我可以使用什么 GEKKO 模型构造或建模策略来固定 MV 在特定时间的 UPPER 值甚至状态属性
What GEKKO model construct(s) or modelling strategy can I use to fix the UPPER value or even status attribute of MV at a specific time
我计划使用 GEKKO 解决动态生产调度问题，涉及 供应、处理、消耗流程，在某些步骤之间具有 hold-up（存储）。我想在 horizon 上最大化一些 objective。但是，在某些时间步，某些单元操作的处理能力可能存在 pre-defined 限制。虽然我可以使用 fix(var,val,pos) 函数将变量完全固定在 pos，但将其限制在 UPPER 一侧更能代表我想要完成的事情，并且在某些情况下可能会产生不同的解决方案.
这是一些玩具问题代码，还没有包含动态）：
"""
Simple toy problem to test flexibillity of limiting MV's at certain time points in the
horizon without fixing them specifically, i.e. leave one bound unconstrained.
"""
from gekko import GEKKO
m=GEKKO(remote=False)
m.time=[0,1,2,3,4] #Use 5 discrete points
m.options.CV_TYPE = 1
supply2=m.Param(5.5) #Supply of stream2 available to a separation unit op
recovery=m.Const(value=0.92) #Extraction efficiency of unit op
feed1=m.MV(20,lb=15,ub=40) #define feed 1 as an independent variable
feed2=m.MV(5,lb=0,ub=10) #define feed 2 as an independent variable
feed1.COST=1 #cost of feed stream 1
feed2.COST=1.5 #cost of feed stream 2
feed1.STATUS=1 #use feed1 in optimisation
feed2.STATUS=1 #use feed2 in optimisation
ovhds=m.CV(30) #define ovhds of unit op as dependent variable
ovhds.STATUS=1 #use in Objective function
ovhds.SPLO=40 #low limit for dependent variable
ovhds.SPHI=50 #high limit for dependenent variable
ovhds.COST=-2 # negative cost (aka profit) from extracted stream
feed1.UPPER=48 #set overall upper limit of 48 for feed1 MV
m.fix(feed1,47,2) #fix feed 1 at a point pos=2 in the horizon
#TODO: add dynamics e.g. differential equations to model inventory volumes.
supply2_flared=m.Intermediate(feed2-supply2) #another independent variable
total_feed=m.Intermediate(feed1+feed2) #the total intake of feed
m.Equation(ovhds==total_feed*recovery) #define relationship between dependent and independent variable
m.options.IMODE=6 #dynamic control, dynamics and dynamic constraints to be added as Equations later.
m.solve()
print("Feed1",feed1.value)
print("Feed2", feed2.value)
print("Product", ovhds.value)
定义一个随时间变化的参数，例如 feed1_ub 并包含一个不等式约束 m.Equation()。
feed1_ub = m.Param([48,48,30,48,48])
m.Equation(feed1<=feed1_ub)
这是一个完整的脚本：
from gekko import GEKKO
m=GEKKO(remote=False)
m.time=[0,1,2,3,4] #Use 5 discrete points
m.options.CV_TYPE = 1
supply2=m.Param(5.5) #Supply of stream2 available to a separation unit op
recovery=m.Const(value=0.92) #Extraction efficiency of unit op
feed1=m.MV(20,lb=15,ub=40) #define feed 1 as an independent variable
feed2=m.MV(5,lb=0,ub=10) #define feed 2 as an independent variable
feed1.COST=1 #cost of feed stream 1
feed2.COST=1.5 #cost of feed stream 2
feed1.STATUS=1 #use feed1 in optimisation
feed2.STATUS=1 #use feed2 in optimisation
ovhds=m.CV(30) #define ovhds of unit op as dependent variable
ovhds.STATUS=1 #use in Objective function
ovhds.SPLO=40 #low limit for dependent variable
ovhds.SPHI=50 #high limit for dependenent variable
ovhds.COST=-2 # negative cost (aka profit) from extracted stream
feed1.UPPER=48 #set overall upper limit of 48 for feed1 MV
supply2_flared=m.Intermediate(feed2-supply2) #another independent variable
total_feed=m.Intermediate(feed1+feed2) #the total intake of feed
m.Equation(ovhds==total_feed*recovery) #define relationship between dependent and independent variable
feed1_ub = m.Param([48,48,30,48,48])
m.Equation(feed1<=feed1_ub)
m.options.IMODE=6 #dynamic control, dynamics and dynamic constraints to be added as Equations later.
m.solve()
print("Feed1",feed1.value)
print("Feed2", feed2.value)
print("Product", ovhds.value)
解决方案
EXIT: Optimal Solution Found.
The solution was found.
The final value of the objective function is -92.03399437705355
---------------------------------------------------
Solver : IPOPT (v3.12)
Solution time : 0.043 sec
Objective : -92.03399324684949
Successful solution
---------------------------------------------------
Feed1 [20.0, 48.0, 30.000000009, 48.0, 48.0]
Feed2 [5.0, 6.3478260396, 10.0, 6.3478260399, 6.3478260394]
Product [30.0, 49.999999965, 36.8000001, 49.999999965, 49.999999965]
我计划使用 GEKKO 解决动态生产调度问题，涉及 供应、处理、消耗流程，在某些步骤之间具有 hold-up（存储）。我想在 horizon 上最大化一些 objective。但是，在某些时间步，某些单元操作的处理能力可能存在 pre-defined 限制。虽然我可以使用 fix(var,val,pos) 函数将变量完全固定在 pos，但将其限制在 UPPER 一侧更能代表我想要完成的事情，并且在某些情况下可能会产生不同的解决方案.
这是一些玩具问题代码，还没有包含动态）：
"""
Simple toy problem to test flexibillity of limiting MV's at certain time points in the
horizon without fixing them specifically, i.e. leave one bound unconstrained.
"""
from gekko import GEKKO
m=GEKKO(remote=False)
m.time=[0,1,2,3,4] #Use 5 discrete points
m.options.CV_TYPE = 1
supply2=m.Param(5.5) #Supply of stream2 available to a separation unit op
recovery=m.Const(value=0.92) #Extraction efficiency of unit op
feed1=m.MV(20,lb=15,ub=40) #define feed 1 as an independent variable
feed2=m.MV(5,lb=0,ub=10) #define feed 2 as an independent variable
feed1.COST=1 #cost of feed stream 1
feed2.COST=1.5 #cost of feed stream 2
feed1.STATUS=1 #use feed1 in optimisation
feed2.STATUS=1 #use feed2 in optimisation
ovhds=m.CV(30) #define ovhds of unit op as dependent variable
ovhds.STATUS=1 #use in Objective function
ovhds.SPLO=40 #low limit for dependent variable
ovhds.SPHI=50 #high limit for dependenent variable
ovhds.COST=-2 # negative cost (aka profit) from extracted stream
feed1.UPPER=48 #set overall upper limit of 48 for feed1 MV
m.fix(feed1,47,2) #fix feed 1 at a point pos=2 in the horizon
#TODO: add dynamics e.g. differential equations to model inventory volumes.
supply2_flared=m.Intermediate(feed2-supply2) #another independent variable
total_feed=m.Intermediate(feed1+feed2) #the total intake of feed
m.Equation(ovhds==total_feed*recovery) #define relationship between dependent and independent variable
m.options.IMODE=6 #dynamic control, dynamics and dynamic constraints to be added as Equations later.
m.solve()
print("Feed1",feed1.value)
print("Feed2", feed2.value)
print("Product", ovhds.value)
定义一个随时间变化的参数，例如 feed1_ub 并包含一个不等式约束 m.Equation()。
feed1_ub = m.Param([48,48,30,48,48])
m.Equation(feed1<=feed1_ub)
这是一个完整的脚本：
from gekko import GEKKO
m=GEKKO(remote=False)
m.time=[0,1,2,3,4] #Use 5 discrete points
m.options.CV_TYPE = 1
supply2=m.Param(5.5) #Supply of stream2 available to a separation unit op
recovery=m.Const(value=0.92) #Extraction efficiency of unit op
feed1=m.MV(20,lb=15,ub=40) #define feed 1 as an independent variable
feed2=m.MV(5,lb=0,ub=10) #define feed 2 as an independent variable
feed1.COST=1 #cost of feed stream 1
feed2.COST=1.5 #cost of feed stream 2
feed1.STATUS=1 #use feed1 in optimisation
feed2.STATUS=1 #use feed2 in optimisation
ovhds=m.CV(30) #define ovhds of unit op as dependent variable
ovhds.STATUS=1 #use in Objective function
ovhds.SPLO=40 #low limit for dependent variable
ovhds.SPHI=50 #high limit for dependenent variable
ovhds.COST=-2 # negative cost (aka profit) from extracted stream
feed1.UPPER=48 #set overall upper limit of 48 for feed1 MV
supply2_flared=m.Intermediate(feed2-supply2) #another independent variable
total_feed=m.Intermediate(feed1+feed2) #the total intake of feed
m.Equation(ovhds==total_feed*recovery) #define relationship between dependent and independent variable
feed1_ub = m.Param([48,48,30,48,48])
m.Equation(feed1<=feed1_ub)
m.options.IMODE=6 #dynamic control, dynamics and dynamic constraints to be added as Equations later.
m.solve()
print("Feed1",feed1.value)
print("Feed2", feed2.value)
print("Product", ovhds.value)
解决方案
EXIT: Optimal Solution Found.
The solution was found.
The final value of the objective function is -92.03399437705355
---------------------------------------------------
Solver : IPOPT (v3.12)
Solution time : 0.043 sec
Objective : -92.03399324684949
Successful solution
---------------------------------------------------
Feed1 [20.0, 48.0, 30.000000009, 48.0, 48.0]
Feed2 [5.0, 6.3478260396, 10.0, 6.3478260399, 6.3478260394]
Product [30.0, 49.999999965, 36.8000001, 49.999999965, 49.999999965]