3

我创建了一个模型来解决 Julia 中的线性程序。我第一次解决了线性程序，但相同的代码不适用于修改后的程序。你能弄清楚发生了什么吗？谢谢！工作正常的代码：

m = Model()
@variable(m, x[1:77] >= 0)
for i in nutrients 
    @constraint(m, dot(data[:, i], x) >= lower[i])
end
@objective(m, Min, sum(dot(c, x)))
status = solve(m)
for i in 1:77
    if getvalue(x[i]) > 0
        println("Take ", getvalue(x[i]), " dollars of ", foods[i], " every day.")
    end
end
println("The total cost for each day is ", getobjectivevalue(m), " dollars.")
println("The total cost for the whole year is ", getobjectivevalue(m) * 365, " dollars.")

这是结果：

Take 0.02951906167648827 dollars of Wheat Flour (Enriched) every day.
Take 0.0018925572907052643 dollars of Liver (Beef) every day.
Take 0.011214435246144865 dollars of Cabbage every day.
Take 0.005007660466725203 dollars of Spinach every day.
Take 0.061028563526693246 dollars of Navy Beans, Dried every day.
The total cost for each day is 0.10866227820675685 dollars.
The total cost for the whole year is 39.66173154546625 dollars.

这是不起作用的代码：

m1 = Model()
for i in 1:77
    if i == a
        @variable(m1, x[i] == 0)
    elseif i == b
        @variable(m1, x[i] == 0)
    else 
        @variable(m1, x[i] >= 0)
    end
end
for i in nutrients 
    @constraint(m1, dot(data[:, i], x) >= lower[i])
end
@objective(m1, Min, sum(dot(c, x)))
status = solve(m1)
for i in 1:77
    if getvalue(x[i]) > 0
        println("Take ", getvalue(x[i]), " dollars of ", foods[i], " every day.")
    end
end
println("The total cost for each day is ", getobjectivevalue(m1), " dollars.")
println("The total cost for the whole year is ", getobjectivevalue(m1) * 365, " dollars.")

这是错误消息：

MethodError: no method matching dot(::NamedArrays.NamedArray{Any,1,Array{Any,1},Tuple{DataStructures.OrderedDict{Any,Int64}}}, ::JuMP.JuMPArray{JuMP.Variable,1,Tuple{Int64}})
Closest candidates are:
  dot(::AbstractArray{T,1}, ::AbstractArray{T,1}) at linalg/generic.jl:302
  dot{T,S,N}(::Array{T,N}, ::JuMP.JuMPArray{S,N,NT<:Tuple{Vararg{T,N}}}) at /Users/yiboliu/.julia/v0.5/JuMP/src/operators.jl:299
  dot{T,S,N}(::JuMP.JuMPArray{T,N,NT<:Tuple{Vararg{T,N}}}, ::JuMP.JuMPArray{S,N,NT<:Tuple{Vararg{T,N}}}) at /Users/yiboliu/.julia/v0.5/JuMP/src/operators.jl:301

 in macro expansion; at /Users/yiboliu/.julia/v0.5/JuMP/src/macros.jl:400 [inlined]
 in macro expansion; at ./In[23]:13 [inlined]
 in anonymous at ./<missing>:?

我知道问题出在限制条件下，但是那部分代码是相同的，并且不起作用。你能告诉我发生了什么吗？

4

2 回答 2

2

错误消息是说调用函数的尝试dot()不符合函数被告知的期望。这可能是因为输入的计数错误，或者它们的类型不是预期的。由于代码在顶层是相同的，我们可以假设是后一种解释，可能类型有问题。

首先检查错误输出以确切了解对 dot 函数的调用（您似乎有两个），然后在调用之前插入一行以输出有关将使用的输入的信息。您可能会使用该typeof()功能。然后将您发送的内容与dot()真正想要的内容进行比较。

于 2017-02-13T07:08:17.107 回答
0

在同一个模型中不能有两个@variable方法调用，因为它们每次都会定义不同的变量。因此，必须将整个第一个for ... end循环m1替换为

@variable(m1, x[1:77] >= 0)
@constraint(m1, x[a] == 0)
@constraint(m1, x[b] == 0)

（我想a并且b已经在别处定义过）

于 2017-02-13T22:10:46.970 回答