Cplex：无法获取决策变量的值
Cplex: cannot get value for decision variables
我正在解决 Cplex 中的医院工作人员调度问题，我是 Cplex 的新手。
但 Cplex 无法配置决策变量的答案。
我想可能是我的模型设计问题。
这是一个长款。
非常感谢能得到帮助。
++型号
--套数：
{字符串} E=...; // 一组经验水平（高级，菜鸟）
{字符串} I=...; // 所有医生的集合
{string} 是=...; //一组经验水平为S的医生
{string} Ir=...;//经验水平为R的医师集合
{字符串} K=...; //轮班类型 (o & t)
{字符串} A1=...; //设置班次o工作（白天，晚上，晚上）
{字符串} A2=...; //设置班次t工作（白天，晚上，晚上1，晚上2）
--参数
int D=...;
range Day=1..D;//规划范围
积分[是]=...; //具有高级经验水平的医生 i 的聘用成本
int Ct[Ir]=...; //具有菜鸟经验水平的医生 i 的聘用成本
int Pku=...; //最大限度。医生执行班次的天数 o
int Pkd=...; //分钟。医生执行班次的天数 o
int Pkuh=...;//最大。医生轮班天数 t
int Pkdh=...;//分钟。医生轮班天数 t
int s=...; //最大限度。连续工作日
int dosd=...; //转移"senior"医生
的日间工作需求
int 剂量=...; //转移 o "senior"医生
的晚间工作需求
int dosn=...; //轮班 "senior" 医生
的夜间工作需求
int dord=...; //转移 "rookie" 医生
的日间工作需求
int dore=...; //转移 o "rookie"医生
的晚间工作需求
int dorn=...; //轮班"rookie"医师
的夜间工作需求
int dtsd=...; //"senior"医生
第t天的工作需求
int dtse=...; //轮班 t 晚间工作需求"senior"医师
int dtsn1=...; //第 t 晚 1 班 "senior" 医生
的工作需求
int dtsn2=...; // shift t night2 "senior"医生
的工作需求
int dtrd=...; //"rookie"医师
的第t天工作需求
int dtre=...; //轮班 t 晚上工作需求"rookie"医生
int dtrn1=...; //第 t night1 班 "rookie" 医生
的工作需求
int dtrn2=...; // shift t night2 "rookie"医生
的工作需求
--二进制判定变量
dvar int x1[I][Day][A1] in 0..1; //1、if senior+ shift o+ 其中一项工作
dvar int x2[I][Day][A2] in 0..1; //1、if senior+ shift t+ 其中一项工作
dvar int y1[I][Day][A1] in 0..1; //1、如果rookie+shift o+其中一项工作
dvar int y2[I][Day][A2] in 0..1; //1、如果rookie+shift t+其中一项工作
--Desicion变量的表达式
dexpr float cost =sum(i in Is, d in Day, a in A1)Co[i]*x1[i][d][a]
+ sum(i in Is, d in Day, a in A2)Co[i]*x2[i][d][a]
+ sum(i in Ir, d in Day, a in A1)Ct[i]*y1[i][d][a]
+ sum(i in Ir, d in Day, a in A2)Ct[i]*y2[i][d][a] ;
--Objective函数
最小化成本；
--约束条件
subject to
{
forall(d in Day, Day in A1)
{
sum(i in Is) x1[i, d ,"Day"] == dosd; //Daily demand:Senior+shift o+day work
}
forall(d in Day, Evening in A1)
{
sum(i in Is ) x1[i, d, "Evening"] == dose; //Daily demand:Senior+shift o+evening work
}
forall(d in Day, Night in A1)
{
sum(i in Is ) x1[i, d, "Night"] == dosn; //Daily demand:Senior+shift o+night work
}
forall(d in Day, Day in A2)
{
sum(i in Is) x2[i, d, "Day"] == dtsd; //Daily demand:Senior+ shift t +day work
}
forall(d in Day, Evening in A2)
{
sum(i in Is) x2[i, d, "Evening"] == dtse; //Daily demand:Senior+shift t+evening work
}
forall(d in Day, Night1 in A2)
{
sum(i in Is) x2[i, d, "Night1"] == dtsn1; //Daily demand:Senior+shift t+night1 work
}
forall(d in Day, Night2 in A2)
{
sum(i in Is) x2[i, d, "Night2"] == dtsn2; //Daily demand:Senior+shift t+night2 work
}
forall(d in Day, Day in A1)
{
sum(i in Ir) y1[i, d, "Day"] == dord; //Daily demand:Rookie+shift o+day work
}
forall(d in Day, Evening in A1)
{
sum(i in Ir) y1[i, d, "Evening"] == dore; //Daily demand:Rookie+shift o+evening work
}
forall(d in Day, Night in A1)
{
sum(i in Ir) y1[i, d, "Night"] == dorn; //Daily demand:Rookie+shift o+night work
}
forall(d in Day, Day in A2)
{
sum(i in Ir) y2[i, d, "Day"] == dtrd; //Daily demand:Rookie+shift t+day work
}
forall(d in Day, Evening in A2)
{
sum(i in Ir) y2[i, d, "Evening"] == dtre; //Daily demand:Rookie+shift t+evening work
}
forall(d in Day, Night1 in A2)
{
sum(i in Ir) y2[i, d, "Night1"] == dtrn1; //Daily demand:Rookier+shift t+night1 work
}
forall(d in Day, Night2 in A2)
{
sum(i in Ir) y2[i, d, "Night2"] == dtrn2; //Daily demand:Rookie+shift t+night2 work
}
forall(i in Is, d in Day:d<D, Evening in A1)
{
x1[i, d, "Evening"]+ x1[i,(d+1), "Day"]<=1; //(Senior)For shift type o,if previous day is evening
work, can't have day work the following day'
}
forall(i in Is, d in Day:d<D, Evening in A2)
{
x2[i, d, "Evening"]+ x2[i,(d+1), "Day"]<=1; //(Senior)For shift type t,if previous day is evening
work, can't have day work the following day'
}
forall(i in Ir, d in Day:d<D, Evening in A1)
{
y1[i, d, "Evening"]+ y1[i,(d+1), "Day"]<=1; //(Rookie)For shift type o,if previous day is evening
work, can't have day work the following day'
}
forall(i in Ir, d in Day:d<D, Evening in A2)
{
y2[i, d, "Evening"]+ y2[i,(d+1), "Day"]<=1; //(Rookie)For shift type t,if previous day is evening
work, can't have day work the following day'
}
forall(i in Is, d in Day:d<D, Night in A1)
{
x1[i, d,"Night"]+ x1[i,(d+1),"Day"]<=1; //(Senior)For shift type o,if previous day is night work,
can't have day work the following day'
}
forall(i in Is, d in Day:d<D, Night1 in A2)
{
x2[i, d,"Night1"]+ x2[i,(d+1),"Day"]<=1; //(Senior)For shift type t,if previous day is night1 work,
can't have day work the following day'
}
forall(i in Is, d in Day:d<D, Night2 in A2)
{
x2[i, d,"Night2"]+ x2[i,(d+1),"Day"]<=1; //(Senior)For shift type t,if previous day is night2 work,
can't have day work the following day'
}
forall(i in Ir, d in Day:d<D, Night in A1)
{
y1[i, d,"Night"]+ y1[i,(d+1),"Day"]<=1; //(Rookie)For shift type o,if previous day is night work,
can't have day work the following day'
}
forall(i in Ir, d in Day:d<D, Night1 in A2)
{
y2[i, d,"Night1"]+ y2[i,(d+1),"Day"]<=1; //(Rookie)For shift type t,if previous day is night1 work,
can't have day work the following day'
}
forall(i in Ir, d in Day:d<D, Night2 in A2)
{
y2[i, d,"Night2"]+ y2[i,(d+1),"Day"]<=1; //(Rookie)For shift type t,if previous day is night2 work,
can't have day work the following day'
}
forall (i in Is, d in Day)
{
sum(a in A1)x1[i,d,a] + sum(a in A2)x2[i,d,a] <=1; //(Senior)each day can only have one work with
one shift
}
forall (i in Ir, d in Day)
{
sum(a in A1)y1[i,d,a] + sum(a in A2)y2[i,d,a] <=1; //(Rookie)each day can only have one work with
one shift
}
forall(i in Is, a in A1)
{
sum(d in Day)x1[i,d,a] <=Pku; //(Senior) can’t have shift type o that more than upper bound. (in
days)
}
forall(i in Ir, a in A1)
{
sum(d in Day)y1[i,d,a] <=Pku; //(Rookie) can’t have shift type o that more than upper bound. (in
days)
}
forall(i in Is, a in A1)
{
sum(d in Day)x1[i,d,a]>=Pkd; //(Senior) can’t have shift type o that less than lower bound. (in
days)
}
forall(i in Ir, a in A1)
{
sum(d in Day)y1[i,d,a]>=Pkd; //(Rookie) can’t have shift type o that less than lower bound. (in
days)
}
forall(i in Is, a in A2)
{
sum(d in Day)x2[i,d,a]<=Pkuh; //(Senior) can’t have shift type t that more than upper bound. (in
days)
}
forall(i in Ir, a in A2)
{
sum(d in Day)y2[i,d,a]<=Pkuh; //(Rookie) can’t have shift type t that more than upper bound. (in
days)
}
forall(i in Is, a in A2)
{
sum(d in Day)x2[i,d,a]>=Pkdh; //(Senior) can’t have shift type t that less than lower bound. (in
days)
}
forall(i in Ir, a in A2)
{
sum(d in Day)y2[i,d,a]>=Pkdh; //(Rookie) can’t have shift type t that less than lower bound. (in
days)
}
forall(i in Is)
{
sum(d in Day:d<=(D-s), a in A1)x1[i,d,a]+sum(d in Day:d<=(D-s), a in A2)x2[i,d,a]<=s; //(Senior)
can’t have consecutive working days (regardless of shift type) that more than upper bound. (in days)
}
forall(i in Ir)
{
sum(d in Day:d<=(D-s), a in A1)y1[i,d,a]+sum(d in Day:d<=(D-s), a in A2)y2[i,d,a]<=s; //(Rookie)
can’t have consecutive working days (regardless of shift type) that more than upper bound. (in days)
}
++数据
E={"Senior","Rookie"};
I={"A","B","C","D","E","F","G", "H","I","J","K","L","M","N","O","P","Q","R","S","T","W","X","Y","Z"};
是={"A","B","C","D","E","F","G", "H"};
Ir={"I","J","K","L","M","N","O", "P","Q","R","S","T","W","X","Y","Z"};
K={"o","t"};
A1={"Day","Evening","Night"};
A2={"Day","Evening","Night1","Night2"};
D=7;
Co=[1200,1150,1100,1050,1000,950,900,850];
Ct=[800,750,750,750,700,700,700,700,700,700,650,650,650,650,650,650];
Pku=10;
Pkd=5;
Pkuh=8;
Pkdh=4;
s=5;
您的模型可能不可行。如果您标记约束，那么 CPLEX 会给您一些松弛和冲突，这将帮助您理解原因。我的建议是标记所有约束条件。
例如你可以改变
forall(d in Day, Day in A1)
{
sum(i in Is) x1[i, d ,"Day"] == dosd; //Daily demand:Senior+shift o+day work
}
进入
forall(d in Day, Day in A1)
ct1:sum(i in Is) x1[i, d ,"Day"] == dosd; //Daily demand:Senior+shift o+day work
我正在解决 Cplex 中的医院工作人员调度问题，我是 Cplex 的新手。
但 Cplex 无法配置决策变量的答案。
我想可能是我的模型设计问题。
这是一个长款。
非常感谢能得到帮助。
++型号
--套数：
{字符串} E=...; // 一组经验水平（高级，菜鸟）
{字符串} I=...; // 所有医生的集合
{string} 是=...; //一组经验水平为S的医生
{string} Ir=...;//经验水平为R的医师集合
{字符串} K=...; //轮班类型 (o & t)
{字符串} A1=...; //设置班次o工作（白天，晚上，晚上）
{字符串} A2=...; //设置班次t工作（白天，晚上，晚上1，晚上2）
--参数
int D=...;
range Day=1..D;//规划范围
积分[是]=...; //具有高级经验水平的医生 i 的聘用成本
int Ct[Ir]=...; //具有菜鸟经验水平的医生 i 的聘用成本
int Pku=...; //最大限度。医生执行班次的天数 o
int Pkd=...; //分钟。医生执行班次的天数 o
int Pkuh=...;//最大。医生轮班天数 t
int Pkdh=...;//分钟。医生轮班天数 t
int s=...; //最大限度。连续工作日
int dosd=...; //转移"senior"医生
的日间工作需求int 剂量=...; //转移 o "senior"医生
的晚间工作需求int dosn=...; //轮班 "senior" 医生
的夜间工作需求int dord=...; //转移 "rookie" 医生
的日间工作需求int dore=...; //转移 o "rookie"医生
的晚间工作需求int dorn=...; //轮班"rookie"医师
的夜间工作需求int dtsd=...; //"senior"医生
第t天的工作需求int dtse=...; //轮班 t 晚间工作需求"senior"医师
int dtsn1=...; //第 t 晚 1 班 "senior" 医生
的工作需求int dtsn2=...; // shift t night2 "senior"医生
的工作需求int dtrd=...; //"rookie"医师
的第t天工作需求int dtre=...; //轮班 t 晚上工作需求"rookie"医生
int dtrn1=...; //第 t night1 班 "rookie" 医生
的工作需求int dtrn2=...; // shift t night2 "rookie"医生
的工作需求--二进制判定变量
dvar int x1[I][Day][A1] in 0..1; //1、if senior+ shift o+ 其中一项工作
dvar int x2[I][Day][A2] in 0..1; //1、if senior+ shift t+ 其中一项工作
dvar int y1[I][Day][A1] in 0..1; //1、如果rookie+shift o+其中一项工作
dvar int y2[I][Day][A2] in 0..1; //1、如果rookie+shift t+其中一项工作
--Desicion变量的表达式
dexpr float cost =sum(i in Is, d in Day, a in A1)Co[i]*x1[i][d][a]
+ sum(i in Is, d in Day, a in A2)Co[i]*x2[i][d][a]
+ sum(i in Ir, d in Day, a in A1)Ct[i]*y1[i][d][a]
+ sum(i in Ir, d in Day, a in A2)Ct[i]*y2[i][d][a] ;
--Objective函数
最小化成本；
--约束条件
subject to
{
forall(d in Day, Day in A1)
{
sum(i in Is) x1[i, d ,"Day"] == dosd; //Daily demand:Senior+shift o+day work
}
forall(d in Day, Evening in A1)
{
sum(i in Is ) x1[i, d, "Evening"] == dose; //Daily demand:Senior+shift o+evening work
}
forall(d in Day, Night in A1)
{
sum(i in Is ) x1[i, d, "Night"] == dosn; //Daily demand:Senior+shift o+night work
}
forall(d in Day, Day in A2)
{
sum(i in Is) x2[i, d, "Day"] == dtsd; //Daily demand:Senior+ shift t +day work
}
forall(d in Day, Evening in A2)
{
sum(i in Is) x2[i, d, "Evening"] == dtse; //Daily demand:Senior+shift t+evening work
}
forall(d in Day, Night1 in A2)
{
sum(i in Is) x2[i, d, "Night1"] == dtsn1; //Daily demand:Senior+shift t+night1 work
}
forall(d in Day, Night2 in A2)
{
sum(i in Is) x2[i, d, "Night2"] == dtsn2; //Daily demand:Senior+shift t+night2 work
}
forall(d in Day, Day in A1)
{
sum(i in Ir) y1[i, d, "Day"] == dord; //Daily demand:Rookie+shift o+day work
}
forall(d in Day, Evening in A1)
{
sum(i in Ir) y1[i, d, "Evening"] == dore; //Daily demand:Rookie+shift o+evening work
}
forall(d in Day, Night in A1)
{
sum(i in Ir) y1[i, d, "Night"] == dorn; //Daily demand:Rookie+shift o+night work
}
forall(d in Day, Day in A2)
{
sum(i in Ir) y2[i, d, "Day"] == dtrd; //Daily demand:Rookie+shift t+day work
}
forall(d in Day, Evening in A2)
{
sum(i in Ir) y2[i, d, "Evening"] == dtre; //Daily demand:Rookie+shift t+evening work
}
forall(d in Day, Night1 in A2)
{
sum(i in Ir) y2[i, d, "Night1"] == dtrn1; //Daily demand:Rookier+shift t+night1 work
}
forall(d in Day, Night2 in A2)
{
sum(i in Ir) y2[i, d, "Night2"] == dtrn2; //Daily demand:Rookie+shift t+night2 work
}
forall(i in Is, d in Day:d<D, Evening in A1)
{
x1[i, d, "Evening"]+ x1[i,(d+1), "Day"]<=1; //(Senior)For shift type o,if previous day is evening
work, can't have day work the following day'
}
forall(i in Is, d in Day:d<D, Evening in A2)
{
x2[i, d, "Evening"]+ x2[i,(d+1), "Day"]<=1; //(Senior)For shift type t,if previous day is evening
work, can't have day work the following day'
}
forall(i in Ir, d in Day:d<D, Evening in A1)
{
y1[i, d, "Evening"]+ y1[i,(d+1), "Day"]<=1; //(Rookie)For shift type o,if previous day is evening
work, can't have day work the following day'
}
forall(i in Ir, d in Day:d<D, Evening in A2)
{
y2[i, d, "Evening"]+ y2[i,(d+1), "Day"]<=1; //(Rookie)For shift type t,if previous day is evening
work, can't have day work the following day'
}
forall(i in Is, d in Day:d<D, Night in A1)
{
x1[i, d,"Night"]+ x1[i,(d+1),"Day"]<=1; //(Senior)For shift type o,if previous day is night work,
can't have day work the following day'
}
forall(i in Is, d in Day:d<D, Night1 in A2)
{
x2[i, d,"Night1"]+ x2[i,(d+1),"Day"]<=1; //(Senior)For shift type t,if previous day is night1 work,
can't have day work the following day'
}
forall(i in Is, d in Day:d<D, Night2 in A2)
{
x2[i, d,"Night2"]+ x2[i,(d+1),"Day"]<=1; //(Senior)For shift type t,if previous day is night2 work,
can't have day work the following day'
}
forall(i in Ir, d in Day:d<D, Night in A1)
{
y1[i, d,"Night"]+ y1[i,(d+1),"Day"]<=1; //(Rookie)For shift type o,if previous day is night work,
can't have day work the following day'
}
forall(i in Ir, d in Day:d<D, Night1 in A2)
{
y2[i, d,"Night1"]+ y2[i,(d+1),"Day"]<=1; //(Rookie)For shift type t,if previous day is night1 work,
can't have day work the following day'
}
forall(i in Ir, d in Day:d<D, Night2 in A2)
{
y2[i, d,"Night2"]+ y2[i,(d+1),"Day"]<=1; //(Rookie)For shift type t,if previous day is night2 work,
can't have day work the following day'
}
forall (i in Is, d in Day)
{
sum(a in A1)x1[i,d,a] + sum(a in A2)x2[i,d,a] <=1; //(Senior)each day can only have one work with
one shift
}
forall (i in Ir, d in Day)
{
sum(a in A1)y1[i,d,a] + sum(a in A2)y2[i,d,a] <=1; //(Rookie)each day can only have one work with
one shift
}
forall(i in Is, a in A1)
{
sum(d in Day)x1[i,d,a] <=Pku; //(Senior) can’t have shift type o that more than upper bound. (in
days)
}
forall(i in Ir, a in A1)
{
sum(d in Day)y1[i,d,a] <=Pku; //(Rookie) can’t have shift type o that more than upper bound. (in
days)
}
forall(i in Is, a in A1)
{
sum(d in Day)x1[i,d,a]>=Pkd; //(Senior) can’t have shift type o that less than lower bound. (in
days)
}
forall(i in Ir, a in A1)
{
sum(d in Day)y1[i,d,a]>=Pkd; //(Rookie) can’t have shift type o that less than lower bound. (in
days)
}
forall(i in Is, a in A2)
{
sum(d in Day)x2[i,d,a]<=Pkuh; //(Senior) can’t have shift type t that more than upper bound. (in
days)
}
forall(i in Ir, a in A2)
{
sum(d in Day)y2[i,d,a]<=Pkuh; //(Rookie) can’t have shift type t that more than upper bound. (in
days)
}
forall(i in Is, a in A2)
{
sum(d in Day)x2[i,d,a]>=Pkdh; //(Senior) can’t have shift type t that less than lower bound. (in
days)
}
forall(i in Ir, a in A2)
{
sum(d in Day)y2[i,d,a]>=Pkdh; //(Rookie) can’t have shift type t that less than lower bound. (in
days)
}
forall(i in Is)
{
sum(d in Day:d<=(D-s), a in A1)x1[i,d,a]+sum(d in Day:d<=(D-s), a in A2)x2[i,d,a]<=s; //(Senior)
can’t have consecutive working days (regardless of shift type) that more than upper bound. (in days)
}
forall(i in Ir)
{
sum(d in Day:d<=(D-s), a in A1)y1[i,d,a]+sum(d in Day:d<=(D-s), a in A2)y2[i,d,a]<=s; //(Rookie)
can’t have consecutive working days (regardless of shift type) that more than upper bound. (in days)
}
++数据
E={"Senior","Rookie"};
I={"A","B","C","D","E","F","G", "H","I","J","K","L","M","N","O","P","Q","R","S","T","W","X","Y","Z"};
是={"A","B","C","D","E","F","G", "H"};
Ir={"I","J","K","L","M","N","O", "P","Q","R","S","T","W","X","Y","Z"};
K={"o","t"};
A1={"Day","Evening","Night"};
A2={"Day","Evening","Night1","Night2"};
D=7;
Co=[1200,1150,1100,1050,1000,950,900,850];
Ct=[800,750,750,750,700,700,700,700,700,700,650,650,650,650,650,650];
Pku=10;
Pkd=5;
Pkuh=8;
Pkdh=4;
s=5;
您的模型可能不可行。如果您标记约束，那么 CPLEX 会给您一些松弛和冲突，这将帮助您理解原因。我的建议是标记所有约束条件。
例如你可以改变
forall(d in Day, Day in A1)
{
sum(i in Is) x1[i, d ,"Day"] == dosd; //Daily demand:Senior+shift o+day work
}
进入
forall(d in Day, Day in A1)
ct1:sum(i in Is) x1[i, d ,"Day"] == dosd; //Daily demand:Senior+shift o+day work