问题如下图：第二个增加的项目它是第一年开始190的，应该不是在第0年吧？我的算法是图片第一个那样？不对在哪里？
选项：
A. 
B. 
C. 
解释：
Debrah_品职答疑助手 · 2020年03月01日
同学你好。关于答案解析，实物扩张期权下投入是在t=1年，产生OCF是在第2-10年，在用计算器计算的时候，是可以用CF0代表第1年，CF1-CF9代表第2-10年的。
关于你的计算步骤，是不对的哦。因为对于期权下的扩张项目，需要分别计算High和low分别在t=1时刻的PVCF1，再经过判断是否选择扩张项目，乘以概率之后再折现到t=0时刻。省略了在t=1时刻的判断，是不对的哦。
拿这道题目来说：
①计算t=1时刻在概率为High的情况下的的NPV，CF0=-190，C01=40，F01=9，I=10，NPV=40.362。计算t=1时刻在概率为low的情况下的的NPV，CF0=-190，C01=20，F01=9，I=10，NPV=-74.82。
②判断：如果需求为high则选择扩张，此时概率为50%。
③求扩张项目的NPV0=（40.362*50%）/(1+10%)=18.346 ④total NPV=-5.663+18.346=12.683。
这道题目是李老师上课讲的原题，老师讲得很细致，如果对中间过程仍有不清楚的，建议可以回听一下李老师上课的讲解。加油。
NO.PZ201601200500000804 请问行权的时候不就是最优价值了吗？为什么最后还要加上没有option的原始NPV呢？谢谢!
NO.PZ201601200500000804 12.68. 31.03. B is correct. Assume we are time = 1. The NPV of the expansion (time 1) if manis \"high\" is NPV=−190+∑t=19401.10t=C$40.361millionNPV=-190+\sum_{t=1}^9\frac{40}{1.10^t}=C\$40.361millionNPV=−190+∑t=191.10t40=C$40.361million The NPV of the expansion (time 1) if manis \"low\" is NPV=−190+∑t=19201.10t=‐C$74.820millionNPV=-190+\sum_{t=1}^9\frac{20}{1.10^t}=‐C\$74.820millionNPV=−190+∑t=191.10t20=‐C$74.820million The optimcision is to expanif manis \"high\" annot expanif \"low.\" Because the expansion option is exerciseonly when its value is positive, whihappens 50 percent of the time, the expectevalue of the expansion project, time zero, is NPV=11.100.50(40.361)=C$18.346millionNPV=\frac1{1.10}0.50(40.361)=C\$18.346millionNPV=1.1010.50(40.361)=C$18.346million The totNPV of the initiprojeanthe expansion projeis NPV = –C$5.663 million + C$18.346 million = C$12.683 million The optionexpansion project, haneoptimally, as sufficient value to make this a positive NPV project.请问老师，40/1.1^t t=9，这个计算器怎么按啊？还是要一个一个按，按9个？
NO.PZ201601200500000804 12.68. 31.03. B is correct. Assume we are time = 1. The NPV of the expansion (time 1) if manis \"high\" is NPV=−190+∑t=19401.10t=C$40.361millionNPV=-190+\sum_{t=1}^9\frac{40}{1.10^t}=C\$40.361millionNPV=−190+∑t=191.10t40=C$40.361million The NPV of the expansion (time 1) if manis \"low\" is NPV=−190+∑t=19201.10t=‐C$74.820millionNPV=-190+\sum_{t=1}^9\frac{20}{1.10^t}=‐C\$74.820millionNPV=−190+∑t=191.10t20=‐C$74.820million The optimcision is to expanif manis \"high\" annot expanif \"low.\" Because the expansion option is exerciseonly when its value is positive, whihappens 50 percent of the time, the expectevalue of the expansion project, time zero, is NPV=11.100.50(40.361)=C$18.346millionNPV=\frac1{1.10}0.50(40.361)=C\$18.346millionNPV=1.1010.50(40.361)=C$18.346million The totNPV of the initiprojeanthe expansion projeis NPV = –C$5.663 million + C$18.346 million = C$12.683 million The optionexpansion project, haneoptimally, as sufficient value to make this a positive NPV project.为何不是在0时刻看，有两种情况 需求低，只投了190，不追加投资，npv为负 追加投资190，需求高，npv为正然后将两种情况各0.5加权求和？现在答案只考虑了第二种情况加权0.5，为何不第一种情况也加权0.5加在一起呢
为什么现金流要乘以0.5呢？即使PROBABILITY是50%，但是这个不是应该假设已经是OPTIMAL了吗，为什么还需要考虑概率。谢谢！
12.68. 31.03. B is correct. Assume we are time = 1. The NPV of the expansion (time 1) if manis \"high\" is NPV=−190+∑t=19401.10t=C$40.361millionNPV=-190+\sum_{t=1}^9\frac{40}{1.10^t}=C\$40.361millionNPV=−190+∑t=191.10t40=C$40.361million The NPV of the expansion (time 1) if manis \"low\" is NPV=−190+∑t=19201.10t=‐C$74.820millionNPV=-190+\sum_{t=1}^9\frac{20}{1.10^t}=‐C\$74.820millionNPV=−190+∑t=191.10t20=‐C$74.820million The optimcision is to expanif manis \"high\" annot expanif \"low.\" Because the expansion option is exerciseonly when its value is positive, whihappens 50 percent of the time, the expectevalue of the expansion project, time zero, is NPV=11.100.50(40.361)=C$18.346millionNPV=\frac1{1.10}0.50(40.361)=C\$18.346millionNPV=1.1010.50(40.361)=C$18.346million The totNPV of the initiprojeanthe expansion projeis NPV = –C$5.663 million + C$18.346 million = C$12.683 million The optionexpansion project, haneoptimally, as sufficient value to make this a positive NPV project.扩张项目的PVCF1已经得出，为什么折现一期的PV就是NPV？能不能用老师说的画图作差法再一下？