2

我需要在 Julia+JuMP 中定义一些常量参数，类似于您在 AMPL 中定义时所做的

set A := a0 a1 a2;

param p :=
a0 1
a1 5
a2 10 ;

我如何在 Julia 中定义类似Aand的东西？p

4

2 回答 2

4

除了 Julia 中可用的之外，JuMP 本身并没有为索引集定义特殊语法。因此，例如，您可以定义

A = [:a0, :a1, :a2]

其中:a0定义了一个符号。

如果你想在这个集合上索引一个变量，那么语法是：

m = Model()
@variable(m, x[A])

JuMP 也不像 AMPL 那样区分数据和模型，因此没有参数的真正概念。相反，您只需在使用数据时提供数据。如果我正确理解你的问题，你可以做类似的事情

p = Dict(:a0 => 1, :a1 => 5, :a2 => 10)
@constraint(m, sum(p[i]*x[i] for i in A) <= 20)

这将添加约束

x[a0] + 5 x[a1] + 10 x[a2] <= 20

我们定义p为 Julia 字典的地方。这里没有特定于 JuMP 的内容，实际上任何 julia 表达式都可以作为系数提供。可以很容易地说

@constraint(m, sum(foo(i)*x[i] for i in A) <= 20)

foo可以执行数据库查找、计算 pi 的位数等的任意 Julia 函数在哪里？

于 2014-07-09T20:44:25.110 回答
0

我无法得到@mlubin 工作的原始答案。此外，网络上的许多示例都使用基于位置的索引，我觉得不太自然，所以我改用字典重写了 GAMS 教程的trnsport.gms示例..感觉更接近 gams/ampl "sets “..

#=
Transposition in JuMP of the basic transport model used in the GAMS tutorial

This problem finds a least cost shipping schedule that meets requirements at markets and supplies at factories.

- Original formulation: Dantzig, G B, Chapter 3.3. In Linear Programming and Extensions.
Princeton University Press, Princeton, New Jersey, 1963.
- Gams implementation: This formulation is described in        detail in:
Rosenthal, R E, Chapter 2: A GAMS Tutorial. In GAMS: A User's Guide.
The Scientific Press, Redwood City, California, 1988.
- JuMP implementation: Antonello Lobianco
=#

using JuMP, DataFrames

# Sets
plants  = ["seattle","san_diego"]          # canning plants
markets = ["new_york","chicago","topeka"]  # markets

# Parameters
a = Dict(              # capacity of plant i in cases
  "seattle"   => 350,
  "san_diego" => 600,
)
b = Dict(              # demand at market j in cases
  "new_york"  => 325,
  "chicago"   => 300,
  "topeka"    => 275,
)

#  distance in thousands of miles
d_table = wsv"""
plants     new_york  chicago  topeka
seattle    2.5       1.7      1.8
san_diego  2.5       1.8      1.4
"""
d = Dict( (r[:plants],m) => r[Symbol(m)] for r in       eachrow(d_table), m in markets)

f = 90 # freight in dollars per case per thousand miles

c = Dict() # transport cost in thousands of dollars per case ;
[ c[p,m] = f * d[p,m] / 1000 for p in plants, m in markets]

# Model declaration
trmodel = Model() # transport model

# Variables
@variables trmodel begin
    x[p in plants, m in markets] >= 0 # shipment quantities in cases
end

# Constraints
@constraints trmodel begin
    supply[p in plants],   # observe supply limit at plant p
        sum(x[p,m] for m in markets)  <=  a[p]
    demand[m in markets],  # satisfy demand at market m
        sum(x[p,m] for p in plants)  >=  b[m]
 end

# Objective
@objective trmodel Min begin
    sum(c[p,m]*x[p,m] for p in plants, m in markets)
end

print(trmodel)

status = solve(trmodel)

if status == :Optimal
    println("Objective value: ", getobjectivevalue(trmodel))
    println("Shipped quantities: ")
    println(getvalue(x))
    println("Shadow prices of supply:")
    [println("$p = $(getdual(supply[p]))") for p in plants]
    println("Shadow prices of demand:")
    [println("$m = $(getdual(demand[m]))") for m in markets]

else
    println("Model didn't solved")
    println(status)
end

# Expected result:
# obj= 153.675
#['seattle','new-york']   = 50
#['seattle','chicago']    = 300
#['seattle','topeka']     = 0
#['san-diego','new-york'] = 275
#['san-diego','chicago']  = 0
#['san-diego','topeka']   = 275

我的相关博客文章中提供了更多评论版本。

于 2017-01-28T17:26:27.100 回答