Problem 1

Determine whether the following statements are true or false, and explain why.

(1) For a standard linear programming problem with n variables and m constraints, the number of basic feasible solutions is (nm)\binom{n}{m}.

  • False. For different bases, we may get the same basic feasible solution. For example, [111123]x=[11],x0\begin{bmatrix} 1 & 1 &1 \\ 1& 2 &3\end{bmatrix}x=\begin{bmatrix}1\\1\end{bmatrix},x\ge0 can get a basic feasible solution (1,0,0)(1,0,0) from {x1,x2}\{x_1,x_2\} and {x1,x3}\{x_1,x_3\}.

(2) The optimal solution to a linear programming problem must be a basic feasible solution.

  • False. If the optimal solution is on a line of the constraint set, the optimal solution may not be basic, as the extreme points are finite and optimal solutions are infinite.

(3) If a linear programming problem has a feasible solution, the feasible region must contain the origin.

  • False. If the origin is not in the constraint set, it can't be in the feasible region.

(4) If a linear programming problem has a feasible solution, then the feasible region of the problem is a convex set.

  • True. Consider F={xAx=b,x0}F=\{x|Ax=b,x\ge0\} is the feasible region of a linear programming problem. Let x,yF,t[0,1]x,y\in F,t\in[0,1], we have A(tx+(1t)y)=tAx+(1t)Ay=bA(tx+(1-t)y)=tAx+(1-t)Ay=b and tx+(1t)yt0+(1t)0=0tx+(1-t)y\ge t\cdot0+(1-t)\cdot 0=0, which means tx+(1t)yFtx+(1-t)y\in F. So FF is a convex set.

Problem 2

Consider the following linear programming problem:

maximizez=CX,AX=b,X0,\operatorname{maximize}z=CX,\enspace AX=b,\enspace X\leq 0,

        where X0X^0 is the optimal solution to the problem. If the objective function coefficient CC is replaced by CC^*, and the optimal solution changes to XX^*, prove that:

(CC)(XX0)0(C^* −C)(X^* −X^0) ≥ 0
  • As X0X^0 is the optimal solution to CC and XX^* is the optimal solution to CC^*, we have CX0CXCX^0\ge CX^* and CXCX0C^*X^*\ge C^*X^0. So CX0+CXCX+CX0CX^0+C^*X^*\ge CX^*+C^*X^0, which means (CC)(XX0)0(C^*-C)(X^*-X^0)\ge0.

Problem 3

Prove: σ<0\sigma < 0\rightarrowThe current basic feasible solution has a higher objective value than all neighboring basic feasible solutions, thus the current solution is optimal.

  • Following the assumption in PPT, we know that we can write the basis chosen as (a1,,am)(a_1,\dots,a_m), and σj<0,j\sigma_j<0,\forall j, which means for all neighboring way to change the basis (a1,,am)(a_1,\dots,a_m) to (a1,,al1,aj,al,,am)(a_1,\dots,a_{l-1},a_{j},a_{l},\dots,a_m), we have z(1)z(0)=θσj<0z^{(1)}-z^{(0)}=\theta\sigma_j<0, thus objective function will decrease. As the bases is corresponding to the basic feasible solutions, we know that the current basic feasible solution has a higher objective value than all neighboring basic feasible solutions. And as the objective function is linear and the feasible region is convex, the local optimum must be a global optimum, thus the current solution is optimal.

Problem 4

For the following linear programming problem, determine all basic solutions, explain whether these basic solutions are basic feasible solutions, and plot each basic feasible solution on the feasible region’s graph.

maximizez=6x1+4x2\operatorname{maximize} z = 6x_1 +4x_2

subject to:

2x1+3x29,x14,x26,x1,x202x_1 +3x_2 \leq 9,\\ x_1 \ge 4,\\ x_2 \leq 6,\\ x_1, x_2 \ge 0
  • Transform this problem to the stardard form, we have that:
maxz=6x1+4x2s.t.{2x1+3x2+s1=9x1s2=4x2+s3=6x1,x2,s1,s2,s30\max z =6x_1+4x_2 \\ \mathrm{s.t.} \left\{\begin{matrix} 2x_1+3x_2+s_1=9 \\ x_1-s_2=4 \\ x_2+s_3=6 \\ x_1, x_2,s_1,s_2,s_3 \ge 0 \end{matrix}\right.

, which can be written as

[231001001001001]x=[946],x0\begin{bmatrix} 2&3&1&0&0\\ 1&0&0&-1&0\\ 0&1&0&0&1 \end{bmatrix}x=\begin{bmatrix}9\\4\\6\end{bmatrix},\enspace x\ge0

By calculating, we can get all basic solutions (4,6,17,0,0)(4,6,-17,0,0), (4.5,6,0,8.5,0)(-4.5,6,0,-8.5,0), (4,1/3,0,0,17/3)(4,1/3,0,0,17/3), (4,0,1,0,6)(4,0,1,0,6), (4.5,0,0,0.5,6)(4.5,0,0,0.5,6), (0,6,9,4,0)(0,6,-9,-4,0), (0,3,0,4,3)(0,3,0,-4,3), (0,0,9,4,6)(0,0,9,-4,6). And as x0x\ge0, (4,1/3,0,0,17/3)(4,1/3,0,0,17/3), (4,0,1,0,6)(4,0,1,0,6) and (4.5,0,0,0.5,6)(4.5,0,0,0.5,6) are three basic feasible solutions. Consider the feasible region in R2\R^2, we have three basic feasible solutions as (4,1/3),(4,0),(4.5,0)(4,1/3),(4,0),(4.5,0).

21a05018a97123bd6c4d7bdbc7e6937c.png

Problem 5

Use the Simplex Method to solve the following linear programming problems, and identify to which class the solution belongs.

maxz=6x1+2x2+10x3+8x4s.t.{5x1+6x24x34x4203x13x2+2x3+8x4254x12x2+x3+3x410x1,x2,x3,x40\max z =6x_1 +2x_2 +10x_3 +8x_4 \\ \mathrm{s.t.} \left\{\begin{matrix} 5x_1 +6x_2 −4x_3 −4x_4 \leq 20 \\ 3x_1 −3x_2 +2x_3 +8x_4 \leq 25 \\ 4x_1 −2x_2 +x_3 +3x_4 \leq 10 \\ x_1, x_2,x_3,x_4 \ge 0 \end{matrix}\right.
  • First convert it into standard form:
maxz=6x1+2x2+10x3+8x4s.t.{5x1+6x24x34x4+c1=203x13x2+2x3+8x4+c2=254x12x2+x3+3x4+c3=10x1,x2,x3,x4,c1,c2,c30\max z =6x_1 +2x_2 +10x_3 +8x_4 \\ \mathrm{s.t.} \left\{\begin{matrix} 5x_1 +6x_2 −4x_3 −4x_4 +c_1= 20 \\ 3x_1 −3x_2 +2x_3 +8x_4 +c_2= 25 \\ 4x_1 −2x_2 +x_3 +3x_4 +c_3= 10 \\ x_1, x_2,x_3,x_4,c_1,c_2,c_3 \ge 0 \end{matrix}\right.

So we have coefficient matrix AA and feasible basis B1B_1:

A=[564410033280104213001]B=[100010001]A= \begin{bmatrix} 5&6&-4&-4&1&0&0\\ 3&-3&2&8&0&1&0\\ 4&-2&1&3&0&0&1 \end{bmatrix} \enspace B=\begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix}

So, we can draw table:

(1)(1) cjc_j\rightarrow c1=6c_1=6 c2=2c_2=2 c3=10c_3=10 c4=8c_4=8 c5=0c_5=0 c6=0c_6=0 c7=0c_7=0 θ\theta
CBC_B XBX_B bb x1x_1 x2x_2 x3x_3 x4x_4 s1s_1 s2s_2 s3s_3
00 s1s_1 2020 55 66 4-4 4-4 11 00 00 -
00 s2s_2 2525 33 3-3 22 88 00 11 00 12.512.5
00 s3s_3 1010 44 2-2 [1][1] 33 00 00 11 1010
σj\sigma_j 66 22 1010\uparrow 88 00 00 00
(2)(2) 00 s1s_1 6060 2121 2-2 00 88 11 00 44 -
00 s2s_2 55 5-5 [1][1] 00 22 00 11 2-2 55
1010 x3x_3 1010 44 2-2 11 33 00 00 11 -
σj\sigma_j 34-34 2222\uparrow 00 22-22 00 00 10-10
(3)(3) 00 s1s_1 7070 1111 00 00 1212 11 22 00
22 x2x_2 55 5-5 11 00 22 00 11 2-2
1010 x3x_3 2020 6-6 00 11 77 00 22 3-3
σj\sigma_j 7676 00 00 66-66 00 22-22 3434

Notice that in (3)(3), σ7>0\sigma_7>0 and a70\mathbf a_7\leq0, which means the problem is unbounded.

Problem 6

Use both the Big M method and the Two-Phase method to solve the following linear programming problems and identify which class the solution belongs to.

maxz=10x1+15x2+12x3s.t.{5x1+3x2+x395x1+6x2+15x3152x1+x2+x35x1,x2,x30\max z =10x_1 +15x_2 +12x_3 \\ \mathrm{s.t.} \left\{\begin{matrix} 5x_1 +3x_2 +x_3 \leq 9 \\ −5x_1 +6x_2 +15x_3 \leq 15 \\ 2x_1 +x_2 +x_3 \ge 5 \\ x_1, x_2,x_3\ge 0 \end{matrix}\right.
  • Big M method: Convert it into M-method form:
maxz=10x1+15x2+12x3Ms4s.t.{5x1+3x2+x3+s1=95x1+6x2+15x3+s2=152x1+x2+x3s3+s4=5x1,x2,x3,s1,s2,s3,s40\max z =10x_1 +15x_2 +12x_3-Ms_4 \\ \mathrm{s.t.} \left\{\begin{matrix} 5x_1 +3x_2 +x_3+s_1 = 9 \\ −5x_1 +6x_2 +15x_3+s_2 = 15 \\ 2x_1 +x_2 +x_3-s_3 +s_4= 5 \\ x_1, x_2,x_3,s_1,s_2,s_3,s_4\ge 0 \end{matrix}\right.

Then, we can draw table:

(1)(1) cjc_j\rightarrow 1010 1515 1212 00 00 00 M-M
CBC_B XBX_B bb x1x_1 x2x_2 x3x_3 s1s_1 s2s_2 s3s_3 s4s_4
00 s1s_1 99 [5][5] 33 11 11 00 00 00
00 s2s_2 1515 5-5 66 1515 00 11 00 00
M-M s4s_4 55 22 11 11 00 00 1-1 11
σj\sigma_j 10+2M10+2M\uparrow 15+M15+M 12+M12+M 00 00 M-M 00
(2)(2) 1010 x1x_1 1.81.8 11 0.60.6 0.20.2 0.20.2 00 00 00
00 s2s_2 2424 00 99 [16][16] 11 11 00 00
M-M s4s_4 1.41.4 00 0.2-0.2 0.60.6 0.4-0.4 00 1-1 11
σj\sigma_j 00 90.2M9-0.2M 10+0.6M10+0.6M\uparrow 20.4M-2-0.4M 00 M-M 00
(3)(3) 1010 x1x_1 1.51.5 11 39/8039/80 00 3/163/16 1/80-1/80 00 00
1212 x3x_3 1.51.5 00 9/169/16 11 1/161/16 1/161/16 00 00
M-M s4s_4 0.50.5 00 43/80-43/80 00 7/16-7/16 3/80-3/80 1-1 11
σj\sigma_j 00 27/843M/8027/8-43M/80 00 20.4M-2-0.4M 5/83M/80-5/8-3M/80 M-M 00

As the optimal solution is x=(1.5,0,1.5,0,0,0,0.5)Tx^*=(1.5,0,1.5,0,0,0,0.5)^T, still having s4s_4, which means that there is no feasible solution for the problem.

  • Two-Phase method:

First phase:

mins4s.t.{5x1+3x2+x3+s1=95x1+6x2+15x3+s2=152x1+x2+x3s3+s4=5x1,x2,x3,s1,s2,s3,s40\min s_4 \\ \mathrm{s.t.} \left\{\begin{matrix} 5x_1 +3x_2 +x_3+s_1 = 9 \\ −5x_1 +6x_2 +15x_3+s_2 = 15 \\ 2x_1 +x_2 +x_3-s_3 +s_4= 5 \\ x_1, x_2,x_3,s_1,s_2,s_3,s_4\ge 0 \end{matrix}\right.
(1)(1) cjc_j\rightarrow 00 00 00 00 00 00 11
CBC_B XBX_B bb x1x_1 x2x_2 x3x_3 s1s_1 s2s_2 s3s_3 s4s_4
00 s1s_1 99 [5][5] 33 11 11 00 00 00
00 s2s_2 1515 5-5 66 1515 00 11 00 00
11 s4s_4 55 22 11 11 00 00 1-1 11
σj\sigma_j 2-2\uparrow 1-1 1-1 00 00 11 00
(2)(2) 00 x1x_1 1.81.8 11 0.60.6 0.20.2 0.20.2 00 00 00
00 s2s_2 2424 00 99 [16][16] 11 11 00 00
11 s4s_4 1.41.4 00 0.2-0.2 0.60.6 0.4-0.4 00 1-1 11
σj\sigma_j 00 0.20.2 0.6-0.6\uparrow 0.40.4 00 11 00
(3)(3) 00 x1x_1 1.51.5 11 39/8039/80 00 3/163/16 1/80-1/80 00 00
00 x3x_3 1.51.5 00 9/169/16 11 1/161/16 1/161/16 00 00
11 s4s_4 0.50.5 00 43/80-43/80 00 7/16-7/16 3/80-3/80 1-1 11
σj\sigma_j 00 43/8043/80 00 0.40.4 3/803/80 00 00

As the optimal solution still has s4s_4, there is no feasible solution for the problem.

Problem 7

Given a linear programming problem that maximizes the objective function and contains two decision variables, its feasible region is shown in the figure below. Write the basic solution of the linear programming problem in vector form as [xs]T[x\enspace s]^T, where xx and ss represent decision variables and slack variables, respectively. Explain why this linear programming problem exhibits degeneration.
(Hint: You can analyze the basic solution written, or explain from the perspective of the graph.)

tmp1919.png

  • Notice that the point (4,2)(4,2) lies on three different lines, which means that it is constrained by three constraints. As we only have two decision variables, the basic solution [420]T[4\enspace 2\enspace 0]^T can be generated by any two of these three constraints, which means that at least one of the slack variables must be 00, thus this linear programming problem exhibits degeneration.