Mathematical Programming: Modelling and Applications September 2009 Sonia Cafieri LIX, École Polytechnique cafieri@lix.polytechnique.fr
Outline � Some basic AMPL useful operations & commands � A modelling problem � Formulation of the mathematical model � The AMPL model � Solution of the problem 2
AMPL basic set operations These operations become useful or necessary in many problems Let A, B, and U be sets � AMPL allows simple set operations: � � Union: U = A union B elements of A or B � Intersection: U = A inter B elements of A and B � Difference: U = A diff B elements of A and not of B Union : Example ampl: set MONTHS := {1,2,3,4}; ampl: set MONTHS0 := {0} union MONTHS; ampl: display MONTHS0; set MONTHS0 := 0 1 2 3 4; 3
AMPL commands AMPL recognizes a lot of commands. Some commands you already know are, e.g.: � model: switch to model mode � data: switch to data mode � exit: exit AMPL � display: print model entities and expressions � let: change data values � …… � Other useful commands: � fix: freeze a variable at its current value � unfix: undo a fix command � delete: delete model entities � purge: delete model entities and their dependents � 4
AMPL commands Note that fix, unfix, delete, purge , .. are used for changing a model : it is possible in AMPL modifying models and data. Fix : Example fix varname • This command instructs AMPL to treat the indicated variable as though fixed at its current value (e.g. in solve command): we have a constant. • If varname is the name of an indexed collection of variables, fix (and unfix) affects all members of the collection. • Fixing a variable we have a further constraint in the problem. 5
AMPL commands: fix ampl: var y{1..4} >=1; ampl: let y[3] := 10; ampl: fix y[3]; ampl: display y[3]; y[3] = 10 ampl: minimize somma: sum{i in 1..4} y[i]; ampl: option solver cplex; ampl: solve; ILOG CPLEX 10.100, licensed to "ecolepolytechnique-palaiseau", options: e m b q use=8 CPLEX 10.1.0: optimal solution; objective 13 0 dual simplex iterations (0 in phase I) ampl: display y; y [*] := 1 1 2 1 3 10 4 1 ; 6
Production planning problem A firm is planning the production of 3 products A1, A2, A3 over a time horizon of 4 months (January to April). We know: • The demand for the products over the 4 months; • Prices, production costs, production quotas, activation costs and minimum batches for each product; Furthermore: • There is a different number of productive days over the 4 months; • The activation status of a production line can be changed every month. • Minimum batches are monthly. • Each product needs to be stored. There are different monthly rates for renting the storage space for each product. • Each product takes the same amount of storage space. The total available volume is given. 7
Production planning problem: data Demand for the products over the months: Demand January February March April A1 5300 1200 7400 5300 A2 4500 5400 6500 7200 A3 4400 6700 12500 13200 Prices, production costs, production quotas, activation costs and minimum batches: Product A1 A2 A3 Selling prices $124 $109 $115 Activation costs $150000 $150000 $100000 Production costs $73.30 $52.90 $65.40 Production quotas 500 450 550 Minimum batches 20 20 16 8
Production planning problem: data Number of productive days over the months: January 23 February 20 March 23 April 22 Monthly rates for storage space: A1 $3.50 A2 $4.00 A3 $3.00 Total available volume: 800 units. Write a mathematical program to maximize the income, and solve it with AMPL. 9
Writing the mathematical model Preliminary observations: • There are some quantities (parameters) defined for each product : selling price, production cost, production quota, activation cost, minimum batch, storage cost for each month: number of production days for each product and each month: maximum demand for each product in each month independently on product/month: storage capacit y. • Furthermore: For each product, there is a different quantity produced, sold, stocked during each month and the activation status of each production line is also dependent on the months. 10
Writing the mathematical model Sets and indices: i ∈ • 3 products � use an index I j ∈ • 4 months � use an index J define parameters / variables indexed on i , j What decisions should we take? quantity produced product , month quantity sold j i quantity stocked also take into account the activation status 11
Mathematical model • Parameters P j : number of production days in month j; d ij : maximum demand for product i in month j; v i : selling price for product i; c i : production cost of product i; q i : maximum production quota of product i; a i : activation cost for production i; b i : minimum batch for production i; s i : storage cost for product i; C : storage capacity in number of units. 12
Mathematical model • Variables x ij : quantity of product i produced during month j; All variables are non w ij : quantity of product i sold during month j; negative z ij : quantity of product i stocked during month j; y ij : activation status for production i during month j; ⎧ 1 if product i is active during month j = ⎨ y ij binary y ij ⎩ 0 otherwise • Objective function maximize the total income total income: 3 products – sum income obtained during each month ⎛ ⎞ ∑ ∑ ∑ ∑ ∑ ⎜ ⎟ − − − max v w c x s z a y ⎜ ⎟ i ij i ij i ij i ij ⎝ ⎠ ∈ ∈ ∈ ∈ ∈ i I j J j J j J j J 13
Mathematical model • Constraints ∀ ∈ ∈ ≤ • demand: i I , j J w d ij ij x ∑ ∀ ∈ ≤ ij • production: j J P j q ∈ i I i ∀ ∈ ∈ + = + • balance: i I , j J z x z w − 1 i , j ij ij ij ∑ ∀ ∈ ij ≤ • capacity: j J z C ∈ i I ∀ ∈ ∈ ≤ • activation: i I , j J x P q y ij j i ij ∀ ∈ ∈ ≥ • minimum batch: i I , j J x b y ij i ij ∀ ∈ 0 = • december: i I z 0 i 14
AMPL model – production planning Starting from the mathematical formulation, try to code the AMPL model � We have 2 indices for many variables/constraints: i (products): A1, A2, A3 j (months): 1,2,3,4 It is useful to define 2 sets: set PRODUCTS; param Months; set MONTHS := 1..Months We also have z i0 -- index 0 set MONTHS0 := MONTHS union {0} 15
AMPL model – production planning � Now we can define the parameters: all parameters are non negative param days{MONTHS} >= 0; param demand { PRODUCTS, MONTHS } >= 0; param price { PRODUCTS } >= 0; param cost { PRODUCTS } >= 0; param quota { PRODUCTS } >= 0; param activation { PRODUCTS } >= 0; param batch { PRODUCTS } >= 0; param storage { PRODUCTS } >= 0; param capacity >= 0; � Variables: var x { PRODUCTS, MONTHS } >= 0; var w { PRODUCTS, MONTHS } >= 0; var z { PRODUCTS, MONTHS0 } >= 0; var y { PRODUCTS, MONTHS } >= 0, binary; 16
AMPL model – production planning � Objective function: maximize revenue: sum {i in PRODUCTS} (price[i] * sum {j in MONTHS} w[i,j] - cost[i] * sum {j in MONTHS} x[i,j] - storage[i] * sum {j in MONTHS} z[i,j] - activation[i] * sum {j in MONTHS} y[i,j]) ; � Constraints: subject to requirement {i in PRODUCTS, j in MONTHS}: w[i,j] <= demand[i,j]; subject to production {j in MONTHS}: sum {i in PRODUCTS} (x[i,j] / quota[i]) <= days[j]; subject to balance {i in PRODUCTS, j in MONTHS}: z[i,j-1] + x[i,j] = z[i,j] + w[i,j]; subject to capacitymag {j in MONTHS}: sum {i in PRODUCTS} z[i,j] <= capacity; subject to active {i in PRODUCTS, j in MONTHS}: x[i,j] <= days[j]*quota[i]*y[i,j]; subject to minbatch {i in PRODUCTS, j in MONTHS}: x[i,j] >= batch[i]*y[i,j]; 17
AMPL dat – production planning set PRODUCTS := A1 A2 A3 ; param Months := 4 ; param days := 1 23 2 20 3 23 4 22 ; param demand: 1 2 3 4 := A1 5300 1200 7400 5300 A2 4500 5400 6500 7200 A3 4400 6700 12500 13200 ; param : price cost quota activation batch storage := A1 124 73.30 500 150000 20 3.5 A2 109 52.90 450 150000 20 4 A3 115 65.40 550 100000 16 3 ; param capacity := 800 ; let {i in PRODUCTS} z[i,0] := 0; fix {i in PRODUCTS} z[i,0]; 18
AMPL run – production planning model productionplanning.mod; data productionplanning.dat; option solver cplex; solve; option display_round 4; display revenue; display x; display y; 19
Solution – production planning ILOG AMPL 10.100, licensed to "ecolepolytechnique-palaiseau". AMPL Version 20060626 (Linux 2.6.9-5.ELsmp) ILOG CPLEX 10.100, licensed to "ecolepolytechnique-palaiseau", options: e m b q use=8 CPLEX 10.1.0: optimal integer solution; objective 1581550 33 MIP simplex iterations 0 branch-and-bound nodes revenue = 1581550.0000 x := y := A1 1 6100.0000 A1 1 1.0000 A1 2 0.0000 A1 2 0.0000 A1 3 0.0000 A1 3 0.0000 A1 4 0.0000 A1 4 0.0000 A2 1 0.0000 A2 1 0.0000 A2 2 3518.1818 A2 2 1.0000 A2 3 0.0000 A2 3 0.0000 A2 4 0.0000 A2 4 0.0000 A3 1 4400.0000 A3 1 1.0000 A3 2 6700.0000 A3 2 1.0000 A3 3 12650.0000 A3 3 1.0000 A3 4 12100.0000 A3 4 1.0000 ; ; 20
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