Rのモンテカルロシミュレーション：ループの戻り値NA

Question

モンテカルロシミュレーション：このコードは、260の要求をシミュレートし、必要な2次シフトの量を計算するためのものです。しかし、私は10回目の実行後にNAの価値に遭遇します。 私はこれを達成しようとしています。 'B' enter image description here

 set.seed(1234) 
     n = 260 
     demand = runif(260, min = 80, max = 130) 
     production_capacity = 100 
     begining_inventory[] = 100 
     post_inventory[] = 0 
     counter = 0 
     for(i in 1:n){ 
      if (i == 1){ 
      begining_inventory[i] = 100 
      ending_inventory[i] = begining_inventory[i] + production_capacity - demand [i] 
      ending_inventory[i] 
      post_inventory[i] = ending_inventory[i] 

      } 
      else{ 
      post_inventory[i] = ending_inventory[i] 
      begining_inventory[i] = post_inventory[i-1] 
      ending_inventory[i] = begining_inventory[i] + production_capacity - demand [i] 
      } 
      if(ending_inventory[i] <= 50){ 
      counter = counter + 1 
      } 
     print(ending_inventory[i]) 
     P.first_shift = (1-counter/n) 
     P.second_shift = 1-P.first_shift 
     } 

Answer 1

から1列目のスタートはそれを考え出した：

set.seed(1234) 
n = 260 
demand = runif(n, min = 80, max = 130) 
production_capacity = 100 
begining_inventory = 100 

counter = 0 
second_shift = rep(0,n) 
ending_inventory = rep(0,n) 

second_shift_production = rep(0,n) 
post_second = rep(0,n) 

for(i in 1:n){ 

    ending_inventory[i] = begining_inventory + production_capacity - demand[i] 

    if(ending_inventory[i] < 50){ 
    print(ending_inventory[i]) 
    second_shift[i] = 1 
    counter = counter + 1 
    }else{ 
    second_shift[i] = 0 
    } 
    second_shift_production[i] = second_shift[i]*100 
    post_second[i] = second_shift_production[i] + ending_inventory[i] 

    if(i > 1){ 
    begining_inventory = post_second[i-1] 
    } 

} 
hist(ending_inventory) 
abline(v = 50, col = "red") 

P.first_shift = (1-counter/n) 
P.second_shift = 1-P.first_shift 

print(paste("Probabiity Only first shift = ", (1-counter/n)*100, "%", sep = " ")) 
print(paste("Probabiity second shift = ", 100-(1-counter/n)*100, "%", sep = " "))