cplex：无法获取决策变量的值

Question

我正在解决Cplex中的医院员工日程安排问题，我是Cplex的新手。

但Cplex无法配置决策html" target="_blank">变量的答案。

我想可能是我的模型设计问题。

这是一个很长的模型。

如果我能得到帮助，我将不胜感激。

模型

--设置：

{string}E=...；//经验等级集（高级，新秀）

{字符串}I=…；//全套医生

{string}为=//具有S级经验的一组医生

{字符串}Ir=//具有R级经验的一组医生

{字符串}K=//班次类型（o

{字符串}A1=//轮班工作（白天、晚上、晚上）

{string}A2=...；//一组shift t work（白天，晚上，晚上1，晚上2）

--参数

int D=...；

范围日=1... D；//规划范围

int Co[Is]=//具有高级经验的医生i的雇佣成本

int Ct[红外]=//具有新手经验的医生i的招聘成本

int北大=//医生执行轮班o的最长天数

int Pkd=//医生轮班的最短天数

int Pkuh=//医生执行t班的最长天数

int Pkdh=//医生执行t班的最短天数

int s=...；//最大限度。连续工作日

int dosd=//“高级”医生的轮班工作需求

int剂量=//轮班o“高级”医生的夜间工作需求

int dos n=...；//“高级”医生夜班需求

int dord=//轮班“新手”医生的日间工作需求

int dore=//轮班o“新手”医生的夜间工作需求

int dorn=//“新手”医生夜班需求

int dtsd=...；//“高级”医生的日间工作需求

int dtse=//t班“高级”医师夜间工作需求

int dtsn1=//t班夜间1“高级”医生工作需求

int dtsn2=//t班夜间2“高级”医生工作需求

int dtrd=...；//“菜鸟”医生的日间工作需求

int dtre=...；//shift t晚间工作对“菜鸟”医生的需求

int dtrn1=...；//shift t night 1对“菜鸟”医生的工作需求

int dtrn2=//t班夜间2“新手”医生工作需求

--二进制十进制变量

dvar int x2[I][Day][A2]in 0...1；//1，如果高级班次t其中一项工作

dvar int y1[I][Day][A1]in 0...1；//1，如果菜鸟换班o其中一项工作

0中的dvar int y2[I][Day][A2]。。1; //1，如果新手轮班

--决策变量的表达式

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

降低成本；

--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={“高级”，“新手”}；

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”、“晚上”}；

A2={“白天”、“晚上”、“晚上1”、“晚上2”}；

D=7；

Co=[12001150110010501000950900850]；

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；

Answer 1

你的模式可能不可行。如果你给你的约束贴上标签，那么CPLEX会给你一些放松和冲突，这会帮助你理解为什么。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 

}

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