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# IE 594 Sample exam question IE 594

sabari Omkar
UIC
GPA 4.0

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Midterm 1 and midterm 2 sample questions and exam questions
COURSE
Distributed Decision Making
PROF.
Mengqi Hu
TYPE
Test Prep (MCAT, SAT...)
PAGES
14
WORDS
CONCEPTS
Distributed Decision Making
KARMA
75 ?

## Popular in Industrial Engineering

This 14 page Test Prep (MCAT, SAT...) was uploaded by sabari Omkar on Monday January 18, 2016. The Test Prep (MCAT, SAT...) belongs to IE 594 at University of Illinois at Chicago taught by Mengqi Hu in Fall 2015. Since its upload, it has received 97 views. For similar materials see Distributed Decision Making in Industrial Engineering at University of Illinois at Chicago.

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Date Created: 01/18/16
Distributed Decision Making IE 594 Exam 1 Review 1. Considering the following game. Player 2 B F Player 1 B 4, 2 1, 0 F 0, 1 2, 4 a. Does this game have a pure strategy Nash equilibrium solution? If yes, find it. b. Find a mixed strategy Nash equilibrium solution Solution: a. (B, B) and (F, F) are pure strategy Nash equilibrium solution b. The probability of player 1 choosing B is p, and F is 1-p The probability of player 2 choosing B is q, and F is 1-q The expected payoff value of player 1 choosing B is 4*q + 1*(1-q) = 1+3q The expected payoff value of player 1 choosing F is 0*q + 2*(1-q) = 2-2q Let 1+3q = 2-2q, we have q=1/5 The expected payoff value of player 2 choosing B is 2*p + 1*(1-p) = 1+p The expected payoff value of player 2 choosing F is 0*p + 4*(1-p) = 4-4p Let 1+p = 4-4p, we have p=3/5 The mixed strategy Nash equilibrium solution is p=3/5 and q=1/5 2. Use min-max algorithm to solve this game. Player 2 A B C D A 8 6 -1 3 Player 1 B 6 -2 1 4 C 3 -5 -1 0 D -6 7 2 -1 Solution: Player 2 A B C D Min A 8 6 -1 3 -1 Player 1 B 6 -2 1 4 -2 C 3 -5 -1 0 -5 D -6 7 2 -1 -6 Max 8 7 2 4 No saddle point 1 Distributed Decision Making IE 594 Exam 2 Review 1. Mike is on the job market to find a job. He has four different alternatives which include Job A, Job B, Job C, and Job D, and uses three criteria to evaluate each choice which include salary, benefit, and location. The compare-wise comparison matrix for each alternative in terms of each criterion, and compare-wise comparison matrix for the criteria are provided. Use analytic hierarchy process to help him selecting a best job. Salary Job A Job B Job C Job D Job A 1 1/2 3 1/8 Job B 2 1 5 1/4 Job C 1/3 1/5 1 1/7 Job D 8 4 7 1 Benefit Job A Job B Job C Job D Job A 1 2 5 2 Job B 1/2 1 6 2 Job C 1/5 1/6 1 1/7 Job D 1/2 1/2 7 1 Location Job A Job B Job C Job D Job A 1 1/2 1/4 1/9 Job B 2 1 2 1/2 Job C 4 1/2 1 1/2 Job D 9 2 2 1 Criterion Salary Benefit Location Salary 1 3 7 Benefit 1/3 1 4 Location 1/7 1/4 1 Solution Relative priority vector for criteria Salary Benefit Location 0.656 0.265 0.080 Relative priority vector for alternatives Salary Benefit Location Job A 0.111 0.413 0.072 Job B 0.207 0.301 0.248 Job C 0.055 0.054 0.201 Job D 0.626 0.232 0.479 1 Rank the priority values Priority Job A 0.188 Job B 0.235 Job C 0.067 Job D 0.510 Decision: Job D 2. Given the following bi-objective linear programming problem Solve it using: a) weighted sum method, and b) bi-objective simplex method Weighted sum method check four extreme points which are (0, 0), (5, 0), (7.5, 2.5), (0, 10) Condition for (0, 0) to be optimal solution for We can find solution such as Condition for (5, 0) to be optimal solution for We can find solution such as Condition for (7.5, 2.5) to be optimal solution for We can find solution such as Condition for (0, 10) to be optimal solution for No solution Using weighted sum method, we can find three efficient solutions which are: (0, 0), (5, 0), (7.5, 2.5) Simplex method 2 Phase I Initial basic feasible solution: (0, 0, 5, 10) Phase II When , the optimal solution is (0, 0, 5, 10) Phase III Iteration 1 RHS 7 3 0 0 0 -10 -1 0 0 0 1 -1 1 0 5 1 1 0 1 10 is the entering variablis the leaving variable when , the efficient solution is (0, 0, 5, 10) Iteration 2 RHS 0 10 -7 -2 -35 0 -11 10 0 50 1 -1 1 0 5 0 2 -1 1 5 3 is the entering variable, is the leaving variable when , the efficient solution is (5, 0, 0, 5) Iteration 3 RHS 0 0 -2 -7 -60 0 0 4.5 5.5 77.5 1 0 0.5 0.5 7.5 0 1 -0.5 0.5 2.5 when , the efficient solution is (7.5, 2.5, 0, 0) Summary Efficient Solution [10/17, 1] (0, 0, 5, 10) 0 0 [11/21, 10/17] (5, 0, 0, 5) 35 50 [0, 11/21] (7.5, 2.5, 0, 0) 60 77.5 3. Using crossover operator to generate two offspring using the following parents a b c d e f g h i j c b a g h i j f d e rd th a) Two point crossover (cut the chromosome at the 3 and 8 letter) a b c g h i j f i j c b a d e f g h d e b) Reorder crossover (cut the chromosome at the 5 and 8 letter) a b c d e g h f i j c b a g h f i j d e 4. Using mutation operator to generate one offspring using the following parent 1 0 0 1 1 0 1 0 a) Bit mutation (mutate the 3 bit) 1 0 1 1 1 0 1 0 rd th b) Inversion (select the 3 and 5 bit) 4 1 0 1 1 0 0 1 0 5. Giving the particle’s position at current iteration t and particles’ best position (pBest) at iteration t-1, update each particle’s best position (pBest) and swarm’s best position at current iteration t Index xt f(xt pBestt-1 f(pBestt-1 1 [0.1009, -1.0519] 1.1167 [0.0351, 0.1949] 0.0392 2 [-1.3696, -0.8214] 2.5505 [-0.3371, -0.3297] 0.2224 3 [-0.7614, 0.2560] 0.6452 [-0.2263, -0.5138] 0.3152 4 [1.6691, -0.8331] 3.4800 [-0.0150, -0.0719] 0.0054 5 [-0.7808, 0.8314] 1.3009 [0.0784, 0.5209] 0.2774 6 [-0.1499, 0.7717] 0.6180 [-0.0215, -0.4401] 0.1941 7 [-0.7316, -1.0195] 1.5746 [0.4543, 0.0289] 0.2073 8 [-1.5481, -1.1060] 3.6198 [0.5614, 0.3493] 0.4371 9 [0.4274, -0.2724] 0.2569 [-0.2361, 0.4727] 0.2792 10 [0.4414, -2.1865] 4.9757 [-0.0311, -0.4862] 0.2374 Solution: Index pBest t f(pBestt gbestt 1 [0.0351, 0.1949] 0.0392 2 [-0.3371, -0.3297] 0.2224 3 [-0.2263, -0.5138] 0.3152 4 [-0.0150, -0.0719] 0.0054 5 [0.0784, 0.5209] 0.2774 6 [-0.0215, -0.4401] 0.1941 [-0.0150, -0.0719] 7 [0.4543, 0.0289] 0.2073 8 [0.5614, 0.3493] 0.4371 9 [0.4274, -0.2724] 0.2569 10 [-0.0311, -0.4862] 0.2374 5 IE 594: Special Topic: Distributed Decision Making Exam 2: November 23, 2015 Due Date: 10am, November 25, 2015 Name: UIN: This exam contains 6 questions. You must submit an electronic copy of your solutions to blackboard, paper copy is not accepted. This exam is worth 100 points. The point value for each question is indicated at the beginning of each question. All work must be shown in order to receive credit. Clearly indicate your final answer (using underline or box). Read each question thoroughly and answer all associated questions. You must complete this exam individually. Working with other students or copying from another student’s exam is academic dishonesty. 1 1. (24 points) Allen wants to buy a house at Chicago. He has four different alternatives which include House A, House B, House C, and House D, and uses three criteria to evaluate each choice which include price, house size, neighborhood, and quality of school system. The compare-wise comparison matrix for each alternative in terms of each criterion, and compare-wise comparison matrix for the criteria are provided. Use analytic hierarchy process to help him selecting a best house. Price House A House B House C House D House A 1 1/5 3 1/5 House B 5 1 6 1/3 House C 1/3 1/6 1 1/8 House D 5 3 8 1 Size House A House B House C House D House A 1 1/4 1/3 1/5 House B 4 1 5 2 House C 3 1/5 1 1/4 House D 5 1/2 4 1 Neighborhood House A House B House C House D House A 1 1/3 5 1/5 House B 3 1 4 1/3 House C 1/5 1/4 1 1/8 House D 5 3 8 1 School House A House B House C House D House A 1 1/5 3 1/4 House B 5 1 6 3 House C 1/3 1/6 1 1/3 House D 4 1/3 3 1 Criterion Price Size Neighborhood School Price 1 5 8 3 Size 1/5 1 5 1/3 Neighborhood 1/8 1/5 1 1/5 School 1/3 3 5 1 2 3 2. (30 points) Given the following bi-objective linear programming problem Solve it using bi-objective simplex method 4 5 3. (18 points) Using the following goal programming procedures to reformulate the following bi-objective linear programming problem a) Find the goal for each objective b) Reformulate the problem using minimax objective goal programming model 6 4. (10 points) Using crossover operator to generate two offspring using the following parents a b c d e f g h i j c b a g h i j f d e a) Reorder crossover (cut the chromosome at the 3 and 7 letter) th b) Uniform crossover Template for parent 1: 2, 2, 1, 1, 1, 2, 1, 2, 1, 2 Template for parent 2: 1, 2, 1, 2, 1, 1, 2, 1, 2, 1 5. (6 points) Using mutation operator to generate one offspring using the following parent 1 0 0 1 1 0 1 0 a) Swap mutation (swap the 3 and 5 bit) th nd th b) Inversion (select the 2 and 6 bit) 7 6. (12 points) Giving the particle’s position at current iteration t and particles’ best position (pBest) at iteration t-1, update each particle’s best position (pBest) and swarm’s best position at current iteration t Index xt f(t) pBestt-1 f(pBestt-1 1 [-1.2933, -1.4007] 3.6346 [0.5070, -1.1566] 1.5947 2 [-0.6253, 1.1099] 1.6228 [0.0199, -0.7357] 0.5417 3 [-0.3139, 0.0529] 0.1013 [0.6535, 0.2515] 0.4903 4 [0.3324, 0.5302] 0.3916 [0.2953, -0.1359] 0.1057 5 [0.8541, 1.8925] 4.3113 [-0.1326, 0.8306] 0.7076 6 [0.5422, -0.0204] 0.2944 [0.0056, -1.3706] 1.8784 7 [0.3454, 0.0674] 0.1239 [0.1454, 0.2674] 0.0927 8 [-0.8709, 0.8152] 1.4231 [-0.8709, -0.3171] 0.8590 9 [0.0278, 1.4849] 2.2056 [-0.7981, 0.3051] 0.7301 10 [0.0339, 2.0627] 4.2559 [-1.5358, 0.2267] 2.4101 8

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