Linear Programming

Is concerned with the optimization (minimization or maximization) of a linear function while satisfying a set of linear equality and/or inequality constraints or restrictions

Basic Definitions Minimize/Maximize: z = c1x1 + c2x2 + ... + cnxn Subject to a11x1 + a12x2 + ... + a1nxn ≥ b1 a21x1 + a22x2 + ... + a2nxn ≥ b2 : : + ... + : : am1x1 + am2x2 + ... + amnxn ≥ bm x1, x2, ..., xn ≥ 0

Objective function Cost coeffiecients Decision variables (activity levels) Constraint (or restriction, or functional, structural or technological constraint) Technological coefficients (or constraint matrix A) Right hand-side vector Nonnegativity constraint Feasible point, or feasible solution Feasible region, or feasible space

Assumptions of Linear Programming Proportionality given a variable xj its contribution to cost is cjxj and its contribution to the ith constraint is aijxj, this means that if xj is doubled, say, so is its contribution to cost and each of the constraints (no savings/extra costs are realized by using more of activity j, that is, there are no economies or returns to scale or discounts. Also, no set up cost for starting the activity is realized) Additivity this assumption guarantees that the total cost is the sum of the individual costs, and that the total contribution to the ith restriction is the sum of the individual contribution of the individual activities (there are no substitution or interaction effects among the activities) Divisibility the assumption ensures that the decision variables can be divided into any fractional levels so that noninteger values for the decision variables are permitted Deterministic the coefficients cj, aij, and bi are all known deterministically. Any probabilistic or stochastic elements inherent in demands, costs, prices, resources availabilities, usages and so on are all assumed to be approximated by these coefficients through some deterministic equivalent

Example 1 Minimize: z = 2x1 + 5x2 Subject to x1 + x2 ≥ 6 x1 + x2 ≤ 18

Linear Programming Modelling Problem Formulation Construct a Mathematical Model Derive a Solution Model Testing, Analysis and (possibly) Restructuring Implementation

Example 2 Produk Kapasitas Yang Digunakan per Unit Ukuran Produksi Kapasitas yang Dapat Digunakan Pabrik 1 2 4 12 3 18 Keuntungan per Unit $3 $5

Maksimumkan: z = 3x1 + 5x2 Berdasarkan Pembatas: x1 ≤ 4 2x2 ≤ 12 3x1 + 2x2 ≤ 18 x1, x2 ≥ 0

Example 3 Masalah Perencanaan Regional Daerah Luas Tanah (Hektar) Alokasi Air Irigasi (m3) 1 400 600 2 800 3 300 375 Data Luas Tanah dan Alokasi Air yang Dapat Digunakan di Daerah 1, 2 dan 3

Jatah Lahan Maksimum (hektar) Hasil Bersih (ribu rp/ha) Jenis Tanaman Jatah Lahan Maksimum (hektar) Konsumsi Air (m3) Hasil Bersih (ribu rp/ha) Tebu 600 3 400 Kapas 500 2 300 Gandum 325 1 100 Data Jatah Lahan Maksimum, Konsumsi Air dan Hasil Bersih Masing-Masing Jenis Tanaman

Berapa hektar tanah yang harus disediakan untuk masing-masing jenis tanaman pada masing-masing daerah, sehingga diperoleh hasil bersih maksimum tanpa melanggar pembatas-pembatas yang telah ditetapkan

Daerah Alokasi (hektar) Tanaman 1 2 3 Tebu X1 X2 X3 Kapas X4 X5 X6 Gandum X7 X8 X9 Variabel-variabel Keputusan

Berdasarkan pembatas: Maksimumkan: z = 400(x1+x2+x3) + 300(x4+x5+x6) + 100(x7+x8+x9) Berdasarkan pembatas: Luas Tanah x1 + x4 + x7 ≤ 400 x2 + x5 + x8 ≤ 600 x3 + x6 + x9 ≤ 300

Air 3x1 + 2x4 + x7 ≤ 600 3x2 + 2x5 + x8 ≤ 800 3x3 + 2x6 + x9 ≤ 375 Jatah Lahan x1 + x2 + x3 ≤ 600 x4 + x5 + x6 ≤ 500 x7 + x8 + x9 ≤ 325

Persetujuan Kepala Daerah (x1+x4+x7)/400 = (x2+x5+x8)/600 (x2+x5+x8)/600 = (x3+x6+x9)/300 (x3+x6+x9)/300 = (x1+x4+x7)/400 Pembatas Nonnegatif x1, x2, ... , x9 ≥ 0