Announcements. Midterm Review Session. Extended TA session on Friday 9am to 12 noon.

Transcription

1 Announcements Extended TA session on Friday 9am to 12 noon. Midterm Review Session The midterm prep set indicates the difficulty level of the midterm and the type of questions that can be asked. Please verify all the calculations shown in solution. Lets take the simplest first! 1) Consider the following probabilities: P(PassExam AND WildParty) = 0.2 P(PassExam AND NOT WildParty) = 0.5 P(NOT PassExam AND WildParty) = 0.2 P(NOT PassExam AND NOT WildParty) = 0.1 a) What is P(PassExam)? (0.7) b) What is P(WildParty)? (0.4) c) What is P(WildParty OR PassExam)? (0.9) d) What is P(PassExam WildParty)? (0.5) e) What is P(PassExam NOT WildParty)? (5/6=0.8333) Hope there are no doubts in this problem. 2) After your yearly checkup, the doctor has bad news and good news. The bad news is that you tested positive for a serious disease, and that the test is 99% accurate (that is, the probability of testing positive given that you have the disease is 0.99, as is the probability of testing negative given that you don t have the disease). The good news is that this is a rare disease, striking only one in 10,000 people. Why is it good news that the disease is rare? What are the chances that you actually have the disease?

2 Let P be the event that you tested positive and D be the event that you actually have the disease. Given: P (D) = 1/10000 = P (P D) = 0.99, P (P NOT D) = 0.01 P (NOT P NOT D) = 0.99 P (NOT P D) = 0.01 To find: P(P) and P (D P) P(P) = P(P ^ D) + P (P ^ NOT D) = P(P D) P (D) + P(P NOT D)P(NOT D) = 0.99 * * = Apply Baye s rule for two variables: P(X Y) = P( Y X) * P (X) P(Y) Thus, P (D P) = P( P D) * P (D) P(P) = 0.99 * = ) a) P (A AND B AND C AND D AND E) = P (E C) P (D B, C) P(C A) P (B A) P (A) = 1.1. ½. 1. ½ = ¼ b) What is P(NOT A AND NOT B AND NOT C AND NOT D AND NOT E)? = P(NOT E NOT C) P(NOT D NOT B< NOT C) P(NOT C NOT A) P(NOT B NOT A) P(NOT A) = ¼. ½. ½ = 0 Infer: P (NOT E C) = 0 ; P (NOT E NOT C) = 1; P (NOT D B AND C) = P(NOT D NOT B AND C) = P (NOT D NOT B AND NOT C) = 0 c) What is P(C A AND B)? = P (C A) = ½ d) Enumerate all 32 combinations:

3 A B C D E T T T T T T T T T F T T T F T T T T F F T T F T T T T F T F T T F F T T T F F F T F T T T T F T T F T F T F T T F T F F T F F T T T F F T F T F F F T T F F F F F T T T T F T T T F F T T F T F T T F F F T F T T F T F T F F T F F T F T F F F F F T T T F F T T F F F T F T F F T F F F F F T T F F F T F F F F F T F F F F F Reduces to 8 All 32 assignments A B C D E T T T T T T T F T F T T F F F F T T T T F T F T F F T F F F F F T T T F F F T F 4) True or false: The amount of memory needed to represent a Bayes net is proportional to the number of nodes times the maximum number of parents of any node? Justify your answer. FALSE Assuming maximum k parents for any node. Thus we have to save maximum 2 k entries in the Conditional Probability Table (CPT) of any node. Hence if there are n nodes it implies that the total memory needed will be proportional to (2 k )*n 5) A used-car buyer can decide to carry out various tests with various costs, and then, depending on the outcome of the tests, decide which car to buy. We will assume that the buyer is deciding whether to buy car c1, that there is time to carry out at most one test, and that t1 is the test of c1 and costs $50. A car can be in good shape (quality q+) or bad shape (quality q-), and the test may help to indicate what shape the car is in. Car c1 costs $1,500, and its market value is $2,000 if it is in good shape; if not, $700 in repairs will be needed to make it in good shape. The buyer s estimate is that c1 has a 70% chance of being in good shape. Event of car being in good shape = q+ Event of car being in bad shape = q- Event of car passing the test = P Calculate the expected net gain from buying c1, given no test. Expected gain from buying will be = P (q+)*profit + P (q-)*loss = 0.7 * ( ) *( ) = 0.7* * (-200) = $290 Expected net gain from buying c1, given no test is $290 b) Tests can be described by the probability that the car will pass or fail given that the car is in good or bad shape. We have the following information: P(pass(c1,t1) q+(c1)) = 0.80 and P(pass(c1,t1) q-(c1)) = 0.35.

4 Ans: Bayes formula: P(X Y) = Given: P (P q+) = 0.8 ; P (P q-) = 0.35 P (q+) = 0.7; P (q-) = 0.3 P( Y X) * P (X) P(Y) c) Calculate the optimal decisions given either a pass or a fail, and their expected net gains. $290 Pass Test Cost $50 $240 Fail (NOT P) To find: P (P) =? P (q+ P) =? P (q+ NOT P) =? P (P) = P ( P q+)*p(q+) + P(P q-)*p(q-) = 0.8 * *0.3 = P (q+ P) = P(P q+) * P( q+) P (P) = 0.8 * = P (q+ NOT P) = P(NOT P q+) * P(q+) P(NOT P) = 0.2 * = but q+ $ $ Don t Buy -50 but q- $42.6 but q+ $ Don t Buy but q- d) Calculate the value of information of the test, and derive an optimal conditional plan for the buyer. - Since we have already paid $50 for the test and the final utility values include this information, $274 will be the gain after paying for the test. Hence the actual utility will = $290 Value of Information = Increase in the utility due to the test = $( ) = $0 < Cost of information. Thus optimal plan would be to just buy the car without testing it. 6) You are an opera singer who is about to take a Chemical Engineering exam. Within the last two days running up to the exam you can be in one of three states of knowledgability about Chemical Engineering: ignorant, competent, or expert. Every day you must decide what you will do today: opera singing (O), or work on your chemical studies (W). If you pass the exam you will receive a one-time payment of $120. Every day that you do opera singing you will earn $25. You earn nothing for working on your chemical studies or failing the exam. As the following figure shows, you can be in 11 states. In six of the states you get to make the O or W decision. For these states, the next state will be a probabilistic choice with the probabilities dependent on your decision. These probabilities are depicted in the figure. You are motivated by money. You want to maximize the expected total (undiscounted) amount of money you will receive.

5 7) Assume that you can produce 3 different kinds of widgets on a machine that you have available for 8 hours. It takes 5 minutes to produce one widget of the first kind, resulting in a profit of $5. It takes 20 minutes to produce one widget of the second kind, resulting in a profit of $15. It takes 10 minutes to produce one widget of the third kind, resulting in a profit of $15. You want to maximize your profit, given the following constraints: a) You can produce at most 100 widgets of the first kind, 20 widgets of the second kind, and 45 widgets of the third kind. b) If you produce any widgets of the first kind, you need to produce at least 20 of them. c) If you produce any widgets of the third kind, you incur a one-time set-up cost of $25. d) For every widget of the third kind that you produce, you need to produce at least one widget of the second kind. Formulate a linear or mixed integer program that models as many of the constraints as is possible. For every constraint explain why a) it cannot be modeled by mixed integer programs, b) it can only be modeled by mixed integer programs but not by linear programs, or c) it can be modeled by both mixed integer programs and linear programs. Let the three widgets be called W1, W2, W3 and let the number widgets produced for each of them in the 8 hours be a, b, c respectively Objective function: Maximize: 5a + 15b + 15c Constraints: Total time available = 8 hours Thus we have 5a + 20b + 10c <= 480 (=8*60) - Time constraint a) a <= 100; b <=20 ; c<= 45 Can be modeled only by Mixed Integer programs as a, b, c are necessarily integers b) a >=20 but a can also be zero. - Cannot be formulated directly in both Mixed integer program and Linear Program using existent variables We have to introduce a variable (say x1) such that x1 = 1 if a > 0 else x1 = 0 Then the constraint becomes a >= {Smallest value of a}*x1 Hence, a >= 20 x1 We also have to make a new constraint a <= {Largest value of a}*x1 Hence a <= 100x1 The 2 nd constraint is necessary because it makes sure that when x1 = 0, we have a <= 0 and a >= 0 which reduces to a = 0. c) If c>0 then objective function becomes Maximize: 5a + 15b + 15c 25 -Such a constraint cannot be directly represented in Mixed programming with existent variables To represent such a constraint we have to introduce another variable (say x2) such that x2 = 1 when c > 0 = 0 when c = 0 So objective function becomes Maximize: 5a + 15b + 15c 25x2 Also we have to introduce another constraint, c <= {Largest value c takes} * x2 which becomes c < 45x2 d) b > c => b c > 0 -Again this constraint can be represented only in Mixed Integer programming as b and c are integers.