三階段抽樣
目錄
基本公式
現用\(\mathbb{E}_3,\mathbb{D}_3\)表示在固定初級單元、二級單元時，對第三階段抽樣求均值和方差；\(\mathbb{E}_2,\mathbb{D}_2\)表示在固定初級單元時，對第二階段求均值和方差；\(\mathbb{E}_1,\mathbb{D}_1\)表示對初級單元求均值和方差。顯然有
\[\mathbb{E}(\hat\theta)=\mathbb{E}_1\mathbb{E}_2\mathbb{E}_3(\hat\theta). \]
於是
\[\begin{aligned} \mathbb{D}(\hat\theta)&=\mathbb{E}(\hat\theta^2)-[\mathbb{E}(\hat\theta)]^2\\ &=\mathbb{E}_1\mathbb{E}_2\mathbb{E}_3(\hat\theta^2)-\{\mathbb{E}_1\mathbb{E}_2\mathbb{E}_3(\hat\theta) \}^2\\ &=\mathbb{E}_1\mathbb{E}_2\mathbb{E}_3(\hat\theta^2)-\{\mathbb{E}_1[\mathbb{E}_2\mathbb{E}_3(\hat\theta)]^2- \mathbb{D}_1[\mathbb{E}_2\mathbb{E}_3(\hat\theta)]\} \\ &=\mathbb{D}_1\mathbb{E}_2\mathbb{E}_3(\hat\theta)+\mathbb{E}_1\mathbb{E}_2\mathbb{E}_3(\hat\theta^2)-\mathbb{E}_1[\mathbb{E}_2\mathbb{E}_3(\hat\theta)]^2\\ &=\mathbb{D}_1\mathbb{E}_2\mathbb{E}_3(\hat\theta)+\mathbb{E}_1[\mathbb{E}_2\mathbb{E}_3(\hat\theta^2)-[\mathbb{E}_2\mathbb{E}_3(\hat\theta)]^2] \end{aligned} \]
注意到\(\mathbb{E}_2\mathbb{E}_3(\hat\theta^2)-[\mathbb{E}_2\mathbb{E}_3(\hat\theta)]^2\)實際上是固定初級單元時，對后面兩個單元合作抽樣的方差，所以
\[\mathbb{E}_2\mathbb{E}_3(\hat\theta^2)-[\mathbb{E}_2\mathbb{E}_3(\hat\theta)]^2=\mathbb{D}_2\mathbb{E}_3(\hat\theta)+\mathbb{E}_2\mathbb{D}_3(\hat\theta), \]
故
\[\mathbb{D}(\hat\theta)=\mathbb{D}_1\mathbb{E}_2\mathbb{E}_3(\hat\theta)+\mathbb{E}_1\mathbb{D}_2\mathbb{E}_3(\hat\theta)+\mathbb{E}_1\mathbb{E}_2\mathbb{D}_3(\hat\theta). \]
等概率三階段抽樣
考慮初級單元中二級單元個數相等，二級單元中三級單元個數相等的情形。第一階段從包含\(N\)個初級單元的總體中以簡單隨機抽樣方式抽取\(n\)個初級單元；第二階段從包含\(M\)個二級單元的總體中以簡單隨機抽樣方式抽取\(m\)個二級單元；第三階段從包含\(K\)個三級單元的總體中以簡單隨機抽樣方式抽取\(k\)個三級單元。
對總體均值的估計為
\[\hat{\bar{\bar {\bar Y}}}=\bar{\bar{\bar y}}=\frac{1}{nmk}\sum_{i=1}^{n}\sum_{j=1}^{m}\sum_{u=1}^{k}y_{iju}. \]
證明\(\bar{\bar{\bar y}}\)的無偏性
可以證明\(\bar{\bar{\bar y}}\)是\(\bar{\bar{\bar Y}}\)的無偏估計，即
\[\mathbb{E}(\bar{\bar{\bar y}})=\bar{\bar{\bar Y}}. \]
此時\(\mathbb{E}(\bar{\bar {\bar y}})=\mathbb{E}_1\mathbb{E}_2\mathbb{E}_3(\bar{\bar {\bar y}})\)，且三個階段均是簡單隨機抽樣。有
\[\begin{aligned} \mathbb{E}(\bar{\bar {\bar y}})&=\mathbb{E}_1\mathbb{E}_2\mathbb{E}_3(\bar{\bar {\bar y}})\\ &=\mathbb{E}_1\mathbb{E}_2\mathbb{E}_3\left(\frac{1}{nm}\sum_{i=1}^{n}\sum_{j=1}^{m}{\bar y}_{ij} \right)\\ &=\mathbb{E}_1\mathbb{E}_2\left(\frac{1}{nm}\sum_{i=1}^{n}\sum_{j=1}^{m}\bar{ Y}_{ij} \right)\\ &=\frac{1}{nm}\mathbb{E}_1\left[\sum_{i=1}^{n}\mathbb{E}_2\left(\sum_{j=1}^{m}\bar{ Y}_{ij} \right)\right] \end{aligned} \]此處，\(\displaystyle{\sum_{j=1}^{m}\bar{Y}_{ij}}\)是第二階段簡單隨機抽樣（將二級單元的總體均值視為抽樣單元）的樣本總值，故\(\displaystyle{\mathbb{E}_2\left(\frac{1}{m}\sum_{j=1}^{m}\bar{Y}_{ij} \right)=\bar {\bar Y}_{i}}\)；接下來，\(\displaystyle{\sum_{i=1}^{n}\bar{\bar Y}_{i}}\)是第一階段簡單隨機抽樣（將一級單元總體均值視為抽樣單元）的樣本總值，故\(\displaystyle{\mathbb{E}_1\left(\frac{1}{n}\sum_{i=1}^{n}\bar{\bar Y_i} \right)}=\bar{\bar{\bar{Y}}}\)。從而
\[\mathbb{E}(\bar{\bar {\bar y}})=\bar{\bar{\bar Y}}. \]
計算\(\bar{\bar{\bar y}}\)的方差
下計算其方差，先給出幾個記號：
\[S_1^2=\frac{1}{N-1}\sum_{i=1}^{N}(\bar{\bar Y}_i-\bar{\bar{\bar Y}})^2,\\ S_{2i}^2=\frac{1}{M-1}\sum_{j=1}^{M}(\bar{Y}_{ij}-\bar{\bar Y}_i)^2,\quad S_{2}^2=\frac{1}{N}\sum_{i=1}^{N}S_{2i}^2, \\ S_{3ij}^2=\frac{1}{K-1}\sum_{u=1}^{K}(Y_{iju}-\bar{Y}_{ij})^2,\quad S_{3}^2=\frac{1}{NM}\sum_{i=1}^{N}\sum_{j=1}^{M}S_{3ij}^2. \]
並記三級抽樣的抽樣比分別為
\[f_1=\frac{n}{N},\quad f_2=\frac{m}{M},\quad f_3=\frac{k}{K}. \]
此時
\[\mathbb{D}(\bar{\bar{\bar y}})=\frac{1-f_1}{n}S_1^2+\frac{1-f_2}{nm}S_2^2+\frac{1-f_3}{nml}S_3^2. \]
有
\[\mathbb{D}(\bar{\bar{\bar y}})=\mathbb{E}_1\mathbb{E}_2\mathbb{D}_3(\bar{\bar{\bar y}})+\mathbb{E}_1\mathbb{D}_2\mathbb{E}_3(\bar{\bar{\bar y}})+\mathbb{D}_1\mathbb{E}_2\mathbb{E}_3(\bar{\bar{\bar y}}). \]逐項計算。
計算第一項要用到此結果：
\[\displaystyle{\mathbb{D}_3(\bar{\bar y}_{ij})=\frac{1-f_3}{k}\frac{1}{K-1}\sum_{u=1}^{K}(Y_{iju}-\bar{\bar{Y}}_{ij})^2}, \]由於\(\displaystyle{\frac{1}{K-1}\sum_{u=1}^{K}(Y_{iju}-\bar{\bar Y}_{ij})^2}=S_{3ij}^2\)，並且\(\displaystyle{S_3^2=\frac{1}{NM}\sum_{i=1}^{N}\sum_{j=1}^{M}S_{3ij}^2}\)，所以
\[\begin{aligned} \mathbb{E}_1\mathbb{E}_2\mathbb{D}_3(\bar{\bar{\bar y}})&=\mathbb{E}_1\mathbb{E}_2\mathbb{D}_3\left(\frac{1}{nm}\sum_{i=1}^{n}\sum_{j=1}^{m}\bar{ y}_{ij} \right)\\ &=\frac{1}{n^2m^2}\mathbb{E}_1\mathbb{E}_2\sum_{i=1}^{n}\sum_{j=1}^{m}\mathbb{D}_3(\bar{ y}_{ij})\\ &=\frac{1-f_3}{nmk}\mathbb{E}_1\mathbb{E}_2\left(\frac{1}{nm}\sum_{i=1}^{n}\sum_{j=1}^{m}S_{3ij}^2 \right)\\ &=\frac{1-f_3}{nmk}\frac{1}{NM}\sum_{i=1}^{N}\sum_{j=1}^{M}S_{3ij}^2\\ &=\frac{1-f_3}{nmk}S_3^2. \end{aligned} \]計算第二項要用到此結果：
\[\mathbb{D}_2\left(\frac{1}{m}\sum_{j=1}^{m}\bar { Y}_{ij} \right)=\frac{1-f_2}{m}\frac{1}{M-1}\sum_{j=1}^{M}(\bar{ Y}_{ij}-\bar{\bar Y}_i)^2 \]由於\(\displaystyle{\frac{1}{M-1}\sum_{j=1}^{M}(\bar{Y}_{ij}-\bar{\bar Y}_i)^2=S_{2i}^2}\)，並記\(\displaystyle{S_2^2=\frac{1}{N}\sum_{i=1}^{N}S_{2i}^2}\)，則第二項為
\[\begin{aligned} \mathbb{E}_1\mathbb{D}_2\mathbb{E}_3(\bar{\bar{\bar y}})&=\mathbb{E}_1\mathbb{D}_2\left(\frac{1}{nm}\sum_{i=1}^{n}\sum_{j=1}^{m}\bar{ Y}_{ij} \right)\\ &=\frac{1}{n^2}\mathbb{E}_1\left[\sum_{i=1}^{n}\mathbb{D}_2\left(\frac{1}{m}\sum_{j=1}^{m}\bar{ Y}_{ij} \right)\right]\\ &=\frac{1}{n^2}\mathbb{E}_1\left(\sum_{i=1}^{n}\frac{1-f_2}{m}S_{2i}^2 \right)\\ &=\frac{1-f_2}{nm}\mathbb{E}_1\left(\frac{1}{n}\sum_{i=1}^{n}S_{2i}^2 \right)\\ &=\frac{1-f_2}{nm}\frac{1}{N}\sum_{i=1}^{N}S_{2i}^2\\ &=\frac{1-f_2}{nm}S_{2}^2. \end{aligned} \]記\(\displaystyle{S_1^2=\frac{1}{N-1}\sum_{i=1}^{N}(\bar{\bar Y}_i-\bar{\bar{\bar Y}})^2}\)，則第三項為
\[\begin{aligned} \mathbb{D}_1\mathbb{E}_2\mathbb{E}_3(\bar{\bar{\bar y}})&=\mathbb{D}_1\left(\frac{1}{n}\sum_{i=1}^{n}\bar{\bar Y}_i \right)\\ &=\frac{1-f_1}{n}\frac{1}{N-1}\sum_{i=1}^{N}(\bar{\bar Y}_i-\bar{\bar{\bar Y}})^2\\ &=\frac{1-f_1}{n}S_1^2. \end{aligned} \]
尋找\(\mathbb{D}(\bar{\bar{\bar y}})\)的無偏估計
先給出以下記號：
\[s_1^2=\frac{1}{n-1}\sum_{i=1}^{n}(\bar{\bar y}_i-\bar{\bar{\bar y}})^2,\\ s_{2i}^2=\frac{1}{m-1}\sum_{j=1}^{m}(\bar{y}_{ij}-\bar{\bar y}_i)^2,\quad s_2^2=\frac{1}{n}\sum_{i=1}^{n}s_{2i}^2,\\ s_{3ij}^2=\frac{1}{k-1}\sum_{u=1}^{k}(y_{ijk}-\bar{y}_{ij})^2,\quad s_3^2=\frac{1}{nm}\sum_{i=1}^{n}\sum_{j=1}^{m}s_{3ij}^2. \]
此時\(\mathbb{D}(\bar{\bar{\bar y}})\)的無偏估計為
\[v(\bar{\bar{\bar y}})=\frac{1-f_1}{n}s_1^2+\frac{f_1(1-f_2)}{nm}s_2^2+\frac{f_1f_2(1-f_3)}{nmk}s_3^2. \]
需要分別計算\(\mathbb{E}(s_1^2),\mathbb{E}(s_2^2),\mathbb{E}(s_3^2)\)，從后向前計算。
\[\begin{aligned} \mathbb{E}(s_3^2)&=\mathbb{E}_1\mathbb{E}_2\mathbb{E}_3(s_3^2)\\ &=\mathbb{E}_1\mathbb{E}_2\mathbb{E}_3\left(\frac{1}{nm(k-1)}\sum_{i=1}^{n}\sum_{j=1}^{m}\sum_{u=1}^{k}(y_{iju}-\bar{y}_{ij})^2 \right)\\ &=\mathbb{E}_1\mathbb{E}_2\left[\frac{1}{nm}\sum_{i=1}^{n}\sum_{j=1}^{m} \mathbb{E}_3\left(\frac{1}{k-1}\sum_{u=1}^{k}(y_{iju}-\bar{y}_{ij})^2 \right) \right]\\ &=\mathbb{E}_1\mathbb{E}_2\left[\frac{1}{nm}\sum_{i=1}^{n}\sum_{j=1}^{m}\frac{1}{K-1}\sum_{u=1}^{K}(Y_{iju}-\bar{Y}_{ij})^2 \right]\\ &=\mathbb{E}_1\left[\frac{1}{n}\sum_{i=1}^{n} \mathbb{E}_2\left(\frac{1}{m}\sum_{j=1}^{m}S_{3ij}^2 \right) \right]\\ &=\mathbb{E}_1\left[\frac{1}{n}\sum_{i=1}^{n}\frac{1}{M}\sum_{j=1}^{M}S_{3ij}^2 \right]\\ &=\frac{1}{NM}\sum_{i=1}^{N}\sum_{j=1}^{M}S_{3ij}^2\\ &=S_3^2. \end{aligned} \]下計算\(\mathbb{E}(s_2^2)\)，可先計算\(\mathbb{E}[(m-1)s_2^2]\)，在此前先計算\(\mathbb{E}_3[(m-1)s_2^2]\)，有
\[\begin{aligned} \mathbb{E}_3[(m-1)s_2^2]&=\mathbb{E}_3\left[\frac{1}{n}\sum_{i=1}^{n}\sum_{j=1}^{m}(\bar y_{ij}-\bar{\bar y}_i)^2 \right]\\ &=\frac{1}{n}\sum_{i=1}^{n}\left[\sum_{j=1}^{m}\mathbb{E}_3(\bar y_{ij}^2)-m\mathbb{E}_3(\bar{\bar y}_i^2) \right]\\ &=\frac{1}{n}\sum_{i=1}^{n}\left[\sum_{j=1}^{m}\{[\mathbb{E}_3(\bar y_{ij})]^2+\mathbb{D}_3(\bar{y}_{ij})\}-m\{[\mathbb{E}_3(\bar{\bar y}_i)]^2+\mathbb{D}_3(\bar{\bar y}_i) \} \right]\\ &=\frac{1}{n}\sum_{i=1}^{n}\left[\sum_{j=1}^{m}\left(\bar Y_{ij}^2+\frac{1-f_3}{k}S_{3ij}^2 \right)-m\left(\frac{1}{m}\sum_{j=1}^{m}\bar Y_{ij} \right)^2-\frac{1-f_3}{mk}\sum_{j=1}^{m}S_{3ij}^2 \right], \end{aligned} \]記\(\displaystyle{\bar{\bar Y}_{m}=\frac{1}{m}\sum_{j=1}^{m}\bar Y_{ij}}\)，則
\[\begin{aligned} \mathbb{E}_3[(m-1)s_2^2]&=\frac{1}{n}\sum_{i=1}^{n}\left[\sum_{j=1}^{m}(\bar Y_{ij}-\bar{\bar Y}_{m})^2+\frac{(1-f_3)(m-1)}{mk}\sum_{j=1}^{m}S_{3ij}^2 \right],\\ \mathbb{E}_2\mathbb{E}_3(s_2^2)&=\frac{1}{n}\sum_{i=1}^{n}\mathbb{E}_2\left[\frac{1}{m-1}\sum_{j=1}^{m}(\bar Y_{ij}-\bar{\bar Y}_{m})^2+\frac{1-f_3}{mk}\sum_{j=1}^{m}S_{3ij}^2 \right]\\ &=\frac{1}{n}\sum_{i=1}^{n}\frac{1}{M-1}\sum_{j=1}^{M}(\bar Y_{ij}-\bar{\bar Y})^2+\frac{1-f_3}{nMk}\sum_{i=1}^{n}\sum_{j=1}^{M}S_{3ij}^2\\ &=\frac{1}{n}\sum_{i=1}^{n}S_{2i}^2+\frac{1-f_3}{nMk}\sum_{i=1}^{n}\sum_{j=1}^{M}S_{3ij}^2, \\ \mathbb{E}(s_2^2)&=\mathbb{E}_1\mathbb{E}_2\mathbb{E}_3(s_2^2)\\ &=\mathbb{E}_1\left[\frac{1}{n}\sum_{i=1}^{n}S_{2i}^2+\frac{1-f_3}{nMk}\sum_{i=1}^{n}\sum_{j=1}^{M}S_{3ij}^2 \right]\\ &=\frac{1}{N}\sum_{i=1}^{n}S_{2i}^2+\frac{1-f_3}{k}\frac{1}{NM}\sum_{i=1}^{N}\sum_{j=1}^{M}S_{3ij}^2\\ &=S_{2}^2+\frac{1-f_3}{k}S_{3}^2. \end{aligned} \]類似處理\(s_1^2\)即可，先記\(\displaystyle{S_{3i}^2=\frac{1}{M}\sum_{j=1}^{M}S_{3ij}^2}\)，\(\displaystyle{\bar{\bar{\bar Y}}_n=\frac{1}{n}\sum_{i=1}^{n}\bar{\bar Y}_i}\)，則有
\[\begin{aligned} &\quad \mathbb{E}_2\mathbb{E}_3[(n-1)s_1^2]\\ &=\mathbb{E}_2\mathbb{E}_3\left[\sum_{i=1}^{n}(\bar{\bar y}_{i}-\bar{\bar{\bar y}})^2\right]\\ &=\sum_{i=1}^{n}\mathbb{E}_2\mathbb{E}_3(\bar{\bar y}_i^2)-n\mathbb{E}_2\mathbb{E}_3(\bar{\bar{\bar y}}^2)\\ &=\sum_{i=1}^{n}\{[\mathbb{E}_2\mathbb{E}_3(\bar{\bar y}_i)]^2+\mathbb{D}_2\mathbb{E}_3(\bar{\bar y}_i)\}-n\{[\mathbb{E}_2\mathbb{E}_3(\bar{\bar{\bar y}})]^2 +\mathbb{D}_2\mathbb{E}_3(\bar{\bar{\bar y}}) \}\\ &=\sum_{i=1}^{n}\bar{\bar Y}_i+\sum_{i=1}^{n}\left[\frac{1-f_2}{m}S_{2i}^2+\frac{1-f_3}{mk}S_{3i}^2 \right]-n\left(\frac{1}{n}\sum_{i=1}^{n}\bar{\bar Y}_i \right)^2\\ &\quad +\frac{1}{n}\sum_{i=1}^{n}\left[\frac{1-f_2}{m}S_{2i}^2+\frac{1-f_3}{mk}S_{3i}^2 \right]\\ &=\sum_{i=1}^{n}(\bar{\bar Y}_i-\bar{\bar{\bar{Y}}}_n)^2+\sum_{i=1}^{n}\left[\frac{(1-f_2)(n-1)}{nm}S_{2i}^2+\frac{(1-f_3)(n-1)}{nmk}S_{3i}^2 \right],\\ &\quad \mathbb{E}(s_1^2)\\ &=\mathbb{E}_1\mathbb{E}_2\mathbb{E}_3(s_1^2)\\ &=\mathbb{E}_1\left\{\frac{1}{n-1}\sum_{i=1}^{n}(\bar{\bar Y}_i-\bar{\bar{\bar Y}}_n)^2+\frac{1-f_2}{nm}\sum_{i=1}^{n}S_{2i}^2+\frac{1-f_3}{nmk}\sum_{i=1}^{n}S_{3i}^2 \right\}\\ &=\frac{1}{N-1}\sum_{i=1}^{N}(\bar{\bar Y}_i-\bar{\bar{\bar Y}})^2+\frac{1-f_2}{m}S_{2}^2+\frac{1-f_3}{mk}S_{3}^2\\ &=S_1^2+\frac{1-f_2}{m}S_2^2+\frac{1-f_3}{mk}S_3^2. \end{aligned} \]最后將上述結論代入，得
\[\begin{aligned} \mathbb{E}[v(\bar{\bar{\bar y}})]&=\frac{1-f_1}{n}\mathbb{E}(s_1^2)+\frac{f_1(1-f_2)}{nm}\mathbb{E}(s_2^2)+\frac{f_1f_2(1-f_3)}{nmk}\mathbb{E}(s_3^2)\\ &=\frac{1-f_1}{n}\left(S_1^2+\frac{1-f_2}{m}S_2^2+\frac{1-f_3}{mk}S_3^3 \right)\\ &\quad +\frac{f_1(1-f_2)}{nm}\left(S_2^2+\frac{1-f_3}{k}S_3^3 \right)\\ &\qquad +\frac{f_1f_2(1-f_3)}{nmk}S_3^2\\ &=\frac{1-f_1}{n}S_1^2+\frac{1-f_2}{nm}S_2^2+\frac{1-f_3}{nmk}S_3^2. \end{aligned} \]
三階段抽樣設計
假設三階段抽樣的費用函數為線性費用函數：\(C_{T}=c_0+c_1n+c_2nm+c_3nmk\)。優化問題為給定\(C_{T}\)極小化方差\(\mathbb{D}(\bar{\bar{\bar y}})\)，或給定\(\mathbb{D}(\bar{\bar{\bar y}})\)極小化\(C_{T}\)，最優的\(k,m\)為
\[k_{\text{opt}}=\frac{S_3}{\sqrt{S_2^2-\frac{S_3^2}{K}}}\sqrt{\frac{c_2}{c_3}},\\ m_{\text{opt}}=\frac{\sqrt{S_2^2-\frac{S_3^2}{K}}}{\sqrt{S_1^2-\frac{S_2^2}{M}}}\sqrt{\frac{c_1}{c_2}}. \]
當確定\(k,m\)的最優值后，如果給定了總費用\(C_{T}\)，則
\[n_{\text{opt}}=\frac{C_{T}-c_0}{c_1+c_2m_{\text{opt}}+c_3m_{\text{opt}}k_{\text{opt}}}; \]
如果給定了估計量方差\(\mathbb{D}(\bar{\bar{\bar y}})=V\)，則
\[n_{\text{opt}}=\dfrac{\dfrac{1-f_1}{S_1^2}+\dfrac{(1-f_2)S_2^2}{m}+\dfrac{(1-f_3)S_3^2}{mk}}{V}. \]
將估計量方差改寫為
\[\begin{aligned} \mathbb{D}(\bar{\bar{\bar y}})&=\frac{1-f_1}{n}S_1^2+\frac{1-f_2}{nm}S_2^2+\frac{1-f_3}{nmk}S_3^2\\ &=\left(\frac{1}{n}-\frac{1}{N} \right)S_1^2+\frac{1}{n}\left(\frac{1}{m}-\frac{1}{M} \right)S_2^2+\frac{1}{nm}\left(\frac{1}{k}-\frac{1}{K} \right)S_3^2\\ &=\frac{1}{n}\left(S_1^2-\frac{S_2^2}{M} \right)+\frac{1}{nm}\left(S_2^2-\frac{S_3^2}{K} \right)+\frac{S_3^2}{nmk}-\frac{S_1^2}{N}, \end{aligned} \]對\(\displaystyle{\left(V+\frac{S_1^2}{N} \right)(C_{T}-c_0)}\)進行極小化，為方便書寫令
\[S_{u}=S_1^2-\frac{S_2^2}{M},\quad S_{v}=S_2^2-\frac{S_3^2}{K}. \]於是有
\[{\left(V+\frac{S_1^2}{N} \right)(C_{T}-c_0)}=\left(S_{u}+\frac{1}{m}S_{v}+\frac{S_3^2}{mk} \right)(c_1+c_2m+c_3mk), \]由柯西不等式，
\[\quad \left(S_{u}+\frac{1}{m}S_{v}+\frac{S_3^2}{mk} \right)(c_1+c_2m+c_3mk) \ge \left(\sqrt{S_{u}c_1}+\sqrt{S_{v}c_2}+\sqrt{S_3^2c_3} \right)^{2}, \]且等號成立當且僅當
\[\frac{S_{u}}{c_1}=\frac{S_{v}}{c_2m^2}=\frac{S_3^2}{c_3m^2k^2}, \]解得
\[m_{\text{opt}}=\sqrt{\frac{S_{v}c_1}{S_{u}c_2}}=\sqrt{\frac{S_2^2-\frac{S_3^2}{K}}{S_1^2-\frac{S_2^2}{M}}}\sqrt{\frac{c_1}{c_2}},\\ k_{\text{opt}}=\sqrt{\frac{S_3^2c_2}{S_vc_3}}=\sqrt{\frac{S_3^2}{S_2^2-\frac{S_3^2}{K}}}\sqrt{\frac{c_2}{c_3}}. \]最后代回即可獲得\(n\)的最優值。