TEST TWO QUANTITATIVE METHODS BSTA 450 March 17, 2003 Name:(Print neatly) Student Number: -Put all your solutions on the paper provided. -make sure you READ the question and provide the solution it asks for. -NOTE: PICK TWO QUESTIONS FROM 4,5 AND 6. DO NOT DO ALL OF THEM, JUST TWO. 1. (20 marks) Given the following table describing the net profits (in thousands of dollars) from 3 different investments, determine the best investment using the following decision criteria Bad economy Average economy Good economy Motel 400 300 3000 Restaurant 100 500 2000 Prison holding tank 250 50 400 a) equal likelihood criterion b) minimax regret criterion c) Hurwicz (α =0.6) d) maximin criterion. e) Given that the probability of an Average economy is 10%, for what probabilities of a Bad economy would you choose the Prison over the restaurant (note: in this part I am NOT asking about the Motel..ignore it). Solutions: a) E(motel) =( 400+300+3000)/3 = 2900,E(restaurant) =( 100+ 3 500+2000)/3 =800,E(prison) =(250+50 400)/3 = 100 Therefore 3 choose Motel. b) 1
Regret Table Bad Average Good Motel 650 200 0 Restaurant 350 0 1000 Prison 0 450 3600 Minimum of Maximum regrets is 650, therefore choose Motel. c) E(motel) =(0.6)3000+(0.4)( 400) = 1640,E(rest) =(0.6)(2000)+ (0.4)( 100) = 1160,E(prison) =(0.6)(250) + (0.4)( 400) = 10.0. Therefore choose Motel. d) E(Motel) = -400, E(restaurant) = -100, E(prison) = -400. Therefore choose Restaurant. e) 100a+50+(0.9 a)(2000) = 250a+5+(0.9 a)( 400) a =. 8018 Therefore the probability of a bad economy would have to be in the interval (80.2%, 90%) 2
(20 marks) 2. Lets say you are a book publisher. The probability for any individual book being successful is only 15%. If it is successful you will make $800,000, but if it is not, then you will lose $300,000 (you always have the option of not producing any individual book ). You can hire an book expert to tell you if any particular book is good or not. He is 95% correct at predicting successful books but only 75% correct at predicting books which do not sell.. Using decision tree analysis,tell me EXACTLY what you should do? Solution: Without any advice E = (0.15)(800000) + 0.85( 300000) = 135000. So do not publish books. p(ps) =(0.15)(.95) + (0.85)(0.25) =. 355,p(pf) =1 0.355 =. 645 p(s ps) =(0.15)(0.95)/.355 =. 4014,p(f ps) =1 0.4014 =. 5986 p(s pf) =(0.15)(0.05)/.645 =.0116,p(f pf) =1.0116 =. 9884 Look at fig 1. You get a positive non-zero expected value if you hire him. So, hire the guy and do what he says. Pay him at most $50246.70. 3
Figure 1: 4
3. (20 marks) Lets say you are a teddy bear producer and you make only two types of bears; happy and sad bears. Happy bears sell for $20 and sad bears sell for $15. Now, Happy bears take 1 hour to produce while sad ones take 2 hours. You have 50 man-hours available each production period. These bears also require stuffing materialofwhichyouhave 60lbs. EachHappy bear requires one pound of stuffing, while each Sad bear requires 3 pounds of stuffing. They also require wool for the body of the bear, of which you have 40 lbs each production period. Happy bears require 2lbs each, while Sad ones require one pound each. How many of each type of bear should you produce to maximize profit?? Solution. Let x represent the Number of happy bears produced. Let y represent the number of sad bears produced. Maximize: z =20x +15y Constraints: x +2y 50 x +3y 60 2x + y 40 and x, y 0 25 20 15 y 10 5 0 0 5 10 x 15 20 25 The points which define the boundary of the region are (0,0), (0,20), (20,0), (12,16) 5
z(0, 0) = 0,z(0, 20) = 300,z(20, 0) = 400,z(12, 16) = 480 Therefore the maximum value is $480 and it occurs when you sell 12 happy and 16 sad bears. 4. (10 marks) Letssayyouproducetwotypesofcandy: peppermintsticksandpackets of mints Each peppermint sells for $1.5 and each packet of mints sells for $2. You produce a bunch of these from you little home factory every month. Every month you ship in 120 lbs of sugar and 80 lbs of weird chemicals to produce your candy. Peppermint sticks require 1 lb of each ingredient, while a packet of mints require 2 lbs of sugar and 1 lb of weird chemicals... a) What is the maximum amount you can increase the price of the peppermint sticks by so that the optimal solution does not change?. b) What is the minimum value you can change the price of the mint packets to, so that the optimal solution does not change?. c) If someone offered you extra free sugar every month, how much would you accept (ie. we do not want wastage here...it all goes toward making candy). Or in other words, what is the maximum increase of sugar so that the shadow price of sugar does not change?. Solution. I didn t ask for optimal point or anything like that, so do not waste time solving for things i didn t ask for, but a quick sketch of the region with slopes will help you solve this. Let x represent the number of peperment sticks produced. Let y represent the number of packets of mints produced. z =1.5x +2y (slope = -0.75) x +2y 120 (slope = -0.5) x + y 80 (slope = -1) 6
80 60 y40 20 0 0 {x + y =80,x+2y =120} The optimal point occurs at the intersecton of the two lines (i do not care what it is...i can tell by the slopes that it is there). a) z = Ax +2y slope = - A. Thus A = 1 A =2. Thus you 2 2 can increase to $2 without changing optimal point.thus the increase allowed is 50 cents. b) z =1.5x + By slope = 1.5 1.5. thus = 1 B =1.5. Thus B B you can decrease the price to $1.50 without changing the optimal point. c) Push the sugar line up till the shadow price is zero. Thus let x+2y = N. Plug in the point (0, 80) N =160 Thus you would accept 40 more pounds of sugar. 20 40 60 x 80 100 120 7
5. ( 10 marks) Arefinery blends four petroleum components into threee grades of gasoline - regular, premium, and low-lead. The maximum quantities available of each component and the cost per barrel are as follows. Component Maximum barrels available/day Cost per barrel 1 5000 $9.00 2 2400 $7.00 3 4000 $12.00 4 1500 $6.00 To ensure that each gasoline grade retains certain essential characteristics, the refinery has put limits on the percentages of the components in each blend. The limits as well as the selling prices for the various grades are as follows. Grade Component Specifications Selling price per Barrel ($) Regular Notlessthat40%of1 Not more than 20% of 2 12.00 Not less than 30% of 3 Premium Not less than 40% of 3 18.00 Low Lead Not more than 50% of 2 Not less than 10% of 1 10.00 The refinery wants to produce at least 3000 barrels of each grade of gasoline. Management wishes to determine the optimal mix of the four components that will maximize profit. FORMULATE A LINEAR PROGRAMMING MODEL FOR THIS PROBLEM Solution: Let x ij represent the number of barrels of component i that go into grade j. Where 1 i 4, and j is either regular, premium or low lead. (12 variables in other words). z =12(x 1r + x 2r + x 3r + x 4r )+18(x 1p + x 2p + x 3p + x 4p )+10(x 1L + x 2L + x 3L + x 4L ) 9(x 1r + x 1p + x 1L ) 7(x 2r + x 2p + x 2L ) 12(x 3r + x 3p + x 3L ) 6(x 4r + x 4p + x 4L ) Constraints: x ij 0 8
x 1r + x 1p + x 1L 5000 x 2r + x 2p + x 2L 2400 x 3r + x 3p + x 3L 4000 x 4r + x 4p + x 4L 1500 x 1r + x 2r + x 3r + x 4r 3000 x 1p + x 2p + x 3p + x 4p 3000 x 1L + x 2L + x 3L + x 4L 3000 x 1r x x 1r +x 2r +x 3r +x 4r 0.4, 2r x x 1r +x 2r +x 3r +x 4r 0.2, 3r x 1r +x 2r +x 3r +x 4r 0.3 x 3p x 1p +x 2p +x 3p +x 4p 0.4 x 2L x x 1L +x 2L +x 3L +x 4L 0.5, 1L x 1L +x 2L +x 3L +x 4L 0.1 6. (10 marks) Answer the following question from the charts on the next page. a) What is the maximum amount that you can increase the coefficient of x1 by so that the optimal point does not change? b) What is the minimum value that the coefficient of x2 can have so that the optimal point does not change? c) If we increased the value of contraint C3 by 3.4, what would the new optimal value be? d) If someone told you that you could increase any of the constraint value, which one would you choose? e) If we decreased the constraint value of C2 by 6, what would the new optimal value be? f) If we decreased the constraint value of C4 by 25.5, what would the new optimal value be? Solution: Here is the type of table you would be given for such a question: 9
Variable Value Reduced Cost Original Value Lower Bound Upper Bound X1 0.69 0 10 3.33 100 X2 6.21 0 5 0.5 15 Constraint Dual Value Slack Original Val Lower Bound Upper Bound Constraint 1 1.551724 0 20 2 20.29 Constraint 2 0 6.034483 15 8.97 Constraint 3 0.6896552 0 10 8.31 27.5 Constraint 4 0 0.3965511 25.5 25.1 Variable Status Value X1 Basic 0.6896552 X2 Basic 6.206897 slack 1 NonBasic 0 slack 2 Basic 6.034482 slack 3 NonBasic 0 slack 4 Basic 0.3965516 Optimal Value (Z) 37.93103 a) Maximum increase is 100-10 = 90 b) 0.5 (just read it off the chart). c) 37.93103 + (3.4)(0.6896552) = 40. 27585 77 d) I would choose C3. The only two which you would consider are C3 and C1. C1 has a higher shadow price BUT you can ONLY increase it by 0.29 (20.29-20). You do not increase maximum very much with this. On the other hand you can increase C3 by 17.5. e) The optimal value would not change.. The shadow price remains zero in this range. z = 37.93 still. f) This is a bit of a trick question...the present value is 25.5. If you decrease by 25.5 this line will go through the origin. (no resource at all left). Thus your profit would be 0. 10