问一道题：NO.PZ2018103102000090

问题如下：

Matt has evaluating the value of Company M of $20 per share by using the two stage model. The company has paid a dividend of $1 per share for the most recent year. The relative highly growth rate is 5% over the first three years. The required rate of return is 8% What`s the long-term growth rate?

选项：

2.00%

2.51%

3.00%

解释：

B is correct.

考点：Two Stage Model

解析：根据公式: $20=\sum_{t=1}^3\frac{1\times\left(1+0.05\right)^t}{\left(1+0.08\right)^t}+\frac{1\times1.05^3\times\left(1+g_L\right)}{\left(0.08-g_L\right)\left(1+0.08\right)^3}$

求得g_L = 2.51%

您好，我想问下难道不是以下的公式么：

20=1.05/1.08+1.05^2/1.08^2+1.05^3/1.08^3+《1.05^3(1+g)/(8%-g)》/(1+8%)^3 么？我怎么觉得答案后半部分给的公式不对啊？

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pourquoi · 2020年07月24日

我也有同样的问题...答案没看懂

2 个答案

maggie_品职助教 · 2020年07月24日

我就是用了一个简单的数字代表了上面复杂公式，上面同学黄色标注的部分和解析中的公式是一样的，请看下面截图：

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maggie_品职助教 · 2019年11月05日

公式没有问题，咱们可以举个简单例子：(2/3)/4=2/12

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pourquoi · 2020年07月23日

这个例子是什么意思？

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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

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

问一道题：NO.PZ2018103102000090

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