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刘哲如 · 2019年02月10日

问一道题:NO.PZ2018103102000090

问题如下图:这个可以用 g=re - D1/P0 吗?

    

选项:

A.

B.

C.

解释:



1 个答案
已采纳答案

maggie_品职助教 · 2019年02月10日

不行啊,你的公式是计算当市场价格等于价值时,通过GGM推出的implied 增长率。现在这道题是两阶段模型,倒推长期稳定阶段的增长率。我建议你先把基础课视频听一下。

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NO.PZ2018103102000090 问题如下 Matt hevaluating the value of Company M of $20 per share using the two stage mol. The company hpaia vinof $1 per share for the most recent year. The relative highly growth rate is 5% over the first three years. The requirerate of return is 8% What`s the long-term growth rate? A.2.00% B.2.51% C.3.00% B is correct.考点Two Stage Mol解析根据公式: 20=∑t=131×(1+0.05)t(1+0.08)t+1×1.053×(1+gL)(0.08−gL)(1+0.08)320=\sum_{t=1}^3\frac{1\times\left(1+0.05\right)^t}{\left(1+0.08\right)^t}+\frac{1\times1.05^3\times\left(1+g_L\right)}{\left(0.08-g_L\right)\left(1+0.08\right)^3}20=∑t=13​(1+0.08)t1×(1+0.05)t​+(0.08−gL​)(1+0.08)31×1.053×(1+gL​)​求得gL = 2.51% 如题

2024-01-17 04:29 1 · 回答

NO.PZ2018103102000090 问题如下 Matt hevaluating the value of Company M of $20 per share using the two stage mol. The company hpaia vinof $1 per share for the most recent year. The relative highly growth rate is 5% over the first three years. The requirerate of return is 8% What`s the long-term growth rate? A.2.00% B.2.51% C.3.00% B is correct.考点Two Stage Mol解析根据公式: 20=∑t=131×(1+0.05)t(1+0.08)t+1×1.053×(1+gL)(0.08−gL)(1+0.08)320=\sum_{t=1}^3\frac{1\times\left(1+0.05\right)^t}{\left(1+0.08\right)^t}+\frac{1\times1.05^3\times\left(1+g_L\right)}{\left(0.08-g_L\right)\left(1+0.08\right)^3}20=∑t=13​(1+0.08)t1×(1+0.05)t​+(0.08−gL​)(1+0.08)31×1.053×(1+gL​)​求得gL = 2.51% =1,=1.05,=1.1025,=1.1576,V3=1.1576*(1+g)/(0.08-g)=20,反算出g=2.09%

2023-09-19 21:32 2 · 回答

NO.PZ2018103102000090问题如下Matt hevaluating the value of Company M of $20 per share using the two stage mol. The company hpaia vinof $1 per share for the most recent year. The relative highly growth rate is 5% over the first three years. The requirerate of return is 8% What`s the long-term growth rate?A.2.00%B.2.51%C.3.00%B is correct.考点Two Stage Mol解析根据公式: 20=∑t=131×(1+0.05)t(1+0.08)t+1×1.053×(1+gL)(0.08−gL)(1+0.08)320=\sum_{t=1}^3\frac{1\times\left(1+0.05\right)^t}{\left(1+0.08\right)^t}+\frac{1\times1.05^3\times\left(1+g_L\right)}{\left(0.08-g_L\right)\left(1+0.08\right)^3}20=∑t=13​(1+0.08)t1×(1+0.05)t​+(0.08−gL​)(1+0.08)31×1.053×(1+gL​)​求得gL = 2.51%这句话我的理解是 v1=1 v2=1.05 v3=1.05^2 这明显和答案不一致 感觉很不太能够理解

2022-07-31 11:14 1 · 回答

2.51% 3.00% B is correct. 考点Two Stage Mol 解析根据公式: 20=∑t=131×(1+0.05)t(1+0.08)t+1×1.053×(1+gL)(0.08−gL)(1+0.08)320=\sum_{t=1}^3\frac{1\times\left(1+0.05\right)^t}{\left(1+0.08\right)^t}+\frac{1\times1.05^3\times\left(1+g_L\right)}{\left(0.08-g_L\right)\left(1+0.08\right)^3}20=∑t=13​(1+0.08)t1×(1+0.05)t​+(0.08−gL​)(1+0.08)31×1.053×(1+gL​)​ 求得gL = 2.51%这个题为什么不用H mol?

2022-06-05 19:58 1 · 回答

NO.PZ2018103102000090 2.51% 3.00% B is correct. 考点Two Stage Mol 解析根据公式: 20=∑t=131×(1+0.05)t(1+0.08)t+1×1.053×(1+gL)(0.08−gL)(1+0.08)320=\sum_{t=1}^3\frac{1\times\left(1+0.05\right)^t}{\left(1+0.08\right)^t}+\frac{1\times1.05^3\times\left(1+g_L\right)}{\left(0.08-g_L\right)\left(1+0.08\right)^3}20=∑t=13​(1+0.08)t1×(1+0.05)t​+(0.08−gL​)(1+0.08)31×1.053×(1+gL​)​ 求得gL = 2.51%请问这道题有用计算器快一点的方法吗?还是只能列出式子之后一步一步算哎

2021-07-11 17:44 2 · 回答