Consider the primal LP problem given below: The optimal tableau is given by: Basic Variables 1. 2 81 21 I₂ x1 0 subject to: 0 1 0 Determine B-¹ X2 0 0 Max = 3r₁ + 2x₂ + 2x3 0 1 2x1 + x₂ + 2x3 ≤ 15 2x1 + x2 + x3 ≤ 12 1₁+1₂ +213 ≤8 11, 12, 13 20 13 -2 1 -1 3 81 0 1 0 0 82 -1 -1 1 -1 83 -1 0 -1 2 RHS 20 3 4 4

Write the dual of the following LPP Minimise Z = 16x₁ + 9x₂ + 21x₃ Subject to the constraints: x₁ + x₂ + x₃ = 16 2x₁ + x₂ + x₃ ≥ 12 x₁ , x₂ ≥ 0 x₃ - unrestricted.

a) Write the dual to the following LPPS: i) Maximize: z= 3x, + 5x. + 4x. ii) Minimize: z= 3x. + 5x. + 2x. subject to: 2x. + X. + 4x 2 subject to: 2x. + 3х 23 3х - 3х+ 2х —8 with: x, x 20 with: x, <0

Write the dual to the following problem. Maximize z = 2x1 + x2  + 3x3 + x4

find: a. The optimal solution for PLE. b. Find the whole solution, rounded. c. How much profit could a company lose if it implemented the rounded solution? Your answer must be accompanied by your calculation memory in pdf file, it will not be valid without this support. Maximize 6X1 + 2X2 + ХЗ r.a: 4X1 + 2X2 + 7X3 > 54 3X1 + 9X2 + 8X3 0 and integers

Construct the dual problem. Minimize  z = – 5x1 – 15x2

Use the simplex method to maximize subject to P = P = 5x₁ + 8x₂ + 2x3 + -X1 + X1 + x₁ ≥ 0 + 7x3 ≤ 12 5x2 + 2x3 ≤ 10 8x2 + 9x3 ≤ 13 X₂ 20 X3 20

Maximize z = 10x₁ + 55x2 + 50x3, subject to 4x₂ + 9x3 ≤ 122 13x2 + 22x3 S 142 6x215x3 S 245 ≥ 0, X3 2 0. X3 $1 16x1 X1 - 8X₁ + 5X1+ x1 ≥ 0, X₂ X2 $2 $3 N

Please use R-programming. Please provide the source code that makes the program run, and then screenshot the output.  Question: Suppose a parallel reliability system consists of cables of wires holding a bridge andthat one of them must have at least 90 wires. If the reliability of a single wire is 0.25 what is a minimal number of wires counted from 90 (or more) should that cable have in order to maintain the reliability of the system to at least 0.95?

Consider two nonnegative numbers x and y where x + y = 5. What is the minimum value of 5x² + 6y ? Enter an exact answer. Provide your answer below: Minimum value =