智能化背景下新能源汽车供应链合作模式

doi:10.6040/j.issn.1671-9352.0.2025.159

Abstract

Abstract: This paper constructs a two-tier supply chain model involving an intelligent system supplier and a new energy vehicle(NEV)manufacturer. Focusing on cooperation modes led by the manufacturer and the supplier, this study employs the Stackelberg game approach to derive the equilibrium strategies and profits of supply chain members, and further examines the selection of cooperation modes in the NEV supply chain. The findings reveal that:(1)When the sales revenue allocation ratio and the R& D cost coefficient exceed a certain threshold, the supplier-led mode results in a higher level of product intelligentization;(2)The R& D cost coefficient and elasticity coefficient exert a significant impact on equilibrium decisions;(3)For both the manufacturer and the supplier, the choice between cooperation modes is influenced by the sales revenue allocation ratio. Under the supplier-led mode, by setting a reasonable sales revenue allocation ratio, both the manufacturer and the supplier can achieve higher profits, thereby realizing a win-win outcome.

Key words: intelligentization, new energy vehicle, supply chain, cooperation mode, sales revenue allocation

CLC Number:

F272

ZHU Mengping, LUO Lijiao, WANG Qifei. Supply chain cooperation modes for new energy vehicles under the background of intelligentization[J].JOURNAL OF SHANDONG UNIVERSITY(NATURAL SCIENCE), 2026, 61(8): 132-144.

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[40] 东方财富网. 赛力斯: 2024年年度报告[EB/OL].(2025-04-01)[2025-04-20]. https://data.eastmoney.com/notices/detail/601127/AN202503311649710345.html. Eastmoney.com. SERES: 2024 Annual Report[EB/OL].(2025-04-01)[2025-04-20]. https://data.eastmoney.com/notices/detail/601127/AN202503311649710345.html. 附录命题1的证明根据逆向归纳法,首先求式(2)关于p的二阶导(ə²πM_m(p,I))/(əp²)=-2 β<0,则πM_m(p,I)是关于p的凹函数。由一阶条件(əπM_m)/(əp)=0,可得p=(α+(c+w)β+γI)/(2 β)。将p=(α+(c+w)β+γI)/(2 β)代入式(1)中,求出式(1)关于w的二阶导(d²πM_s(w))/(dw²)=-β<0,则πM_s(w)是关于w的凹函数,由一阶条件(dπM_s(w))/(dw)=0,可得w=(2α-2c β+k βI²+2γI)/(4 β)。将w=(2α-2c β+k βI²+2γI)/(4 β)代入p=(α+(c+w)β+γI)/(2 β)中,得到p=(6α+2c β+k βI²+6γI)/(8β)。将w=(2α-2c β+k βI²+2γI)/(4 β)和p=(α+(c+w)β+γI)/(2 β)代入式(2),得到πM_m(I)=([2α-2c β+2γI-k βI²] ²)/(64 β)。在实践中制造商的边际利润为正,需要保证p-w-c=(2α-2c β+2γΙ-k βΙ²)/(8β)>0,即2α-2c β+2γΙ-k βΙ²>0。因此,最大化πM_m(I)即等价于最大化2α-2c β+2γI-k βI²。定义f(I)=2α-2c β+2γI-k βI²,可得(df(I))/(dI)=2γ-2k βI且(d²f(I))/(dI²)=-2k β<0。可知f(I)为关于I的凹函数,根据上述一阶条件,可得I ^{M^*}=γ/(k β)。将I^{M^*}=γ/(k β)代入w=(2α-2c β+k βI²+2γI)/(4 β)和p=(6α+2c β+k βI²+6γI)/(8β)中,可得w^{M^*}=(2k β(α-c β)+3γ²)/(4k β²), p^{M^*}=1/8(2c+(6kα β+7γ²)/(k β²))。将I^{M^*}、w^{M^*}、 p^{M^*}代入式(1)、(2),可得π^{M^*}_s=([2k β(α-c β)+γ²] ²)/(32k² β³), π^{M^*}_m=([2k β(α-c β)+γ²] ²)/(64k² β³)。命题2的证明根据逆向归纳法,首先求式(4)关于p的二阶导(d²πS_m(p))/(dp²)=-2 β(1-σ)<0,可知π^S_m(p)是关于p的凹函数。由一阶条件(dπ^S_m(p))/(dp)=0,可得p=1/2((α+γI)/β+c/(1-σ))。将p=1/2((α+γI)/β+c/(1-σ))代入式(3)中,得到利润 πS_s(I)=([c β+α(-1+σ)+γΙ(-1+σ)] {k βΙ²-[α+c β+Ι(γ+k βΙ)] σ+(α+γΙ)σ²})/(4 β(-1+σ)²)。(A1) 式(A1)是关于I的三次函数,基于一阶条件(dπ^S_s(I))/(dI)=0,可得2个驻点I₁=(Z+β A)/(3k βγ(1-σ))或I₂=(Z-β A)/(3k βγ(1-σ))。对于这2个驻点来说,只有当二阶导数小于0,利润函数取得最大值;否则,利润函数取得最小值。式(3)关于I的二阶导(d²π^S_s(I))/(dI²)=1/2[(γ²σ)/β-k(α+3γI)+(ck β)/(1-σ)] ,将I₁、I₂分别代入以上表达式,可得I₁对应的二阶导数为 -((k²[c β+α(-1+σ)] ²-(2k γ²[c β-2α(-1+σ)] (-1+σ)σ)/β+(γ⁴(-1+σ)²σ²)/(β²))^1/2)/(2(1-σ))<0, 对应π^S_s(I)的最大值;I₂对应的二阶导数为 ((k²[c β+α(-1+σ)] ²-(2k γ²[c β-2α(-1+σ)] (-1+σ)σ)/β+(γ⁴(-1+σ)²σ²)/(β²))^1/2)/(2(1-σ))>0, 对应πS_s(I)的最小值。因此I^{S^*}=I₁=(Z+β A)/(3k βγ(1-σ))。将I^{S^*}=(Z+β A)/(3k βγ(1-σ))代入p=1/2((α+γI)/β+c/(1-σ)),可得 p^{S^*}=(γ²(1-σ)σ+2k β(α+2c β-ασ)+β A)/(6k β²(1-σ))=(Y+β A)/(6k β²(1-σ))。将I^{S^*}、 p^{S^*}代入式(3)和式(4),可得供应链各成员的利润分别为 π^{S^*}_s=(3σ(X+β A)(Y+β A))/(108k² β³(1-σ)²)-((X+β A)(Z+β A)²)/(108k² β³(1-σ)³γ²), πS^*_m=({A β+2k β[-c β+α(1-σ)] +γ²(1-σ)σ}²)/(36k² β³(1-σ))=((A β+X)²)/(36k² β³(1-σ))。注 A=(k²[α(1-σ)-c β] ²+(γ⁴(1-σ)²σ²)/(β²)+(2k γ²(1-σ)σ(2α+c β-2ασ))/β)^1/2, X=2k β[α(1-σ)-c β] +γ²(1-σ)σ,Y=γ²(1-σ)σ+2k β(α+2c β-ασ), Z=k β[α(1-σ)-c β] +γ²(1-σ)σ。定理1的证明 ΔI=I ^M*-I ^{S^*}=(A β+k β[c β-α(1-σ)] - γ²(3-σ)(1-σ))/(3k βγ(σ-1)),由于(ə(ΔI))/(əA)=1/(3k γ(σ-1))<0,所以ΔI关于A单调递减,关于A的临界值为:(~overA)₁=(γ²(3-σ)(1-σ)+k β(α-c β-ασ))/β。因此,当A>(~overA)₁=(γ²(3-σ)(1-σ)+k β(α-c β-ασ))/β时,I ^{M^*}S^*;反之I ^{M^*}≥I ^{S^*}。注意到,A>(~overA)₁可以等价于A²>(~overA)²₁, A²-(~overA)²₁=(3γ²(-1+σ){2k β[α(1-σ)²-c β] +γ²(1-σ)(3-2σ)})/(β²)。 (1)在供应商模式下,供应商有较大的渠道权力确保其销售收入分配不低于一定比例。因此,α(1-σ)²-c β<0是一个合理的设定。在此情形下,A²-(~overA)²₁关于k单调递增,且临界条件为(~overk)=(γ²(1-σ)(3-2σ))/(2 β[c β-α(1-σ)²] )。当k>(~overk)时,有A²-(~overA)²₁>0,此时 I ^{M^*}S^*;否则,I ^{M^*}≥I ^{S^*}。 (2)当α(1-σ)²-c β>0时,I^{M^*}>I ^{S^*}恒成立。综合以上两种情况且进一步化简,可以得到当σ>1-((c β)/α)^1/2且k>(γ²(1-σ)(3-2σ))/(2 β[c β-α(1-σ)²] )时,I^{M^*}S^*;否则,I^{M^*}≥I ^{S^*}。定理2的证明 Δp=p^{M^*}-p^{S^*}=(4A β-γ²(1-σ)(21-4σ)+2k β[5α(-1+σ)+c β(5+3σ)] )/(24k β²(-1+σ)),(ə(Δp))/(əA)=1/(6k β(σ-1))<0,因此Δp关于A单调递减,关于A的临界值为(~overA)₂=(γ²(1-σ)(21-4σ)-2k β[c β(5+3σ)-5α(1-σ)] )/(4 β)。因此,当A>(~overA)₂=(γ²(1-σ)(21-4σ)-2k β[c β(5+3σ)-5α(1-σ)] )/(4 β)时,p^{M^*}S^*;反之,p^{M^*}≥p^{S^*}。定理3的证明 ΔD=D^{M^*}-D^{S^*}=(4A β+γ²[(7-4σ)σ-3] -2k β[c β(1+3σ)-α(1-σ)] )/(24k β(-1+σ)),(ə(ΔD))/(əA)=1/(6k(σ-1))<0,因此ΔD关于A单调递减,关于A的临界值为(~overA)₃=(γ²(1-σ)(3-4σ)+2k β[c β(1+3σ)-α(1-σ)] )/(4 β)。当A>(~overA)₃=(γ²(1-σ)(3-4σ)+2k β[c β(1+3σ)-α(1-σ)] )/(4 β)时,D^{M^*}S^*;反之,D^{M^*}≥D^{S^*}。定理4的证明 Δπm=π^{M^*}_m-π^{S^*}_m=([2k β(α-c β)+γ²] ²)/(64k² β³)-((A β+X)²)/(36k² β³(1-σ)),(ə(Δπm))/(əA)=(X+A β)/(18k² β²(σ-1))<0,因此,Δπm关于A单调递减,关于A的临界值为(~overA)₄=(3[2k β(α-c β)+γ²] (1-σ)^1/2)/(4 β)- X/β。因此,当A>(~overA)₄=(3[2k β(α-c β)+γ²] (1-σ)^1/2)/(4 β)- X/β时,π^{M^*}_m<π^{S^*}_m;反之,π^{M^*}_m≥π^{S^*}_m。

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