lecture2 - Overview of Lecture 15.053 February 3, 2011 ! ...

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Unformatted text preview: Overview of Lecture 15.053 February 3, 2011 !  Goals –  get practice in recognizing and modeling linear constraints and objectives More Linear and Non-linear Programming Models –  and non-linear objectives –  to see a broader use of models in practice plus applications to radiation therapy Note: Read tutorials 01 and 02 on the website. They cover material not covered in this class. 1 2 Overview on modeling Quotes for today Reality is merely an illusion, albeit a very persistent one. !  !  Applications to workforce scheduling !  Everything should be made as simple as possible, but not one bit simpler. Modeling as an art form !  Albert Einstein Modeling as a mathematical skill Applications to radiation therapy Albert Einstein, (attributed) 3 4 Formulating as an LP Scheduling Postal Workers !  !  Let s see if we can come up with what the decision variables should be. !  Day Demand !  Don t look ahead. Discuss with your neighbor how one might formulate this problem as an LP. Each postal worker works for 5 consecutive days, followed by 2 days off, repeated weekly. !  Mon Tues 17 Wed Thurs Fri 13 15 19 14 Sat Sun 16 11 Minimize the number of postal workers (for the time being, we will permit fractional workers on each day.) 5 6 For students who have a new response card today. The linear program Day Demand Minimize subject to Mon Tues 17 Wed Thurs 13 15 19 Fri Sat Sun 14 16 11 z = x1 + x2 + x3 + x4 + x5 + x6 + x7 x1 + x 4 + x5 + x6 + x7 ! 17 Mon. x5 + x6 + x7 ! 13 x1 + x 2 + Tues. x6 + x7 ! 15 Wed. x1 + x 2 + x 3 + x1 + x 2 + x 3 + x 4 + x1 + x 2 + x 3 + x 4 + x 5 x 2 + x3 + x4 + x5 + x6 x7 ! 19 ! 14 ! 16 x3 + x4 + x5 + x6 + x7 ! 11 The two digit code for our receiver is 10. Thurs. Fri. Red: Green: Yellow: Yellow: Sat. Sun. Response not received. Response was received. (Multiple flash) In the process of sending (Single flash) Polling not open xj ! 0 for j = 1 to 7 7 8 On the selection of decision variables !  A Modifications of the Model A choice of decision variables that doesn t work !  –  Let yj be the number of workers on day j. Suppose that there was a pay differential. The cost of each worker who works on day j is cj. The new objective is to minimize the total cost. –  No. of Workers on day j is at least dj. (easy to formulate) What is the objective coefficient for the shift that starts on Monday for the new problem? –  Each worker works 5 days on followed by 2 days off (hard). !  Conclusion: sometimes the decision variables incorporate constraints of the problem. 1.  2.  –  We will see more of this in integer programming. c1 +c2 +c3 +c4 +c5 3.  –  Hard to do this well, but worth keeping in mind c1 c1 +c4 +c5 +c6 +c7 9 A Different Modification of the Model !  10 Model 2 Suppose that there is a penalty for understaffing and penalty for understaffing. If you hire k too few workers on day j, the penalty is 5 k2. If you hire k too many workers on day j, then the penalty is k2. How can we model this? Minimize 5 ! 7 i =1 d i2 + ! i =1 ei2 x1 + 7 x4 + x5 + x6 + x7 + d1 – e1 = 17 x5 + x6 + x7 + d2 – e2 = 13 x6 + x7 + d3 – e3 = 15 x7 + d4 – e4 = 19 + d5 – e5 = 14 + d6 – e6 = 16 x3 + x4 + x5 + x6 + x7 + d7 – e1 = 11 x1 + x 2 + x1 + x 2 + x 3 + x1 + x 2 + x 3 + x 4 + Step 1. Create new decision variables. x1 + x 2 + x 3 + x 4 + x 5 Let ej = excess workers on day j x 2 + x3 + x4 + x5 + x6 Let di = deficit workers on day j xj ! 0, dj ! 0, ej ! 0 for j = 1 to 7 11 What is wrong with this model, other than the fact that variables should be required to be integer valued? 12 More Comments on Model 2. What is wrong with Model 2? Difficulty: The feasible region permits feasible solutions that do not correctly model our intended constraints. Let us call these bad feasible solutions. The good feasible solutions are ones in which d1 = 0 or e1 = 0 or both. They correctly model the scenario. 1.  The constraints should have inequalities. 2.  The constraints don t make sense. 3.  The objective is incorrect. (Note: it is OK that it is nonlinear) Resolution: All optimal solutions are good. 4.  It s possible that ej and dj are both positive. Illustration of why it works: 5.  Nothing is wrong. 10 + 10 + 0 + 0 + 0 + d1 – e1 = 17 e1 = 4 and d1 = 1 is a bad feasible solution. e1 = 3 and d1 = 0 are good feasible solution. For every bad feasible solution, there is a good feasible solution whose objective is better. 13 More on the model 14 Non-linear Programs !  Usually: every feasible solution to our math model (LP) corresponds to a feasible solution to our managerial model. Such solutions are good. !  An optimization problem with a single objective and multiple constraints. !  Occasionally, we will permit bad feasible solutions provided that the LP solver will never select one. The LP solver won t select it because for every bad feasible solution, there is a good solution with lower cost. !  Linear programs are a special case. !  We will see this technique more in this lecture, and in other lectures as well. 15 16 Examples of Nonlinear Objective Functions ! Min Max Min ! !  " 7 ( x j )2 j =1 5 Cos (e j ) 7 j =1 ! ! dj 7 j =1 On Nonlinear Programs Examples of Nonlinear Constraints 7 j =1 But they usually can be solved if the objective is to minimize a convex function, and the constraints are linear. 5 Cos (e j ) 7 j =1 " xj ( x j )2 ! 30 In general, nonlinear programs are incredibly hard to solve. Sometimes they are impossible to solve. dj 7 j =1 = 13.76 t righ he the for t d Fin rithm lem o alg t prob righ x j ! 13 17 Convex functions of one variable www.natasafety1st.org/posters/2001_Mission_Im... 18 Which functions are convex? A function f(x) is convex if for all x and y, the line segment on the curve joining (x, f(x)) to (y, f(y) lies on or above the curve. f(x) = x2 25 f(x) = x3 for x ! 0 f(x) = x.5 Step Function whatever 20 f(x) 15 10 5 f(x) = |x| 0 0 5 x 10 19 Yes No 20 Other enhancements Time for a mental break 21 Math Programming and Radiation Therapy !  Math Programming and Radiation Therapy An important application area for optimization !  22 Thanks to Rob Freund and Peng Sung for some of the following slides !  High doses of radiation (energy/unit mass) can kill cells and/or prevent them from growing and dividing –  True for cancer cells and normal cells http://www.youtube.com/watch? v=GMPaArG4CcM&feature=related !  23 Radiation is attractive because the repair mechanisms for cancer cells is less efficient than for normal cells 24 Radiation Imaging !  Radiation Delivery Recent improvements in imaging !  Improvements of delivery of radiation !  Relatively new field: tomotherapy –  MRI –  CT Scan –  other IMRT www.ottawahospital.on.ca/sc/cancer Optimizing the Delivery of Radiation Therapy to Cancer patients, by Shepard, Ferris, Olivera, and Mackie, SIAM Review, Vol 41, pp 721-744, 1999. cherrypit.princeton.edu/mri.gif www.spiralock.com/images/Mri-machine%201.JPG 25 http://www.psl.wisc.edu/wp-content/themes/default/images/tomo.gif 26 Conventional Radiotherapy Use of Multi-leaf Collimaters !  http://pages.cs.wisc.edu/ ~ferris/papers/sirev-cancer.pdf multi-leaf collimator –  blocks radiation –  turns a large beam into a focused beam welchcancercenter.org Relative Intensity of Dose Delivered 27 28 Conventional Radiotherapy Conventional Radiotherapy !  In conventional radiotherapy –  3 to 7 beams of radiation –  radiation oncologist and physicist work together to determine a set of beam angles and beam intensities –  determined by manual trial-and-error process Relative Intensity of Dose Delivered 29 30 Optimization approach: the decision variables Goal: maximize the dose to the tumor while minimizing dose to the critical area, whatever that means. !  First, discretize the space –  Divide up region into a 2D (or 3D) grid of pixels Critical Area Tumor area wp = intensity weight assigned to beamlet p for p = 1 to n; 31 32 Some linear constraints !  Create the beamlet data for each of p = 1, ..., n possible beamlets. Dijp = unit dose delivered to pixel (i, j) by beamlet p. Determine objective function(s) and constraints that limit the solution. Dij = dosage delivered to pixel (i, j) What suggestions do you have on how the objective might be formulated. Should we consider additional constraints? Dij = ! p=1 Dijp w p n wp ! 0 Or what questions do you have? Imagine that there is an expert on radiation therapy that can answer any question. for all p 33 34 Lower bounds on radiation level for tumors. upper bounds on radiation level for critical region. Questions and Suggestions –  dosage over the tumor area will be at least a target level !L . –  dosage over the critical area will be at most a target level !U. Dij " ! L for ( i , j ) # T Dij " ! U for ( i , j ) # C Difficulty: these LPs are rarely feasible. 35 In an example reported in the 1999 paper, there were more than 63,000 variables, and more than 94,000 constraints (excluding upper/lower bounds) 36 Results from LP optimization Dij ! .9 for ( i , j ) "T More LP Results Max beam weight / min beam weight " 5 .9 ! Dij ! 1.1 for ( i , j ) "T 37 What happens if the model is infeasible? minimize ! y Dij ( i , j ) ij Dij = ! p=1 Dijp w p n 38 An even better model Allow the constraint for pixil (i,j) to be violated by an amount yij, and then minimize the violations. minimize ! jD ! ( i (,(iji,),j()) yyijijij) 2 Dij = ! p=1 Dijp w p minimize the sum of squared violations. n DDij " ij L ! L forf( i , j ) #jT# T or ( i , ) ij + y ! " DDij " ij L ! L forf( i , j ) #jT# T or ( i , ) ij + y ! " Dij ij " ! U ! U for f(or j()i#jC$ C D " yij # i, , ) Dij ij " ! U ! U for f(or j()i#jC$ C D " yij # i, , ) wp ! 0 yij ! 0 wp ! 0 yij ! 0 for all p for all ( i , j ) 39 Least squares for all p for all ( i , j ) This is a nonlinear program (NLP). This one can be solved efficiently. 40 Optimal Solution with 7 beams Optimal Solution with 15 Beams 41 Optimal Solution with 71 Beams 42 Is that the end of the story on modeling? !  !  43 No. Models can almost always be enhanced. But this was a huge step over doctors determining the beam angles. 44 Modeling as a design process A Closer Look at the Constraint Matrix x1 x2 x3 x4 x5 x6 x7 1 0 0 1 1 1 1 ! 17 Mon 1 1 0 0 1 1 1 ! 13 Tues 1 1 1 0 0 1 1 ! 15 Wed 1 1 1 1 0 0 1 ! 19 Thurs 1 1 1 1 1 0 0 ! 14 Fri 0 0 1 0 1 1 1 1 1 1 1 1 0 1 ! ! 16 11 Sat Sun It is cyclically repeating in both rows and columns. 45 46 ...
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