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Aimee🌸 · 2022年07月31日

a dividend of $1 per share for the most recent year

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:

A.

2.00%

B.

2.51%

C.

3.00%

解释:

B is correct.

考点:Two Stage Model

解析:根据公式: 20=t=131×(1+0.05)t(1+0.08)t+1×1.053×(1+gL)(0.08gL)(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}

求得gL = 2.51%

这句话我的理解是 div1=1 div2=1.05 div3=1.05^2 这明显和答案不一致 感觉很不太能够理解

1 个答案

王园圆_品职助教 · 2022年07月31日

嗨,从没放弃的小努力你好:


同学根据上一题的回答,你应该已经知道了,D0 = 1而不是D1=1

所以你后面的D1,D2,D3需要重新计算哦~

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就算太阳没有迎着我们而来,我们正在朝着它而去,加油!

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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

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