Ground transport
Description
| Problem name and type, features | Difficulty | Related examples |
| E‑1 | Car rental: Transport problem | *** | transport_graph.mos |
|---|
| data preprocessing, set operations, sqrt and ^2, if-then-elif |
| E‑2 | Choosing the mode of transport: Minimum cost flow | ** | mincostflow_graph.mos |
|---|
| formulation with extra nodes for modes of transport; encoding of arcs, finalize, union of sets, nodes labeled with strings |
| E‑3 | Depot location: Facility location problem | *** | facilityloc_graph.mos |
|---|
| modeling flows as fractions, definition of model cuts |
| E‑4 | Heating oil delivery: Vehicle routing problem (VRP) | **** | vrp_graph.mos |
|---|
| elimination of inadmissible subtours, cuts; selection with `|', definition of model cuts |
| E‑5 | Combining different modes of transport | *** | |
|---|
| modeling implications, weak and strong formulation of bounding constraints; triple indices |
| E‑6 | Fleet planning for vans | *** | |
|---|
| maxlist, minlist, max, min |
Further explanation of this example:
'Applications of optimization with Xpress-MP', Chapter 10: Ground transport
Source Files
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Data Files
e1carrent.mos
(!******************************************************
Mosel Example Problems
======================
file e1carrent.mos
``````````````````
Fleet management in car rental
A rental company has 94 cars at 10 locations. Assume the
distance between each location is 1.3 times the Euclidean
distance. Given the transport cost per km, determine how
many cars to move between locations so that morning demand
is met while minimizing total cost.
Here the data is checked to confirm there are the number
of cars required, exiting if stock does not meet demand.
Then the locations are sorted into 'NEED' or 'EXCESS'
locations with 'if-then-elif' to determine start (source) and
destination (sink) locations. To calculate distance, this
problem introduces the function 'sqrt' and the exponential
operator '^'.
(c) 2008-2022 Fair Isaac Corporation
author: S. Heipcke, Mar. 2002, rev. Mar. 2022
*******************************************************!)
model "E-1 Car rental"
uses "mmxprs"
declarations
AGENTS = 1..10 ! Car rental agencies
REQ: array(AGENTS) of integer ! Required number of cars
STOCK: array(AGENTS) of integer ! Number of cars in stock
X,Y: array(AGENTS) of integer ! Coordinates of rental agencies
COST: real ! Cost per km of moving a car
NEED: set of integer ! Agencies needing more cars
EXCESS: set of integer ! Agencies with too many cars
end-declarations
initializations from 'e1carrent.dat'
REQ STOCK X Y COST
end-initializations
if sum(a in AGENTS) (STOCK(a)-REQ(a)) <> 0 then
writeln("Problem is infeasible")
exit(0)
end-if
! Calculate sets of agencies with excess or deficit of cars
forall(a in AGENTS)
if STOCK(a) - REQ(a) < 0 then
NEED += {a}
elif STOCK(a) - REQ(a) > 0 then
EXCESS += {a}
end-if
finalize(NEED); finalize(EXCESS)
declarations
DIST: array(EXCESS,NEED) of real ! Distance between agencies
move: array(EXCESS,NEED) of mpvar ! Cars exchanged between agencies
end-declarations
! Calculate distances between agencies
forall(a in EXCESS,b in NEED)
DIST(a,b):= 1.3*sqrt((X(a)-X(b))^2 + (Y(a)-Y(b))^2)
! Objective: total transport cost
Cost:= sum(a in EXCESS,b in NEED) COST*DIST(a,b)*move(a,b)
! Agencies with excess availability
forall(a in EXCESS) sum(b in NEED) move(a,b) = STOCK(a) - REQ(a)
! Agencies in need of cars
forall(b in NEED) sum(a in EXCESS) move(a,b) = REQ(b) - STOCK(b)
forall(a in EXCESS,b in NEED) move(a,b) is_integer
! Solve the problem
minimize(Cost)
! Solution printing
writeln("Total cost: ", getobjval)
write(" ->")
forall(b in NEED) write(strfmt(b,3))
writeln
setparam("realfmt","%3g")
forall(a in EXCESS) do
write(a," ")
forall(b in NEED) write(getsol(move(a,b)))
writeln
end-do
end-model
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