# Stackelberg Game for Product Renewal in Supply Chain

Authors: Yong Luo; Yamin Lou; Shizhao Wang
Journal: Discrete Dynamics in Nature and Society
 (2013)

---

## Abstract

The paper studied the process of product renewal in a supply chain, which is composed of one manufacturer and one retailer. There are original product and renewal product in the supply chain. A market share shift model for renewal product was firstly built on a increment function and a shift function. Based on the model, the decision-making plane consisting of two variables was divided into four areas. Since the process of product renewal was divided into two stages, Stackelberg-Nash game model and Stackelberg-merger game model could be built to describe this process. The optimal solutions of product pricing strategy of two games were obtained. The relationships between renewal rate, cost, pricing strategy, and profits were got by numerical simulation. Some insights were obtained from this paper. Higher renewal rate will make participants’ profits and total profit increase at the same margin cost. What is more important, the way of the optimal decision making of the SC was that RP comes onto the market with a great price differential between OP and RP.


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## Body

## 1. Introduction

With the development of IT, the speed of product renewal process becomes faster and faster. This situation brings some new problems to traditional supply chain (SC), such as the dynamic nature of an SC and the relationship coordination between participants of an SC. The dynamic nature of an SC includes the dynamic variety of products; that is to say, the market requests SC to satisfy diverse needs of customer with the fastest speed, as well as the best quality, which calls for SC improving performance to adapt to the product variety. A valid way for SC enterprises to follow the variety of market is to renew their existing products, which can make full use of enterprise’s existing resources and supply/distribution outlet of SC. Thus, SC will adapt to a variational market economically and quickly. Product renewal is an effective method that strengthens a SC enterprise’s core competitiveness, and it will even impact on the survival of a enterprise. There are quite a few cases that enterprises collapse because of the mistakes of product renewal decision making, such as the failure of WANGAN Computer Corp.Now, it is necessary to differentiate “renewal product (RP)” from “innovated product (IP).” IP’s structure and principle are different from original product (OP), while RP’s main structure and principle are the same as OP. Furthermore, RP has some appended components or upgraded functions. Thereby, RP is the renewal of existing product. At present, most researches are mainly focused on complete new product innovation, including management of product innovation process, optimization of product innovation investment, promotion of new product, and design of product innovation drive mechanism. Dereli et al. proposed a framework of the rapid response for innovative product development using reverse engineering approach [1]. Bourreau et al. studied the effects of modular design on firms’ product innovation strategies and postinnovation competition in digital market [2]. Ju and Xiao built a model of product lifecycle evaluation based on context knowledge [3]. Wan et al. built a new product investment model [4]. Researches about product innovation based on SC frame have also emerged. Huo has studied the R&D strategy in SC using game theory [5].At present, the researches concerning product renewal have been carried out. Bass P. I. and Bass F. M. had proposed a new product expansion model and studied the process of multigeneration renewal products coming onto market [6, 7]. Wu examined the reuse/redesign, quality, speed-to-market, and marketing decisions for two consecutive generations of a multicomponent modular product, utilizing stylized models [8]. Druehl et al.  developed a model to gain insight into which factors drive the pace of product updates [9]. Huang and Ling studied the tracks of product update of ICT enterprises [10].  Koca et al. studied the product rollover strategy decision, where a firm decides whether to phase out an old generation product to be replaced by a new with either a dual or single roll [11]. Quan analyzed the relationship of the optimal launch time with parameters in the process of introducing a renewal product [12].The other focus of product renewal is on the marketing process of renewal product, especially pricing decision. Luo and Tu studied a price decision problem for product renewal supply chain based on Nash game [13]. Wei studied the optimal decision of inventory and pricing in a supply chain, in which there are original products and renewal products [14]. But the work of product renewal is far less than enough.The RP and OP are virtually differentiated products. When they are on sale at the same market, they can substitute for each other. Compared with common differentiated products, the substitution of RP for OP is unidirectional and partial. Therefore, the coming of RP onto market will cause some influences on OP market. At the same time, each entity in SC has independent profit. Therefore, the coordination of SC will become much more complex if there are RP and OP in the SC. A potent tool to coordinate SC is price. The fluctuation of price will make supply and demand tend to an equilibrium. On the other hand, price’s fluctuation can make the profit of each entity in SC distributed reasonably, so as to achieve the aim of coordination of SC. Therefore pricing strategy is the main content of the paper.The SC including OP and RP is called product renewal SC. The change of market share caused by product renewal in the SC is studied in this paper. When there is a RP in SC, the enterprises in SC will make strategies for both of OP and RP to achieve an equilibrium of enterprises’ profits. Compared with other researches, this paper firstly studied Stackelberg game for product renewal, aiming at solving complex decision-making question in actual renewal SC, and obtained some novel insights.The rest of this paper is organized as follows. Product renewal model is formulated in Section2. In Section 3, we provide a Stackelberg-Nash Game based on product renewal model. Stackelberg-merger game is played in Section 4, and numerical examples are provided in Section 5. Finally, we conclude our paper in Section 6.


## 2. Model Description

### 2.1. Assumption

Consider a two-stage SC which is composed of one manufacturer and one retailer. First, suppose the manufacturer produces productA and sells it in market via the retailer. After a period of renewal process, the manufacturer develops product A’s renewal product, B, and sells it with product A in market via the retailer.In order to fully describe the model of the SC, we state the following hypotheses.(i)
The retailer is in a monopoly position; that is, only the retailer sellsA and B in market.(ii)
Suppose that the consumer’s repurchase is possible  and each time one customer only buys one product. When there is an RP, some customers of OP may shift to buying RP in other words, there is repetition purchase for the products studied in this paper, such as toys, and small electric appliance equipment.(iii)
There is no stochastic demand in the market, so the demand is only decided by price.(iv)
Inventory is not considered.(v)
The manufacturer’s productivity is sufficient enough to satisfy demand.(vi)
There is no fixed cost for unit product, and margin production cost is an invariant for manufacturer.(vii)
The product renewal generally causes its performance promotion.The SC model of this paper is as follows: there is one manufacturer and one retailer in SC. Suppose that the manufacturer produces productsA and B at marginal costs CA and CB, respectively, and then distributes them at the wholesale price of fA and fB to retailer. Based on the wholesale price, a price markup gA and gB is, respectively, added by retailer to product A and B to form the market retail price: Pi=gi+fi. Retailer’s selling cost is c per unit. The product renewal SC including one manufacturer and one retailer that sells combination products is shown in Figure 1.Figure 1
The structure of renewal products SC.


### 2.2. Market Model

Product renewal rate is used to measure the change of product performance caused by product renewal. Combined with price and performance price ratio(Eij=kPj/Pi), product renewal rate can be used to forecast the market share change. In fact, what is focused on is not performance value, but the ratio of OP’s performance value to RP’s [15], namely renewal rate k(k>1).The RP marketing process can be divided into two stages:In Stage 1. there is only OP in market. Supposing the OP price ofA is PA, according to [15], the market demand for A is
(1)T=a-bPA.In expression (1), a means market scale, the absolute value of b denotes price flexibility, and b>0.In Stage 2. RP comes onto the market and competes with OP. If RP comes onto the market, it will cause several changes in the market:(1) RP will stimulate new consumption demand because of its appended or upgraded functions, which makes the latent customer quantity of the total products market enlarge, and the increment is ΔT. ΔT is related to original market demand T and a decreasing function of RP price, while an increasing function of renewal rate k. (2) Customers have the tendency to purchase RP. In addition, the performance price ratio of RP is generally higher than that of OP. So some OP’s customers will shift to buying RP. The number of shift customers is marked as Ts. (3) This will cause OP’s price to be adjusted to market.Definition 1.
There exists an OPA and its RP B in the market. Suppose that the demand of A is T, and the product renewal rate is k.Then ΔT satisfies the following function:
(2)ΔT=w1T(k-w2PB),
where w1 is the influence coefficient of T and 0<w1<1. w2 is the influence coefficient of RP’s price on its market increment, and w2≪1.FromΔT=0 and T=0, PB1=k/w2 and PA1=a/b can be got. If PA>PA1, OP’s demand will be zero, which leads ΔT to be zero. If PB>PB1, no matter how great OP’s demand is, demand increment ΔT will be equal to zero.Theorem 2.
The market shift functionTs relates to OP price, RP price, and OP demand. Ts has the following features.(i)
IfPB≤PA, Ts=T, that is, all of the OP demand will shift to RP. So OP will have to exit from the market, and RP will hold the market.(ii)
ExistingPB0>PA, if PB≥PB0, Ts=0. PB0 can be treated as the zero point of market shift.(iii)
IfPA<PB<PB0, Ts will be a monotone decreasing function of PB, namely, ∂Ts/∂PB<0.(iv)
IfPA<PB<PB0, Ts is to be a monotone increasing function of PA, namely, ∂Ts/∂PA>0.(v)
With the increase ofk, PB0 increases correspondingly. Ts is an increasing function of k, namely, ∂Ts/∂k>0.(vi)
LetΔT=0, and the corresponding price is PB1, which is the upper limit of RP price.(vii)
LetTs=0, and there is still demand increment caused by RP; that is to say, on any condition, PB0<PB1.(viii)
Ts is a direct proportion increasing function with T.From Theorem2, given PA, the laws of changes of Ts and ΔT can be got with the variety of PB, as shown in Figure 2. The specific shape of this figure is changing with the variety of PA.Figure 2
The changes ofTs and ΔT with the variety of PB.According to Figure2, the shift function Ts in [0,PA] is
(3)Ts1=T.Definition 3.
According to Figure2, in [PA,PB0], there exist an OP A and its RP B; suppose the prices of A, B are PA, PB, respectively, and the demand of A is T. Then, Ts2 satisfies the following function:
(4)Ts2(PA,PB)=w3T(kw4PA-PB),
where w3 is the influence coefficient of PA and PB, 0<w1<1, w4 is the influence coefficient of PA, and w4>1/k.The functions (3) and (4) meet eight features listed previewed, and can be used as the paper’s model base. From the previous analysis, the shift function includes two portions, so it can be expressed as follows:(5)Ts={TPB≤PA,[w3T(kw4PA-PB)]+PB>PA.LettingTs=0 and solve (5), the zero point of market shift function PB0 can be got:
(6)PB0=kw4PA.
With the increase of k, PB0 obviously becomes higher. Therefore, when RP comes onto the market, the total demand will change. The market share of RP B is
(7)TB=ΔT+Ts,
and OP A is
(8)TA=T-Ts.In addition, product renewal will make marginal production cost change fromCA to CB.Figure2 is a variety chart of shift demand Ts in the situation that PA is given, so the shift demand Ts actually depends on PA and PB simultaneously. Accordingly, a three-dimensional coordinate can be built, PA-PB-Ts, as shown in Figure 3, and axis Ts is shift demand. Plane PA-PB is divided into the following four areas by six lines in Figure 3:Area I: the area surrounded by axisPA and lines PA=PB and PA=a/b,Area II: the area surrounded by linesPA=PB, PA=a/b and PB=kw4PA,Area III: the area surrounded by axisPB and lines PB=kw4PA, PA=a/b and PB=k/w2,Area IV: the areas excepting area I, II, and III.Figure 3
The change ofTs with the variety of PB and PA.In area I, LettingPA=0, Ts’s maximum value a can be obtained from (1) and (2). Given PA, the line Ts1 will be parallel to plane PA-PB, and it is clear that the slope of line Ts1 is kTs1=0. From (3), with the increasing of PA, T will become small. Meanwhile, Ts1 will become small. If PA=a/b, T=Ts1=0. Therefore, Ts will be a sloped plane in area I.In area II, givenPA, kTs2=-w3T<0 can be obtained from (5). The line Ts2 will become smoother and smoother with the increase of PA and kTs2. If PA=a/b,  kTs2=Ts2=0. Letting Ts2=0 in (4), a line l1:PB=kw4PA in PA-PB plane can be got. The line linking maximal Ts point (0,0,a) to point (a/b,a/b,0) is a Ts line named l2. The lines linking the points at which line l1 or l2 intersects with Ts2 constitute Ts complex curving surface; in other words, Ts in area II is a complex curving surface.In area III,PB>kw4PA, Ts=0 can be got from (5).ConsideringTs function’s feature (vii), it is obvious that market increment ΔT>0 in area I, II, and III.In area IV,PA>a/b or PB>k/w2, ΔT=Ts=0. OP and RP do not come onto the market in this area.The shift functions and increment functions in different areas are shown in Table1.Table 1
The table of area’s shift function and increment function.
T
s
(
P
A
,
P
B
)
Δ
T
(
P
A
,
P
B
)
Area I
T
(
P
A
)
w
1
T
(
P
A
)
(
k
-
w
2
P
B
)
Area II
w
3
T
(
P
A
)
(
k
w
4
P
A
-
P
B
)
w
1
T
(
P
A
)
(
k
-
w
2
P
B
)
Area III
0
w
1
T
(
P
A
)
(
k
-
w
2
P
B
)
Area IV
0
0According to the feature (vii) of shift function, the zero point of increment function is bigger than that of shift function, namely,PB1>PB0=kw4PA. From expression (1), it is known that the maximum of PA is a/b. So it requires b>aw2w4 to meet the seventh feature of shift function. Based on the previous  analysis, the retailer’s and manufacturer’s profits including the profits of OP and RP are, respectively, as follows:
(9)πr=(Ts+ΔT)(gB-c)+(T-Ts)(gA-c),πm=(ΔT+Ts)(fB-CB)+(T-Ts)(fA-CA)-Cg.
Where Cg is the fixed cost of the manufacturer. According to the price relationship between RP and OP in different areas, the profit functions can be shown as follows.(i) In area I,PB≤PA, ΔT>0 and Ts=T>0. The market share of RP B is ΔT+Ts, and OP A is zero. The profits of retailer and manufacturer are
(10)πr(gA,gB)=T(1+w1(k-w2PB))(gB-c)=(a-b(fA+gA))×(1+w1(k-w2(fB+gB)))(gB-c),πm(fA,fB)=T(1+w1(k-w2PB))(fB-CB)-Cg=(a-b(fA+gA))×(1+w1(k-w2(fB+gB)))(fB-CB)-Cg.(ii) In area II,PA≤PB≤PB0, ΔT>0 and Ts>T>0. The market share of RP B is ΔT+Ts, and OP A is T-Ts. The profits of retailer and manufacturer are(11)πr(gA,gB)=T{(gA-c)(w1(k-w2PB)+w3(kw4PA-PB))×(gB-c)+(1-w3(kw4PA-PB))×(gA-c)((w1(k-w2PB)+w3(kw4PA-PB))}=(a-b(fA+gA))×{(w1(k-w2(fB+gB))+w3(kw4(fA+gA)-(fB+gB)))(gB-c)+(1-w3(kw4(fA+gA)-(fB+gB)))×(gA-c)(},πm(fA,fB)=T{(w1(k-w2PB)+w3(kw4PA-PB))(fB-CB)+(1-w3(kw4PA-PB))(fA-CA)}-Cg=(a-b(fA+gA))×{(fA-CA)(w1(k-w2(fB+gB))+w3(kw4(fA+gA)-(fB+gB)))(fB-CB)+(1-w3(kw4(fA+gA)-(fB+gB)))×(fA-CA)(}-Cg.(iii) In area III,PB0≤PB≤PB1, ΔT>0 and Ts=0. RP B’s market share is ΔT, and OP A’s market share is T. The profits of retailer and manufacturer are
(12)πr(gA,gB)=T{w1(k-w2PB)(gB-c)+gA-c}=(a-b(fA+gA))×{w1(k-w2(fB+gB))(gB-c)+gA-c},πm(fA,fB)=T{w1(k-w2PB)(fB-CB)+fA-CA}-Cg=(a-b(fA+gA))×{w1(k-w2(fB+gB))×(fB-CB)+fA-CA}-Cg.(iv) In area IV,PB>PB1 or PA>PA1, ΔT=Ts=0. The prices of OP and RP are so high that it could not come onto the market. It is not necessary to consider this situation.


## 2.1. Assumption

Consider a two-stage SC which is composed of one manufacturer and one retailer. First, suppose the manufacturer produces productA and sells it in market via the retailer. After a period of renewal process, the manufacturer develops product A’s renewal product, B, and sells it with product A in market via the retailer.In order to fully describe the model of the SC, we state the following hypotheses.(i)
The retailer is in a monopoly position; that is, only the retailer sellsA and B in market.(ii)
Suppose that the consumer’s repurchase is possible  and each time one customer only buys one product. When there is an RP, some customers of OP may shift to buying RP in other words, there is repetition purchase for the products studied in this paper, such as toys, and small electric appliance equipment.(iii)
There is no stochastic demand in the market, so the demand is only decided by price.(iv)
Inventory is not considered.(v)
The manufacturer’s productivity is sufficient enough to satisfy demand.(vi)
There is no fixed cost for unit product, and margin production cost is an invariant for manufacturer.(vii)
The product renewal generally causes its performance promotion.The SC model of this paper is as follows: there is one manufacturer and one retailer in SC. Suppose that the manufacturer produces productsA and B at marginal costs CA and CB, respectively, and then distributes them at the wholesale price of fA and fB to retailer. Based on the wholesale price, a price markup gA and gB is, respectively, added by retailer to product A and B to form the market retail price: Pi=gi+fi. Retailer’s selling cost is c per unit. The product renewal SC including one manufacturer and one retailer that sells combination products is shown in Figure 1.Figure 1
The structure of renewal products SC.


## 2.2. Market Model

Product renewal rate is used to measure the change of product performance caused by product renewal. Combined with price and performance price ratio(Eij=kPj/Pi), product renewal rate can be used to forecast the market share change. In fact, what is focused on is not performance value, but the ratio of OP’s performance value to RP’s [15], namely renewal rate k(k>1).The RP marketing process can be divided into two stages:In Stage 1. there is only OP in market. Supposing the OP price ofA is PA, according to [15], the market demand for A is
(1)T=a-bPA.In expression (1), a means market scale, the absolute value of b denotes price flexibility, and b>0.In Stage 2. RP comes onto the market and competes with OP. If RP comes onto the market, it will cause several changes in the market:(1) RP will stimulate new consumption demand because of its appended or upgraded functions, which makes the latent customer quantity of the total products market enlarge, and the increment is ΔT. ΔT is related to original market demand T and a decreasing function of RP price, while an increasing function of renewal rate k. (2) Customers have the tendency to purchase RP. In addition, the performance price ratio of RP is generally higher than that of OP. So some OP’s customers will shift to buying RP. The number of shift customers is marked as Ts. (3) This will cause OP’s price to be adjusted to market.Definition 1.
There exists an OPA and its RP B in the market. Suppose that the demand of A is T, and the product renewal rate is k.Then ΔT satisfies the following function:
(2)ΔT=w1T(k-w2PB),
where w1 is the influence coefficient of T and 0<w1<1. w2 is the influence coefficient of RP’s price on its market increment, and w2≪1.FromΔT=0 and T=0, PB1=k/w2 and PA1=a/b can be got. If PA>PA1, OP’s demand will be zero, which leads ΔT to be zero. If PB>PB1, no matter how great OP’s demand is, demand increment ΔT will be equal to zero.Theorem 2.
The market shift functionTs relates to OP price, RP price, and OP demand. Ts has the following features.(i)
IfPB≤PA, Ts=T, that is, all of the OP demand will shift to RP. So OP will have to exit from the market, and RP will hold the market.(ii)
ExistingPB0>PA, if PB≥PB0, Ts=0. PB0 can be treated as the zero point of market shift.(iii)
IfPA<PB<PB0, Ts will be a monotone decreasing function of PB, namely, ∂Ts/∂PB<0.(iv)
IfPA<PB<PB0, Ts is to be a monotone increasing function of PA, namely, ∂Ts/∂PA>0.(v)
With the increase ofk, PB0 increases correspondingly. Ts is an increasing function of k, namely, ∂Ts/∂k>0.(vi)
LetΔT=0, and the corresponding price is PB1, which is the upper limit of RP price.(vii)
LetTs=0, and there is still demand increment caused by RP; that is to say, on any condition, PB0<PB1.(viii)
Ts is a direct proportion increasing function with T.From Theorem2, given PA, the laws of changes of Ts and ΔT can be got with the variety of PB, as shown in Figure 2. The specific shape of this figure is changing with the variety of PA.Figure 2
The changes ofTs and ΔT with the variety of PB.According to Figure2, the shift function Ts in [0,PA] is
(3)Ts1=T.Definition 3.
According to Figure2, in [PA,PB0], there exist an OP A and its RP B; suppose the prices of A, B are PA, PB, respectively, and the demand of A is T. Then, Ts2 satisfies the following function:
(4)Ts2(PA,PB)=w3T(kw4PA-PB),
where w3 is the influence coefficient of PA and PB, 0<w1<1, w4 is the influence coefficient of PA, and w4>1/k.The functions (3) and (4) meet eight features listed previewed, and can be used as the paper’s model base. From the previous analysis, the shift function includes two portions, so it can be expressed as follows:(5)Ts={TPB≤PA,[w3T(kw4PA-PB)]+PB>PA.LettingTs=0 and solve (5), the zero point of market shift function PB0 can be got:
(6)PB0=kw4PA.
With the increase of k, PB0 obviously becomes higher. Therefore, when RP comes onto the market, the total demand will change. The market share of RP B is
(7)TB=ΔT+Ts,
and OP A is
(8)TA=T-Ts.In addition, product renewal will make marginal production cost change fromCA to CB.Figure2 is a variety chart of shift demand Ts in the situation that PA is given, so the shift demand Ts actually depends on PA and PB simultaneously. Accordingly, a three-dimensional coordinate can be built, PA-PB-Ts, as shown in Figure 3, and axis Ts is shift demand. Plane PA-PB is divided into the following four areas by six lines in Figure 3:Area I: the area surrounded by axisPA and lines PA=PB and PA=a/b,Area II: the area surrounded by linesPA=PB, PA=a/b and PB=kw4PA,Area III: the area surrounded by axisPB and lines PB=kw4PA, PA=a/b and PB=k/w2,Area IV: the areas excepting area I, II, and III.Figure 3
The change ofTs with the variety of PB and PA.In area I, LettingPA=0, Ts’s maximum value a can be obtained from (1) and (2). Given PA, the line Ts1 will be parallel to plane PA-PB, and it is clear that the slope of line Ts1 is kTs1=0. From (3), with the increasing of PA, T will become small. Meanwhile, Ts1 will become small. If PA=a/b, T=Ts1=0. Therefore, Ts will be a sloped plane in area I.In area II, givenPA, kTs2=-w3T<0 can be obtained from (5). The line Ts2 will become smoother and smoother with the increase of PA and kTs2. If PA=a/b,  kTs2=Ts2=0. Letting Ts2=0 in (4), a line l1:PB=kw4PA in PA-PB plane can be got. The line linking maximal Ts point (0,0,a) to point (a/b,a/b,0) is a Ts line named l2. The lines linking the points at which line l1 or l2 intersects with Ts2 constitute Ts complex curving surface; in other words, Ts in area II is a complex curving surface.In area III,PB>kw4PA, Ts=0 can be got from (5).ConsideringTs function’s feature (vii), it is obvious that market increment ΔT>0 in area I, II, and III.In area IV,PA>a/b or PB>k/w2, ΔT=Ts=0. OP and RP do not come onto the market in this area.The shift functions and increment functions in different areas are shown in Table1.Table 1
The table of area’s shift function and increment function.
T
s
(
P
A
,
P
B
)
Δ
T
(
P
A
,
P
B
)
Area I
T
(
P
A
)
w
1
T
(
P
A
)
(
k
-
w
2
P
B
)
Area II
w
3
T
(
P
A
)
(
k
w
4
P
A
-
P
B
)
w
1
T
(
P
A
)
(
k
-
w
2
P
B
)
Area III
0
w
1
T
(
P
A
)
(
k
-
w
2
P
B
)
Area IV
0
0According to the feature (vii) of shift function, the zero point of increment function is bigger than that of shift function, namely,PB1>PB0=kw4PA. From expression (1), it is known that the maximum of PA is a/b. So it requires b>aw2w4 to meet the seventh feature of shift function. Based on the previous  analysis, the retailer’s and manufacturer’s profits including the profits of OP and RP are, respectively, as follows:
(9)πr=(Ts+ΔT)(gB-c)+(T-Ts)(gA-c),πm=(ΔT+Ts)(fB-CB)+(T-Ts)(fA-CA)-Cg.
Where Cg is the fixed cost of the manufacturer. According to the price relationship between RP and OP in different areas, the profit functions can be shown as follows.(i) In area I,PB≤PA, ΔT>0 and Ts=T>0. The market share of RP B is ΔT+Ts, and OP A is zero. The profits of retailer and manufacturer are
(10)πr(gA,gB)=T(1+w1(k-w2PB))(gB-c)=(a-b(fA+gA))×(1+w1(k-w2(fB+gB)))(gB-c),πm(fA,fB)=T(1+w1(k-w2PB))(fB-CB)-Cg=(a-b(fA+gA))×(1+w1(k-w2(fB+gB)))(fB-CB)-Cg.(ii) In area II,PA≤PB≤PB0, ΔT>0 and Ts>T>0. The market share of RP B is ΔT+Ts, and OP A is T-Ts. The profits of retailer and manufacturer are(11)πr(gA,gB)=T{(gA-c)(w1(k-w2PB)+w3(kw4PA-PB))×(gB-c)+(1-w3(kw4PA-PB))×(gA-c)((w1(k-w2PB)+w3(kw4PA-PB))}=(a-b(fA+gA))×{(w1(k-w2(fB+gB))+w3(kw4(fA+gA)-(fB+gB)))(gB-c)+(1-w3(kw4(fA+gA)-(fB+gB)))×(gA-c)(},πm(fA,fB)=T{(w1(k-w2PB)+w3(kw4PA-PB))(fB-CB)+(1-w3(kw4PA-PB))(fA-CA)}-Cg=(a-b(fA+gA))×{(fA-CA)(w1(k-w2(fB+gB))+w3(kw4(fA+gA)-(fB+gB)))(fB-CB)+(1-w3(kw4(fA+gA)-(fB+gB)))×(fA-CA)(}-Cg.(iii) In area III,PB0≤PB≤PB1, ΔT>0 and Ts=0. RP B’s market share is ΔT, and OP A’s market share is T. The profits of retailer and manufacturer are
(12)πr(gA,gB)=T{w1(k-w2PB)(gB-c)+gA-c}=(a-b(fA+gA))×{w1(k-w2(fB+gB))(gB-c)+gA-c},πm(fA,fB)=T{w1(k-w2PB)(fB-CB)+fA-CA}-Cg=(a-b(fA+gA))×{w1(k-w2(fB+gB))×(fB-CB)+fA-CA}-Cg.(iv) In area IV,PB>PB1 or PA>PA1, ΔT=Ts=0. The prices of OP and RP are so high that it could not come onto the market. It is not necessary to consider this situation.


## 3. Stackelberg-Nash Game

Product renewal process can be regarded as two stages. First, manufacturer and retailer make their optimal decisions for OPA, respectively, and synchronously. After OP A comes onto market, RP B is developed by the manufacturer and comes onto market, too. In this condition, both OP A and RP B are sold in the same market. Then the manufacturer and retailer make their optimal decisions for RP B, respectively, and synchronously. The optimal decision-making process is also divided into two stages. The two-stage games constitute a Stackelberg-Nash game, which is the real process of product renewal game.In Stage 1: There is a decision for productA. The manufacturer and the retailer play a Nash game. The manufacturer makes a strategic decision of fA, and the retailer makes a strategic decision of gA.In Stage 2: There is a decision for productB. The manufacturer and the retailer also play a Nash game. The manufacturer makes a strategic decision of fB, and the retailer makes a strategic decision of gB.The game solving process is divided into three steps. Following three steps, the optimal solution of product renewal Stackelberg-Nash game can be derived.In Step 1, givenfA and gA, according to the Nash equilibrium of the manufacturer and retailer profit functions, the optimal reaction function of fB(fA,gA), gB(fA,gA) or the optimal value of gB and fB can be obtained from ∂πr/∂gB=0, ∂πm/∂fB=0.In Step 2, put the optimal reaction function or the optimal value into the profit functions of Step 1, and the optimal solutionfA* and gA* can be obtained from the Nash equilibrium of the manufacturer and retailer profit functions in Step 1.In Step 3, lettingfA* and gA* substitute fA and gA in the fB(fA,gA) and gB(fA,gA) of Step 2, the optimal value of fB* and gB* can be obtained.Theorem 4.
In Stackelberg-Nash game, in areas I, II, and III, givenfA and gA,  πr is a convex function of gB, and πm is convex functions of fB.Proof.
In area I(13)∂πr(gA,gB)∂gB=(a-b(fA+gA))×(-2w1w2gB+w1w2(c-fB)+w1k+1),∂πm(fA,fB)∂fB=(a-b(fA+gA))×(-2w1w2fB+w1w2(CB-gB)+w1k+1),∂πr2(gA,gB)∂gB2=-2w1w2(a-b(fA+gA))<0,∂πm2(fA,fB)∂fB2=-2w1w2(a-b(fA+gA))<0.
Therefore,πr is a convex function of gB, and πm is a convex function of fB. The proofs in II and III are the same as area I.Theorem 5.
The Stackelberg-Nash decision-making process can be divided into the following three situations.(i)
In area I, there is no Stackelberg-Nash equilibrium.(ii)
In area II, there exists only one optimal pricing strategy for Stackelberg-Nash game.(iii)
In area III, there exists only one optimal pricing strategy for Stackelberg-Nash game.Proof.
(i) In area I, according to (10), based on Theorem 4 and Step 1, the optimal value of gB and fB can be derived as follows:
(14)gB*=(1+w1k+3w1w2(2c-CB))(3w1w2),fB*=(1+w1k+3w1w2(2CB-c))(3w1w2).
From the supply chain’s structure, it is obvious thatgB*+fB*=PB*. Because PB* is greater than PA1 (from Theorem 6), considering the condition of area I, PB≤PA, the conclusion that PB* is out of area I can be drawn. Accordingly, there is no Nash equilibrium in area I.
(ii) In area II,PA≤PB≤PB0. According to (11), based on Theorem 4 and Step 1, the optimal reaction (fB(fA,gA),gB(fA,gA)) of the manufacturer and retailer can be got as follows:
(15)fB(fA,gA)=(w3(kw4+2)fA+w3(kw4-1)gA+w1k+2w3(CB-CA)+w1w2(2CB-c))/(3w3+3w1w2),gB(fA,gA)=(w3(kw4-1)fA+w3(kw4+2)gA+w1k+w3(CA-CB)+w1w2(2c-CB))/(3w3+3w1w2).
From Step 2, two binary cubic equations can be got. Because the analytic solution of the equations cannot be found, the numerical method can be used to find the optimal approximate solution(fA*,gA*). From Step 3, the optimal of (fB*,gB*) can be got.
(iii) In area III,PB0≤PB≤PB1. According to (12), based on Theorem 4 and Step 1, the optimal values of gB and fB of πr and πm are as follows:
(16)gB*=(k+w2(2c-CB))(3w2),fB*=(k+w2(2CB-c))(3w2).
From Step 2, the optimal values gA* and fA* of πr and πm can be got as follows:
(17)gA*=-(w1b(k-w2CB)2+2w1w2bc(w2CB-1)+w1bc2w22+9bw2(CA-2c)-9000w2(k-w2CB)2)/(27w1b),fA*=-(w1b(k-w2CB)2+2w1w2bc(w2CB-1)+w1bc2w22+9bw2(c-2CA)-9000w2(k-w2CB)2)/(27w1b).Theorem 6.
The value of Nah equilibrium for area I,PB*, is out of area I.Proof.
From the supply chain structure, it is obvious that:gB*+fB*=PB*:
(18)PB*=gB*+fB*=(2+2w1k)(3w1w2)+CB-c.
Because of 0<w1<1 and k>1,
(19)PB*≥(2+2k)(3w2)+CB-c≥4(3w2)+CB-c,
whereas w2≪1,
(20)PB*≥PA1=ab.
It is clear that the value of Nah equilibrium, PB*, is greater than PA1. Since PA1 is the maximum of PA in area I, considering the condition of area I, PB≤PA, the value of Nah equilibrium, PB*, does not satisfy this condition. So PB* is out of area I.


## 4. Stackelberg-Merger Game

In the decision-making process of Stackelberg-Nash game mentioned previously, if the manufacturer and retailer constitute a manufacturing and sales league to maximize the SC profit, that is, there is no retailer, and manufacturer sells its product in market directly, the product renewal Stackelberg-Nash decision-making process will become a Stackelberg-merger decision-making process. This process can be divided into two stages, and this process will play a Stackelberg-merger game. In this condition, the league which consists of the manufacturer and the retailer can be regarded as a enterprise, because the league’s decision variables are the retail pricePi of OP and RP (fi and gi can be regarded as the transfer price of the manufacturer and retailer profits distribution). In Stage 1, the league sells OPA and makes a strategic decision of PA.In Stage 2, the league sells OPA and RP B at the same time, and the league makes strategic decisions of PB and PA.The solving process of Stackelberg-merger game is divided into three steps, and by three steps the optimal solution of product renewal Stackelberg-merger game can be derived.In Step 1, OP retail pricePA is given; the optimal reaction function  PB(PA) or the optimal value of PB can be obtained from ∂π/∂PB=0.In Step 2, put the optimal reaction functionPB(PA) or the optimal value PB* into the first stage; the optimal pricing PA* can be obtained from ∂π/∂PA=0.In Step 3, putPA* into the optimal reaction function PB(PA); the optimal pricing PB* can be derived.Theorem 7.
In Stackelberg-merger game, in areas I, II, and III, givenPA, π  is a convex function of PB.Proof.
In area I(21)∂π(PA,PB)∂PB=(a-bPA)×(-2w1w2PB+w1w2(CB+c)+w1k+1),(22)∂π2(PA,PB)∂PB2=-2w1w2(a-bPA)<0.
Therefore, π  is a convex function of PB. The proofs in II and III are the same as those of area I.Theorem 8.
The process of the Stackelberg-merger game is divided into three situations according to areas I, II, III. There exists only one optimal pricing strategy for Stackelberg-merger game in each situation.Proof.
(i) In area I,PB≤PA. the total profit of the SC can be got from the sum of (10):
(23)π(PA,PB)=(a-bPA)(w1(k-w2PB)+1)×(PB-CB-c)-Cg.
Base on Theorem7 and Step 1, the optimal value PB of π can be obtained as follows:
(24)PB*=(1+w1k)(2w1w2)+(CB+c)2.
The value ofw2 is so small that the value of PB* is greater than the maximum value of PA1. Considering PB≤PA and Theorem 7, the maximum value of π can be obtained in the condition that PB=PA, and then (21) can be expressed as follows:
(25)π(PA)=(a-bPA)(w1(k-w2PA)+1)(PA-CB-c)-Cg.
From∂π/∂PA=0, the two values of PA can be got. One solution is so small that it cannot match the situation, and it is cast out. The other is the optimal value of PA. So there is only one solution for area I (from Theorem 9).
(ii) In area II, the total profit of SC can be got from the sum of (11):
(26)π(PA,PB)=(a-bPA)×{)(w3(kw4PA-PB)+w1(k-w2PB))×(PB-CB-c)+(1-w3(kw4PA-PB))(PA-CA-c)}-Cg.
Based on Theorem7 and Step 1, the optimal reaction PB(PA) of π can be obtained as follows:
(27)PB(PA)=(w3(kw4+1)PA+w1w2(CB+c)+w1k+w3CA)2(w1w2+w3).
From Step 2,PA* can be got, and the expression of PA* is so long that it could not be written down here.
From Step 3, the optimal value ofPB for π can be obtained.
(iii) In area III, the total profit of SC can be derived from the sum of (12):
(28)π(PA,PB)=(a-bPA)×(w1(k-w2PB)(PB-CB-c)+PA-CA-c).
Based on Theorem7 and Step 1, the optimal value of PB for π  can be obtained as follows:
(29)PB*=k2w2+(CB+c)2.
From Step 2, the optimal ofPA for π can be obtained as follows:
(30)PA*=a2b-(w1(k-(CB+c))(k-w2(CB+c))-4(CA+c))8.Theorem 9.
In area I, the optimal value ofPA is the only one valid solution, which matches the real situation.Proof.
In area I, from∂π(PA)/∂PA=0,
(31)3bw1w2PA2-(2w1w2a+b(CB+c)(w1w2))+(w1k+1)(a-b(CB+c))+w1w2a(CB+c)=0.
Two PA can be got from (31):
(32)PA=((2w1w2a+b(CB+c)(w1w2-1))±F)(6bw1w2),
where
(33)F=(2w1w2a+b(CB+c)(w1w2))2-12bw1w2{(w1k+1)(a-b(CB+c))+w1w2a(CB+c)}.
In (32), PA with the negative value of sqrt F is
(34)PA1=((2w1w2a+b(CB+c)(w1w2-1))-F)(6bw1w2)≈a(3b)+(CB+c)6-1(6w1w2)-F(6bw1w2),
where
(35)F(6bw1w2)≈a(3b)+(CB+c)6-1(6w1w2).
Substituting (35) in (36),
(36)PA1≈0<CB+c.
It is obvious that the solution cannot match the real situation, and it was cast out. Therefore, the optimal value of PA is the only one valid solution, which matches the real situation.
The key of the Stackelberg-merger game is how to distribute the total profit between manufacturer and retailer. Using Nash bargaining model can coordinate profit of each entity [16], and the study of this field needs to go further.


## 5. Numerical Simulation

### 5.1. Stackelberg-Nash Game Simulation

Look at mobile phone, for example. There are two types of mobile phoneA and B in the same market, and B is the renewal product of A. The related parameters’ values are as follows: a=1000, b=2, w1=0.8, w2=0.001, w3=0.008, w4=1.2, CA=200. Table 2 shows the effect of the change of RP cost on price, demand, and profits of both products in the condition k=1.2. Table 3 shows the effect in the condition k=1.4. From Tables 2 and 3, the effect of renewal rate k on price, demand, and profits can be concluded.Table 2
The effect of RP cost changing in Stackelberg-Nash game (k=1.2).
Case result
k
=
1.2 CB=220
k
=
1.2 CB=250
P
B
≤
P
A
P
A
<
P
B
<
P
B
0
P
B
0
<
P
B
<
P
B
1
P
B
≤
P
A
P
A
<
P
B
<
P
B
0
P
B
0
<
P
B
<
P
B
1
f
A
284.3316
266.0267
286.2994
267.7067
g
A
104.3316
86.0267
106.2994
87.7067
f
B
371.5171
540.0000
393.8308
560.0000
g
B
171.5171
340.0000
163.8308
330.0000
P
A
365.9493
388.6632
352.0533
381.8052
392.5987
355.4133
P
B
365.9493
543.0343
880.0000
381.8052
557.6615
890.0000
T
268.1014
222.6737
295.8933
236.3896
214.8025
289.1733
Δ
T
178.8881
117.0312
75.7487
154.7302
110.3807
71.7150
T
s
268.1014
29.6436
0
236.3896
13.1986
0
T
A
0
193.0301
295.8933
0
201.6040
289.1733
T
B
536.2027
252.3172
295.8933
472.7791
228.0011
289.1733
π
r
38502
43776
35173
41811
π
m
28502
33776
25173
31811
π
46298
67005
77553
33729
60346
73621Table 3
The effect of RP cost changing in Stackelberg-Nash decision (k=1.4).
Case result
k
=
1.4 CB=220
k
=
1.4 CB=250
P
B
≤
P
A
P
A
<
P
B
<
P
B
0
P
B
0
<
P
B
<
P
B
1
P
B
≤
P
A
P
A
<
P
B
<
P
B
0
P
B
0
<
P
B
<
P
B
1
f
A
279.2893
253.4637
281.5733
255.4993
g
A
99.2893
73.4637
101.5733
75.4993
f
B
399.1822
606.6667
422.1999
626.6667
g
B
199.1822
406.6667
192.1999
396.6667
P
A
366.3029
378.5785
326.9000
382.0862
383.1467
331.0000
P
B
366.3029
598.3644
1013.3000
382.0862
614.3999
1023.3000
T
267.3943
242.8429
346.1452
232.8276
233.7066
338.0030
Δ
T
221.1238
155.7372
107.0742
192.0417
146.8800
101.8516
T
s
267.3943
73.1396
0
235.8276
54.7557
00
T
A
0
169.7034
346.1452
0
178.9509
338.0030
T
B
534.7886
315.9825
346.1452
471.6552
288.4624
338.0030
π
r
54466
59910
49319
57120
π
m
44466
49910
39319
47120
π
51701
98933
109820
37958
88639
104250From numerical simulation in Tables2 and 3, the following conclusions can be drawn.(1)
There is no Stackelberg-Nash equilibrium in area I, and so Stackelberg-merger is used to simulate this situation. The total profit in this situation is the optimal profit of area I. Obviously, the optimal total profit of area I is less than those of areas II and III. Therefore, area I(PB=PA) is not the optimal area of Stackelberg-Nash game. The rational manufacturer and retailer could not select area I.(2)
For all parameter conditions, the profits of manufacturer, retailer, and the total profit of SC in area III are greater than those of area II (in area II, there is a little difference between the price of RP and OP). Because the manufacturer and retailer are rational, the final decision-making result(PA,PB) of the whole SC should be in area III. In this condition, OP is set at a low price (break-even sales), while RP is priced at a high level (the price differential between RP and OP is huge); that is to say, RP comes onto the market at a high price, and OP comes onto the market at a low price, which leads to larger market scale and wider influence on product. In this case, even if RP is set at a high price, the shift demand and the increment demand of market are proportional to the size of OP scale, so high price has less influence on market demand. But the profit of unit renewal product rises substantially with its price increase. In this situation, RP comes onto the market with a great price differential, and so the profits of all products are the highest; that is, area III is the optimal pricing area of supply chain.(3)
The profit of retailer in areas II and III is greater than that of manufacturer, mainly because the retailer is a direct participator of market, and it reacts very rapidly when the demand of market changes. Therefore, the retailer can adjust the process of decision making in the fastest time to increase revenue and reduce losses. The manufacturer’s reaction to the market relies on the retailer decision making information’s transmission, and there is a delay in the transmission process. Accordingly, the manufacturer’s profit is less than that of the retailer.(4)
With the cost of RP increasing, the profits of the manufacturer, retailer, and total will decrease.(5)
If the cost of RP is constant, with product renewal ratek increasing, the profits of the manufacturer, retailer and total will increase.


### 5.2. Stackelberg-Merger Game Simulation

The parameters are the same as Section5.1. With the changing of product renewal rate or the cost of RP, the variety of price, demand, and profits are shown in Tables 4 and 5.Table 4
The effect of RP cost changing in Stackelberg-merger model (k=1.2).
Case result
k
=
1.2 CB=220
k
=
1.2 CB=250
P
B
≤
P
A
P
A
<
P
B
<
P
B
0
P
B
0
<
P
B
<
P
B
1
P
B
≤
P
A
P
A
<
P
B
<
P
B
0
P
B
0
<
P
B
<
P
B
1
P
A
365.9493
346.2141
267.8400
381.8052
346.2141
273.5100
P
B
365.9493
458.5284
720.0000
381.8052
473.5284
735.0000
T
268.1014
307.5718
464.3200
236.3896
307.5718
452.9800
Δ
T
178.8881
182.4446
178.2989
154.7302
178.7538
168.5086
T
s
268.1014
98.4720
0
236.3896
61.5634
0
T
A
0
209.0998
464.3200
0
246.0084
452.9800
T
B
536.2027
406.0438
464.3200
472.7791
369.1352
452.9800
π
46298
87780
107800
33729
79981
102600Table 5
The effect of RP cost changing in Stackelberg-merger model (k=1.4).
Case result
k
=
1.4 CB=220
k
=
1.4 CB=250
P
B
≤
P
A
P
A
<
P
B
<
P
B
0
P
B
0
<
P
B
<
P
B
1
P
B
≤
P
A
P
A
<
P
B
<
P
B
0
P
B
0
<
P
B
<
P
B
1
P
A
366.3029
346.2141
225.4400
382.0862
346.2141
232.3100
P
B
366.3029
505.3881
820.0000
382.0862
520.3881
835.0000
T
267.3943
307.5718
549.1200
232.8276
307.5718
535.3800
Δ
T
221.1238
220.1259
254.7917
192.0417
216.4351
241.9918
T
s
267.3943
187.6227
0
235.8276
150.7141
0
T
A
0
119.491
549.1200
0
156.8577
535.3800
T
B
534.7886
495.1945
549.1200
471.6552
458.2859
535.3800
π
51701
123350
150770
37958
111730
143320From numerical simulation in Tables4 and 5, the following conclusions can be drawn:(i)
The profits in areas II and III of Stackelberg-merger game are greater than those in area I, so the Stackelberg-merger pricing strategy would not be in area I. The situation is consistent with Stackelberg-Nash game.(ii)
If OP adopts low-price strategy, and RP adopts high-price strategy, the total profit will be maximum value; in other words, area III is the optimal merger pricing area for the SC. The situation and its reason are the same as those of Stackelberg-Nash game.(iii)
If the cost of RP is constant, with the product renewal rate increasing, the total profit will increase obviously.(iv)
With the cost of RP increasing, the total profit will decrease.From the analysis of Sections5.1 and 5.2, the following can be drawn: (i) Both two kinds of decision-making processes are consistent with each other in profits, cost, and product renewal rate changing (with the cost of RP increasing, profits decrease, with product renewal rate increasing, profits increase). (ii) Both cases obtain the optimal profits in area III; in other words, the RP which comes onto the market has a great price differential with OP. (iii) Compared with Stackelberg-Nash game, the price in Stackelberg-merger game is lower while the profits are more, so Stackelberg-merger decision making is virtually a kind of win-win situation for enterprises and market.


## 5.1. Stackelberg-Nash Game Simulation

Look at mobile phone, for example. There are two types of mobile phoneA and B in the same market, and B is the renewal product of A. The related parameters’ values are as follows: a=1000, b=2, w1=0.8, w2=0.001, w3=0.008, w4=1.2, CA=200. Table 2 shows the effect of the change of RP cost on price, demand, and profits of both products in the condition k=1.2. Table 3 shows the effect in the condition k=1.4. From Tables 2 and 3, the effect of renewal rate k on price, demand, and profits can be concluded.Table 2
The effect of RP cost changing in Stackelberg-Nash game (k=1.2).
Case result
k
=
1.2 CB=220
k
=
1.2 CB=250
P
B
≤
P
A
P
A
<
P
B
<
P
B
0
P
B
0
<
P
B
<
P
B
1
P
B
≤
P
A
P
A
<
P
B
<
P
B
0
P
B
0
<
P
B
<
P
B
1
f
A
284.3316
266.0267
286.2994
267.7067
g
A
104.3316
86.0267
106.2994
87.7067
f
B
371.5171
540.0000
393.8308
560.0000
g
B
171.5171
340.0000
163.8308
330.0000
P
A
365.9493
388.6632
352.0533
381.8052
392.5987
355.4133
P
B
365.9493
543.0343
880.0000
381.8052
557.6615
890.0000
T
268.1014
222.6737
295.8933
236.3896
214.8025
289.1733
Δ
T
178.8881
117.0312
75.7487
154.7302
110.3807
71.7150
T
s
268.1014
29.6436
0
236.3896
13.1986
0
T
A
0
193.0301
295.8933
0
201.6040
289.1733
T
B
536.2027
252.3172
295.8933
472.7791
228.0011
289.1733
π
r
38502
43776
35173
41811
π
m
28502
33776
25173
31811
π
46298
67005
77553
33729
60346
73621Table 3
The effect of RP cost changing in Stackelberg-Nash decision (k=1.4).
Case result
k
=
1.4 CB=220
k
=
1.4 CB=250
P
B
≤
P
A
P
A
<
P
B
<
P
B
0
P
B
0
<
P
B
<
P
B
1
P
B
≤
P
A
P
A
<
P
B
<
P
B
0
P
B
0
<
P
B
<
P
B
1
f
A
279.2893
253.4637
281.5733
255.4993
g
A
99.2893
73.4637
101.5733
75.4993
f
B
399.1822
606.6667
422.1999
626.6667
g
B
199.1822
406.6667
192.1999
396.6667
P
A
366.3029
378.5785
326.9000
382.0862
383.1467
331.0000
P
B
366.3029
598.3644
1013.3000
382.0862
614.3999
1023.3000
T
267.3943
242.8429
346.1452
232.8276
233.7066
338.0030
Δ
T
221.1238
155.7372
107.0742
192.0417
146.8800
101.8516
T
s
267.3943
73.1396
0
235.8276
54.7557
00
T
A
0
169.7034
346.1452
0
178.9509
338.0030
T
B
534.7886
315.9825
346.1452
471.6552
288.4624
338.0030
π
r
54466
59910
49319
57120
π
m
44466
49910
39319
47120
π
51701
98933
109820
37958
88639
104250From numerical simulation in Tables2 and 3, the following conclusions can be drawn.(1)
There is no Stackelberg-Nash equilibrium in area I, and so Stackelberg-merger is used to simulate this situation. The total profit in this situation is the optimal profit of area I. Obviously, the optimal total profit of area I is less than those of areas II and III. Therefore, area I(PB=PA) is not the optimal area of Stackelberg-Nash game. The rational manufacturer and retailer could not select area I.(2)
For all parameter conditions, the profits of manufacturer, retailer, and the total profit of SC in area III are greater than those of area II (in area II, there is a little difference between the price of RP and OP). Because the manufacturer and retailer are rational, the final decision-making result(PA,PB) of the whole SC should be in area III. In this condition, OP is set at a low price (break-even sales), while RP is priced at a high level (the price differential between RP and OP is huge); that is to say, RP comes onto the market at a high price, and OP comes onto the market at a low price, which leads to larger market scale and wider influence on product. In this case, even if RP is set at a high price, the shift demand and the increment demand of market are proportional to the size of OP scale, so high price has less influence on market demand. But the profit of unit renewal product rises substantially with its price increase. In this situation, RP comes onto the market with a great price differential, and so the profits of all products are the highest; that is, area III is the optimal pricing area of supply chain.(3)
The profit of retailer in areas II and III is greater than that of manufacturer, mainly because the retailer is a direct participator of market, and it reacts very rapidly when the demand of market changes. Therefore, the retailer can adjust the process of decision making in the fastest time to increase revenue and reduce losses. The manufacturer’s reaction to the market relies on the retailer decision making information’s transmission, and there is a delay in the transmission process. Accordingly, the manufacturer’s profit is less than that of the retailer.(4)
With the cost of RP increasing, the profits of the manufacturer, retailer, and total will decrease.(5)
If the cost of RP is constant, with product renewal ratek increasing, the profits of the manufacturer, retailer and total will increase.


## 5.2. Stackelberg-Merger Game Simulation

The parameters are the same as Section5.1. With the changing of product renewal rate or the cost of RP, the variety of price, demand, and profits are shown in Tables 4 and 5.Table 4
The effect of RP cost changing in Stackelberg-merger model (k=1.2).
Case result
k
=
1.2 CB=220
k
=
1.2 CB=250
P
B
≤
P
A
P
A
<
P
B
<
P
B
0
P
B
0
<
P
B
<
P
B
1
P
B
≤
P
A
P
A
<
P
B
<
P
B
0
P
B
0
<
P
B
<
P
B
1
P
A
365.9493
346.2141
267.8400
381.8052
346.2141
273.5100
P
B
365.9493
458.5284
720.0000
381.8052
473.5284
735.0000
T
268.1014
307.5718
464.3200
236.3896
307.5718
452.9800
Δ
T
178.8881
182.4446
178.2989
154.7302
178.7538
168.5086
T
s
268.1014
98.4720
0
236.3896
61.5634
0
T
A
0
209.0998
464.3200
0
246.0084
452.9800
T
B
536.2027
406.0438
464.3200
472.7791
369.1352
452.9800
π
46298
87780
107800
33729
79981
102600Table 5
The effect of RP cost changing in Stackelberg-merger model (k=1.4).
Case result
k
=
1.4 CB=220
k
=
1.4 CB=250
P
B
≤
P
A
P
A
<
P
B
<
P
B
0
P
B
0
<
P
B
<
P
B
1
P
B
≤
P
A
P
A
<
P
B
<
P
B
0
P
B
0
<
P
B
<
P
B
1
P
A
366.3029
346.2141
225.4400
382.0862
346.2141
232.3100
P
B
366.3029
505.3881
820.0000
382.0862
520.3881
835.0000
T
267.3943
307.5718
549.1200
232.8276
307.5718
535.3800
Δ
T
221.1238
220.1259
254.7917
192.0417
216.4351
241.9918
T
s
267.3943
187.6227
0
235.8276
150.7141
0
T
A
0
119.491
549.1200
0
156.8577
535.3800
T
B
534.7886
495.1945
549.1200
471.6552
458.2859
535.3800
π
51701
123350
150770
37958
111730
143320From numerical simulation in Tables4 and 5, the following conclusions can be drawn:(i)
The profits in areas II and III of Stackelberg-merger game are greater than those in area I, so the Stackelberg-merger pricing strategy would not be in area I. The situation is consistent with Stackelberg-Nash game.(ii)
If OP adopts low-price strategy, and RP adopts high-price strategy, the total profit will be maximum value; in other words, area III is the optimal merger pricing area for the SC. The situation and its reason are the same as those of Stackelberg-Nash game.(iii)
If the cost of RP is constant, with the product renewal rate increasing, the total profit will increase obviously.(iv)
With the cost of RP increasing, the total profit will decrease.From the analysis of Sections5.1 and 5.2, the following can be drawn: (i) Both two kinds of decision-making processes are consistent with each other in profits, cost, and product renewal rate changing (with the cost of RP increasing, profits decrease, with product renewal rate increasing, profits increase). (ii) Both cases obtain the optimal profits in area III; in other words, the RP which comes onto the market has a great price differential with OP. (iii) Compared with Stackelberg-Nash game, the price in Stackelberg-merger game is lower while the profits are more, so Stackelberg-merger decision making is virtually a kind of win-win situation for enterprises and market.


## 6. Conclusion

This paper studied the process of product renewal. A market shift model of RP was built on incremental function and shift function. Based on the model, Stackelberg-Nash game model and Stackelberg-merger game model for RP in SC were built, and their theoretical analysis was carried out. The following conclusions can be drawn. (i) In both two models, the increase of RP cost will make participants’ profits and total profit decrease, while higher renewal rate will make the profits increase at the same margin cost. (ii) Manufacturer and retailer obtain the optimal profits in area III; in other words, the way of the optimal decision making in SC is that RP comes onto the market with a great price differential with OP. (iii) Compared with Stackelberg-Nash game model, Stackelberg-merger game model’s pricing is lower and the profits are higher, which is actually a kind of win-win situation for enterprises and market.In this paper, a part of premise conditions of the model was built on the basis of some rational hypothesis, and the model parameters need to be confirmed by real statistics data. These problems need to be further studied.



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*Source: 726536-2013-03-13.xml*