问题如下图：
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老师，计算机从CF0到结束能重投说一遍吗一直没按对
NO.PZ2015122802000166 问题如下 analyst gathers or estimates the following information about a stock:Baseon a vinscount mol, the stois most likely: A.unrvalue B.fairly value C.overvalue is correct.The current priof €22.56 is less ththe intrinsic value (V0) of €24.64; therefore, the stoappears to currently unrvalue Accorng to the two-stage vinscount mol:V0=∑t=1n(1+gs)t(1+r)t+Vn(1+r)nV_0=\sum_{t=1}^n\frac{{\splaystyle 0(1+g_s)}^t}{{\splaystyle(1+r)}^t}+\frac{V_n}{{(1+r)}^n}V0=∑t=1n(1+r)t(1+gs)t+(1+r)nVn anVn=+1r−gLV_n=\frac{{n+1}}{r-g_L}Vn=r−gL+1+1 = (1+gS)n(1+gL) = €1.60 × 1.09 = €1.744 = €1.60 × (1.09)2 = €1.901 = €1.60 × (1.09)3 = €2.072 = €1.60 × (1.09)4 = €2.259 = [€1.60 × (1.09)4](1.04) = €2.349V4 = €2.349/(0.12 – 0.04) = €29.363V0=1.744(1.12)1+1.901(1.12)2+2.072(1.12)3+2.259(1.12)4+29.363(1.12)4V_0=\frac{1.744}{{(1.12)}^1}+\frac{1.901}{{(1.12)}^2}+\frac{2.072}{{(1.12)}^3}+\frac{2.259}{{(1.12)}^4}+\frac{29.363}{{(1.12)}^4}V0=(1.12)11.744+(1.12)21.901+(1.12)32.072+(1.12)42.259+(1.12)429.363 = 1.557+1.515+1.475+1.436+18.661 = €24.64(whiis greater ththe current priof €22.56)考点Multi-stage Mol计算器具体步奏如下CFO=0, C01=1.744,F01=1,C02=1.901,F02=1,C03=2.072,F03=1,C04=2.259+29.363,F04=1,NPV , I=12,CPT NPV 此题V4是用什么公式算的？为什么不用讲义P495页求P的公式？求V4,我下方的算法有问题吗？这一类题的解题思路，李老师根本没讲清楚。
NO.PZ2015122802000166 问题如下 analyst gathers or estimates the following information about a stock:Baseon a vinscount mol, the stois most likely: A.unrvalue B.fairly value C.overvalue is correct.The current priof €22.56 is less ththe intrinsic value (V0) of €24.64; therefore, the stoappears to currently unrvalue Accorng to the two-stage vinscount mol:V0=∑t=1n(1+gs)t(1+r)t+Vn(1+r)nV_0=\sum_{t=1}^n\frac{{\splaystyle 0(1+g_s)}^t}{{\splaystyle(1+r)}^t}+\frac{V_n}{{(1+r)}^n}V0=∑t=1n(1+r)t(1+gs)t+(1+r)nVn anVn=+1r−gLV_n=\frac{{n+1}}{r-g_L}Vn=r−gL+1+1 = (1+gS)n(1+gL) = €1.60 × 1.09 = €1.744 = €1.60 × (1.09)2 = €1.901 = €1.60 × (1.09)3 = €2.072 = €1.60 × (1.09)4 = €2.259 = [€1.60 × (1.09)4](1.04) = €2.349V4 = €2.349/(0.12 – 0.04) = €29.363V0=1.744(1.12)1+1.901(1.12)2+2.072(1.12)3+2.259(1.12)4+29.363(1.12)4V_0=\frac{1.744}{{(1.12)}^1}+\frac{1.901}{{(1.12)}^2}+\frac{2.072}{{(1.12)}^3}+\frac{2.259}{{(1.12)}^4}+\frac{29.363}{{(1.12)}^4}V0=(1.12)11.744+(1.12)21.901+(1.12)32.072+(1.12)42.259+(1.12)429.363 = 1.557+1.515+1.475+1.436+18.661 = €24.64(whiis greater ththe current priof €22.56)考点Multi-stage Mol计算器具体步奏如下CFO=0, C01=1.744,F01=1,C02=1.901,F02=1,C03=2.072,F03=1,C04=2.259+29.363,F04=1,NPV , I=12,CPT NPV 请问V4的分母是应该是的数值还是（1+4%）？
NO.PZ2015122802000166问题如下analyst gathers or estimates the following information about a stock:Baseon a vinscount mol, the stois most likely:A.unrvalueB.fairly valueC.overvalue is correct.The current priof €22.56 is less ththe intrinsic value (V0) of €24.64; therefore, the stoappears to currently unrvalue Accorng to the two-stage vinscount mol:V0=∑t=1n(1+gs)t(1+r)t+Vn(1+r)nV_0=\sum_{t=1}^n\frac{{\splaystyle 0(1+g_s)}^t}{{\splaystyle(1+r)}^t}+\frac{V_n}{{(1+r)}^n}V0=∑t=1n(1+r)t(1+gs)t+(1+r)nVn anVn=+1r−gLV_n=\frac{{n+1}}{r-g_L}Vn=r−gL+1+1 = (1+gS)n(1+gL) = €1.60 × 1.09 = €1.744 = €1.60 × (1.09)2 = €1.901 = €1.60 × (1.09)3 = €2.072 = €1.60 × (1.09)4 = €2.259 = [€1.60 × (1.09)4](1.04) = €2.349V4 = €2.349/(0.12 – 0.04) = €29.363V0=1.744(1.12)1+1.901(1.12)2+2.072(1.12)3+2.259(1.12)4+29.363(1.12)4V_0=\frac{1.744}{{(1.12)}^1}+\frac{1.901}{{(1.12)}^2}+\frac{2.072}{{(1.12)}^3}+\frac{2.259}{{(1.12)}^4}+\frac{29.363}{{(1.12)}^4}V0=(1.12)11.744+(1.12)21.901+(1.12)32.072+(1.12)42.259+(1.12)429.363 = 1.557+1.515+1.475+1.436+18.661 = €24.64(whiis greater ththe current priof €22.56)考点Multi-stage Mol计算器具体步奏如下CFO=0, C01=1.744,F01=1,C02=1.901,F02=1,C03=2.072,F03=1,C04=2.259+29.363,F04=1,NPV , I=12,CPT NPV 老师请问怎么判断题目给的1.6是还是啊？
NO.PZ2015122802000166 老师，最后第五年的计算不太明白，课堂上的也没听明白，能不能麻烦讲一下，谢谢！
NO.PZ2015122802000166 fairly value overvalue A is correct. The current priof €22.56 is less ththe intrinsic value (V0) of €24.64; therefore, the stoappears to currently unrvalue Accorng to the two-stage vinscount mol: V0=∑t=1n(1+gs)t(1+r)t+Vn(1+r)nV_0=\sum_{t=1}^n\frac{{\splaystyle 0(1+g_s)}^t}{{\splaystyle(1+r)}^t}+\frac{V_n}{{(1+r)}^n}V0=∑t=1n(1+r)t(1+gs)t+(1+r)nVn anVn=+1r−gLV_n=\frac{{n+1}}{r-g_L}Vn=r−gL+1 +1 = (1+gS)n(1+gL) = €1.60 × 1.09 = €1.744 = €1.60 × (1.09)2 = €1.901 = €1.60 × (1.09)3 = €2.072 = €1.60 × (1.09)4 = €2.259 = [€1.60 × (1.09)4](1.04) = €2.349 V4 = €2.349/(0.12 – 0.04) = €29.363 V0=1.744(1.12)1+1.901(1.12)2+2.072(1.12)3+2.259(1.12)4+29.363(1.12)4V_0=\frac{1.744}{{(1.12)}^1}+\frac{1.901}{{(1.12)}^2}+\frac{2.072}{{(1.12)}^3}+\frac{2.259}{{(1.12)}^4}+\frac{29.363}{{(1.12)}^4}V0=(1.12)11.744+(1.12)21.901+(1.12)32.072+(1.12)42.259+(1.12)429.363 = 1.557+1.515+1.475+1.436+18.661 = €24.64(whiis greater ththe current priof €22.56) 考点Multi-stage Mol 计算器具体步奏如下 CFO=0, C01=1.744,F01=1,C02=1.901,F02=1,C03=2.072,F03=1,C04=2.259+29.363,F04=1,NPV , I=12,CPT NPV 可以算到➕P5么
