Question

Solve the following model using linear programming (allow for continuous values and determine the values of the decision variables and objective function. Then, round the decision variables values down to the nearest integer and determine the value of the decision variables and objective function, this is an approximate answer to solving the model using integer programming. Observe if the rounding provides a "feasible solution, all constraints are satisfied. Finally, solve the model using integer programming and determine the values of the decision variables and the objective function using integer constraints. Observe the differences in the answers using the three methods. Maximize Z = 4x1 + 5x2 + 5x3 subject to 1x1 + 1x2 + 1x3 3 8 -2x1 + 1x2 + -1x33 -7 - 1x1 + - 1x2 + 1x3 5 x1, x2, x320 LP Method Round Down Method x1 = IP Method

Answer 1

Solution LP Method Max 124 +62 +243 sit 121 + 1 2 +123 cis -2x, + 1 X 2 -1x ₂ <=-4 -12 -2 tllze=56 found FS : 19. 33333 0.000000 2 131 Global optimal solution Obiective value: infeasibiuties : Total solver iterations: Eloupsed runtime Seconds Model class : Total variables : Nonu near variables: integer variables o Total constraints : 4 Nonine ar constraints o Total non geros: 12 Nonlinear monzeros : 0

variable value Reduced cost 3333333 2.666667 0.000000 0.000000 0.000000 0.6666667 Slack or Surplus 19.33333 0.000000 Dual price 1.000000 4.333333 11666667 0.000000 11.00000 0.000000 X1 = 3.33 X 2 = 267 x=0 z = 19.33 Round down method X1=3 X2=2 X=0 Z = 1x3 +6x2 + 2x0 215 using these values, constraint 1 = 1x3+12271x0 -52-6, therefore, constraint is satisfied.

constraint o 0-2x3+1X2 +-Ixo - -44--4, therefore, constraint & is satistied constraint3: -1x3 - 1X2 + 1x0 = -52-5) i constraint 3 is satisfied IP method and solution using LINGO is 1p model as follows Mpx 17-тьх, -10х3 -24 + 1X2 -1X3 45-4 -1X1 - 1X2 + 1 X ₂ L=5 END GIN XI GIN X2 GIN X3 Global optimal solution found Objeetive value: 18.00000 Objective bound: 180 in feasibilities: 0.000000 Extended Solve steps: 0 Total solver iterations : 2 Elapsed runtime seconds: 0.45

Model class PILP Total variables : 3 Nonlinear variables :o Integer variables :3 Total constraints 4 Non linear constraints o Total rion zeros i 12 Nonlinear nonze ros: 0 variable value Reduced cost 2.000000 - 1.000000 X2 2.000000 - 6.000000 2.000000 - 2.000000 Row dual poke 1.000000 0.000000 0.000000 0.00000 da J slack or Surplus 18.00000 0.000000 0.000000 7.000000 x = 2 x2 = 2 Az = 2 Z1X2 + 6x2 + 2x2 18