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标题:
SAP ABAP 使用GENIOS求解线性规划问题的简单例子
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作者:
熊熊出没
时间:
2023-7-22 01:31
标题:
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<br>lr_objective = lr_model->create_objective( if_genios_model_c=>gc_obj_minimization ).<br>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' ).
复制代码
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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