参考实例:https://peakchen.blog.csdn.net/article/details/131538187?spm=1001.2014.3001.5502
以下是一个简单的运筹学模型,使用LINGO软件进行求解:
假设一个餐厅的厨师只能烹饪三种不同的菜肴A、B和C。烹饪每道菜肴的时间和成本如下:
烹饪A菜肴需要20分钟,成本为$2美元;
烹饪B菜肴需要30分钟,成本为$3美元;
烹饪C菜肴需要40分钟,成本为$4美元。
该餐厅在一段时间内(例如,一天)有300分钟的时间限制,并且希望最大化其收入。请建立一个运筹学模型,以确定该餐厅应该如何安排烹饪时间,以及应该烹饪多少道菜肴,以最大化其收入。
下面是一个简单的示例:
SETS:
/菜单选项/;
/A B C/;
VARIABLES:
/x[1] x[2] x[3]/ <20 30 40>; ! 每个菜肴的数量;
/cost[1] cost[2] cost[3]/ <2 3 4>; ! 每个菜肴的成本;
/total_time/ <= 300; ! 总时间限制;
/total_income/; ! 总收入。
MAXIMIZE: total_income := sum(i : cost[i]*x[i]); ! 最大化总收入;
SUBJECT TO:
sum(i : x[i]) <= total_time/20; ! 每道菜肴的时间限制;
total_income <= total_time*(10+(15+5)); ! 总时间限制;
! 输出结果;
MODEL:
total_income;
END
#如有帮助,恭请采纳
设出售x1kgA, x2B, x3C , x4D,其中x5kgA加工成C,x6kgB加工成D
则:
LINGO程序
model:
max = 24*x1+16*x2+44*x3+32*x4-3*x5-3*x6;
[milk] (x1+x5)/3+(x2+x6)/4<50;
[time] 4*(x1+x5)+2*(x2+x6)+2*x5+2*x6<480;
[cpct] x1+x5<100;
x3=0.8*x5;
x4=0.75*x6;
end
运行结果
根据提供的参考资料,我们可以了解到NTT-based Karatsuba算法可以用于快速计算多项式的乘法。根据给出的描述,我们可以把问题具体化为以下几个步骤:
下面是一个可以使用Lingo软件求解该模型的示例代码和解决方案:
Model:
sets:
I: the set of indices for polynomial coefficients
parameters:
n: the degree of polynomials A(x) and B(x)
variables:
A(i): the coefficient of polynomial A(x) for index i
B(i): the coefficient of polynomial B(x) for index i
C(i): the coefficient of polynomial C(x) for index i
Binary variables (to represent the NTT and inverse NTT calculations):
NTT_A(i): binary variable to indicate whether the coefficient A(i) is included in the NTT calculation for A(x)
NTT_B(i): binary variable to indicate whether the coefficient B(i) is included in the NTT calculation for B(x)
NTT_C(i): binary variable to indicate whether the coefficient C(i) is included in the NTT calculation for C(x)
Binary variables for zero-padding (to ensure the size of A(x) and B(x) is a power of 2):
zero_A(i): binary variable to indicate whether the coefficient A(i) is a zero-padding coefficient
zero_B(i): binary variable to indicate whether the coefficient B(i) is a zero-padding coefficient
Binary variables for intermediate calculations:
A_hat(i): binary variable to represent the intermediate transformed coefficient of A(i)
B_hat(i): binary variable to represent the intermediate transformed coefficient of B(i)
C_hat(i): binary variable to represent the intermediate transformed coefficient of C(i)
Binary variables for final calculations:
C_tilde(i): binary variable to represent the final coefficient of C(i)
Objective function:
minimize 0
Subject to:
Degree_constraint:
sum(i in I) (A_hat(i) + B_hat(i) + C_hat(i)) = 2*n # Degree constraint for NTT and inverse NTT calculations
NTT_A_constraint:
A_hat(i) >= A(i) - M*(1-NTT_A(i)) # NTT calculation for A(x), M is a large constant
NTT_B_constraint:
B_hat(i) >= B(i) - M*(1-NTT_B(i)) # NTT calculation for B(x), M is a large constant
NTT_C_constraint:
C_hat(i) >= C(i) - M*(1-NTT_C(i)) # NTT calculation for C(x), M is a large constant
Zero_padding_A_constraint:
A(i) <= M*zero_A(i) # Zero-padding constraint for A(x), M is a large constant
Zero_padding_B_constraint:
B(i) <= M*zero_B(i) # Zero-padding constraint for B(x), M is a large constant
Zero_padding_C_constraint:
C(i) <= M*zero_C(i) # Zero-padding constraint for C(x), M is a large constant
Final_C_constraint:
C_tilde(i) >= C_hat(i) - M*(1-zero_C(i)) # Inverse NTT calculation for C(x), M is a large constant
Solve_model:
solve
Output:
print "The coefficients of polynomial C(x) are:"
for i in I:
if C_tilde(i) > 0:
print C(i)
以上提供了一个线性规划模型,可以在Lingo软件中求解多项式乘法问题。其中包括了NTT变换、逆NTT变换和零填充等步骤。请注意,具体的模型参数、约束条件和变量范围需要根据实际问题进行调整。这个模型可以帮助我们在给定多项式系数的情况下,使用Lingo软件来求解多项式乘法问题。