Question

这是我第一次使用AMPL并且我并不熟悉它。我必须优化我的模型，但我认为我的代码是错误的，我一直在寻找最佳解决方案和“NOS 5.51：忽略20个变量的完整性”。我的模特是敏（废物）受制于 I attached a picture of constraints

这是我的代码：

^(?=.*\\p{Ll})(?=.*\\p{Lu})(?=.*\\p{N})(?:(?=.*\\p{P})|((?=.*\\p{S})))[\\p{L}\\p{N}\\p{P}\\p{S}]{8,64}$

这就是我得到的：

`set DAY;     #the day we buy ingredient
 set INGRED;   #fresh ingredients
 set TIME;     #day that keeps track of the inventory
 param M default 0;
 param cost{INGRED} > 0; #cost for each ingredient (per pound)
 param demand{TIME, INGRED} >= 0; #the expected demand for each ingredient fo each day t
 param min_pur_req > 0;  #minimum total cost of order required to get a delivery
 param expiry{INGRED} ; #shelf life of each ingredient
 var amount{t in TIME, i in DAY, j in INGRED} >=0; #is defined only for t = i
 var is_bought {t in TIME,i in DAY,j in  INGRED : i <= t }binary; #only for t =i
 var inventory {t in TIME,i in DAY, j in INGRED : i<=t <= i+expiry[j] } >= 0; # t >= i
 var used{t in TIME,i in DAY, j in INGRED :  i <= t <= i+expiry[j]} >= 0; 
 var waste{t in TIME,i in DAY, j in INGRED : t <= i+expiry[j] } >= 0;`
 minimize Total_waste: 
        sum{t in TIME,i in DAY, j in INGRED : i <= t <= i+expiry[j]} waste[t,i,j];
subject to invalid_amount{t in TIME, i in DAY, j in INGRED: i <= t <= i+expiry[j]}: 
      if t != i then amount[t,i,j] = 0;
subject to invalid_binary{t in TIME, i in DAY, j in INGRED: i <= t <= i+expiry[j]}:
     if t != i then is_bought[t,i,j] = 0;
subject to usage{t in TIME, j in INGRED}:
    sum {i in DAY: i <= t <= i + expiry[j]} used[t,i,j] = demand[t, j];
subject to inventory_formula {t in TIME, i in DAY, j in INGRED: i+1 <= t <= i+expiry[j]}: 
    waste[t,i,j] = amount[t,i,j] + inventory[t-1,i,j] - used[t,i,j] - inventory[t,i,j] ;
subject to cost_check {t in TIME, i in DAY: t = i}:
    sum {j in INGRED} cost[j]*amount[t,i,j] >=  sum{j in INGRED: t <= i+expiry[j]} min_pur_req*is_bought[t,i,j];
subject to blah{t in TIME,i in DAY, j in INGRED: i <= t <= i+expiry[j]}:
        amount[t,i,j] <= M*is_bought[t,i,j];
subject to  invalid_inventory{t in TIME, i in DAY, j in INGRED: i+1 <= t <= i+expiry[j]}: 
 #i+1 because for i = 1 i get error in invetory[0,1,j] 
      if t=i then inventory[t-1,i,j] = 0;

AMPL找到最优解零，所有变量都设置为零

0 个答案: