<\/p>\n
An MIP Example<\/strong><\/h4>\nMaximize the function f(x, y) = x + 10<\/em> subject to the following constraints:<\/p>\n\n\n\nx + 7 y \u2264 17.5<\/td>\n<\/tr>\n \nx \u2264 3.5<\/td>\n<\/tr>\n \nx \u2265 0<\/td>\n<\/tr>\n \ny \u2265 0<\/td>\n<\/tr>\n \nx,\u00a0y\u00a0integers<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nSolution<\/strong><\/h5>\nFor clarity, we would solve this problem using the simple 5-step linear programming approach. Then we would modify the program to use an MIP approach.<\/p>\n
Step 1 – Create your solver<\/strong><\/p>\nYou’ll need to import the ortools.linear_solver from pywrap and create a solver variable. the code below does that<\/p>\n
# IMPORT THE SOLVER<\/span>\r\nfrom<\/span> ortools.linear_solver<\/span> import<\/span> pywraplp\r\nsolver =<\/span> pywraplp.<\/span>Solver.<\/span>CreateSolver('GLOP'<\/span>)\r\n<\/pre>\n <\/p>\n
Step 2 – Declare the variables<\/strong><\/p>\nWe do this using the solver’s NumVar<\/em> method and specify: the upper bound, the lower bound and the variable name.<\/p>\n# DECLARE THE VARIABLES<\/span>\r\nx =<\/span> solver.<\/span>NumVar(0<\/span>, 3.5<\/span>, 'x'<\/span>)\r\ny =<\/span> solver.<\/span>NumVar(0<\/span>, solver.<\/span>infinity(), 'y'<\/span>)\r\n<\/pre>\n <\/p>\n
Step 3 – Create the Constraints<\/strong><\/p>\nConstraints are of the form 0 <= f(x,y) <= 17.5<\/em>. We would create only this one constraints because the other constraints have been implicitly created with the variable declaration.<\/p>\n# CREATE THE CONSTRAINT 0 <= f(x,y) <= 17.5<\/span>\r\nct =<\/span> solver.<\/span>Constraint(0<\/span>, 17.5<\/span>) # specifies upper and lower bound<\/span>\r\nct.<\/span>SetCoefficient(x, 1<\/span>) # specify the x coefficient<\/span>\r\nct.<\/span>SetCoefficient(y, 7<\/span>) # specify the y coefficient<\/span>\r\n<\/pre>\n <\/p>\n
Step 4 – Define the Cost function<\/strong><\/p>\nWe simply declare the objective variable and then set the coefficient of x and y in the function.<\/p>\n
# DEFINE THE OBJECTIVE FUNCTION x + 10y<\/span>\r\nobj =<\/span> solver.<\/span>Objective()\r\nobj.<\/span>SetCoefficient(x, 1<\/span>)\r\nobj.<\/span>SetCoefficient(y, 10<\/span>)\r\nobj.<\/span>SetMaximization() # set the problem goal as maximization<\/span>\r\n<\/pre>\n <\/p>\n
Step 5 – Call the Solver and display the result<\/strong><\/p>\nYou must say solver.Solve(). The value of the objective function is in the Value() method of the objective variable. Similarly, the value of each variable if in the solution_value() method of each of the variables. The code below displays the results.<\/p>\n
# CALL THE SOLVER AND SHOW THE RESULT<\/span>\r\nsolver.<\/span>Solve()\r\nprint<\/span>('Objective value = '<\/span>, obj.<\/span>Value())\r\nprint<\/span>('x = '<\/span>, x.<\/span>solution_value())\r\nprint<\/span>('y = '<\/span>, y.<\/span>solution_value())\r\n<\/pre>\nThe result is given below:<\/p>\n
Objective value = 25.0\r\nx = 0.0\r\ny = 2.5\r\n<\/pre>\n <\/p>\n
Solving Using MIP Approach<\/strong><\/h4>\nTo solve using MIP you need to make just a few changes.<\/p>\n
\n- in the variable declaration, instead of using NumVar(), you use IntVar()<\/li>\n
- in creating the solver, you need to use ‘SCIP’ instead of ‘GLOP”<\/li>\n<\/ul>\n
If you make this changes, then you will have the output:
\n<\/p>\n
Objective value = 23.0\r\nx = 3.0\r\ny = 2.0\r\n<\/pre>\n <\/p>\n
| x + 7 y \u2264 17.5<\/td>\n<\/tr>\n |
| x \u2264 3.5<\/td>\n<\/tr>\n |
| x \u2265 0<\/td>\n<\/tr>\n |
| y \u2265 0<\/td>\n<\/tr>\n |
x,\u00a0y\u00a0integers<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nSolution<\/strong><\/h5>\nYou’ll need to import the ortools.linear_solver from pywrap and create a solver variable. the code below does that<\/p>\n # IMPORT THE SOLVER<\/span>\r\nfrom<\/span> ortools.linear_solver<\/span> import<\/span> pywraplp\r\nsolver =<\/span> pywraplp.<\/span>Solver.<\/span>CreateSolver('GLOP'<\/span>)\r\n<\/pre>\n |
![\"Solution](\"https:\/\/www.kindsonthegenius.com\/data-science\/wp-content\/uploads\/sites\/12\/2021\/01\/Screenshot-2021-01-20-at-12.04.29.png\")
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