问一道题：NO.PZ2018103102000118

NO.PZ2018103102000118 问题如下 Matt is concting a FCFF sensitivity analysis for Company M`s value. The most recent FCFF of Company M is $100 million. Assuming the trate is 35% ana stable capitstructure of 40% an60% equity, the company`s value is least sensitive to: A.cost of equity B.growth rate C.after-tcost of C is correct.考点Sensitivity Analysis in FCFF anFCFE解析1. Cost of equity敏感性分析保持r= 5.5%，g = 5.0%当re = 8.0%时，WA= 0.4*5.5% + 0.6*8.0% = 7%Firm  Value=FCFF0×(1+g)WACC−g=100×(1+5%)7%−5%=5250Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+5\%\right)}{7\%-5\%}=5250FirmValue=WACC−gFCFF0​×(1+g)​=7%−5%100×(1+5%)​=5250当re = 12.0%时，WA= 0.4*5.5% + 0.6*12.0% = 9.4%Firm  Value=FCFF0×(1+g)WACC−g=100×(1+5%)9.4%−5%=2386.36Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+5\%\right)}{9.4\%-5\%}=2386.36FirmValue=WACC−gFCFF0​×(1+g)​=9.4%−5%100×(1+5%)​=2386.36Range = 5250 – 2386.36 = 2863.642. After-tcost of bt敏感性分析保持re = 10%，g = 5%当r= 5.0%时，WA= 0.4*5.0% + 0.6*10.0% = 8%Firm  Value=FCFF0×(1+g)WACC−g=100×(1+5%)8%−5%=3500Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+5\%\right)}{8\%-5\%}=3500FirmValue=WACC−gFCFF0​×(1+g)​=8%−5%100×(1+5%)​=3500当r= 6.0%时，WA= 0.4*6.0% + 0.6*10.0% = 8.4%Firm  Value=FCFF0×(1+g)WACC−g=100×(1+5%)8.4%−5%=3088.24Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+5\%\right)}{8.4\%-5\%}=3088.24FirmValue=WACC−gFCFF0​×(1+g)​=8.4%−5%100×(1+5%)​=3088.24Range = 3500 – 3088.24 = 411.763. Growth rate敏感性分析保持re = 10%，r= 5.5%WA= 0.4*5.5% + 0.6*10.0% = 8.2%当g = 4%时, Firm  Value=FCFF0×(1+g)WACC−g=100×(1+4%)8.2%−4%=2476.2Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+4\%\right)}{8.2\%-4\%}=2476.2FirmValue=WACC−gFCFF0​×(1+g)​=8.2%−4%100×(1+4%)​=2476.2当g = 6%时, Firm  Value=FCFF0×(1+g)WACC−g=100×(1+6%)8.2%−6%=4818.2Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+6\%\right)}{8.2\%-6\%}=4818.2FirmValue=WACC−gFCFF0​×(1+g)​=8.2%−6%100×(1+6%)​=4818.2Range =4818.2 - 2476.2 = 2342三者相比，敏感性最弱的是after-tcost of bt。 印象中，sensitivity analysis是某个因素单位变化所引起的结果变化分析。这题中，三个因素的变化幅度不同，high和low的结果直接相减就OK吗？不用考虑每个因素的变化幅度？

NO.PZ2018103102000118 问题如下 Matt is concting a FCFF sensitivity analysis for Company M`s value. The most recent FCFF of Company M is $100 million. Assuming the trate is 35% ana stable capitstructure of 40% an60% equity, the company`s value is least sensitive to: A.cost of equity B.growth rate C.after-tcost of C is correct.考点Sensitivity Analysis in FCFF anFCFE解析1. Cost of equity敏感性分析保持r= 5.5%，g = 5.0%当re = 8.0%时，WA= 0.4*5.5% + 0.6*8.0% = 7%Firm  Value=FCFF0×(1+g)WACC−g=100×(1+5%)7%−5%=5250Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+5\%\right)}{7\%-5\%}=5250FirmValue=WACC−gFCFF0​×(1+g)​=7%−5%100×(1+5%)​=5250当re = 12.0%时，WA= 0.4*5.5% + 0.6*12.0% = 9.4%Firm  Value=FCFF0×(1+g)WACC−g=100×(1+5%)9.4%−5%=2386.36Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+5\%\right)}{9.4\%-5\%}=2386.36FirmValue=WACC−gFCFF0​×(1+g)​=9.4%−5%100×(1+5%)​=2386.36Range = 5250 – 2386.36 = 2863.642. After-tcost of bt敏感性分析保持re = 10%，g = 5%当r= 5.0%时，WA= 0.4*5.0% + 0.6*10.0% = 8%Firm  Value=FCFF0×(1+g)WACC−g=100×(1+5%)8%−5%=3500Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+5\%\right)}{8\%-5\%}=3500FirmValue=WACC−gFCFF0​×(1+g)​=8%−5%100×(1+5%)​=3500当r= 6.0%时，WA= 0.4*6.0% + 0.6*10.0% = 8.4%Firm  Value=FCFF0×(1+g)WACC−g=100×(1+5%)8.4%−5%=3088.24Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+5\%\right)}{8.4\%-5\%}=3088.24FirmValue=WACC−gFCFF0​×(1+g)​=8.4%−5%100×(1+5%)​=3088.24Range = 3500 – 3088.24 = 411.763. Growth rate敏感性分析保持re = 10%，r= 5.5%WA= 0.4*5.5% + 0.6*10.0% = 8.2%当g = 4%时, Firm  Value=FCFF0×(1+g)WACC−g=100×(1+4%)8.2%−4%=2476.2Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+4\%\right)}{8.2\%-4\%}=2476.2FirmValue=WACC−gFCFF0​×(1+g)​=8.2%−4%100×(1+4%)​=2476.2当g = 6%时, Firm  Value=FCFF0×(1+g)WACC−g=100×(1+6%)8.2%−6%=4818.2Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+6\%\right)}{8.2\%-6\%}=4818.2FirmValue=WACC−gFCFF0​×(1+g)​=8.2%−6%100×(1+6%)​=4818.2Range =4818.2 - 2476.2 = 2342三者相比，敏感性最弱的是after-tcost of bt。 虽然计算结果对选择答案也是一样，但是给了35%这个条件总得要用吧

NO.PZ2018103102000118问题如下 Matt is concting a FCFF sensitivity analysis for Company M`s value. The most recent FCFF of Company M is $100 million. Assuming the trate is 35% ana stable capitstructure of 40% an60% equity, the company`s value is least sensitive to: A.cost of equityB.growth rateC.after-tcost of btC is correct.考点Sensitivity Analysis in FCFF anFCFE解析1. Cost of equity敏感性分析保持r= 5.5%，g = 5.0%当re = 8.0%时，WA= 0.4*5.5% + 0.6*8.0% = 7%Firm  Value=FCFF0×(1+g)WACC−g=100×(1+5%)7%−5%=5250Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+5\%\right)}{7\%-5\%}=5250FirmValue=WACC−gFCFF0​×(1+g)​=7%−5%100×(1+5%)​=5250当re = 12.0%时，WA= 0.4*5.5% + 0.6*12.0% = 9.4%Firm  Value=FCFF0×(1+g)WACC−g=100×(1+5%)9.4%−5%=2386.36Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+5\%\right)}{9.4\%-5\%}=2386.36FirmValue=WACC−gFCFF0​×(1+g)​=9.4%−5%100×(1+5%)​=2386.36Range = 5250 – 2386.36 = 2863.642. After-tcost of bt敏感性分析保持re = 10%，g = 5%当r= 5.0%时，WA= 0.4*5.0% + 0.6*10.0% = 8%Firm  Value=FCFF0×(1+g)WACC−g=100×(1+5%)8%−5%=3500Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+5\%\right)}{8\%-5\%}=3500FirmValue=WACC−gFCFF0​×(1+g)​=8%−5%100×(1+5%)​=3500当r= 6.0%时，WA= 0.4*6.0% + 0.6*10.0% = 8.4%Firm  Value=FCFF0×(1+g)WACC−g=100×(1+5%)8.4%−5%=3088.24Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+5\%\right)}{8.4\%-5\%}=3088.24FirmValue=WACC−gFCFF0​×(1+g)​=8.4%−5%100×(1+5%)​=3088.24Range = 3500 – 3088.24 = 411.763. Growth rate敏感性分析保持re = 10%，r= 5.5%WA= 0.4*5.5% + 0.6*10.0% = 8.2%当g = 4%时, Firm  Value=FCFF0×(1+g)WACC−g=100×(1+4%)8.2%−4%=2476.2Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+4\%\right)}{8.2\%-4\%}=2476.2FirmValue=WACC−gFCFF0​×(1+g)​=8.2%−4%100×(1+4%)​=2476.2当g = 6%时, Firm  Value=FCFF0×(1+g)WACC−g=100×(1+6%)8.2%−6%=4818.2Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+6\%\right)}{8.2\%-6\%}=4818.2FirmValue=WACC−gFCFF0​×(1+g)​=8.2%−6%100×(1+6%)​=4818.2Range =4818.2 - 2476.2 = 2342三者相比，敏感性最弱的是after-tcost of bt。能否求导数来做，这计算量…没心情算了😐

NO.PZ2018103102000118问题如下 Matt is concting a FCFF sensitivity analysis for Company M`s value. The most recent FCFF of Company M is $100 million. Assuming the trate is 35% ana stable capitstructure of 40% an60% equity, the company`s value is least sensitive to: A.cost of equityB.growth rateC.after-tcost of btC is correct.考点Sensitivity Analysis in FCFF anFCFE解析1. Cost of equity敏感性分析保持r= 5.5%，g = 5.0%当re = 8.0%时，WA= 0.4*5.5% + 0.6*8.0% = 7%Firm  Value=FCFF0×(1+g)WACC−g=100×(1+5%)7%−5%=5250Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+5\%\right)}{7\%-5\%}=5250FirmValue=WACC−gFCFF0​×(1+g)​=7%−5%100×(1+5%)​=5250当re = 12.0%时，WA= 0.4*5.5% + 0.6*12.0% = 9.4%Firm  Value=FCFF0×(1+g)WACC−g=100×(1+5%)9.4%−5%=2386.36Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+5\%\right)}{9.4\%-5\%}=2386.36FirmValue=WACC−gFCFF0​×(1+g)​=9.4%−5%100×(1+5%)​=2386.36Range = 5250 – 2386.36 = 2863.642. After-tcost of bt敏感性分析保持re = 10%，g = 5%当r= 5.0%时，WA= 0.4*5.0% + 0.6*10.0% = 8%Firm  Value=FCFF0×(1+g)WACC−g=100×(1+5%)8%−5%=3500Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+5\%\right)}{8\%-5\%}=3500FirmValue=WACC−gFCFF0​×(1+g)​=8%−5%100×(1+5%)​=3500当r= 6.0%时，WA= 0.4*6.0% + 0.6*10.0% = 8.4%Firm  Value=FCFF0×(1+g)WACC−g=100×(1+5%)8.4%−5%=3088.24Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+5\%\right)}{8.4\%-5\%}=3088.24FirmValue=WACC−gFCFF0​×(1+g)​=8.4%−5%100×(1+5%)​=3088.24Range = 3500 – 3088.24 = 411.763. Growth rate敏感性分析保持re = 10%，r= 5.5%WA= 0.4*5.5% + 0.6*10.0% = 8.2%当g = 4%时, Firm  Value=FCFF0×(1+g)WACC−g=100×(1+4%)8.2%−4%=2476.2Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+4\%\right)}{8.2\%-4\%}=2476.2FirmValue=WACC−gFCFF0​×(1+g)​=8.2%−4%100×(1+4%)​=2476.2当g = 6%时, Firm  Value=FCFF0×(1+g)WACC−g=100×(1+6%)8.2%−6%=4818.2Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+6\%\right)}{8.2\%-6\%}=4818.2FirmValue=WACC−gFCFF0​×(1+g)​=8.2%−6%100×(1+6%)​=4818.2Range =4818.2 - 2476.2 = 2342三者相比，敏感性最弱的是after-tcost of bt。考试时都要这样计算么 还是可以有简便方法

NO.PZ2018103102000118 问题如下 Matt is concting a FCFF sensitivity analysis for Company M`s value. The most recent FCFF of Company M is $100 million. Assuming the trate is 35% ana stable capitstructure of 40% an60% equity, the company`s value is least sensitive to: A.cost of equity B.growth rate C.after-tcost of C is correct.考点Sensitivity Analysis in FCFF anFCFE解析1. Cost of equity敏感性分析保持r= 5.5%，g = 5.0%当re = 8.0%时，WA= 0.4*5.5% + 0.6*8.0% = 7%Firm  Value=FCFF0×(1+g)WACC−g=100×(1+5%)7%−5%=5250Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+5\%\right)}{7\%-5\%}=5250FirmValue=WACC−gFCFF0​×(1+g)​=7%−5%100×(1+5%)​=5250当re = 12.0%时，WA= 0.4*5.5% + 0.6*12.0% = 9.4%Firm  Value=FCFF0×(1+g)WACC−g=100×(1+5%)9.4%−5%=2386.36Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+5\%\right)}{9.4\%-5\%}=2386.36FirmValue=WACC−gFCFF0​×(1+g)​=9.4%−5%100×(1+5%)​=2386.36Range = 5250 – 2386.36 = 2863.642. After-tcost of bt敏感性分析保持re = 10%，g = 5%当r= 5.0%时，WA= 0.4*5.0% + 0.6*10.0% = 8%Firm  Value=FCFF0×(1+g)WACC−g=100×(1+5%)8%−5%=3500Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+5\%\right)}{8\%-5\%}=3500FirmValue=WACC−gFCFF0​×(1+g)​=8%−5%100×(1+5%)​=3500当r= 6.0%时，WA= 0.4*6.0% + 0.6*10.0% = 8.4%Firm  Value=FCFF0×(1+g)WACC−g=100×(1+5%)8.4%−5%=3088.24Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+5\%\right)}{8.4\%-5\%}=3088.24FirmValue=WACC−gFCFF0​×(1+g)​=8.4%−5%100×(1+5%)​=3088.24Range = 3500 – 3088.24 = 411.763. Growth rate敏感性分析保持re = 10%，r= 5.5%WA= 0.4*5.5% + 0.6*10.0% = 8.2%当g = 4%时, Firm  Value=FCFF0×(1+g)WACC−g=100×(1+4%)8.2%−4%=2476.2Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+4\%\right)}{8.2\%-4\%}=2476.2FirmValue=WACC−gFCFF0​×(1+g)​=8.2%−4%100×(1+4%)​=2476.2当g = 6%时, Firm  Value=FCFF0×(1+g)WACC−g=100×(1+6%)8.2%−6%=4818.2Firm\;Value=\frac{FCFF_0\times\left(1+g\right)}{WACC-g}=\frac{100\times\left(1+6\%\right)}{8.2\%-6\%}=4818.2FirmValue=WACC−gFCFF0​×(1+g)​=8.2%−6%100×(1+6%)​=4818.2Range =4818.2 - 2476.2 = 2342三者相比，敏感性最弱的是after-tcost of bt。 re是波动了（-2%，+2%）， r（1-T）波动了（-0.5%，+0.5%），g波动了（-1%，+1%）如果把算出来的波动结果标准化，针对re（5250-2386.3626）/4针对r（1-T)（3500-3088.2353)/1针对g（4818.1818-2476.1905）/2