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附录 A.1 纯标准产品策略(PS策略) 在临界值vs这点,购买商品的效用等于不购买商品的效用.得出 vsQ+λ(vs+1)2-ps=λvs2(A1)
当 vs=ps-λ2Q, 式(A1)成立.利润函数变为 π1=ps(1-ps-λ2Q).通过π1对ps求一阶导,得到下面最优解:
dπdps=1-2ps-λ2Q(A2) p*s1=Q2+λ4(A3) vs1=ps-λ2Q(A4) 联立 (A2)~(A4),得到临界值 vs1 和最优利润π*1:
vs1=Q2-λ4Q(A5) π*1=Q2+λ42Q(A6)
A.2 纯定制产品策略(PC策略) 和PS策略相同, 通过下式求解临界值 vc : vc(Q+x)+λ(vc+1)2-pc=λvc2(A7)
当vc=pc-λ2Q+x, 式(A7)成立.利润函数变为π2=pc(1-pc-λ2Q+x)-ax2-bx.π2对pc求一阶导,得到下面最优解:
p*c2=Q2+λ4+x2(A8)
vc2=Q2-λ4+x2Q+x(A9) π*2=Q2+λ4+x22Q+x-ax2-bx(A10) 该利润函数的二阶导数也满足条件. A.3 无法自由定制的混合产品策略(MN策略) 在这种情况下,计算两个平衡点vs3和vc3.财富水平为vs3的消费者购买标准产品和不购买任何产品来获得的效用相同;当消费者财富水平为vc3时,消费者获得的标准产品效用和定制产品效用没有区别.根据均衡条件,建立以下方程组:
vc3Q+λ(vs3+vc3)2-ps=vc3(Q+x)+λ(vc3+1)2-pc(A11)
vs3Q+λ(vs3+vc3)2-ps=λvs32(A12) 该方程关于vs3,vc3的解集为
vs3=4xps+2λps+λ2-2λpc4xQ+λ2(A13) vc3=2λps-4Qps-2λQ+4Qpc4xQ+λ2(A14) vc-vs=4Qpc-4Qps-4xps-2λQ-λ2+2λpc4xQ+λ2(A15) 1-vc=4xQ+2λQ+λ2-2λps+4Qps-4Qpc4xQ+λ2(A16) π3=ps(vc-vs)+pc(1-vc)-ax2-bx(A17) 该种产品策略下的利润函数为 π3=ps(vc-vs)+pc(1-vc)-ax2-bx. 联立 (A15)~(A17),得出公司关于ps和pc的利润函数如下:
π3=ps4Qpc-4Qps-4xps-2λQ-λ2+2λpc4xQ+λ2+pc4xQ+2λQ+λ2-2λps+4Qps-4Qpc4xQ+λ2-ax2-bx(A18)
式(A18)是ps和pc的二元函数.通过对ps和pc分别取偏导数,得到以下公式:
π3ps=8Qpc-8Qps-8xps-2λQ-λ24xQ+λ2=0(A19)
π3pc=8Qps-8Qpc+4xQ+2λQ+λ24xQ+λ2=0(A20)
联立 (A19) 和 (A20),有 p*s3=Q2(A21) p*c3=Q2+λ4+x2+λ28Q(A22) 将(A21)和(A22)代入(A18),得 π*3=Q4+λ4+x4+λ216Q-ax2-bx(A23) 命题3.3证明 MN与PS策略的利润公式分别为
π*3=Q4+λ4+x4+λ216Q-ax2-bx(A24) π*1=Q4+λQ4+λ216(A25) 为了讨论产品策略的选择,不妨假设π*3>π*1,则有 ax2+bx-x4-(λ216Q+λ4-λQ4-λ216)<0(A26)
令G=λ216Q+λ4-λQ4-λ216,改写为G=λ216Q-λ216+λ4-λQ4.当Q>1时,λ216Q-λ216<0,λ4-λQ4<0,故G<0;同理可得:当Q<1时,G>0.
当Q>1时,有G>0,有ax2+bx-x4-G<0,解该一元二次不等式: 当G<0,解得0因为当Q>1时,总有G>0,当G>(b-14)24a时,该不等式无解,故π*3>π*1无解,得到命题3.3③; 当Q>1,G<(b-14)24a,两个解14-b± (b-1/4)2+4aG2a均小于0,由于解代表定制程度必须大于0,故在定义域内无解,得到命题3.3④; 当Q>1,G<(b-14)24a,b<1/4时,求得定义域内两解为14-b± (b-1/4)2+4aG2a,可得到不等式的解集为14-b- (b-1/4)2+4aG2a A.4 可自由选择定制程度的混合产品策略(MF策略)
π4=∫kx+vsvsQ2+λ4+M(vi-vs)kdvi+(1-kx-vs)Q2+λ4+Mx-(ax2+bx)(A27)
化简为 π4=Q4+λ4+λ216Q+Mx2+Mxλ4Q-kMx22-ax2-bx(A28)
式(A28)对x求一阶导,得到以下结果: dπ4dx=M2+Mλ4Q-kMx-2ax-b=0(A29)
x*4=2QM-λM-4Qb8Qa+4QMk(A30)
命题3.6证明 x*4=2QM-λM-4Qb8Qa+4QMk,x*4对M求导得到 dx*4M=(2Q-λ)(8Qa+4QMk)-4Qk(2QM-λM-4Qb)(8Qa+4QMk)2(A31)
化简为
dx*4M=16Q2a-8λQa+16Q2bk(8Qa+4QMk)2=8Qa2vs3+16Q2bk(8Qa+4QMk)2>0(0 得证.
命题4.1证明 改写需求函数公式:
D1=c+142+λ2-p1-(c+1)c4(A33)
令m=2+λ2-p1,化简后得:D1=14cm+m-(c+1)2c4,为了求得需求函数随价格差的变化关系,对需求函数求一阶导,令一阶导等于0,求得c=± 4+48m-46,根据函数导数判断单调性,得到命题4.1.
命题4.2证明 公司1的利润函数如下:
π1=2cp1+λ2cp1-cp21-cp1(c+1)c4+2p1+λ2p1-p21-(c+1)c4p14(A34) 同样为了求得公司1利润与价格差c的关系,将利润函数对c求导: π1c=142p1+λ2p1-p21-3c2p1+4cp1+p14(A35)
令π1c=0,即 3c2p1+4cp1+4p21-7p1-2λp1=0(A36) 令n=4p21-7p1-2λp1;将式(A36)改写为3c2p1+4cp1+n=0.求得 c=-4p1± 16p21-12p1n6p1(A37) 通过对根号内的数值正负性加以考虑,进行分类讨论: ①当p1<2512+λ2时,16p21-12p1n>0,可正常求解得到命题4.2①和4.2②; ②当p1>2512+λ2时,16p21-12p1n<0,π1c<0,得到命题4.2③.
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