4. Given the deformation field of x3 2X1 + X2 + 3X3 1 Calculate (a) The...
2x1 − x2 − 3x3 − 2x4 = 1 x1 − x2 − 4x3 − 2x4 = 5 3x1 − x2 − x3 − 3x4 = −2 x1 + 2x3 − x4 = −4
Consider the mathematical program max 3x1 x2 +3x3 s.t. 2X1 + X2 + X3 +X4-2 x1 + 2x2 + 3x3 + 2xs 5 2x1 + 2x2 + x3 + 3x6 = 6 Three feasible solutions ((a) through (c)) are listed below. (b) xo) (0.9, 0, 0, 0.2,2.05, 1.4) (c) xo) (0.3, 0.1, 0.4, 0.9, 1.65, 1.6) Please choose one appropriate interior point from the list, and use the Karmarkar's Method at the interior point and determine the optimal solution.
Problem 1 (20 pts) Consider the mathematical program max 3x1+x2 +3x3 s.t. 2x1 +x2 + x3 +x2 x1 + 2x2 + 3x3 +2xs 5 2x 2x2 +x3 +3x6-6 Xy X2, X3, X4, Xs, X620 Three feasible solutions ((a) through (c)) are listed below. (0.3, 0.1, 0.4, 0.9, 1.65, 1.6) (c) x Please choose one appropriate interior point from the list, and use the Karmarkar's Method at the interior point and determine the optimal solution. 25
Problem 1 (20 pts) Consider the mathematical program max 3x1+x2 +3x3 s.t. 2x1 +x2 + x3 +x2 x1 + 2x2 + 3x3 +2xs 5 2x 2x2 +x3 +3x6-6 Xy X2, X3, X4, Xs, X620 Three feasible solutions ((a) through (c)) are listed below. (0.3, 0.1, 0.4, 0.9, 1.65, 1.6) (c) x Please choose one appropriate interior point from the list, and use the Karmarkar's Method at the interior point and determine the optimal solution. 25
Problem 1 (20 pts) Consider...
3. For the infinitesimal deformation defined by the following displacement field: u; = a*(X, + X3) Uz = a?(X2 + x3) Uz = -a(X,X2) a) Given that a is a small constant, find the normal strain at point (0.2.-1) along the 18:-1:43 direction b) Find the change in the right angle between infinitesimal material fibers at point (0.2.-1) lying along the directions (8:-1:43 and 4:4-7)
Solve the problems: 1. x2 x33. 2. 2x, + 2x1 + 4x2-3x3 → min. 8x1 3x2 + 3x3 3 40, x2 20. 3. xi + X2 = 1, x120, x120.
Solve the problems: 1. x2 x33. 2. 2x, + 2x1 + 4x2-3x3 → min. 8x1 3x2 + 3x3 3 40, x2 20. 3. xi + X2 = 1, x120, x120.
Consider the following LPP: Maximize z = 50x1 + 20x2 + 30x3 subject to 2x1 + x2 + 3x3 + 90 (Resource A) x1 + 2x2 + x3 + 50 (Resource B) x1 + x2 + x3 + 80 (Resource C) x1, x2 , x3 > 0 The final simplex table is Basis cj x1 x2 x3 s1 s2 s3 Solution 50 20 30 0 0 0 x1 50 1 -1 0 1 -1 0 40 x3 30 0...
Solve the system X1 + 2x2 – 3x3 = 5 2x1 + x2 – 3x3 = 13 - X1 + X2 = -8 1 x= 1), sec 0 a. b. N x=s3 sec -5 0 X=S SEC -1 O O d. X=t 1 1 1 tec Oe. x= -1, sec 0
Solve the system X1 + 2x2 – 3x3 = 5 2x1 + x2 – 3x3 = 13 - X1 + x2 = -8 [1 X=t1 tec 1 a. b. SEC Oc. 1 - -- 1. Jeee -2 -0. x=t0 O d. -1 , SEC e. SEC o f. X=S 2 3 ], sec -5
Given the LPP: Max z=-2x1+x2-x3 St: x1+x2+x3<=6 -x1+2x2<=4 x1,x2<=0 What is the new optimal, if any, when the a) RHS is replaced by [3 4] b) Column a2 is changed from[1 2] to [2 5] c) Column a1 is changed from[1 -1] to [0 -1] d) First constraint is changed to x2-x3<=6 ? e) New activity x6>=0 having c6=1 and a6=[-1 2] is introduced ?