Online Appendix to Measuring the Bullwhip Effect: Discrepancy and Alignment between Information and Material Flows

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1 Online Appendix to Measuring the Bullwhip Effect: Discrepancy and Alignment between Information and Material Flows i Chen Wei uo Kevin Shang S.C. Johnson Graduate School of Management, Cornell University, Ithaca, New York 4853, USA IESE Business School, University of Navarra, Av. Pearson 2, 834 Barcelona, Spain Fuqua School of Business, Duke University, Durham, North Carolina 2778, USA li.chen@cornell.edu wluo@iese.edu khshang@duke.edu Rest of Proofs Proof Proposition 2 et ϕt denote the standard normal density function, and define Φ x = x t xϕtdt. et Gs, = VM t] VM t] = 2E s D t t D t t s ]. When demand is i.i.d. normal, we can write Gs, as where Īs ξ = Gs, = 2 sξ Īs ξ Bs ξ 2π e s ξ u 2π e 2 uµ 2 ξµ 2 dξ. 2 du = Φ s ξ µ where the last equality follows from a change of variable operation. Similarly, we can show that Bs ξ = u s ξ 2π e 2 uµ 2 s ξ µ du = Φ. Thus, we have sξ Gs, = 2 2 Φ s ξ µ s ξ µ Φ 2π e 2 ξµ The optimal base-stock level under normal demand distribution is s = µ z, where z is the safety factor. Substituting this into the above expression and also making a change of, 2 dξ. variable based on t = ξ µ/, we have Gs, = Gz, = 2 2 Φ z t Φ z t ϕtdt. A

2 Now let x = t z, i.e., t = x z /, then we have et fx = Φ xφ x, then Gz, = 22 dgz, dz = 22 = 22 Φ xφ xϕ x z dx. x z fxdϕ x z f xϕ dx. It is easy to show f x > for x <, f x < for x >, and f x = f x. When z =, dgz, = 22 ] f xϕxdx f xϕxdx =. dz When z >, we have dgz, dz = 22 = 22 = 22 = 22 x z f xϕ dx f xϕ x z dx x z f xϕ dx x z f x ϕ ϕ ] x z f xϕ dx f xϕ ] x z f xϕ dx x z ] dx. ] x z dx It is straightforward to verify that ϕ xz ϕ xz for all x. Thus, we have dgz, /dz < for z >. Symmetrically, we can show that dgz, /dz > for z <. Therefore, Gz, is unimodal in z, with the peak attained at z =. Or, equivalently, Gs, is unimodal in s, with the peak attained at s = µ. With this result, we conclude that the gap between r M and r O is monotonically increasing in s when s µ, and is monotonically decreasing in s when s µ. Proof Proposition 3 et fx = Φ xφ x, where Φ x = t xϕtdt and ϕt is x the standard normal density function. Also, let Φx denote the cumulative density function of the standard normal distribution. It is easy to verify that f x = ϕx2φx 2xΦx Φx f x = xϕx2φx 2ϕx 2 2Φx Φx. From the equation A in the proof of Proposition 2, we have Gz, = 2 2 f t z ϕtdt. 2

3 et x = t z, i.e., t = x z /, then we have dgz, = 2 f t z t z ϕtdt d 2 = f z x z x x ϕ dx. et us focus on the integration term in the above expression: f z x z x x ϕ dx = = = f xxϕ f xxϕ x z dx x z dx f xx f x ] ϕ x z dx. f x x z ] x z ϕ dx f xdϕ ] x z Thus, to show dgz, /d <, it suffices to show that hx = f xx f x < for all x. Note that f x = f x and f x = f x. Therefore, hx = f xx f x = f xx f x = hx. Thus, to show hx < for all x, it suffices to show hx < for all x >. Substituting the expressions of f x and f x into hx, we obtain hx = xf x f x = 2xϕx2Φx 2ϕx 2 2 x 2 Φx Φx. From this, we have h = /π /2 <, and h =. Thus, to show hx < for x >, it suffices to show h x >. Note h x = 4ϕx2Φx 4xΦx Φx and h = h =. Thus, it suffices to show that h > and h x is unimodal. To do that, it suffices to show h x = 8ϕx 2 4Φx Φx crosses zero once note we have h = 4/π > and h =. Now note that h x = 4ϕx4xϕx 2Φx. et gx = 4xϕx 2Φx. Then g x = 2ϕx 2x 2. Clearly, g x crosses zero once at x = 2/2, with g x > for < x < 2/2 and g x < for x > 2/2. Therefore, gx is unimodal. Also note that g = and g = <. Thus, gx crosses zero just once in 2/2,. Suppose the zero-crossing point of gx is x, then we have h x < for < x < x and h x > for x > x. Because h =, we have h x < for x > x. Because h > and h x < for < x < x, we have h x only crosses zero once in the range < x < x. Therefore, we conclude that hx < 3

4 for x >. In other words, dgz, /d <. It follows that the gap between r M and r O is monotonically decreasing in. Proof Proposition 4 et Gρ = VD t VM t]. When ρ =, from we have G =Vε t ε t V ε t ε t ε t z ε t z ] = V ε t z ε t z ] 2E ε t ε t z ε t ε t z ] = V ε t z ε t z ] Now, let X = ε t ε t z ε t z. Then we have VM t] = VX 2ρE ε t ε t z ] Vρε t Vρ 2 D t2, VD t = Vε t Vρε t Vρ 2 D t2. Recall from Proposition that Vε t VX = 2E z ε t ] Eε t z ]. Thus Gρ = 2E z ε t ] Eε t z ] 2ρE ε t ε t z ]. It is easy to verify that G because E z ε t ] Eε t z ]. It is also easy to verify that Eε t ε t z ] is decreasing in z and approaches zero when z goes to infinity. It follows that Eε t ε t z ] for any z, hence G ρ. Therefore, Gρ crosses zero only once from positive to negative at ρ = E z ε t ] Eε t z ]/E ε t ε t z ]. Because VM t ] = VO t ] in the single-stage system, it immediately follows that r M > r O if ρ < ρ, and r M r O otherwise. Proof Proposition 5 For >, to compare VD t ] and VM t ], it is sufficient to ] compare V k= ρk ε tk and V k= ρk ρ ε tk k k= ρ ε tk z k= When ρ =, recall from Proposition that V ε t k= ε ] tk z k= ε tk z Vε t ]. ] ρ k ρ ε tk z. Thus, from continuity, there exists a thresholds ρ, such that, if ρ ρ, VM t] < VD t ]. When 4

5 ρ approaches, we have V k= ε tk k= k ε ] tk z k= k ε tk z ] = V k= ε tk V k= k ε ] tk z k= k ε tk z { 2E k= k ε tk z k= ε tk } k= k ε tk z k= ε tk ] = V k= ε tk V k= k ε ] tk z k= k ε tk z ] V k= ε tk. Thus, from continuity, there exits a threshold ρ such that, if ρ ρ <, VM t] VD t ]. Proof Proposition 6 Define X t = D t t D t t 2 s 2 and Y t = D t t D t t 2 s 2. Adding Y t to both sides of 8, we have M t Y t = X t X t s X t s. Since X t is identically distributed, using a similar sample-path argument as in the proof of Proposition 7 we can show that VM t Y t ] VX t ] = VD t Y t ]. Therefore, to show VM t] VD t ], it suffices to show CovM t, Y t ]. CovM t, Y t ] =CovX t s X t s, Y t ] A2 =E { X t s X t s } ]Y t EXt s X t s ]EY t ] { Xt =E s X t s ] D t t D t t 2 s 2 ]} A3 ] ] =E X t s D t t E X t s D t t { Xt E s X t s ] D t t 2 s 2 } { =E X t s Dt t D t t 2 s 2 s ] D t t 2 s 2 } A4 { =E X t s D t t Dt 2 t 2 s 2 s ] D t t 2 s 2 }, where A2 follows from the independence of D t and Y t ; A3 is because X t is identically distributed; A4 is because D t t is independence from D t, D t, D t t 2, D t t 2, and D t is i.i.d.; A5 is because D t and D t 2 are i.i.d.; A6 follows from D t 2 t 2 s 2 D t 2 t 2 s 2. A5 A6 5

6 Proof Proposition 7 Because M t is a stationary process, we can compare VM t] and VM t] without affecting the conclusion. We first prove the base case with 2 = and =, i.e., to show VM t] VM t] under the condition s 2 s. Consider any cycle t, t,..., tτ such that B t =, B t >, B t >,..., B t τ >, B t τ =, we calculate M t i for i =,..., τ in the following three steps. a For i =, since B t = and B t >, we have D t s D t2 s 2 s s 2, hence, M t = s D t s 2 = s. b For i = 2,..., τ, since B t i 2 > and B t i >, we have M t i = D ti2 D ti3 s 2 D ti2 s 2 = M t i. c For i = τ, since B tτ > and B tτ =, we have D tτ s D tτ s 2 s s 2, and M t τ = D tτ D tτ D tτ2 s 2 s. Combining results a and b, we have M 2 t i M 2 t i = M 2 t τ M 2 t τ s 2 M 2 t τ ], A7 where from 9, 2 and result c above, we have M t τ = D tτ D tτ2 s 2 D tτ s 2, M t τ = D tτ D tτ s 2, M t τ = D tτ D tτ D tτ2 s 2 s. As shown above, it is not immediately clear whether M t is more variable than M t. To further simplify A7, we discuss the following four cases. Case. When D tτ2 s 2 and D tτ s 2, from B t τ > we have D tτ > s D tτ2 s 2 = s. Together with D tτ s as in result c, equation A7 becomes M 2 t i M 2 t i = Dtτ 2 Dtτ 2 s 2 D tτ D tτ s 2] = 2D tτ s s D tτ. Case 2. When D tτ2 s 2 and D tτ > s 2, from B t τ = we have D tτ s D tτ s 2 = s s 2 D tτ. Therefore, equation A7 becomes M 2 t i M 2 t i = s 2 2 D tτ D tτ s 2 2 s 2 D tτ D tτ s 2] = 2s 2 s s 2 s D tτ D tτ. 6

7 Case 3. When D tτ2 > s 2 and D tτ s 2, from B t τ > we have D tτ D tτ2 > s s 2. Together with D tτ s as in result c, equation A7 becomes M 2 t i M 2 t i = D tτ2 D tτ s 2 2 Dtτ 2 s 2 D tτ2 D tτ D tτ s 2 s 2] = 2s D tτ D tτ2 D tτ s 2 s. Case 4. When D tτ2 > s 2 and D tτ > s 2, from B t τ = we have D tτ s D tτ s 2 = s s 2 D tτ. Together with D tτ2 > s 2 s, equation A7 becomes M 2 t i M 2 t i = Dtτ2 2 D tτ D tτ s 2 2 s 2 D tτ2 D tτ D tτ s 2 s 2] = 2D tτ2 s s 2 s D tτ D tτ. Since the above four cases include all possible demand sample paths, we can conclude that τ M 2t i τ M 2t i. Note that when B t 2 >, period t is the last period of the previous cycle; When B t 2 = B t =, period t is not included in any cycle, in which case we have D t s s 2 and D t2 s s 2, hence M t = M t = D t. This holds for any period outside the cycle. Also, it is straightforward to verify that τ M t i = τ M t i. Because B t is a stationary process, by the ergodic property, we conclude that EM 2 t] EM 2 t ] and EM t] = EM t ]. Therefore, VM t] VM t ]. Next, we prove the case with general lead time. Define s = s D t2 t, then 8 becomes M t = D t D t 2 D t 2 2 s 2 s Dt D t 2 s 2 s Given s, the above expression has the same structure as in the case of = and 2 =. From the condition s 2 s, it immediately follows that s 2 s. Thus, following the above analysis, we conclude that VM t] VM t s ] for any given s. Also, from flow conservation, we have EM t s ] = ED t for all s. Therefore, we have VM t] EVM t s ]] = VM t ], which implies that r M r O. Proof Proposition 8 Suppose that the demand distribution has positive support in the range b, d, where b and d. When s b, the material flow dynamics equations 7 7

8 and 8 become M t = D t D t2 t 2 s 2 D t t 2 s 2, M t = D t D t 2 t 2 s 2 D t t 2 2 s. Thus, we have VM t] = VM t]. Therefore, the result holds trivially with s 2 =. When s > b, let us consider two cases. Case : d =. In this case, we note that, if s 2 =, the material flow dynamics equations 7 and 8 become the following: M t = D t, M t = D t D t t s D t t s. The above expressions have the same structure as the material flow equations 5 and 6 in the single-stage system. Thus, the result of Proposition can be applied to this case. Moreover, because d =, for any given s > b, we have s VM t] VM t] = 2E D t t D t t s ] >. Thus, by continuity, there exists a threshold s 2 s, such that for any s 2 s 2 s, VM t] VM t], implying r M r O. Case 2: d <. et us define s 2 = s 2 D t 2 t 2, then 8 becomes M t = D t D t t D t 2 s 2 s Dt t D t s 2 s. Since D t 2, D t, and D t 2 t 2 VM t] = V = V = V D t are mutually independent, from equation 7 we have ] D t2 D t2 t 2 s 2 D t D t2 t 2 s 2 ] D t 2 D t 2 t 2 s 2 D t D t 2 t 2 s 2 D t D t D t 2 s 2 Dt s 2 ]. Given s 2 and s, VM t s 2 ] and VM t s 2 ] have the same structure as VM t] and VM t] in the case of 2 =. Now let s 2 s = s 2 d. Thus, the condition s 2 s 2 s ensures s 2 s for any demand realization. Thus, from Proposition 7, we have VM t s 2 ] VM t s 2 ]. Also, from flow conservation, we have EM t s 2 ] = EM t s 2 ] = ED t for all s 2. Therefore, we conclude that VM t] VM t], implying that r M r O. Combining the two cases, we arrive at the desired result. 8

9 Proof Proposition 9 Suppose that the demand distribution has positive support in the range b, d, where b and d. If b >, let s 2 s = 2 b >. We have, for any s 2 s 2 s, the material flow dynamics equations 7 and 8 become: M t = D t2, M t = D t D t t 2 s D t t 2 s. Thus, the result of Proposition can be applied to this case. implying r M r O. We have VM t] VM t], On the other hand, if b =, consider two cases. Case : When s =, the material flow dynamics equations 7 and 8 become M t = D t D t2 t 2 s 2 D t t 2 s 2, M t = D t D t 2 t 2 s 2 D t t 2 2 s. Thus, we have VM t] = VM t]. Therefore, the result holds trivially with s 2 =. Case 2: When s >, we know that, if s 2 =, the material flow dynamics equations 7 and 8 become: M t = D t2, M t = D t D t t 2 s D t t 2 s. Thus, the result of Proposition can be applied to this case. Because < s < 2 d, we have s VM t] VM t] = 2E D t t D t t s ] >. Thus, by continuity, there exists a threshold s 2 >, such that for any s 2 s 2 VM t] VM t], implying r M r O. 9