简单的优化模型 |
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步骤一:找到决策变量 设 C i C_i Ci类包装箱厚度 t i t_i ti厘米,重量 w i w_i wi公斤,件数为 n i n_i ni件。 设第一辆车装载 C i C_i Ci类包装箱 x i x_i xi件,第二辆装载 C i C_i Ci类包装箱 y i y_i yi件( i i i=1,2,…,7)。 (决策变量是 x i x_i xi , y i y_i yi) 步骤二:找到目标函数 第一辆车剩余空间 Z 1 = 1020 = ∑ i = 1 7 t i ⋅ x i Z_1 = 1020 = \sum_{i=1}^7t_i\cdot x_i Z1=1020=∑i=17ti⋅xi 第二辆车剩余空间 Z 2 = 1020 = ∑ i = 1 7 t i ⋅ y i Z_2 = 1020 = \sum_{i=1}^7t_i\cdot y_i Z2=1020=∑i=17ti⋅yi 总剩余空间最小为目标,则目标函数为: m i n Z = Z 1 + Z 2 minZ = Z_1 + Z_2 minZ=Z1+Z2 步骤三:找到约束条件 件数满足 x i + y i ≤ n i , i = 1 , 2 , … , 7 x_i + y_i \leq n_i, i = 1,2,\dots,7 xi+yi≤ni,i=1,2,…,7 ,且 x i x_i xi , y i y_i yi 为整数。 各辆车载重为40吨,有: ∑ i = 1 7 w i ⋅ x i ≤ 40000 \sum_{i=1}^7w_i\cdot x_i \leq 40000 ∑i=17wi⋅xi≤40000 , ∑ i = 1 7 w i ⋅ y i ≤ 40000 \sum_{i=1}^7w_i\cdot y_i \leq 40000 ∑i=17wi⋅yi≤40000 各辆车有10.2米(1020厘米)的地方用来装包装箱,有: ∑ i = 1 7 t i ⋅ x i ≤ 1020 \sum_{i=1}^7t_i\cdot x_i \leq 1020 ∑i=17ti⋅xi≤1020 , ∑ i = 1 7 t i ⋅ y i ≤ 1020 \sum_{i=1}^7t_i\cdot y_i \leq 1020 ∑i=17ti⋅yi≤1020 C 5 C_5 C5, C 6 C_6 C6, C 7 C_7 C7 类包装箱的厚度不能超过302.7厘米,有: ∑ i = 5 7 t i ⋅ ( x i + y i ) ≤ 302.7 \sum_{i=5}^7t_i\cdot (x_i + y_i) \leq 302.7 ∑i=57ti⋅(xi+yi)≤302.7 总的线性规划模型: m i n Z = ( 1020 − ∑ i = 1 7 t i ⋅ x i ) + ( 1020 − ∑ i = 1 7 t i ⋅ y i ) { ∑ i = 1 7 t i ⋅ x i ≤ 1020 ∑ i = 1 7 t i ⋅ y i ≤ 1020 x i + y i ≤ n i , i = 1 , 2 , … , 7 ∑ i = 1 7 w i ⋅ x i ≤ 40000 ∑ i = 1 7 w i ⋅ y i ≤ 40000 ∑ i = 5 7 t i ⋅ ( x i + y i ) ≤ 302.7 x i , y i 为 非 负 整 数 minZ = (1020-\sum_{i=1}^7t_i\cdot x_i) + (1020-\sum_{i=1}^7t_i\cdot y_i) \\ \\ \begin {cases} \sum_{i=1}^7t_i\cdot x_i \leq 1020 \\ \\ \sum_{i=1}^7t_i\cdot y_i \leq 1020 \\ \\ x_i + y_i \leq n_i, i = 1,2,\dots,7 \\ \\ \sum_{i=1}^7w_i\cdot x_i \leq 40000 \\ \\ \sum_{i=1}^7w_i\cdot y_i \leq 40000 \\ \\ \sum_{i=5}^7t_i\cdot (x_i + y_i) \leq 302.7 \\ \\ x_i, y_i为非负整数 \end {cases} minZ=(1020−i=1∑7ti⋅xi)+(1020−i=1∑7ti⋅yi)⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧∑i=17ti⋅xi≤1020∑i=17ti⋅yi≤1020xi+yi≤ni,i=1,2,…,7∑i=17wi⋅xi≤40000∑i=17wi⋅yi≤40000∑i=57ti⋅(xi+yi)≤302.7xi,yi为非负整数 步骤四:编写程序求解 得到的解为(老师课件中的解): x 1 = 6 , x 2 = 2 , x 3 = 9 , x 4 = 1 , x 5 = 0 , x 6 = 0 , x 7 = 0 x_1 = 6, \quad x_2 = 2, \quad x_3 = 9, \quad x_4 = 1, \quad x_5 = 0, \quad x_6 = 0, \quad x_7 = 0 x1=6,x2=2,x3=9,x4=1,x5=0,x6=0,x7=0 y 1 = 2 , y 2 = 5 , y 3 = 0 , y 4 = 5 , y 5 = 3 , y 6 = 3 , y 7 = 0 y_1 = 2, \quad y_2 = 5, \quad y_3 = 0, \quad y_4 = 5, \quad y_5 = 3, \quad y_6 = 3, \quad y_7 = 0 y1=2,y2=5,y3=0,y4=5,y5=3,y6=3,y7=0 目标值 Z = 0.6 Z = 0.6 Z=0.6 ( x i x_i xi, y i y_i yi可以有不同的结果,但是目标值是唯一的) 其中第一辆平板车剩余空间为0.1厘米;第二辆平板车剩余空间为0.5厘米; 我运行的LINGO(LINGO 12.0)程序得出的解为: x 1 = 5 , x 2 = 1 , x 3 = 5 , x 4 = 3 , x 5 = 2 , x 6 = 2 , x 7 = 0 x_1 = 5, \quad x_2 = 1, \quad x_3 = 5, \quad x_4 = 3, \quad x_5 = 2, \quad x_6 = 2, \quad x_7 = 0 x1=5,x2=1,x3=5,x4=3,x5=2,x6=2,x7=0 y 1 = 3 , y 2 = 6 , y 3 = 4 , y 4 = 3 , y 5 = 1 , y 6 = 1 , y 7 = 0 y_1 = 3, \quad y_2 = 6, \quad y_3 = 4, \quad y_4 = 3, \quad y_5 = 1, \quad y_6 = 1, \quad y_7 = 0 y1=3,y2=6,y3=4,y4=3,y5=1,y6=1,y7=0 目标值 Z = 0.6 Z = 0.6 Z=0.6 其中第一辆平板车剩余空间为0.6厘米;第二辆平板车剩余空间为0厘米; LINGO程序代码: |
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