# Lumber transportation problem (J. Reeb and S. Leavengood)¶

Millco has three wood mills and is planning three new logging sites. Each mill has a maximum capacity and each logging site can harvest a certain number of truckloads of lumber per day. The cost of a haul is \$2/mile of distance. If distances from logging sites to mills are given below, how should the hauls be routed to minimize hauling costs while meeting all demands? Logging Site Mill A Mill B Mill C Max loads per day 1 8 15 50 20 2 10 17 20 30 3 30 26 15 45 Mill demand 30 35 30 ### Solution¶ In : # this solution uses NamedArrays, which are arrays indexed over sets for both x and y dimensions. using JuMP, Clp, NamedArrays sites = [ 1, 2, 3] mills = [:A, :B, :C] cost_per_haul = 4 # don't forget the return trip! dist = NamedArray( [8 15 50; 10 17 20; 30 26 15], (sites,mills), ("Sites","Mills") ) supply = Dict(zip( sites, [20 30 45] )) demand = Dict(zip( mills, [30 35 30] )) m = Model(solver=ClpSolver()) @variable(m, x[sites,mills] >= 0) # x[i,j] is lumber shipped from site i to mill j. @constraint(m, sup[i in sites], sum(x[i,j] for j in mills) == supply[i] ) # supply constraint @constraint(m, dem[j in mills], sum(x[i,j] for i in sites) == demand[j] ) # demand constraint @objective(m, Min, cost_per_haul*sum( x[i,j]*dist[i,j] for i in sites, j in mills ) ) # minimize transportation cost status = solve(m) println(status) # nicely formatted solution solution = NamedArray( Int[getvalue(x[i,j]) for i in sites, j in mills], (sites,mills), ("Sites","Mills") ) println( solution ) println() println("Total cost will be \$", getobjectivevalue(m))

Optimal
3×3 Named Array{Int64,2}
Sites ╲ Mills │ :A  :B  :C
──────────────┼───────────
1             │ 20   0   0
2             │ 10  20   0
3             │  0  15  30

Total cost will be \$5760.0


### Compact version of the problem¶

In :
using JuMP, Clp

m = Model(solver=ClpSolver())

# incidence matrix:
A = [ 1  1  1  0  0  0  0  0  0
0  0  0  1  1  1  0  0  0
0  0  0  0  0  0  1  1  1
-1  0  0 -1  0  0 -1  0  0
0 -1  0  0 -1  0  0 -1  0
0  0 -1  0  0 -1  0  0 -1 ]

# supply and demand
b = [ 20, 30, 45, -30, -35, -30 ]

# distances
c = [ 8, 15, 50, 10, 17, 20, 30, 26, 15, ]

@variable(m, x[1:9] >= 0)
@constraint(m, A*x .== b)
@objective(m, Min, 4*dot(c,x))

solve(m)
xsol = getvalue(x)
display( reshape(xsol,3,3)' )

3×3 Array{Float64,2}:
20.0   0.0   0.0
10.0  20.0   0.0
0.0  15.0  30.0
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