Introduction to the course Linear programming CHEM-E7155 Production Planning and Control 19/04/2012
Lecture 1 Course Introduction Operations Research, Hillier, Lieberman Ch. 1 and 2 Introduction to Linear Programming Hillier, Lieberman Ch. 3
Course introduction Pre-requisites: CHEM-E7140 Process Automation Lecturer: Dr. Alexey Zakharov Email: alexey.zakharov@aalto.fi Lectures: Tuesday 10-12 Ke4, Thursday 10-12 Ke4 Exercises: Wednesday 14-16 class1, Friday 8-10 class 1
Content of the course Basic concepts of production planning and control Linear programming, Simplex method (lectures 1-4) Transportation problems Dynamic programming Integer programming Nonlinear programming Scheduling Inventory systems
Content of the Course Lectures, exercises, two assignments Learn the standard types of mathematical models Study examples of small-size problems which can be fit to these models Learn to build mathematical models from text description of the problems Basic solving algorithms (key concepts, geometric interpretations, implementation) 5
Content of the Course A Neste Oil case study A practical view on the topic 6
Course Material Hillier, Lieberman Introduction to Operations Research Buffa, Sarlin Scheduling Lecture notes
Course Performance Exam
II Operations Research
Operation research Operations research is a scientific approach to the management of organizations Operations research is applied to problems that concern how to conduct and coordinate operations and activities within an organization Operations research has been applied extensively in financial planning, manufacturing, the military, public services, health care, etc
General approach The process begins by carefully observing and formulating the problem Gathering the relevant data The next step is to construct a scientific model that attempts to abstract the essence of the real problem Finding the optimal solution of the problem under consideration
General approach: formulating the problem Operations research is also concerned with the practical management of the entire organization Finding solutions optimal for the whole organization rather than suboptimal solutions best for its components Attempt to resolve the conflicts of interest among the components of the organizations as a whole Selecting appropriate objectives: Long-run profit maximization Combining satisfactory level of profits with other objectives, like increasing share of the market, providing product diversification, increase company prestige, etc.
Operation research approach: constructing a mathematical model Models are idealized representation of the problems Approximations and simplifying assumptions are needed to make the model tractable (capable of being solved) A trade-off between the precision and tractability of the model The related decisions to be made a represented as decision variables (,,, ) The measure of performance is called the objective function Various restrictions (for example available resources) are expressed mathematically as constraints 13
Operation research approach: finding an optimal solution Standard algorithms are normally applied on a computer If time or cost required to find the optimal solution is too large, heuristic procedures (simple procedures to find a good suboptimal solution) are used Post-optimality analysis What-if analysis: what would happen if different assumptions are made about future conditions Sensitivity analysis: which parameters of the model are the most critical to determine the solution
Effects of Operations Research
Effects of Operations Research
Linear Programming Hillier, Lieberman Chapter 3
Linear Programming (LP) Contents Lecture 1 Prototype example LP-model, assumptions of LP More examples Two examples from industry Lecture 2. Simplex method Lecture 3. Revised simplex (simplex in matrix form) Lecture 4. Dual LP, other algorithms to solve LP
Linear Programming. Prototype example Wyndor Glass CO has 3 plants Plant 1. Aluminium frames and hardware Plant 2. Wood frames Plant 3. Produces glass and assembles the products Releasing production capacity to launch two new products having large sales potential: Product 1: 8-ft. glass door with aluminium framing. Product 2: 4x6 foot double-hung wood-framed window
Linear Programming. Prototype example The company could sell as much of either product as could be produced by these plants Because both products would be competing for the same production capacity in plant 3, it is not clear which mix of the two products would be most profitable
Linear Programming. The relevant data for the example Number of hours of production time available in each plant per week Number of hours of production time used in each plant for one batch of each new product Profit per batch of each new product Product mix type linear programming problem
Linear Programming. Prototype example Necessary data is summarized in the table: Plant Product 1 Product 2 per week 1 1 0 4 2 0 2 12 3 3 2 18 profit/ $3 000 $5 000 batch
Linear Programming. Prototype example as a linear programming problem Formulation as a LP problem x 1 and x 2 numbers of batches produced / week (the decision variables for the model) Z total profit / week Maximize: Z = 3 x 1 +5 x 2 = max! The restrictions: x 1 4 2x 2 12 3x 1 + 2x 2 18 x i 0, i=1,2.
Linear Programming. Prototype example Graphical solution: 2 decision variables => 2 dimensions => graphical solution can be used to solve it Identify the values of (x 1,x 2 ) that are permitted by restrictions (feasible region)
Linear Programming. Prototype example 6 6 x 1 4 2x 2 12 3x 1 + 2x 2 18 0, i=1,2. x i 1 2 3x 1 +2x 2 =18 6 (2,6) 4 4 = the feasible region 3 Z=36=3x 1 +5x 2 1 = x 1 and x 2 feasible region 2 = set of permissible values 3 = maximizes the value Z=10=3x 1 +5x 2 Z=20=3x 1 +5x 2 4
Linear Programming. Prototype example Pick out the point that maximizes the value of Z Drawing a family of parallel lines containing at least one point in the feasible region Selecting the line that corresponds to the largest value of Z. Solution: products 1 : 2 batches / week products 2 : 6 batches / week Total profit = $36 000/week.
Linear Programming (LP). Contents Lecture 1 Prototype example LP-model, assumptions of LP More examples Examples from industry Lecture 2. Simplex method Lecture 3. Revised simplex (simplex in matrix form) Lecture 4. Dual LP, other algorithms to solve LP
LP model: terminology The key terms are resources and activities Resources can be used to do activities. Resources are limited => allocating resources to activities Prototype example Production capacities of plants 3 plants Production of products 2 products Production rate of product j, (x j ) ProfitZ General LP problem Resources m resources Activities n activities level of activity j, (x j ) Overall measure of performance Z Determine this allocation involves choosing the levels of the activities that achieve the best possible value of the Z
LP model: terminology A feasible solution = a solution for which all the constrains are satisfied A feasible region = the collection of all feasible solutions Optimal solution = a feasible solution that has the most favourable value of objective function
A standard form of a LP model Describe the resource allocation problem Astandard form of LP is: Maximize the objective function: Z= c 1 x 1 + c 2 x 2 + +c n x n All functional constrains with a less-than or equal-to inequality: a i1 x 1 + a i2 x 2 + + a in x n b i The non-negativity constrains for all decision values: 0, for all values of j 30
The general form of a LP model Other forms are: Minimizing the objective function: Z= c 1 x 1 + c 2 x 2 + +c n x n Some functional constrains with a larger-than-or equal-to inequality: a i1 x 1 + a i2 x 2 + + a in x n b i, for some values of i Some functional constrains in the equality form: a i1 x 1 + a i2 x 2 + + a in x n =b i,for some values of i Deleting the non-negativity constrains for some decision values: unrestricted in sing, for some values of j
Assumptions of LP Proportionality The contribution of each activity to the objective function Z is proportional to the level of activity x j Similarly, the contribution of each activity to the left hand-side side of each functional constraint is proportional to the level of the activity x j, as represented by the a ij x j term in the constraint Proportionality fails because of Start-up costs Decreasing marginal returns If proportionality assumption fails, nonlinear programming or integer programming must be used in most cases
LP model Assumptions of LP Additivity Every function in a linear programming model is the sum of the individual contributions of the respective activities Object function and restrictions If additivity fails, nonlinear programming can be used Divisibility Decision variables in a linear programming model are allowed to have any values, including non-integer value If it fails, integer programming can be used Certainty The value assigned to each parameter of a linear programming model is assumed to be constant and precisely known
Linear Programming (LP). Contents Lecture 1 Prototype example LP-model, assumptions of LP More examples Design of radiation therapy Regional planning Controlling air pollution Examples from industry Lecture 2. Simplex method Lecture 3. Revised simplex (simplex in matrix form) Lecture 4. Dual LP, other algorithms to solve LP
Radiation Therapy Example Mary has just been diagnosed as having a cancer at a fairly advanced stage Radiation therapy damaging both cancerous and healthy tissues Aggregate dose to critical tissues must not exceed established tolerance levels, in order to prevent complications that can be more serious than the disease itself The total dose to the entire healthy anatomy must be minimized
Radiation Therapy Example Fraction of absorbed entry dose by area Beam 1 Beam 2 Healthy anatomy Critical tissues Tumor region 0,4 0,5 minimize 0,3 0,1 2,7 0,5 0,5 = 6 Center of tumor 0,6 0,4 6
Radiation Therapy Example 1 and 2 are the decision variables to the dose for beam 1 and 2 Because the total dosage reaching the healthy anatomy is to be minimized, let Z denote this quantity The data from the table can be used directly to formulate the linear programming model
Radiation Therapy Example Formulation as LP Minimize Subject to = 0.4 + 0.5 0.3 + 0.1 2.7 0.5 + 0.5 =6 0.6 + 0.6 6, 0
Radiation Therapy Example Graphical solution
Regional Planning Example A group of three kibbutz is planning agricultural production for the coming year Each kibbutz is limited by both the amount of available irrigable land and the quantity of water Kibbutz Usable land Water allocation (Acres) (acre / feet) 1 400 600 2 600 800 3 300 375
Regional Planning Example: The activities The crops suited for this region include sugar beets, cotton and sorghum These crops differ primarily in their expected net return per acre and their consumption of water The Ministry of Agricultural has set a maximum quota for the total acreage that can be devoted to each of these crops Maximum Water Net Return Crop Quata (Acres) Consumption ($ / acre) Sugar beets 600 3 1000 Cotton 500 2 750 Sorghum 325 1 250
Regional planning Example: Decision Variables The quantities to be decided are the numbers of acres to devote to each of three crops at each of three farms Kibbutz 1 2 3 Crop Acres Sugar beets x1 x2 x3 Cotton x4 x5 x6 Sorghum x7 x8 x9
Regional planning LP formulation Maximize the total profit: = 1000 + + + 750 + + + 250( + + ) Restrictions on usable land for each farm: + + 400 + + 600 + + 300 Restrictions on water allocation for each farm : 3 +2 + 600 3 +2 + 800 3 +2 + 375
Regional planning LP formulation Restrictions on total acreage of each crop (the quota): + + 600 + + 500 + + 325 Equal proportion of land planted: + + = + + 400 600 + + = + + 600 300 All are nonnegative 44
Regional planning Optimal solution ( 9 x, x, x, x, x, x, x, x, x ) = 1 2 3 4 5 6 7 8 (1331/ 3,100,25,100,250,150,0,0,0)
Controlling Air Pollution Example The steel company has to reduce its annual emission of pollutants Sources of pollution Blast furnaces Open-heart furnaces The most effective types of abatement methods Increasing the height of the smokestacks Using filter devices Cleaner fuels for the furnaces No single method by itself could achieve all the required reductions. A combination of methods is needed
Controlling Air Pollution Example The target emission example Pollutant Required reduction in annual emission rates particulates 60 sulfur oxides 150 hydrocarbons 125
Controlling Air Pollution Example The target emission reduction
Controlling Air Pollution Example The total annual cost
Controlling Air Pollution Example The decision variables
Controlling Air Pollution Formulation of LP problem
Linear Programming (LP). Contents Lecture 1 Prototype example LP-model, assumptions of LP More examples Examples from industry Grinding plant optimization Pulp Mill optimization Lecture 2. Simplex method Lecture 3. Revised simplex (simplex in matrix form) Lecture 4. Dual LP, other algorithms to solve LP
Optimization of a Grinding Plant with Linear Programming Industrial example 1
Grinding plant optimization: the process description The aim is to decrease the particle size before the flotation stage, where the minerals are separated from the gangue u 1 y 4 u 2 y 1, y 2 y 3 u 1 : the rod-mill feed (t/h) u 2 : the pump sump water addition (m 3 /h) y 1 : the hydrocyclone overflow density (% solids) y 2 : the fraction of particles smaller than 47Am in the product (%) y 3 : the tonnage through the ball mill (t/h) y 4 : the pump sump level (%)
Grinding plant optimization: the process constraints y 1 - the hydrocyclone overflow density: 48% < y 1 < 52% (the product density must be kept in this range to fit the requirements of the following stage of the process) y 2 - the fraction of particles < 47Am (%): y 2 = 48% (the target product fineness) y 3 - the tonnage through the ball mill (t/h): y 3 < 820 t/h (not overloading the ball mill) y 4 - the pump sump level (%): 15% < y 4 < 85% (the pump sump box must not dry out or overflow)
Grinding plant optimization: the dynamic process model The process output is defined as follows: = + 59.5 75 1520 930 where is the transfer function matrix:
Grinding plant optimization: the scheme of the process control The throughput through the Rod Mill must be maximized under the presented constraints
Grinding plant optimization: a steady-state process model The dynamic model is needed for process control The optimizer needs a steady-state model instead: 59.5 = + 75, 1520 930 where the steady input-output gain matrix is obtained from the transfer function: 0.0255 0.14 = 0 = 0.2 0.012 13.8 4.2 5.749 1.962
Grinding plant optimization: the LP constraints q The first output (overflow density) + > 48 59.5 + < 52 59.5 q The second output (particle size) + = 48 75 q The third output (ball mill throughput) + < 820 1520) q The fourth output (the pump box level) + > 15 930) + < 85 930) q The coefficients are elements of the input output gain matrix
Grinding plant optimization: graphical solution The throughput through the Rod Mill must be maximized: The optimal solution The feasible interval
Grinding plant optimization: Literature Lestage R., Pomerleau A., Hodouin D. (2002), Constrained real-time optimization of a grinding circuit using steady-state linear programming supervisory control, Powder Technology, 124, pp. 254-263
Maximization of the profitability of a Pulp Mill with Linear Programming Industrial example 2
Optimization of a Pulp Mill: The process description The primary goal of a pulp mill is to produce pulp of a given Kappa no. (amount of the lignin remaining in the fibers) and brightness while minimizing energy costs, utilities and chemical make-up streams Pulp mills can be divided in two major areas: fiber line and chemical recovery
Optimization of a Pulp Mill: The process description Fiberline Removes lignin with NaOH and NaSH, O 2, ClO 2 Recovery line Regenerates the white liquor from the extracted liquors from the digester and brown stock washer
Optimization of a Pulp Mill: the process control strategy Economical efficiency optimization Decision variables Model Predictive Control MPC inputs SISO Control Loops SISO inputs Free inputs Pulp Mill Process Measurements
Optimization of a Pulp Mill: List of the decision variables (V) Short name Variable Name Related to Lower bound Upper bound U1 D2 production rate MPC1 629,979 630 U2 Digester Kappa MPC1 12,5 27,5 U3 Digester upper EA MPC1 2,399 16,799 U4 Digester lower EA MPC1 2,25 15,75 U5 Upper extract conductivity MPC1 17,46 12,224 U6 Lower extract conductivity MPC1 12,11 84,822 U7 O kappa MPC1 5 15 U8 E kappa MPC2 2,4 2,6 U9 E washer [OH] MPC2 0,00044999 0,0012499 U10 D2 brightness MPC2 0,8 0,84 U11 Black liquor solids MPC3 0,625 0,675 U12 Slaker temperature MPC3 669,012 679,012 U13 Kiln O2 excess % MPC4 0,0149 0,0549 U14 Kiln CaCO3 residual % MPC4 0,02487 0,02487 U15 Digester liquor temperature SISO 425,5 445,5 U16 O tower temperature SISO 368 373,55 U17 D1 tower temperature SISO 329 347,4999 U18 E0 tower temperature SISO 344 353,2499 U19 D2 tower temperature SISO 343 352,2499 U20 O tower consistency SISO 0,05 0,1425 U21 Recaust CaOH2 concentr. SISO 3 30,75 U22 WL temperature SISO 358 376,4999 U23 WL NaOH concentration SISO 99,1498 150 U24 Steam flow 3 (digester) FREE 0,0373 0,18025 U25 Excess WL split (digester) FREE 0,9 1,1 U26 D1 water flow FREE 68,5125 330,75 U27 E back-flush flow FREE 240 847,105 U28 D2 caustic flow FREE 0,07325 0,4797 U29 Split fraction 6 (brown stock) FREE 0,2936 0,3408 U30 Split fraction 8 (brown stock) FREE 0,7176 0,83304 U31 Split fraction 9 (brown stock) FREE 0,38 0,417 U32 Split fraction 1 (MEE) FREE 0,7202 1,575 U33 Effect 3 steam flow FREE 821,92 952,44 U34 Filter lower flow FREE 0,54 1,508 U35 Kiln primary air flow FREE 62,5 437,5 U36 Effect 5 exit flow FREE 75 525 U37 Effect 4 exit flow FREE 75 525 U38 Effect 3 exit flow FREE 75 525 U39 Effect 6 exit flow FREE 75 525 U40 Effect 2 exit flow FREE 75 525 U41 Effect 1a exit flow FREE 75 525 U42 Effect 1b exit flow FREE 75 525 U43 Effect 1c exit flow FREE 75 525 U44 Coolant 4 flow (post Slaker) FREE 2,5 17,5
Optimization of a Pulp Mill: The economically significant variables (ESV) The economically significant variables are the variables needed to compute the plant profitability The economically significant variables are most of the manipulated variables of the plant, the pulp production rate and the steam production rate The reasons to consider these variables are: The variables are involved to the objective (plant profitability) The constraints imposed on these variables
Optimization of a Pulp Mill: The list of ESV (U) Short name Variable name Related to Lower bound Upper bound V1 Production rate MPC1 629,97 630 V2 Steam flow 1 SISO 0 0,398 V3 Steam flow 2 MPC1 0 0,5306 V4 Steam flow 4 MPC1 0 0,3172 V5 Steam flow 5 MPC1 0 0,3184 V6 Water flow SISO 0 1 324,60 V7 O steam flow 1 SISO 0 134 V8 O steam flow 2 SISO 0 134 V9 O steam flow 3 SISO 0 392 V10 D1 ClO2 flow MPC2 0 3 316 V11 D1 wash water SISO 0 1 237 V12 D1 steam flow SISO 0 990,6 V13 E caustic flow MPC2 0 536 V14 E steam flow SISO 0 505,6 V15 D2 ClO2 flow MPC2 0 1 904 V16 D2 wash water SISO 0 6263,1 V17 Coolant flow MPC1 0 1264,1 V18 Oxygen flow SISO 0 50 V19 Salt-cake flow SISO 0 168,5 V20 Coolant 1 flow MPC3 0 8 V21 Wash stream 1 SISO 0 0,1 V22 Fresh lime flow SISO 0 8 V23 Coolant 2 flow SISO 0 2 V24 Caustic flow 3 SISO 0 0,068 V25 Wash stream 2 SISO 0 4,466 V26 Wash stream 3 SISO 0 1,508 V27 Wash stream 4 SISO 0 0,298 V28 Wash stream 4 SISO 0 0,454 V29 Kiln fuel flow MPC4 0 74 V30 Water flow SISO 0 0,944 V31 Steam production FREE V32 Wood chips flow MPC1 0 5,2286 V33 O caustic flow MPC1 0 87 V34 Effect 1 steam flow MPC3 0 1847,2 V35 Effect 2 steam flow MPC4 92,36 1385,4 V36 Steam flow 3 Free 0 0,206 V37 D1 water flow Free 0 378 V38 E back-flush flow Free 0,06 968,06 V39 D2 caustic flow Free 0 0,586 V40 Effect 3 steam flow Free 0 1763,8
Optimization of a Pulp Mill: The objective function A typical entry in the objective function is a mass or volumetric flow rate (denoted as u) of a raw material, multiplied by the unit price and the concentration or density of the active ingredient if necessary. The resulting expression for the objective function f (in terms of USD / min) is linear in terms of the economic significant variables u: = + + +
Optimization of a Pulp Mill: Modeling of ESVs Economically significant variables can be considered as functions of the decision variables (plant setpoints) The optimization-relevant model is identified using a number of plant tests and setpoint changes in the control structure of the pulp mill benchmark problem For this purpose the 44 decision variables (denoted as u) are stepped up and down 5% from their nominal values and the change in the steady state values of the economical variables v are determined:
Optimization of a Pulp Mill: Modeling of ESVs The ESV model can be written using 40 linear equality constraints 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1,,,,,,, here = ( ),,,,,,,, ( ), =
Optimization of a Pulp Mill: Formulation of a LP problem The objective: max The constraints: =,,,,
Optimization of a Pulp Mill: optimization results A 12% increase in profit has been achieved The profitability at the nominal conditions is 106.7 USD/min The profitability at the optimal point (according to the linear model) is 126.4 USD/min The profitability at the optimal point (according to the simulations) is 118.8 USD/min Due to model inaccuracy, the simulations do not demonstrate as good result as it was promised by the model
Optimization of a Pulp Mill: optimization results
Optimization of a Pulp Mill: Literature Castro J., Doyle F., (2004), A Pulp Mill benchmark problem for control: problem description, Journal of Process Control, 14, pp. 17-29 Mercangoz M., Doyle F., (2008), Real-time optimization of the pulp mill benchmark problem, Computers and Chemical Engineering, 32, pp. 789-804
Questions How would you briefly describe a general linear programming problem? What the decision variables/objective/constraints are? What issues prevents a problem to be described with a LP model? What is the role of the process models in the industrial examples presented in the lecture?