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1、?27?3?20126?JOURNAL OF SYSTEMS ENGINEERINGVol.27 No.3Jun.2012?,?(?,?400044)?:?,?Stackelberg?,?.?,?,?.?,?.?,?.?:?;?;?;?:C931?:A?:10005781(2012)03034407Price discount model for coordination of dual-channelsupply chain under e-commerceXU Guang-ye,DAN Bin(College of Economics and Business Administration
2、,Chongqing University,Chongqing 400044,China)Abstract:For the dual-channel mode of the coexistence of traditional retail channel and electronic directchannel under e-commerce,this paper establishes a price discount model to coordinate dual-channel supplychain when the manufacturer is a leader in Sta
3、ckelberg game.Under the price discount coordination mech-anism,it is proved that the profits of the dual-channel supply chain and retailer are increased,but the profitof the manufacturer is reduced.With regards to this,a transfer payment mechanism is designed to enable themanufacturer and retailer t
4、o achieve a win-win situation.Finally,a numerical example is used to verify theeffectiveness of the designed coordination mechanisms.Key words:dual-channel;supply chain coordination;price discount;e-commerce1?,?,?IBM?1.?,?“?”?(double marginalization)2,?3,4.?,?,?,?,?,?5.?,?,?,?.?:20091222;?:20100506.
5、?:?(70972056);?(CDJXS11022202).?3?:?345?,?.Chiang?6?,?,?,?.?,Yan?7?,?,?,?.Tsay?8?,?.Kurata?9?,?.?10?Stackelberg?Bertrand?,?,?,?Pareto?.?,?.?,?.?,Seong?11?,?,?.?12?,?,?,?.?13?,?;Chiang14?,?;?Boyaci5?,?.?5,13,14?,?,?,?.?13?,?;?5,14?,?.?,?,?,?,?.?.2?,?,?.?,?.?,?1?,?2?.?I?,?PD?,?NPD?,?*?.?:1?;2?;d1?;d2?
6、;w?;c?;p1?,p1 w;p2?,p2 w.?,?.?d1=sa p1+bp2,(1)d2=(1 s)a p2+bp1,(2)?a(a 0)?,s(0 s 1)?,b(0 b 0,(1 s)a p2 0,?.3?,?.?,?.?,?II=(p1 c)(sa p1+bp2)+(p2 c)(1 s)a p2+bp1).(3)?(3)?,?,?(3)?p1?p2?0,?pI1=(s+(1 s)b)a+(1 b2)c2(1 b2),(4)pI2=(1 s)+sb)a+(1 b2)c2(1 b2).(5)?,?I=?12(1 s)(1 b)sb?a2(1 b2)ca+(b3 b+1)c22(1 b
7、2).(6)?,?,?.4?,?,?,?Stackelberg?(leader),?(follower).?w(p2)=c+k(p2 c).(7)?k?,?0 k 1,?w(p2)?c w(p2)0,?sa c sa pPD1=sa p1 0,?pPD1 wPD 0,?.?.?1?,?,?.?,?PD1=(sa (1 b)c)24,(21)PD2=(1 s+sb)a (1 b2)c)24(1 b2).(22)?(21)?(22)?PD=PD1+PD2=?12(1 s)(1 b)sb?a2(1 b2)ca+(b3 b+1)c22(1 b2)=I.(23)?,?,?.?,?,?.?,?,?;?,?
8、,?.?,?348?27?Pareto?,?,?,?.?,?.5?Pareto?,?,?NPD1=(p1 w)(sa p1+bp2),(24)NPD2=(w c)(sa p1+bp2)+(p2 c)(1 s)a p2+bp1).(25)?,?(24)?(25)?,?(24)?(25),?,?NPD1=(sa+bc c)216,(26)NPD2=(s2b2+4(1 s)sb+3s2 4s+2)a28(1 b2)2(1 b2)(2 s+bs)ca+(1 b2)(b2 2b+3)c28(1 b2),(27)NPD=(s2b2+8(1 s)sb+7s2 8s+4)a28(1 b2)2(1 b2)(4
9、s+bs)ca+(1 b2)(b2+6b 7)c28(1 b2).(28)?,?,?2.?2?,?.?(23)?(28)?,?=(sa+bc c)2/16 0,(29)?(21)?(26),?11=3(sa+bc c)2/16 0,(30)?(22)?(27),?22=(sa+bc c)2/8 0.(31)?(29)?(30)?(31)?.?.?2?,?.?Stackelberg?,?.?,?,?Pareto?,?3.?3?,?T?T 2,1?,?Pareto?.?(30)?(31)?3,?.?3?,?,?.?,?,?3?:?349?,?.?,?,?,?.6?,?.?a=200,?s=0.4,
10、c=10,1=1 T,2=2+T.?k?b,?.?1b?,T?,?1?.?1?4?7?10?,?T?,?b?;?b?;?b?.?,?T?,?,?,?,?,?b?.?,?1?,?b=0.8?,?,?,?T=750?2=760?3.?,?,?T.?1?,?3,?,?Pareto?.?1T=750?Table 1 The optimal result of T=750bPD1NPD11PD2NPD22PDNPD0.3582.25333.06249.195 749.455 665.5883.876 331.705 998.64333.060.4619.00342.25276.756 887.196 8
11、21.6965.507 506.197 163.94342.250.5656.25351.56304.698 502.088 455.2146.879 158.338 806.77351.560.6694.00361.00333.0010 951.0010 923.0028.0011 645.0011 284.00361.000.7732.25370.56361.6915 067.0615 058.198.8715 799.3115 428.75370.560.8771.00380.25390.7523 350.1123 360.61 10.5024 121.1123 740.86380.25
12、?2T?,b?,?2?.?2?4?7?10?,?b?,?T?;?T?;?,?.?b?,?,?T?,?.?,?2?2?3?4?,?T=600,650,700?,?,?,?T=600,650,700?2=703.125?3.?,?,?T.?2?4,?.?2b=0.5?Table 2 The optimal result of b=0.5TPD1NPD11PD2NPD22PDNPD600806.25351.56454.698352.088455.2103.139158.338806.77351.56650756.25351.56404.698402.088455.2153.139158.338806
13、.77351.56700706.25351.56354.698452.088455.21 3.139158.338806.77351.56750656.25351.56304.698502.088455.2146.879158.338806.77351.56800606.25351.56254.698552.088455.2196.879158.338806.77351.56?,?T?(?T=100),?,?Stackelberg?,?,?350?27?,?.7?.?,?,?.?,?,?.?,?,?,?.?,?.?:1 Wilder C.HPs online push N.Informatio
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