《高等运筹学》复习题手写解答 Advanced Operations Research: Final Exam:Review Exercises
文章目錄
- Nonlinear Program 非線性規劃
- KKT condition KKT條件
- Golden section method 黃金分割法
- Newton's method 牛頓法
- Gradient steepest descent/ ascent method 梯度下降/上升法
- Integer Programming 整數規劃
- Branch and bound 分支定界法
- Gomory cutting plane algorithm 割平面法/ Column generation 列生成算法
- Dynamic programming 動態規劃
- IP 有整數約束
- LP 無整數約束
- Linear programming 線性規劃
- Central path 中心路徑法(內點法)
- Karmarkar算法(內點法)
- Eliposoid method 橢球法(外點法)
- Transportation problem & unimodular matrix 運輸問題與幺模矩陣
- Graph Theory 圖論
- Maximum flow 最大流
- 附原題
Nonlinear Program 非線性規劃
Problem 29.
Suppose p<1,p≠0p<1,p\neq 0p<1,p?=0 . Show that the function of
f(x)=(∑i=1nxip)1/pf(x) ={( \sum_{i=1}^nx_i^p)}^{1/p} f(x)=(i=1∑n?xip?)1/p with domf=R++ndomf = R^n_{++}domf=R++n? is concave.
Solution
Problem 18.
Let ??? be a subset of EnE^nEn and let f∈C2f \in C^2f∈C2 be a function on ???. If x?x^?x? is a relative minimum point of fff over ???, then for any d∈End\in E^nd∈En that is a feasible direction at x?x^?x? we have
Solution
Problem 4.
Find the minimum number of ccc, such that any local optimal solution is a global optimal solution for the following nonlinear program of P4P4P4.
P4)max?f(x)=?c?x12+x1x2+2x1?12x22s.t.x1≤2P4) \max \ f(x) = ?c?x^2_1 + x_1x_2 + 2x_1 ? \frac{1}{2}x_2^2 \\ s.t. \ \ x_1 ≤ 2 P4)max?f(x)=?c?x12?+x1?x2?+2x1??21?x22?s.t.??x1?≤2
Solution
KKT condition KKT條件
Problem 3.
Show the KKT condition for the nonlinear program of P3P3P3, and try to solve it with the KKT conditions.
P3)max??2x12?2x1x2?x22+10x1+10x2s.t.x12+x22≤53x1+x2≤6P3) \ \max{ \ ?2x^2_1 ?2x_1x_2 ?x^2_2 + 10x_1 + 10x_2} \\ s.t. \ \ \ x^2_1 + x^2_2 ≤ 5 \\ \ \ \ \ \ \ \ 3x_1 + x_2 ≤ 6 P3)?max??2x12??2x1?x2??x22?+10x1?+10x2?s.t.???x12?+x22?≤5???????3x1?+x2?≤6
Solution
Golden section method 黃金分割法
Problem 16.
For the problem min?f(x)\min{f(x)}minf(x) over x∈[a,b]x\in [a,b]x∈[a,b], we use golden section method to solve it. Please prove the convergence ratio is 0.6180.6180.618.
Solution
Newton’s method 牛頓法
Problem 17.
For one dimension nonlinear problem min?f(x)\min f(x)minf(x) over x∈[a,b]x\in [a,b]x∈[a,b], we use Newton’s Method to solve it. Please prove the convergence ratio of Newton’s method is at least two.
Solution
Gradient steepest descent/ ascent method 梯度下降/上升法
*Problem 4.
max?f(x)=?12x12+x1x2+2x1?12x22\max \ f(x) = ?\frac{1}{2}x^2_1 + x_1x_2 + 2x_1 ? \frac{1}{2}x_2^2 \\ max?f(x)=?21?x12?+x1?x2?+2x1??21?x22? Use the gradient steepest ascent method to find the optimal solution of above nonlinear program, with the start point x0=(0,4)Tx_0 = (0,4)^Tx0?=(0,4)T.
Solution
Integer Programming 整數規劃
Branch and bound 分支定界法
Problem 1.
Use branch and bound method to solve P1)P1)P1).
Note that at each branching node, you can use graphic method to get the (local) upper bound for each node.
Draw the branch and bound tree, and point out the Node Program, GLB and LUB clearly.
P1)max?13x1+5x2s.t.3x1+2x2≤6.5?2x1+3x2≤8x1,x2≥0,integersP1) \max{\ \ \ 13x_1+5x_2}\\ s.t. \ \ 3x_1+2x_2\leq 6.5\\ -2x_1+3x_2\leq 8\\ x_1,x_2\geq0,integers P1)max???13x1?+5x2?s.t.??3x1?+2x2?≤6.5?2x1?+3x2?≤8x1?,x2?≥0,integers
Solution
branch and bound tree
Gomory cutting plane algorithm 割平面法/ Column generation 列生成算法
Problem 26.
min?cTxs.t.Ax=bx∈integer\min \textbf{\textit{c}}^T\textbf{\textit{x}}\\ s.t. \textbf{\textit{Ax}}=\textbf{\textit{b}}\\ \textbf{\textit{x}}\in integer mincTxs.t.Ax=bx∈integerFor above IPIPIP, suppose now we have a basic feasible solution corresponding to basis matrix B\textbf{\textit{B}}B and Non-basis N\textbf{\textit{N}}N, i.e., A=(B,N)\textbf{\textit{A}} = (\textbf{\textit{B}},\textbf{\textit{N}})A=(B,N), and let aij=(B?1Aj)ia_{ij} = (\textbf{\textit{B}}^{?1}\textbf{\textit{A}}_j)_iaij?=(B?1Aj?)i?and ai0=(B?1b)ia_{i0} = (\textbf{\textit{B}}^{?1}\textbf{\textit{b}})_iai0?=(B?1b)i?. Prove:
xi+∑j∈N?aij?xj≤?ai0?x_i+\sum_{j\in \textbf{\textit{N}}}\left \lfloor a_{ij} \right \rfloor x_{j}\leq \left \lfloor a_{i0} \right \rfloor xi?+j∈N∑??aij??xj?≤?ai0??
Solution
Problem 12.
Use the Gomory cutting plane algorithm to solve the following integer linear programming.
min?x1?2x2s.t.?4x1+6x2≤9x1+x2≤4x1,x2≥0,integers\min \ \ x_1 ?2x_2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ s.t. \ \ ?4x_1 + 6x_2 ≤ 9 \\ x_1 + x_2 ≤ 4 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ x_1,x_2 ≥ 0, integers min??x1??2x2????????????????s.t.???4x1?+6x2?≤9x1?+x2?≤4?????????????x1?,x2?≥0,integers
Solution
Problem 28.
IP)min?cTxs.t.Ax=bx≥0and∈integersIP)\min{c^Tx} \\ s.t.\ \ Ax = b \\ x ≥ 0 \ and \in integers IP)mincTxs.t.??Ax=bx≥0?and∈integersFor above IPIPIP, denote A=[a1,a2,???,an]A = [a_1,a_2,··· ,a_n]A=[a1?,a2?,???,an?] by columns. Let k<nk<nk<n, denote Ak=[a1,a2,???,ak]A_k = [a_1,a_2,··· ,a_k]Ak?=[a1?,a2?,???,ak?].
IPR)min?cTxs.t.Ax=bx≥0IPR)\min{c^Tx} \\ s.t.\ \ Ax = b \\ \ \ \ \ \ \ x ≥ 0 IPR)mincTxs.t.??Ax=b??????x≥0IPRk)min?cTxs.t.Akx=bx≥0IPR_k)\min{c^Tx} \\ s.t.\ \ A_kx = b \\ \ \ \ \ \ \ x ≥ 0 IPRk?)mincTxs.t.??Ak?x=b??????x≥0Suppose xk?x_k^*xk??, and x?x^?x? is the optimal solution for subproblem IPRk)IPR_k)IPRk?) and IPR)IPR)IPR) respectively, and objectives of OBJIPRkOBJ_{IPR_k}OBJIPRk??,OBJIPROBJ_{IPR}OBJIPR? correspondingly. Let S={a1,a2,???,an}S = \{a_1,a_2,··· ,a_n\}S={a1?,a2?,???,an?}, BBB is the basis related to xk?x^?_kxk?? for IPRkIPR_kIPRk?.
PP)min?ca?cBB?1as.t.a∈SPP)\min c_a ?c_BB^{?1}a\\ s.t.\ \ a \in S PP)minca??cB?B?1as.t.??a∈SPlease prove: If OBJPP≥0OBJ_{PP} ≥ 0OBJPP?≥0, then OBJIPRk=OBJIPROBJ_{IPR_k} = OBJ_{IPR}OBJIPRk??=OBJIPR?
Solution
Dynamic programming 動態規劃
IP 有整數約束
Problem 8.
Use dynamic programming method to solve the following integer programming.
max?2x1+3x2+x3+2x4s.t.5x1+7x2+6x3+5x4≤14x1,x2,x3,x4≥0,integers\max {\ \ 2x_1 + 3x_2 + x_3 + 2x_4}\\ \ \ \ \ \ \ \ \ \ \ \ \ s.t. \ \ \ 5x_1 + 7x_2 + 6x_3 + 5x_4 ≤ 14\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_1,x_2,x_3,x_4 ≥ 0, integers max??2x1?+3x2?+x3?+2x4?????????????s.t.???5x1?+7x2?+6x3?+5x4?≤14?????????????????x1?,x2?,x3?,x4?≥0,integersNote that you need to point out recursive formula clearly, and write down calculation steps clearly.
Solution
LP 無整數約束
*Problem 8.
Use dynamic programming method to solve the following linear programming.
max?2x1+3x2+x3+2x4s.t.5x1+7x2+6x3+5x4≤14x1,x2,x3,x4≥0,\max {\ \ 2x_1 + 3x_2 + x_3 + 2x_4}\\ \ \ \ \ \ \ \ \ \ \ \ \ s.t. \ \ \ 5x_1 + 7x_2 + 6x_3 + 5x_4 ≤ 14\\ \ \ \ \ x_1,x_2,x_3,x_4 ≥ 0, max??2x1?+3x2?+x3?+2x4?????????????s.t.???5x1?+7x2?+6x3?+5x4?≤14????x1?,x2?,x3?,x4?≥0,Note that you need to point out recursive formula clearly, and write down calculation steps clearly.
Solution
Linear programming 線性規劃
Central path 中心路徑法(內點法)
Problem 25.
Let (x(μ),y(μ),s(μ))(x(\mu),y(\mu),s(\mu))(x(μ),y(μ),s(μ)) be the central path of
x?s=μeAx=bATy+s=cx≥0,s≥0x?s = \mu e\\ Ax = b\\ A^Ty + s = c\\ x ≥ 0,s ≥ 0 x?s=μeAx=bATy+s=cx≥0,s≥0 Then prove:
(a) The central path point (x(μ),y(μ),s(μ))(x(\mu),y(\mu),s(\mu))(x(μ),y(μ),s(μ)) is bounded for 0<μ≤μ00 < \mu ≤ \mu_00<μ≤μ0? and any given 0<μ<∞0 < \mu < ∞0<μ<∞.
(b) For 0<μ′<μ0 < {\mu}'< \mu0<μ′<μ,
cTx(μ′)≤cTx(μ)andbTy(μ′)≥bTy(μ)c^Tx( {\mu}') ≤ c^Tx(\mu)\ \ and\ \ b^Ty( {\mu}') ≥ b^Ty(μ) cTx(μ′)≤cTx(μ)??and??bTy(μ′)≥bTy(μ)Furthermore, if x(μ′)≠x(μ)x( {\mu}')\neq x(\mu)x(μ′)?=x(μ) and y(μ′)≠y(μ)y( {\mu}')\neq y(\mu)y(μ′)?=y(μ).
cTx(μ′)<cTx(μ)andbTy(μ′)>bTy(μ)c^Tx( {\mu}') < c^Tx(\mu)\ \ and\ \ b^Ty( {\mu}') > b^Ty(μ) cTx(μ′)<cTx(μ)??and??bTy(μ′)>bTy(μ)
Solution
Problem 11.
Compute the central path for the following linear programming.
min?x1s.t.x1+x2+x3=2\min{\ x_1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ s.t. \ \ x_1 + x_2 + x_3 = 2 min?x1??????????????????????????s.t.??x1?+x2?+x3?=2
Solution
Karmarkar算法(內點法)
Problem 6.
What are the differences between Karmarkar method and Simplex method for linear programming? Please show the logics of them clearly. Possibly you can use figures to show what your idea.
Solution
Eliposoid method 橢球法(外點法)
Problem 7.
Use ellipsoid method solve:
max?0s.t.x1+5x2≤7x1+2x2≥6x1,x2≥0.\max \ \ 0 \\ s.t. \ \ x_1 + 5x_2 ≤ 7 \\ \ \ \ \ \ \ \ \ x_1 + 2x_2 ≥ 6 \\ \ \ \ \ x_1,x_2 ≥ 0 . max??0s.t.??x1?+5x2?≤7????????x1?+2x2?≥6????x1?,x2?≥0.The initial Ellipsoid is taken to be E(0,100×I2×2)E(0,100×I_{2×2})E(0,100×I2×2?).
You only need to give THREE STEPS
Solution
Transportation problem & unimodular matrix 運輸問題與幺模矩陣
Problem 20.
For transportation problem stated as follows.
There are m origins that contain various amounts of a commodity that must be shipped to nnn destinations to meet demand requirements. Specially, origin iii contains an amount aia_iai?, and destination jjj has a requirement of amount bib_ibi?. It is assumed that the system is balanced in the sense that total supply equals total demand. There is unit cost cijc_{ij}cij? associated with the shipping of the commodity from origin iii to destination jjj. The problem is to find the shipping pattern between origins and destinations that satisfies all the requirements and minimized the total shipping cost.
(1) build an mixed integer linear programming model for above problem.
(2) If the row and column sums of a transportation problem are integers, then the basic variables in any basic solution are integers. 如果運輸問題的行和列的和是整數,證明所有基可行解的基變量為整數。
Problem 21.
A matrix A\textbf{\textit{A}}A is said to be totally unimodular if the determinant of every square submatrix formed from it has value 0,+10,+10,+1 or ?1?1?1. (完全幺模矩陣的各階子式均為0,1或-1)
(1) Show that the matrix A\textbf{\textit{A}}A defining the equality constraints of a transportation prolblem is totally unimodular. 證明運輸問題的系數矩陣是完全幺模的。
(2) In the system of equations Ax=b\textbf{\textit{Ax}}=\textbf{\textit{b}}Ax=b, assume that A\textbf{\textit{A}}A is totally unimodular and that all elements of A\textbf{\textit{A}}A and b\textbf{\textit{b}}b are integers. Show that all basic solutions have integer components.
與Problem20(2)的區別在于:此時A\textbf{\textit{A}}A是任意矩陣,秩不再是m+n?1m+n-1m+n?1
Solution
Graph Theory 圖論
Maximum flow 最大流
Problem 9.
Use the Ford-Fulkerson method to solve decide the maximum flow for the following network problem.
Solution
附原題
總結
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