问题如下图：
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解释：
这道题为什么要强调term structure is flat, 如果不是会有什么影响吗，这个条件的作用我没能想到
NO.PZ2016082402000006 问题如下 A portfolio manager ha bonposition worth US100 million. The position ha mofieration of eight years ana convexity of 150 years. Assume ththe term structure is flat. how mues the value of the position change if interest rates increase 25 basis points? US-2,046,875 US-2,187,500 US-1,953,125 US-1,906,250 ANSWER: CThe change in priis given △P=−[×P](△y)+12[C×P](△y)2=−(8×100)×\;\bigtriangleup P=-{\lbraast\times P\rbrack}{(\bigtriangleup y)}+\frac12{\lbraC\times P\rbrack}{(\bigtriangleup y)}^2=-{(8\times100)}\times△P=−[×P](△y)+21[C×P](△y)2=−(8×100)× (0.0025)+0.5(150×100)×(0.0025)2=−2.000000+0.046875=−1.953125.{(0.0025)}+0.5{(150\times100)}\times{(0.0025)}^2=-2.000000+0.046875=-1.953125.(0.0025)+0.5(150×100)×(0.0025)2=−2.000000+0.046875=−1.953125. The fomulshoulbe:△P=−[×P](△y)+1/2[C×CONVESITY ](△y)2,right?
NO.PZ2016082402000006 A portfolio manager ha bonposition worth US100 million. The position ha mofieration of eight years ana convexity of 150 years. Assume ththe term structure is flat. how mues the value of the position change if interest rates increase 25 basis points? US-2,046,875 US-2,187,500 US-1,953,125 US-1,906,250 ANSWER: C The change in priis given △P=−[×P](△y)+12[C×P](△y)2=−(8×100)×\;\bigtriangleup P=-{\lbraast\times P\rbrack}{(\bigtriangleup y)}+\frac12{\lbraC\times P\rbrack}{(\bigtriangleup y)}^2=-{(8\times100)}\times△P=−[×P](△y)+21[C×P](△y)2=−(8×100)× (0.0025)+0.5(150×100)×(0.0025)2=−2.000000+0.046875=−1.953125.{(0.0025)}+0.5{(150\times100)}\times{(0.0025)}^2=-2.000000+0.046875=-1.953125.(0.0025)+0.5(150×100)×(0.0025)2=−2.000000+0.046875=−1.953125. 请问求出来-1.953125之后，怎么得出来A
NO.PZ2016082402000006 US-2,187,500 US-1,953,125 US-1,906,250 ANSWER: C The change in priis given △P=−[×P](△y)+12[C×P](△y)2=−(8×100)×\;\bigtriangleup P=-{\lbraast\times P\rbrack}{(\bigtriangleup y)}+\frac12{\lbraC\times P\rbrack}{(\bigtriangleup y)}^2=-{(8\times100)}\times△P=−[×P](△y)+21[C×P](△y)2=−(8×100)× (0.0025)+0.5(150×100)×(0.0025)2=−2.000000+0.046875=−1.953125.{(0.0025)}+0.5{(150\times100)}\times{(0.0025)}^2=-2.000000+0.046875=-1.953125.(0.0025)+0.5(150×100)×(0.0025)2=−2.000000+0.046875=−1.953125.用的2021年教材；没找到公式啊！公式在教材哪页？
不懂，这是哪个知识点
