Problem 29.
Suppose p < 1 , p ≠ 0 p<1,p\neq 0 p<1,p=0 . Show that the function of
f ( x ) = ( ∑ i = 1 n x i p ) 1 / p f(x) ={( \sum_{i=1}^nx_i^p)}^{1/p} f(x)=(i=1∑nxip)1/p with d o m f = R + + n domf = R^n_{++} domf=R++n is concave.
Solution
Problem 18.
Let Ω Ω Ω be a subset of E n E^n En and let f ∈ C 2 f \in C^2 f∈C2 be a function on Ω Ω Ω. If x ∗ x^∗ x∗ is a relative minimum point of f f f over Ω Ω Ω, then for any d ∈ E n d\in E^n d∈En that is a feasible direction at x ∗ x^∗ x∗ we have
Solution
Problem 4.
Find the minimum number of c c c, such that any local optimal solution is a global optimal solution for the following nonlinear program of P 4 P4 P4.
P 4 ) max f ( x ) = − c ∗ x 1 2 + x 1 x 2 + 2 x 1 − 1 2 x 2 2 s . t . x 1 ≤ 2 P4) \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+x1x2+2x1−21x22s.t. x1≤2
Solution
Problem 3.
Show the KKT condition for the nonlinear program of P 3 P3 P3, and try to solve it with the KKT conditions.
P 3 ) max − 2 x 1 2 − 2 x 1 x 2 − x 2 2 + 10 x 1 + 10 x 2 s . t . x 1 2 + x 2 2 ≤ 5 3 x 1 + x 2 ≤ 6 P3) \ \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 −2x12−2x1x2−x22+10x1+10x2s.t. x12+x22≤5 3x1+x2≤6
Solution
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 0.618 0.618 0.618.
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.
*Problem 4.
max f ( x ) = − 1 2 x 1 2 + x 1 x 2 + 2 x 1 − 1 2 x 2 2 \max \ f(x) = −\frac{1}{2}x^2_1 + x_1x_2 + 2x_1 − \frac{1}{2}x_2^2 \\ max f(x)=−21x12+x1x2+2x1−21x22 Use the gradient steepest ascent method to find the optimal solution of above nonlinear program, with the start point x 0 = ( 0 , 4 ) T x_0 = (0,4)^T x0=(0,4)T.
Problem 1.
Use branch and bound method to solve P 1 ) 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.
P 1 ) max 13 x 1 + 5 x 2 s . t . 3 x 1 + 2 x 2 ≤ 6.5 − 2 x 1 + 3 x 2 ≤ 8 x 1 , x 2 ≥ 0 , i n t e g e r s P1) \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+5x2s.t. 3x1+2x2≤6.5−2x1+3x2≤8x1,x2≥0,integers
Solution
branch and bound tree
Problem 26.
min c T x s . t . Ax = b x ∈ i n t e g e r \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 I P IP IP, 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 a i j = ( B − 1 A j ) i a_{ij} = (\textbf{\textit{B}}^{−1}\textbf{\textit{A}}_j)_i aij=(B−1Aj)iand a i 0 = ( B − 1 b ) i a_{i0} = (\textbf{\textit{B}}^{−1}\textbf{\textit{b}})_i ai0=(B−1b)i. Prove:
x i + ∑ j ∈ N ⌊ a i j ⌋ x j ≤ ⌊ a i 0 ⌋ 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 x 1 − 2 x 2 s . t . − 4 x 1 + 6 x 2 ≤ 9 x 1 + x 2 ≤ 4 x 1 , x 2 ≥ 0 , i n t e g e r s \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.
I P ) min c T x s . t . A x = b x ≥ 0 a n d ∈ i n t e g e r s IP)\min{c^Tx} \\ s.t.\ \ Ax = b \\ x ≥ 0 \ and \in integers IP)mincTxs.t. Ax=bx≥0 and∈integersFor above I P IP IP, denote A = [ a 1 , a 2 , ⋅ ⋅ ⋅ , a n ] A = [a_1,a_2,··· ,a_n] A=[a1,a2,⋅⋅⋅,an] by columns. Let k < n k
I P R ) min c T x s . t . A x = b x ≥ 0 IPR)\min{c^Tx} \\ s.t.\ \ Ax = b \\ \ \ \ \ \ \ x ≥ 0 IPR)mincTxs.t. Ax=b x≥0 I P R k ) min c T x s . t . A k x = b x ≥ 0 IPR_k)\min{c^Tx} \\ s.t.\ \ A_kx = b \\ \ \ \ \ \ \ x ≥ 0 IPRk)mincTxs.t. Akx=b x≥0Suppose x k ∗ x_k^* xk∗, and x ∗ x^∗ x∗ is the optimal solution for subproblem I P R k ) IPR_k) IPRk) and I P R ) IPR) IPR) respectively, and objectives of O B J I P R k OBJ_{IPR_k} OBJIPRk, O B J I P R OBJ_{IPR} OBJIPR correspondingly. Let S = { a 1 , a 2 , ⋅ ⋅ ⋅ , a n } S = \{a_1,a_2,··· ,a_n\} S={a1,a2,⋅⋅⋅,an}, B B B is the basis related to x k ∗ x^∗_k xk∗ for I P R k IPR_k IPRk.
P P ) min c a − c B B − 1 a s . t . a ∈ S PP)\min c_a −c_BB^{−1}a\\ s.t.\ \ a \in S PP)minca−cBB−1as.t. a∈SPlease prove: If O B J P P ≥ 0 OBJ_{PP} ≥ 0 OBJPP≥0, then O B J I P R k = O B J I P R OBJ_{IPR_k} = OBJ_{IPR} OBJIPRk=OBJIPR
Problem 8.
Use dynamic programming method to solve the following integer programming.
max 2 x 1 + 3 x 2 + x 3 + 2 x 4 s . t . 5 x 1 + 7 x 2 + 6 x 3 + 5 x 4 ≤ 14 x 1 , x 2 , x 3 , x 4 ≥ 0 , i n t e g e r s \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.
*Problem 8.
Use dynamic programming method to solve the following linear programming.
max 2 x 1 + 3 x 2 + x 3 + 2 x 4 s . t . 5 x 1 + 7 x 2 + 6 x 3 + 5 x 4 ≤ 14 x 1 , x 2 , x 3 , x 4 ≥ 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.
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 = μ e A x = b A T y + s = c x ≥ 0 , s ≥ 0 x◦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 ≤ μ 0 0 < \mu ≤ \mu_0 0<μ≤μ0 and any given 0 < μ < ∞ 0 < \mu < ∞ 0<μ<∞.
(b) For 0 < μ ′ < μ 0 < {\mu}'< \mu 0<μ′<μ,
c T x ( μ ′ ) ≤ c T x ( μ ) a n d b T y ( μ ′ ) ≥ b T y ( 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).
c T x ( μ ′ ) < c T x ( μ ) a n d b T y ( μ ′ ) > b T y ( 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 x 1 s . t . x 1 + x 2 + x 3 = 2 \min{\ x_1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ s.t. \ \ x_1 + x_2 + x_3 = 2 min x1 s.t. x1+x2+x3=2
Solution
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.
Problem 7.
Use ellipsoid method solve:
max 0 s . t . x 1 + 5 x 2 ≤ 7 x 1 + 2 x 2 ≥ 6 x 1 , x 2 ≥ 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 × I 2 × 2 ) E(0,100×I_{2×2}) E(0,100×I2×2).
You only need to give THREE STEPS
Problem 20.
For transportation problem stated as follows.
There are m origins that contain various amounts of a commodity that must be shipped to n n n destinations to meet demand requirements. Specially, origin i i i contains an amount a i a_i ai, and destination j j j has a requirement of amount b i b_i bi. It is assumed that the system is balanced in the sense that total supply equals total demand. There is unit cost c i j c_{ij} cij associated with the shipping of the commodity from origin i i i to destination j j j. 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. 如果运输问题的行和列的和是整数,证明所有基可行解的基变量为整数。
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 , + 1 0,+1 0,+1 or − 1 −1 −1. (完全幺模矩阵的各阶子式均为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. 证明运输问题的系数矩阵是完全幺模的。
(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.
与Problem20(2)的区别在于:此时 A \textbf{\textit{A}} A是任意矩阵,秩不再是 m + n − 1 m+n-1 m+n−1
Problem 9.
Use the Ford-Fulkerson method to solve decide the maximum flow for the following network problem.