模擬測試解法

Python 的供應鏈分析

Aaren Stubberfield

Supply Chain Analytics Mgr.

注意事項

需要很久才能解的問題，不適合用 LP 或 IP

注意圖示，男子滑倒

整體概念

整體概念：

對你選的關鍵輸入加入隨機雜訊
反覆求解模型
觀察分佈

為什麼嘗試

為什麼要做：

輸入多半是估計，可能不準確。
先前的敏感度分析一次只改一個輸入。

情境說明

情境－玻璃公司－資源規劃：

Resource	Prod. A	Prod. B	Prod. C
Profit $US	$500	$450	$600

限制：

有需求、生產能力與倉儲容量的限制

風險：

利潤估計可能不準確

# Initialize Class, & Define Variables
model = LpProblem("Max Glass Co. Profits", LpMaximize)
A = LpVariable('A', lowBound=0)
B = LpVariable('B', lowBound=0)
C = LpVariable('C', lowBound=0)

# Define Objective Function
model += 500 * A + 450 * B + 600 * C

# Define Constraints & Solve
model += 6 * A + 5 * B + 8 * C <= 60
model += 10.5 * A + 20 * B + 10 * C <= 150
model += A  <= 8
model.solve()

程式範例－步驟 2

a, b, c = normalvariate(0,25), 
          normalvariate(0,25),
          normalvariate(0,25)

# Define Objective Function
model += (500+a)*A + (450+b)*B + (600+c)*C

# Initialize Class, & Define Variables
model = LpProblem("Max Glass Co. Profits", 
                   LpMaximize)

A = LpVariable('A', lowBound=0)
B = LpVariable('B', lowBound=0)
C = LpVariable('C', lowBound=0)
a, b, c = normalvariate(0,25), 
          normalvariate(0,25), 
          normalvariate(0,25)
# Define Objective Function
model += (500+a)*A + (450+b)*B + (600+c)*C

# Define Constraints & Solve
model += 6 * A + 5 * B + 8 * C <= 60
model += 10.5 * A + 20 * B + 10 * C <= 150
model += A  <= 8
model.solve()

def run_pulp_model():
    # Initialize Class
    model = LpProblem("Max Glass Co. Profits", LpMaximize)
    A = LpVariable('A', lowBound=0)
    B = LpVariable('B', lowBound=0)
    C = LpVariable('C', lowBound=0)
    a, b, c = normalvariate(0,25), normalvariate(0,25), normalvariate(0,25)

    # Define Objective Function
    model += (500+a)*A + (450+b)*B + (600+c)*C

    # Define Constraints & Solve
    model += 6 * A + 5 * B + 8 * C <= 60
    model += 10.5 * A + 20 * B + 10 * C <= 150
    model += A  <= 8
    model.solve()
    o = {'A':A.varValue, 'B':B.varValue, 'C':C.varValue, 'Obj':value(model.objective)}
    return(o)

程式範例－步驟 4

def run_pulp_model():
    # Initialize Class
    model = LpProblem("Max Glass Co. Profits",
                       LpMaximize)
    A = LpVariable('A', lowBound=0)
    B = LpVariable('B', lowBound=0)
    C = LpVariable('C', lowBound=0)
    a, b, c = normalvariate(0,25), 
              normalvariate(0,25),
              normalvariate(0,25)

    # Define Objective Function
    model += (500+a)*A + (450+b)*B + 
             (600+c)*C

    # Define Constraints & Solve
    model += 6 * A + 5 * B + 8 * C <= 60
    model += 10.5 * A + 20 * B + 10 * C <= 150
    model += A  <= 8
    model.solve()
    o = {'A':A.varValue, 'B':B.varValue, 
         'C':C.varValue, 
         'Obj':value(model.objective)}
    return(o)

for i in range(100):
    output.append(run_pulp_model())
df = pd.DataFrame(output)

程式範例－步驟 5

print(df['A'].value_counts())
print(df['B'].value_counts())
print(df['C'].value_counts())

輸出：（結果可能不同）

6.666667    73
0.000000    14
8.000000    13
Name: A, dtype: int64

4.000000    73
5.454546    14
2.400000    13
Name: B, dtype: int64
0.000000    86
4.090909    14
Name: C, dtype: int64

用直方圖視覺化

產品 A：產品 A、B、C 與目標的直方圖

產品 B：產品 B 的直方圖

產品 C：產品 C 的直方圖

目標值：目標值的直方圖

重點總結

不適合用在需要很久才能解的問題
好處
- 觀察隨輸入改變，最佳結果如何變化
步驟
1. 以標準 PuLP 模型程式碼開始
2. 用 Python 的 normalvariate 對關鍵輸入加雜訊
3. 將 PuLP 模型程式包成回傳輸出的函式
4. 建立迴圈呼叫新函式並將結果存成 DataFrame
5. 視覺化結果的 DataFrame

一起來練習吧！

Python 的供應鏈分析