SAP ABAP 使用GENIOS求解线性规划问题的简单例子

主要内容来自Operations Research & ABAP ,结合我遇到的需求,做了一些修改。

需求:有BOX1和BOX2两种箱子,分别能包装不同数量的A物料和B物料,给出若干数量的A, B物料,怎样包装可以使箱子数最少?

线性规划有助于解决类似问题。

以下是一个示例程序,包含必要的注释,

*&---------------------------------------------------------------------*
*& Report YTEST_LP1
*&---------------------------------------------------------------------*
*&
*&---------------------------------------------------------------------*
REPORT ytest_lp1.

*BOX1 箱子可以装2个A物料 1个B物料, BOX2箱子可以装1个A物料 3个B物料, 现有10A 11B, 如何包装箱子最少?
*假设最终结果用到了X个 BOX1箱子, Y个BOX2箱子,问题可以写成如下形式
* minimize X + Y
* subject to : 2X + 1Y >= 10
*              X + 3Y >= 11
*              X >= 0
*              Y >= 0
DATA:
  lr_model       TYPE REF TO cl_genios_model, "problem instance
  lr_objective   TYPE REF TO cl_genios_objective, "objective function
  lr_environment TYPE REF TO cl_genios_environment,
  lr_x           TYPE REF TO cl_genios_variable, "variables
  lr_y           TYPE REF TO cl_genios_variable,
  lr_constraint  TYPE REF TO cl_genios_linearconstraint,
  lr_solver      TYPE REF TO cl_genios_solver,
  ls_result      TYPE genioss_solver_result,
  ls_variable    TYPE genioss_variable,
  lt_variables   TYPE geniost_variable,
  lv_value       TYPE genios_float,
  lv_name        TYPE string. "变量名

lr_environment = cl_genios_environment=>get_environment( ).
lr_model = lr_environment->create_model( 'PRIORIZATION' ). 
*定义离散变量,因为箱子是离散的定义连续变量,因为SIMP只支持连续变量
lr_x = lr_model->create_variable( iv_name = 'X'
  iv_type = if_genios_model_c=>gc_var_continuous ).
lr_y = lr_model->create_variable( iv_name = 'Y'
  iv_type = if_genios_model_c=>gc_var_continuous ).
"下面设置目标函数,取满足条件的(X+Y)最小值,现实中也可能有不同的系数,比如X的价格是2,Y是1,那么在这里做相应调整
lr_objective = lr_model->create_objective( if_genios_model_c=>gc_obj_minimization ).
lr_objective->add_monom( io_variable = lr_x iv_coefficient = 1 ). lr_objective->add_monom( io_variable = lr_y iv_coefficient = 1 ).
* 定义线性约束c1: 2X + 1Y >= 10 lr_constraint = lr_model->create_linearconstraint( iv_name = 'c1' iv_type = if_genios_model_c=>gc_con_greaterorequal iv_righthandside = 10 ). lr_constraint->add_monom( io_variable = lr_x iv_coefficient = 2 ). lr_constraint->add_monom( io_variable = lr_y iv_coefficient = 1 ). * 定义线性约束c2: 1X + 3Y >= 11 lr_constraint = lr_model->create_linearconstraint( iv_name = 'c2' iv_type = if_genios_model_c=>gc_con_greaterorequal iv_righthandside = 11 ). lr_constraint->add_monom( io_variable = lr_x iv_coefficient = 1 ). lr_constraint->add_monom( io_variable = lr_y iv_coefficient = 3 ). * 定义线性约束c3: X >= 0 lr_constraint = lr_model->create_linearconstraint( iv_name = 'c3' iv_type = if_genios_model_c=>gc_con_greaterorequal iv_righthandside = 0 ). lr_constraint->add_monom( io_variable = lr_x iv_coefficient = 1 ). * 定义线性约束c4: Y >= 0 lr_constraint = lr_model->create_linearconstraint( iv_name = 'c4' iv_type = if_genios_model_c=>gc_con_greaterorequal iv_righthandside = 0 ). lr_constraint->add_monom( io_variable = lr_y iv_coefficient = 1 ). lr_solver = lr_environment->create_solver( 'SIMP' ). lr_solver->load_model( lr_model ). ls_result = lr_solver->solve( ). * Get the result IF ls_result-solution_status = if_genios_solver_result_c=>gc_optimal OR ls_result-solution_status = if_genios_solver_result_c=>gc_abortfeasible. lt_variables = lr_model->get_variables( ). LOOP AT lt_variables INTO ls_variable. lv_name = ls_variable-variable_ref->gv_name. lv_value = ls_variable-variable_ref->get_primalvalue( ). DATA: lv_int TYPE P LENGTH 10 DECIMALS 2. lv_int = lv_value. WRITE: /,lv_name,' = ',lv_int. ENDLOOP. ENDIF. lr_environment->destroy_solver( 'SIMP' ). lr_environment->destroy_model( 'PRIORIZATION' ).

 

运行程序,可以看到结果

X = 3.80
Y = 2.40

示例中的Solver SIMP 仅支持连续变量,对这个需求显然不合适,因为箱子不应该为小数。如果想使用离散变量,就需要换成其他Solver比如MILP,可惜我的系统无法使用MILP,所以不能测试。据说SAP APO或SAP SCM Optimizer可以满足更复杂的需求。

 

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