F T (c q, c u, x, q, u) is convex in (c q, c u ). Since convexiy is preserved under maximizaion, F T (c q, c u, x) is convex in (c q, c u ). (c) When c u =, he sysem can be viewed as one wih only he allowance consrain. Thus, he second sage problem can be formulaed by opimaliy equaion (3), and he firs sage problem can be formulaed as max { c qq + v T (q)}. Le F (c q, q) = c q q + v T (q) and q le K be he smalles q such ha v T (q) =. Then for q < K, v T (q) >, and for q K, v T (q) =. This implies ha q T (,, ) = K. Noe ha v T (z) is wice differeniable. Thus we have v T (K) =. As a consequence, q T (c q,, ) c q = cq= Proof of Proposiion 8: Since c u 2 qc q F (cq, qt (c q,, )) 2 F (c q 2 q, qt (c q,, )) 1 = cq= v T (q T (c q,, )) =. cq= =, he sysem can be viewed as one wih only he allowance consrain. Le Q (x, q) denoe he expeced cumulaive amoun produced, under he opimal policy, from period o he end of he planning horizon, wih saring invenory level x and remaining allowance q. Then, u T (w, x) = Q T (x,c T (w,x)) c T (w,x). Noe ha Q (x, q), for = T,..., 1, 6

7 can be compued recursively as EQ 1 ((x D) +, q), Q (x, q) = y (x, q) x + EQ 1 ((y (x, q) D) +, x + c y (x, c)), if x y (x, c), oherwise, and Q (x, q) =. We can hen show by inducion ha, for 1 T, v (z) < when z < y b, and v (z) = when z y b. This implies ha q T (,, x) = (T yb x) +. We can also show by inducion ha q Q (x, q) = when x + q y b. As a consequence, η T (c q,, ) c q = q T (,, ) q Q (, q T (,, )) c q qt (,, ) Q (, q T (,, )) c q qt (,, ) cq= (qt (,, ))2 = T yb q Q (, T y b ) Q (, T y b ) (T y b ) 2 c q q T (,, ) =. Proof of Theorem 9: Le ḡ (x, z) = v(x, z x) cx, f (y, z, d) = f (y, z y, d), and L(x) = ( p + αc)x h(x) + b( x) +. Then we can rewrie he opimaliy equaions in (5) and (6) as (7) and (8). In he proof of Lemma 1, we show ha xḡ(x, z) + z ḡ(x, z) p c and F (y, z, l) = L(y l) + py + αḡ 1 (y l, z l) is concave in l. Le l (y, z) = argmax { L(y l) + l z py + αḡ 1 (y l, z l)}. When l y, we have l F (y, z, l) = p αc + h α( xḡ 1(y l, z l) + z ḡ 1(y l, z l)). This implies ha l (y, z) = argmax{( p + αc)(y l) b(l y) + py + αḡ 1 (y l, z l)}. Le y l z w (q) = argmax{(p αc b)w+αḡ 1 ( w, q w)}. Then l (y, z) = y+w (z y). Given invenory w q level y, remaining allowance q 1, and realized demand d, he opimal amoun of demand o fulfill equals min(d, y +w (q 1 )). By propery (c) of Lemma 2 of Huh and Janakiraman (21), we have d dq w (q) 1. For (c), noe ha w (q) = argmax{(p αc b)w + αḡ 1 ( w, q w)}. If we can show w q xḡ(x, z) + z ḡ(x, z) xḡ 1(x, z) + z ḡ 1(x, z), hen we have w (q) w+1 (q). Nex we show by inducion ha xḡ(x, z) + z ḡ(x, z) xḡ+1(x, z) + z ḡ+1(x, z) and z ḡ(x, z) z ḡ+1(x, z). Clearly, xḡ1(x, z)+ z ḡ1(x, z) = xḡ(x, z)+ z ḡ(x, z) and z ḡ1(x, z) = z ḡ(x, z). Now, suppose xḡ 1(x, z)+ z ḡ 1(x, z) xḡ(x, z)+ z ḡ(x, z) and z ḡ 1(x, z) z ḡ(x, z). Then, we have w (q) w+1 (q). By he srucure of he opimal fulfillmen policy, Ḡ (y, z) = cy y y+w y [( p + αc)(y ξ) h(y ξ) + py + αḡ 1 (y ξ, z ξ)]φ(ξ)dξ y+w [( p + αc)(y ξ) b(ξ y) + py + αḡ 1 (y ξ, z ξ)]φ(ξ)dξ [( p + αc)( w ) bw + py + αḡ 1 ( w, z y w )]φ(ξ)dξ, 7

8 where w = w (z y). Hence, y Ḡ(y, z) + z Ḡ(y, z) =p c (h + b)φ(y) (p αc b)φ(y + w ) + α y+w [ x v 1(y ξ, z ξ) + z v 1(y ξ, z ξ)]φ(ξ)dξ. By he definiion of w, x v 1(y w, z w) + z v 1(y w, z w) (p αc b) for any w [, w ] and x v 1(y w, z w) + z v 1(y w, z w) (p αc b) for any w [w, z]. This implies ha y Ḡ(y, z) + z Ḡ(y, z) p c (h + b)φ(y) (p αc b)φ(y + w+1) + α y+w +1 [ x v 1(y ξ, z ξ) + z v 1(y ξ, z ξ)]φ(ξ)dξ = p c (h + b)φ(y) (p αc b)φ(y + w +1) + α y+w +1 [ x v (y ξ, z ξ) + z v (y ξ, z ξ)]φ(ξ)dξ = y Ḡ+1(y, z) + z Ḡ+1(y, z). Similarly, we can show ha z Ḡ(y, z) z Ḡ+1(y, z). I can be verified ha, for n = 1,..., T, xḡn(x, z) + z ḡn(x, z) = ( y Ḡn(y, z) + z Ḡn(y, z)) y=yn(x,z), where y n (x, z) is he opimal order-up-o level in period n. Le ȳ (z) = argmax{ cy + y z E f (y, z, D)}. If ȳ (z) ȳ +1 (z), hen y (x, z) y +1 (x, z) and xḡ(x, z) + z ḡ(x, z) ( y Ḡ(y, z) + z Ḡ(y, z)) y=y+1 (x,z) ( y Ḡ+1(y, z) + z Ḡ+1(y, z)) = xḡ+1(x, z) + z ḡ+1(x, z), where he firs inequaliy is due o he L -convexiy of Ḡ(y, z). y=y+1 (x,z) If ȳ (z) < ȳ +1 (z), hen y (x, z) y +1 (x, z). I can be verified ha xḡ(x, z) xḡ+1(x, z). Moreover, z ḡ(x, z) = z Ḡ(y, z) y=y(x,z) z Ḡ(y, z) y=y+1 (x,z) z Ḡ+1(y, z)) = y=y+1 (x,z) z ḡ+1(x, z), where he firs inequaliy is due o he supermodulariy of Ḡ (y, z). Therefore, xḡ(x, z) + z ḡ(x, z) xḡ+1(x, z) + z ḡ+1(x, z). Similarly, we can show z ḡ(x, z) z ḡ+1(x, z). This complees he inducion and he proof. Proof of Proposiion 1: Noe ha w (q) = argmax{( p + αc b)w + αḡ 1 ( w, q w). w q When p αc b, we have p αc b α( xḡ 1( w, q w)+ z ḡ 1( w, q w)) p αc b. This implies ha w (q) =. When (1 αc)p b, we have p αc b α( xḡ 1( w, q w) + 8

9 z ḡ 1( w, q w)) p αc b α(p c). This implies ha w (q) = q. (a) and (b) are proved. For (c), noe ha given x and q, i is opimal o back order a mos w (x + q y (x, q )) unis of demand, where y (x, q ) is defined in he proof of Theorem 2. Since w (q) is nondecreasing in q and x + q y (x, q) is nondecreasing in q, w (x + q y (x, q )) is nondecreasing in q. Clearly, w (x + q y (x, q )) > when q is sufficienly large. Le q (x ) be he larges q such ha w (x + q y (x, q )) =. Then w (x + q y (x, q )) = when q q (x ) and w (x + q y (x, q )) > when q > q (x ). References Huh WT, Janakiraman G (21). On he Opimal Policy Srucure in Serial Invenory Sysems wih Los Sales. Operaions Research. 58(2): Poreus E (22). Foundaions of Sochasic Invenory Theory. (Sanford Universiy Press, Redwood Ciy, CA) Zipkin P (28). On he Srucure of Los-Sales Invenory Models. Operaions Research. 56(4):