我正在 Cplex 中解决医院工作人员调度问题，我是 Cplex 的新手。
但是 Cplex 无法配置决策变量的答案。
我想也许这与我的模型设计问题有关。
这是一个很长的模型。
如果我能得到帮助，非常感谢。
++模型
--集：
{string} E=...;//一组经验水平(高级、新秀(
{string} I=...;//所有医生的集合
{string} Is=...;//一组具有经验水平 S 的医生
{string} Ir=...;一组具有R级经验的医生
{string} K=...;//shift type of shifts (o & t(
{string} A1=...;//set of shift o work (day， night， night(
{string} A2=...;//班次 t 工作集(白天、晚上、晚上 1、晚上 2(
--参数
int D=...;
范围 日=1..D;//计划区间
范围int Co[Is]=...;//具有高级经验水平的医生的招聘成本
int Ct[Ir]=...;//具有新手经验水平的医生 i 的招聘成本
int Pku=...;//医生执行轮班的最长天数
oint Pkd=...;//医生执行轮班的最小天数
int Pkuh=...;医生执行轮班 T 的最长天数
int pkdh=...;医生执行轮班 T 的最短天数
int s=...;//最大连续工作日
int dosd=...;//shift o "高级"医生的日常工作需求
int dose=...;//shift o 晚间工作对"高级"医师的需求
int dosn=...;//shift o "高级"医生的夜间工作需求
int dord=...;//shift o "菜鸟"医生的日常工作需求
int dore=...;//shift o 晚间工作对"菜鸟"医生的需求
int dorn=...;//shift o "菜鸟"医生的夜间工作需求
int dtsd=...;//shift t "高级"医师的日常工作需求
int dtse=...;//shift t 晚间工作对"高级"医师的需求
int dtsn1=...;//shift t night1 "高级"医师的工作需求
int dtsn2=...;//shift t night2 "高级"医师的工作需求
int dtrd=...;//shift t "菜鸟"医生的日常工作需求
int dtre=...;//shift t 晚间工作对"菜鸟"医生的需求
int dtrn1=...;//shift t night1 "菜鸟"医生的工作需求
int dtrn2=...;//shift t night2 "菜鸟"医生的工作需求
--二进制十进制变量
dvar int x1[I][Day][A1] in 0..1;//1， 如果 senior+ shift o+ 工作之一
dvar int x2[I][Day][A2] in 0..1;//1， if senior+ shift t+ 工作之一
dvar int y1[I][Day][A1] in 0..1;//1， if rookie+ shift o+ 其中一项工作
dvar int y2[I][Day][A2] in 0..1;//1， if+ shift t+ 其中一项工作
--干燥变量的表达式
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] ;
--目标函数
最小化成本;
--约束
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={"高级"，"新秀"};
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"};
Is={"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={"白天"，"晚上"，"夜晚"};
A2={"白天"，"晚上"，"晚上1"，"晚上2"};
D=7;
Co=[1200,1150,1100,1050,1000,950,900,850];
ct=[
800,750,750,750,700,700,700,700,700,700,650,650,650,650,650,650,650,650];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，则 CPLEX 将为您提供一些放松和冲突，以帮助您理解原因。我的建议是标记所有约束。
例如，您可以更改
forall(d in Day, Day in A1)
{
sum(i in Is) x1[i, d ,"Day"] == dosd; //Daily demand:Senior+shift o+day work
}
到
forall(d in Day, Day in A1)
ct1:sum(i in Is) x1[i, d ,"Day"] == dosd; //Daily demand:Senior+shift o+day work