Price discount model for coordination of dual-channel supply chain under e-commerce

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½ 27 ½ 3 2012 6 JOURNAL OF SYSTEMS ENGINEERING Vol.27 No.3 Jun. 2012 ô Î ÆÆ î º žâ, Ê ( ï Ä Ò, ï 400044) ý : ô íûđ Î, ë Ǒ à Stackelberg, ÅÍÅÆÆÎ î º. ÝÅ îææ ë,ǒ ÍÅÇ Î, ðë ëä.ǒ, ÇÅè ëë ÍÅ ÎÁ., Ä Ù Å Ç ÆÆ ë. ³ : Î ; ÆÆ; î; : C931 ÞÁ : A : 1000 5781(2012)03 0344 07 Price discount model for coordination of dual-channel supply chain under e-commerce XU Guang-ye, DAN Bin ( College of Economics and Business Administration, Chongqing University, Chongqing 400044, China ) Abstract: For the dual-channel mode of the coexistence of traditional retail channel and electronic direct channel under e-commerce, this paper establishes a price discount model to coordinate dual-channel supply chain when the manufacturer is a leader in Stackelberg game. Under the price discount coordination mechanism, it is proved that the profits of the dual-channel supply chain and retailer are increased, but the profit of the manufacturer is reduced. With regards to this, a transfer payment mechanism is designed to enable the manufacturer and retailer to achieve a win-win situation. Finally, a numerical example is used to verify the effectiveness of the designed coordination mechanisms. Key words: dual-channel; supply chain coordination; price discount; e-commerce 1 ² Ù, íûđ Î Ǒë â í, íµ IBM  Đòë â æåî [1]. ô,ǒåî Å Ù Î ËÞ Ã( double marginalization ) [2],Á Ë Â Å ÆÆ ë [3,4]., Ù Î ë Éð Õð û, Î ÉĐ Î ËÞ ÃÕĐ ü Ã, è, ½Ù þ à, ßÆÆ èë è Î [5]., îî Î ËÞ ü, ÝÎ ÆÆ, ðî â Ã. þú : 2009 12 22; ú : 2010 05 06. á : Á á (70972056); ÅÔ â á (CDJXS11022202).

½ 3 žâ : ô Î ÆÆ î º 345,³ÙÎ ³ Ù Ñ ù û ÆÆ. Chiang [6] Ä Î í û, ÙÝë Ñ ½Ù, û Óð ±, Ð ÆÆ., Yan [7] Ǒë Ñ Ö ÂÔ, ä üþ, û Ä Å èù Ë, ÆÆÎ. Tsay [8] ÙÝë Ä í ½ó Æ, Ö Å Î. Kurata [9] û ô, ÇÅ ë ÅÆÆ.À [10] Ç Å Stackelberg Bertrand Î, Ǒ èù ½Ù Ë, ÑĐ ü, û ÓĐ ÆÆ ëè ù Ù è ð Pareto. ÞàðĐ Å Ñ û ÆÆ, Ú Đ Ñ û ÆÆ. û ð, ÙÎ îî Î ËÞ Ã ü à ±. à, Seong [11] Î ë î¹ý Í, Ð ÝÎ ÆÆ, ðè àåûú ÆÑ. ÝÎ [12] Î,² ÅÎ ÆÆà, Ñ ÆÆ Ñ Æ Æ, ûú Æ ÆÆ ë. ïù [13] Å ÝÎ ÆÆ î º,û ÄÆ î Îà ÎÁ; Chiang [14] ÇÅè Đ à í þ ë, Î ÅÆÆ; î Boyaci [5] Ǒ à í î, Î ÆÆ. Þ [5,13,14]Ǒ ÇÅ ÆÆ ë Î ÆÆ, ð, Ç ÆÆ ëðð Æ Ù, û ÝÆÆ ë Ë Æð ½. Þ[13] Đ ðë, ûú Đ ; Þ[5,14] Ç ÆÆ ë Í ÝÎà ÎÁ, çææ ë Ë ð ½. Ù, Î, Ð ³ Ù, Å ÆÆÎ, í ³ î º. û è Ä ë ë îææ ÎÁ. 2 º Đ è è ë Ìë È Î, ë ǑöÜ Ò, Þ Ã ½. ë Ǒ à, ǑÐà., Á 1ãÄ, Á 2ãÄë.Á IÄ Ǒ,Á PDÄ î ë,á NPDÄ ê î ë,á *ãä.û º ³ õþ :π 1 Ǒ þ; π 2 Ǒë þ; d 1 Ǒ Á; d 2 Ǒ Á; wǒ ½Ù ; cǒë ; p 1 Ǒ, p 1 > w ;p 2 Ǒ í, p 2 > w. Ùë íî å à í î å à ù è ð, à Ôû ³. Áø» Ǒ d 1 = sa p 1 + bp 2, (1) d 2 = (1 s)a p 2 + bp 1, (2) a(a > 0)Ǒù ÓÁ, s(0 < s < 1)ðÎ ù, b(0 < b < 1)ð

346 ½ 27 Ñ, sa p 1 > 0,(1 s)a p 2 > 0, ûë Đ à. 3 º ǑǑÈ í ë È ù ëë, à  ΠÞ. ì», ð., ì» Î ø π I Ǒ π I = (p 1 c)(sa p 1 + bp 2 ) + (p 2 c)((1 s)a p 2 + bp 1 ). (3) Ð (3)ðĐ Î ø ð³ù í ø,, (3) dz٠p 1 p 2 è û¹ Ù 0, µ í ÇǑ è, Î Ǒ π I = p I 1 = (s + (1 s)b)a + (1 b2 )c, (4) 2 = ((1 s) + sb)a + (1 b2 )c. (5) p I ( ) 1 (1 s) (1 b)sb 2 a 2 (1 b 2 )ca + (b 3 b + 1)c 2. (6) Ô Å Â, íµ, îææ ë. 4 Î ÆÆ î º ô Î îææ ëî, ùë ô Î ÆÆ Á, Í ÆÆ, Ǒ ë Ǒ Stackelberg à( leader ), Ǒ Ðà( follower ). ë Ù ½Ù ù» w (p 2 ) = c + k (p 2 c). (7) kǒ î, Ú0 < k < 1, ݽ٠w (p 2 ) Úc < w (p 2 ) < p 2, ë ø ÇǑ π PD 1 = (p 1 w (p 2 ))d 1 = (p 1 kp 2 (1 k)c)(sa p 1 + bp 2 ), (8) π PD 2 = (w (p 2 ) c)d 1 + (p 2 c)d 2 = (p 1 kp 2 (1 k)c)(sa p 1 + bp 2 ) + (p 2 c)((1 s)a p 2 + bp 1 ). (9), Ä Ð Ò (8) (9)µ, í ÇǑ 1 = (3sk2 + 2(1 (1 + b)s)k (sb 2 2(1 s)b 4s))a + 4(2 + k 2 b 2 2bk) ( ) (1 + 3b) k 2 2(1 + b) 2 k + 4 b 3 b 2 + 2b c, (10) 4(2 + k 2 b 2 2bk)

½ 3 žâ : ô Î ÆÆ î º 347 2 = (2(1 s) + (b + k) s)a + (2k2 (1 + 3b) k + 2 b 2 + b) c. (11) 2(2 + k 2 b 2 2bk) óã 1 ë  î k = bî, Î ÆÆ. Ý ǑÅ ÝÎ ÆÆ, î ë Ç, Á Ú (4) (10) Çã (12), µ k 1 = b, 1 = p I 1, (12) 2 = p I 2. (13) k 2 = ( sb3 + 3sb + 2(1 s)) a (1 + b) 2 (1 b)(2 b)c. (15) (3sk 2 + 2(1 s)b s)a + (1 b 2 )(1 3b)c (5) (11) Çã (13), µ k 3 = b, (16) k 4 = ((1 s)b + s)a (1 b2 )c (sb + (1 s))a (1 b 2 )c. (17) (14) (15) (16)(17), î Ǒ k = b. (18) è, (11) (18)ã (7), ë  ½Ù (10) (19) ½³, (14) w PD = (s + (1 s)b)ba + (1 b2 ) (2 b)c. (19) 1 w PD = (sa c) + b((1 s)a + bc). (20) 2(1 + b) sa p 1 > 0, Đ sa c > sa 1 = sa p 1 > 0, 1 w PD > 0, Ú½Ù Ù. ݺ. óã 1 ÙÝè Ýæ, îë  î ð Ñ, Å î ëê ß ð½. î ë Âã ³ ø, Đ ë ÇǑ ¾ (21) (22) ÆÆ Î π PD = π PD 1 + π PD 2 = π PD (sa (1 b)c)2 1 =, 4 (21) π PD 2 = ((1 s + sb) a (1 b2 ) c) 2. 4(1 b 2 ) (22) ( 1 (1 s) (1 b)sb 2 ) a 2 (1 b 2 )ca + (b 3 b + 1)c 2 = π I. (23) ë Ä î ë ¹Ý, Î ÝÆÆ, Î Ó Þ., Ù Ò, Å Đ. Ùë,à ÆÆ ÙÆÆî,ë  î ë; Ù,à ÆÆ ÙÆÆ î, û ë.ǒ ð, Ý Ì ë Å

348 ½ 27 ParetoÍ, à Ý ÅÎÁìõ, ÍÅ ë ûâ û ë,íå»ææî., ÆÆ ë Þìõ. 5 ÆÆ Þ ParetoÍ îææ ë, ë Þ, à ø ë ø ÇǑ π NPD 1 = (p 1 w)(sa p 1 + bp 2 ), (24) π NPD 2 = (w c)(sa p 1 + bp 2 ) + (p 2 c)((1 s)a p 2 + bp 1 ). (25), Ä Ð Ò (24) (25)µ,û ½Ù Çã (24) (25), îææ ë, ë Î ÇǑ π NPD 1 = (sa + bc c)2, (26) 16 π NPD 2 = (s2 b 2 + 4(1 s) sb + 3s 2 4s + 2)a 2 (2 s + bs) ca + (1 b 2 ) (b 2 2b + 3)c 2, (27) π NPD = (s2 b 2 + 8(1 s) sb + 7s 2 8s + 4)a 2 (4 s + bs) ca + (1 b 2 ) (b 2 + 6b 7)c 2. (28) è,đ ë ÆÆ Þ, à óã 2. óã 2 ÆÆ Î Ç, ë. Ý ³ (23) (28),ÆÆ ÆÆîÎ «πǒ π = (sa + bc c) 2 /16 > 0, (29) ³ (21) (26), ÆÆ «π 1 Ǒ π 1 = 3(sa + bc c) 2 /16 > 0, (30) ³ (22) (27),ë ÆÆ «π 2 Ǒ π 2 = (sa + bc c) 2 /8 < 0. (31) (29) (30)(31)óà Ý. ݺ. óã 2 Å î ëǒ ÍÅÇ Î, ð ë. Ùë Ǒ Stackelberg à, ë å  îææ ë.,ǒå Ý îææ ë ½, ë Çè ë Îà ÍÅ ParetoÍ,à óã 3. óã 3 îææ ë, åë T Ú T [ π 2, π 1 ]î, Ý Î Å ParetoÍ. (30) (31)ðÝóà 3,Ý. óã 3 Å ÝÎ ÆÆ î, ÍÅ Úèù, à ë ÎÁ. Ù îææ, Û Ç, û

½ 3 žâ : ô Î ÆÆ î º 349 ë, Å ë ½., Ä ÒÆ, Æ Æ Î Ç Îà ½¾, æù ÃíÙÎà Í. 6 ǑÅ è ÇÆÆ ë, Ô Ä ½ ³. Î ù ÓÁ a = 200, è¹ s = 0.4,c = 10, π 1 = π 1 T, π 2 = π 2 + T. Ù î k Ù Ñ b, Ô Ç Ä ì» ½ ³. ì» 1 b Þ,T ù, ÄÇ µ Ä 1 Â. ÄÄ 1 ½ 4 710 Ù Ý, T ù,ææ Ç á b Ç Ç ;ÆÆ ë Ç á b Ç ;ÆÆ Î Ç á b Ç Ç. Å î ë, T ù, Ñ,, ë,û Ç Ùë, ð ǑÆÆ ½Ù á b Ç. è, ÄÄ 1 è½, b = 0.8î,ÆÆ Ǒ Ç, ð ë, ð Ǒ T = 750 Ù π 2 = 760 Úóà 3.,ë ǑÅ ÝÆÆ, ÌÂÔ T. ÙÄ 1 ½, Ù Úóà 3, ÆÆ ë Ç, Ý ParetoÍ. Ä 1 T = 750î  Table 1 The optimal result of T = 750 b π1 PD π1 NPD π 1 π2 PD π2 NPD π 2 π PD π NPD π 0.3 582.25 333.06 249.19 5 749.45 5 665.58 83.87 6 331.70 5 998.64 333.06 0.4 619.00 342.25 276.75 6 887.19 6 821.69 65.50 7 506.19 7 163.94 342.25 0.5 656.25 351.56 304.69 8 502.08 8 455.21 46.87 9 158.33 8 806.77 351.56 0.6 694.00 361.00 333.00 10 951.00 10 923.00 28.00 11 645.00 11 284.00 361.00 0.7 732.25 370.56 361.69 15 067.06 15 058.19 8.87 15 799.31 15 428.75 370.56 0.8 771.00 380.25 390.75 23 350.11 23 360.61 10.50 24 121.11 23 740.86 380.25 ì» 2 T Þ,b ù, ÄÇ µ Ä 2 Â. ÄÄ 2 ½ 4 710 Ù Ý, b ù,ææ Ç á T Ç ;ÆÆ ë Ç á T Ç Ç ; ÆÆ Î Ç ÙµÍ, ë Ç. ð Ǒ b ùî,ææë ù, îææ ë à T ³,Î Ù Î. è, ÄÄ 2 ½ 2 34½, T = 600,650,700î, ÆÆ Ǒ Ç, ðë, ð Ǒ Ù ù T = 600,650,700 Ù π 2 = 703.125 Úóà 3.,ë ǑÅ ÝÆÆ, ÌÂÔ T. ÙÄ 2 ½ Úóà 4, ÆÆ ë Ç. Ä 2 b = 0.5î  Table 2 The optimal result of b = 0.5 T π1 PD π1 NPD π 1 π2 PD π2 NPD π 2 π PD π NPD π 600 806.25 351.56 454.69 8 352.08 8 455.21 03.13 9 158.33 8 806.77 351.56 650 756.25 351.56 404.69 8 402.08 8 455.21 53.13 9 158.33 8 806.77 351.56 700 706.25 351.56 354.69 8 452.08 8 455.21 3.13 9 158.33 8 806.77 351.56 750 656.25 351.56 304.69 8 502.08 8 455.21 46.87 9 158.33 8 806.77 351.56 800 606.25 351.56 254.69 8 552.08 8 455.21 96.87 9 158.33 8 806.77 351.56 Ä ì» Ǒ ÙÝ, T Ôî(³ T = 100 ), î ë ø, Ù Stackelberg Ǒ à ë,à Ý ÆÆ Ç,è ì»

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