...Foundations of Operational Research and Business Analysis 1 Assignment 2013/14 Author: Thibaut Achard de Leluardière Abstract: Looking through the infinite number of theories and models developed in organisations, this assignment aims at finding out the founding principles of a good OR/MS model and general issues encountered in the setting-up of OR interventions. To try out and compare the insights presented, this assignment proposes to study a specific case about OR modelling in Fishery management. Fishery Management is related to the preservation of fish resources and optimisation of catch and profit of this industry, in a context of high-yield practices and increasingly more complex environmental issues present. This case applies to a large and complex system linked to today’s topics issues of sustainable development. In addition, a personal experience of analytical project related to an internship position as assistant project manager in a leading oil company is proposed to illustrate this essay. This essay concludes by giving recommendations about what could be the characteristics of an ideal portrait of OR/MS model. Introduction: In a letter addressed to English universities after the Second World War, general Pile, a popular British officer who commended the Anti-Aircraft Command, claims for men of sciences ‘able to quickly understand complex issues and to find them simple’. Thus, supporting the fast economic growth in Europe after war, operational...
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...Homework 1 Q1. The SKKU company, one of social enterprise produces chairs and tables from two resources – labor and wood. The company has 80 hours of labor and 36 pounds of wood available each day. Demand for chairs is limited to 6 per day. Each chair requires 8 hours of labor and 2 pounds of wood, whereas a table requires 10 hours of labor and 6 pounds of wood. The profit derived from each chair is $400 and from each table, $100. The company wants to determine the number of chairs and tables to produce each day in order to maximize profit A. Formulate a linear programming model for this problem B. Solve the model by using computer sw. Q2. In problem 1, how much labor and wood will be unused if the optimal numbers of chairs and table produced? Q3. In problem1, explain the effect on the optimal solution of changing the profit on the table from $100 to $500. * Solve the model by using computer SW. Q4. RPR has been conducted to do survey following an election primary in New Hampshire. The firm must assign interview to conduct the survey by telephone or in person. One person can conduct 80 telephone interviews or 40 personal interviews in a single day. The following criteria have been established by the firm to ensure a representative survey: An A. Formulate a linear programming model for this problem B. Solve the linear programming model by using computer sw(Excel or DS for...
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...DERY CYRIL DOMEYELLE EBA LEVEL: 300 QUANTITATIVE BUSINESS METHODS ASSIGNMENT (A) Let x = number of units of product X y = number of units of product Y z = number of units of product Z Maximize 20x + 18y + 16z (Objective function) Subject to 5x + 3y + 6z ≤ 3,000 (Machine hours constraint) 2x + 5y + 3z ≤ 2,500 (Labour hours constraint) 8x + 10y + 3z ≤ 10,000 (Materials constraint) Changing the inequalities into equalities and adding a slack variable Maximize 20x + 18y + 16z (Objective function) Subject to 5x + 3y + 6z + S1 = 3,000 2x + 5y + 3z + S2 = 2,500 8x + 10y + 3z + S3 = 10,000 Initial Simplex Tableau Solution Variables Products X Y Z Slack Variables S1 S2 S3 S1 5 3 6 1 0 0 3,000 S2 2 5 3 0 1 0 2,500 S3 8 10 3 0 0 0 10,000 Z 20 18 16 0 0 0 0 (B) Initial Simplex Tableau Solution Variables Products X Y Z Slack Variables S₁ S₂ S₃ S₁ 5 3 6 1 0 0 3,000 S₂ 2 5 3 0 ...
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...Yağmur GÜVEN IE-202 PROJECT PART 1 LP: Minimize 4*X121+8*X122+7*X131+14*X132+4*X211+8*X212+10*X231+20*X232+13*X241+26*X242+7*X311+14*X312+10*X321+20*X322+13*X421+26*X422+4*Y121+8*Y122+7*Y131+14*Y132+4*Y211+8*Y212+10*Y231+20*Y232+13*Y241+26*Y242+7*Y311+14*Y312+10*Y321+20*Y322+13*Y421+26*Y422 s.t. X111+X112<=0 X121+X122<=175 X131+X132<=175 X141+X142<=0 X211+X212<=175 X222+X221<=0 X231+X232<=175 X241+X242<=175 X311+X312<=175 X321+X322<=175 X331+X332<=0 X341+X342<=0 X411+X412<=0 X421+X422<=175 X431+X432<=0 X441+X442<=0 Y111+Y112<=0 Y121+X122<=175 Y131+Y132<=175 Y141+Y142<=0 Y211+Y212<=175 Y221+Y222<=0 Y231+Y232<=175 Y241+Y242<=175 Y311+Y312<=175 Y321+Y322<=56 Y331+Y332<=0 Y341+Y342<=0 Y411+Y412<=0 Y421+Y422<=175 Y431+Y432<=0 Y441+Y442<=0 (X211+X311+X411)-(Y121+Y131+Y141)>=0 (X121+X321+X421)-(Y211+Y231+Y241)>=50 (X131+X231+X431)-(Y311+Y321+Y341)>=130 (X141+X241+X341)-(Y411+Y421+Y431)>=0 (X212+X312+X412)-(Y122+Y132+Y142)>=75 (X122+X322+X422)-(Y212+Y232+Y242)>=0 (X132+X232+X432)-(Y312+Y322+Y342)>=0 (X142+X242+X342)-(Y412+Y422+Y432)>=100 Y211+Y311+Y411>=0 Y121+Y321+Y421>=75 Y131+Y231+Y431>=80 Y141+Y241+Y341>=0 Y212+Y312+Y412>=110 Y122+Y322+Y422>=0 Y132+Y232+Y432>=0 Y142+Y242+Y342>=80 | | Son | Azaltılmış | Hedef | İzin Verilen | İzin Verilen | Hücre | Ad...
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...Solutions 56:171 Operations Research Homework #3 Solutions – Fall 2002 1. Revised Simplex Method Consider the LP problem Maximize subject to z = 3 x1 − x2 + 2 x3 x1 + x2 + x3 ≤ 15 2 x1 − x2 + x3 ≤ 2 − x1 + x2 + x3 ≤ 4 x j ≥ 0, j = 1, 2,3 a. Let x4 , x5 , &, x6 denote the slack variables for the three constraints, and write the LP with equality constraints. Answer: Maximize z = 3 x1 − x2 + 2 x3 subject to x1 + x2 + x3 + x4 = 15 2 x1 − x2 + x3 + x5 = 2 − x1 + x2 + x3 + x6 = 4 x j ≥ 0, j = 1, 2,3, 4,5, 6 After several iterations of the revised simplex method, 1 0 the basis B={4,3,2} and the basis inverse matrix is ( AB ) −1 = 0 1 2 0 − 1 2 −1 1 . 2 1 2 b. Proceed with one iteration of the revised simplex method, by i. Computing the simplex multiplier vector π Answer: 1 0 −1 B −1 0 1 1 = 0, 3 , 1 π = CB ( A ) = [0 2 −1] 2 2 2 2 0 − 1 1 2 2 = [ 0, 1.5, 0.5] ii. “pricing”, i.e., computing the “relative profits”, of the non-basic columns. Answer: 56:171 O.R. -- HW #3 Solutions Fall 2002 page 1 of 8 Solutions 1 0 0 C = [3 0 0 ] , A = 2 1 0 −1 0 1 N N N −3 −1 C = C −π A = 1 2 2 2 The relative profits for non-basic variables are C1 = 0.5 , C5 = −1.5 , C6 = −0.5 . iii. Selecting the column to enter the basis. Answer: Only the relative profit of X 1 is positive and the problem is Max problem, and so X 1 should enter the basic. iv. Computing the substitution rates of the entering column. Answer: The substitution...
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...|Commerce | | | | | | |Honours Bachelor of Commerce (4-year) | | | | | | | | |DEGREE CHECKLIST | | |TOTAL: 120 credits | |HBComm (4yr) |Required courses (63 credits) | | |Students must obtain a minimum grade of 60% on all required courses and in each of the 21 COMM | | |credits at the 4000 level. | | |COMM 1006 E/F - Foundations of the Management of Organizations I | | |COMM 1007 E/F - Foundations...
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...R Tools for Portfolio Optimization Guy Yollin Quantitative Research Analyst Rotella Capital Management Bellevue, Washington Backgrounder Rotella Capital Management Quantitative Research Analyst Systematic CTA hedge fund trading 80+ global futures and foreign exchange markets Insightful Corporation Director of Financial Engineering Developers of S-PLUS®, S+FinMetrics®, and S+NuOPT® J.E. Moody, LLC Financial Engineer Futures Trading, Risk Management, Business Development OGI School of Engineering at Oregon Health & Science University Adjunct Instructor Statistical Computing & Financial Time Series Analysis Electro Scientific Industries, Inc Director of Engineering, Vision Products Division Machine Vision and Pattern Recognition Started Using R in 1999 R Tools for Portfolio Optimization 2 Introduction DJIA: 12/02/2008 - 04/15/2009 100 GM C 80 IBM annualized return (%) JPM 60 INTC HD DD GE PG BAC R-SIG-FINANCE QUESTION: stock price 90 40 MMM 20 KFT XOM PFE CVX JNJ VZ HPQ MCD KO UTX WMT T CAT MRK AXP BA Can I do < fill in the blank > portfolio optimization in R? MSFT DIS AA 0 IBM: 12/02/2008 - 04/15/2009 0 100 5 10 conditional value-at-risk (%) 15 20 95 80 85 Maximum Drawdown Jan Mar ANSWER: drawdown (%) IBM Underwater Graph 0 -5 P/L Distribution -10 -15 Yes! (98% confidence level) 0.12 0.14 Jan Mar 0.10 VaR 0.08 Density CVaR...
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...After doing extensive researches us the research team of DooDads Inc. has determined that if we are operating at our most efficient level. The optimal level of widget production to maximize our profit will be to produce 1,763 widgets. We acquired this number by using the information we already have and creating a demand function. The steps we took to create a demand function are the following 1. Using the number of widgets demand and their prices at two intervals and finding the slope.2. Once the slope is found from these two intervals we chose one interval and plug it into the point slope form, giving us our demand function. This allows us to acquire our revenue function, which is the demand function time the number of widgets.3. We create our cost function “price per widget time number of widgets plus fixed cost”.4. Of course (revenue – cost=profit) so now all is left to do is to solve the equation. So, now we have determined from this solving this equation the optimal level of widgets sold is 1,763 however we still need to know the maximum profit. This is determined by plugging optimal level of widgets sold into the profit function; we came out with an optimal profit of $21,170.24. The average cost of each widget (1763plugged into our cost function divided by 1763) is $7.84 per widget. The average revenue per widget (21170.24/1763) is $12.01.Average profit per widget (1,763 plugged into profit function) is $4.17. Since our breakeven point ( where cost = revenue or where...
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...VOLCANO Application Review Relative costs associated with VOLCANO stemmed mainly from the $100,000 grants given to researchers at MIT. Considering the vast amount of savings already incurred, the cost to UPS is miniscule from an R&D and implementation perspective. Actual costs associated with integration of VOLCANO into all UPS processes and planning designs are unknown. References Abrams, B. (2003). Bill. Linear Modeling to the Max: An Interview with UPS Operations Research Manager Keith Ware . Retrieved September 9, 2009, from http://www.secondmoment.org/articles/ups.php Armacost, A. P., Barnhart, C., & Ware, K. A. (2002). Composite Variable Formulations for Express Shipment Service Network Design [Electronic version]. Transportation Science, 36(1), 1-20. Armacost, A. P., Barnhart, C., Ware, K. A., & Wilson, A. M. (2004). UPS Optimizes Its Air Network [Electronic version]. Interfaces, 34(1), 15-25. UPS Optimizes Its Air Network (Video) ECU Blackboard http://core.ecu.edu/dsci/swartw/EDELMAN/UPS.M4V Walkey, F. H. (1998). Composite variable analysis: A simple and transparent alternative to factor analysis [Electronic version]. Personality and Individual Differences, 22(5), 757-767. doi:10.1016/S0191-8869(96)00238-3 Youssef, A. (n.d.). Retrieved September 11, 2009, from http://www.seas.gwu.edu/~ayoussef/cs212/branchandbound.html VOLCANO Assessment Quiz TRUE 2). What was wrong with the integer programming model used before VOLCANO?. Constant...
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...terms of an objective function to be maximized or minimized subject to a set of constraints. • find feasible solutions for maximization and minimization linear programming problems using the graphical method of solution. • solve maximization linear programming problems using the simplex method. • construct the Dual of a linear programming problem. • solve minimization linear programming problems by maximizing their Dual. 0.1.2 Introduction One of the major applications of linear algebra involving systems of linear equations is in finding the maximum or minimum of some quantity, such as profit or cost. In mathematics the process of finding an extreme value (maximum or minimum) of a quantity (normally called a function) is known as optimization . Linear programming (LP) is a branch of Mathematics which deals with modeling a decision problem and subsequently solving it by mathematical techniques. The problem is presented in a form of a linear function which is to be optimized (i.e maximized or minimized) subject to a set of linear constraints. The function to be optimized is known as the objective function . Linear programming finds many uses in the business and industry, where a decision maker may want to utilize limited available resources in the best possible manner. The limited resources may include material, money, manpower, space and time. Linear Programming provides various methods of solving such problems. In this unit, we present the basic concepts of linear programming...
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...Simplex method for the Paint Company MAT/205 Finite Mathematics December, 20, 2011 The world of economics for a business can be a challenging area for the owners and operators to keep control of. The process of balancing cost of production to the profit of the item has to be constantly balanced. There are methods that can help a business owner to make the balancing process easier. The simplex method is an algebraic method that can help an individual solve a problem that can have large amounts of variables and problem constraints, and the method also puts inequalities in a format that is usable by computers (Raymond A. Barnett, Michael R. Ziegler, and Karl E. Byleen, 2011). A paint company uses the simplex method to help determine the amount of paint that needs to be produced while keeping cost at a minimum and maximizing profit. The simplex method will be broken down to allow for thorough understanding of the process with the results displayed for the business. The benefits of the method will be discussed so the paint company can utilize the information. A paint company has two plants that produce paint and primer. The A plant produces 20 gallons of paint and 10 gallons of primer per hour. The B plant produces 5 gallons of paint and 25 gallons of primer per hour. The C plant produces 15 gallons of paint and 15 gallons of primer per hour. The price of operating A plant is $80 per hour, the price to operate B plant is $70 per hour, and C plant...
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...92 (2) Let be the number of grade 9 oranges O.J. uses for bags, be the number of grade 9 oranges O.J. uses for juice, be the number of grade 6 oranges O.J. uses for bags, and be the number of grade 6 oranges O.J. uses for juice. The objective is to maximize the profit. For juice, the profit will be ($1.50$1.05)*( + ), and for bags, the profit will be ($0.50-$0.20)*( + ). Thus, the total profit will be . The objective function is: Maximize The constraints are: Constraint 1: The total number of grade 9 oranges is 100,000 lb. That is, Constraint 2: The total number of grade 6 oranges is 120,000 lb. That is, Constraint 3: The average quality of juice is at least 8. That is, This gives us: Constraint 4: The average quality of juice is at least 8. That is, This gives us: Constraint 5: All variables must be nonnegative. That is, Using Excel, we can compute the optimal numbers of (checking the box Make Unconstrained Variables Non-Negative to satisfy the Constraint 5): Variables Coefficients Objective Function Constraints Constraint 1 Constraint 2 Constraint 3 Constraint 4 x11 x21 x12 x22 46666.66667 93333.33333 53333.33333 26666.66667 0.3 78000 Total 100000 120000 0 0 0.3 0.45 0.45 1 0 2 0 0 1 -1 0 1 0 0 1 0 1 0 -2 = 100000 120000 0 0 98 (4) To make product 3 from product 1, there are 2 methods. First, we can directly process product 1 into product 3 (1 to 3), which will cost us 3 hours of labor and $2. Besides, we can process product 1 into 2, then process 2 into...
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...MBA662 HW1 Dr. Nuo Xu (Adapted from Lawrence and Pasternack, Chapter 2 Problem 5) Anderson & Blount (A&B) Woodworks makes tables and chairs from 30-inch wide mahogany sheets that it purchases the linear foot. It can purchase whatever mahogany it desires for $10 per linear foot up to 2250 linear feet per week. Each table requires 9 linear feet and each chair 3 linear feet (including waste). Each chair also utilizes a soft cushion. Up to 500 cushions can be purchased each week for $25 each. Other required hardware (supports, braces, nuts, bolts, etc.) averages $45 for each table and $25 for each chair. A&B sells the tables to retailers for $300 each and each chair for $150 each. The 10 craftsmen employed by A&B are salaried workers. Their wages of $800 each per week as well as the $5000 per week in rent, insurance and utility costs are all considered fixed costs. To produce a table requires 1 hour of a craftsmen's time, whereas each chair requires only 36 minutes. Each craftsman averages 37.5 productive work-hours per week. Company policy mandates that the ratio of chairs to tables must be between 4 to 1 and 6 to 1. a. Develop a linear programming model for A&B. The objective function should maximize its gross weekly profit (gross revenue less the variable costs of wood, cushions and other materials). Express the feasible region by the non-negativity constraints and a set of five functional constraints (wood and cushion availability, the minimum and maximum chair to table ratios...
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...Julia's Food Booth Part A (Formulate) | | | | | Step 1: | Define the decision variables: | | | | | | x1 | = | How many hot dogs to produce to maximize profit | | | x2 | = | How many BBQ Sandwiches to produce to maximize profit | | x3 | = | How many Cheese Pizza slices to sell to maximize profit | | | | | | | | | | | Step 2: | Define the objective function. | | | | | | Maximize the profits of Julia's booth. | | | | | | | | | | | | | | | | | | Cost | Selling Price | Profit | | | | | Profit of Pizza = | $ 0.75 | $ 1.50 | $ 0.75 | | | | | Profit of BBQ = | $ 0.90 | $ 2.25 | $ 1.35 | | | | | Profit of Hot Dog = | $ 0.45 | $ 1.50 | $ 1.05 | | | | | | | | | | | | | | Maximize Z = | $1.05x1 | + | $1.35x2 | + | $.75x3 | | | | | | | | | | | Step 3: | Define the constraints: | | | | | | Size of the shelves in the warming oven (space is a constraint). | | | | | | | | | | | | | | Budget Constraint = | $0.45x1 + $0.90x2 + $0.75x3 <=$1,500 | | | | | | | | | | | | | Space Constraint = | Total Space in Oven = | 192 | Sq Ft | | | | | | | | 27648 | in sq | | | | | She is refilling at half time = | 55296 | in sq | | | | | | | | | | | | | | Space required for a pizza = | 196 | in sq | | | | | for a slice of pizza...
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...Answer 1 a. We have to make a decision, how much space should we lease and for how long. Constraint is the minimum space required. The objective of this function is to minimize b. Decision Variables: Let Xij = the space leased in month (i) for the period of (j) months, for i = 1, 2, …, 5 and j = 1, …, 6. Objective Function: Minimize Z = 65(X11 + X21 + X31 + X41 + X51) + (100X12 + X22 + X32 + X42) + 135(X13 + X23 + X33) +160(X14 + X24) + 190X15 Constraints: X11 + X12 + X13 + X14 + X15 >= 30,000 Month 1 Req. X12 + X13 + X14 + X15 + X21 + X22 + X23 + X24 >= 20,000 Month 2 Req. X13 + X14 + X15 + X22 + X23 + X24 + X31 + X32 + X33 > = 40,000 Month 3 Req. X14 + X15 + X23 + X24 + X32 + X33 + X41 + X42 >= 10,000 Month 4 Req. X15 + X24 + X33 + X42 + X51 >= 50,000 Month 5 Req. Answer 2: b. Decision Variables: Let F1 = Number of full-time consultants working the morning shift (8 AM – 4 PM) Let F2 = Number of full-time consultants working the evening shift (4 PM – 12AM) Let P1 = Number of part-time consultants working the 1st shift (8 AM – 12 PM) Let P2 = Number of part-time consultants working the 2nd shift (12 PM – 4 PM) Let P3 = Number of part-time consultants working the 3rd shift (4 PM – 8 PM) Let P4 = Number of part-time consultants working the 4th shift (8 PM – 12AM) Objective Function: Minimize Z = 14*8* (F1 + F2 ) + 12*4 * (P1 + P2 + P3 + P4 ) Constraints: Minimum number of consultant requirement F1 + P1 >= 4, F1 +...
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