Solution: x + y 8 (6 x) + (9 y) 11 (6 x) 0 (9 y) 0 x 0, y 0

Transcription

1 MA Test 3 Justify your answers and show all relevant work. The exam paper will not be graded, put all your work in the blue book provided, including graph paper. Problem ( points) A car dealer with warehouses in New York and Baltimore receives orders from dealers in Philadelphia and Trenton. The dealer in Philadelphia needs 6 cars and the dealer in Trenton needs 9. The New York warehouse has 8 cars and the Baltimore warehouse has. The cost of shipping cars from Baltimore to Philadelphia is $0 per car, from Baltimore to Trenton $ 00 per car, from New York to Philadelphia $00 per car, and from New York to Trenton $80 per car. Find the number of cars to be shipped from each warehouse to each dealer so that the shipping cost is minimized. The diagram represents this situation: NY and Balt are labeled with the number of cars they have, and the dealers in Philly and Trenton are labeled with the number of cars they need. Let x and y represent the number of cars shipped from NY to Philly and NY to Trenton, respectively. Then since the total number of cars shipped to Philly must be 6, we must ship 6 x cars from Baltimore to Philly. Similarly we must ship 9 y cars from Baltimore to Trenton. Since we only have 8 cars in NY, the total car outflow from NY is 8 and x+y 8. Then the total car outflow from Baltimore is, so (6 x) + (9 y). Since the number of cars shipped out from the warehouses must be nonnegative, we have x, y, (6 x), and (9 y) 0. So we have the set of constraints x + y 8 (6 x) + (9 y) (6 x) 0 (9 y) 0 x 0, y 0

2 Here are the constraints in standard form: y x + 8 y x + 4 x 6 y 9 x 0, y 0 The objective we wish to minimize is the total shipping cost. The shipping cost for each leg is (the number of cars) (the cost per car); for example, the shipping cost for the leg from NY to Philly is x(00) = 00x. We add the costs for the four legs together to obtain the total shipping cost: 00x + 80y + 0(6 x) + 00(9 y) = 560 0x 0y We ll graph the feasible set for the constraints above and check the value of this objective at each vertex to find the pair (x, y) that will give us the solution. Here is the graph of the feasible set (Note the inequality y 9 plays no role in the graph): We check the value of the objective at all five vertices: Vertex Cost = 560 0x 0y A = (0, 8) 560 0(0) 0(8) = 400 B = (6, ) 560 0(6) 0() = 460 C = (6, 0) 560 0(6) 0(0) = 500 D = (4, 0) 560 0(4) 0(0) = 50 E = (0, 4) 560 0(0) 0(4) = 480 The minimum value for the objective of 400 is reached at A with values x = 0, y = 8. So we should ship no cars from NY to Philly, 6-0 = 6 cars from Balt to Philly, 8 cars from NY to Trenton, and 9-8 = cars from Balt to Trenton.

3 Note: if we assign x and y to represent the cars shipped from Baltimore to Philly and Baltimore to Trenton, respectively, then our objective function will be x + 0y and our family of constraints is x + y (6 x) + (9 y) 8 (6 x) 0 (9 y) 0 x 0, y 0 Problem The two matrices P = [ ] 0., Q = [ ] represent transition matrices for Markov processes with states and. (a) Draw transition diagrams for P and Q.

4 (b) Find P and Q. P = P P = [ [ () + 0.(0) (0.) + 0.(0.8) 0() + 0.8(0) 0(0.) + 0.8(0.8) Q = Q Q = [ [ 0(0) + 0.() 0(0.) + 0.(0.8) (0) + 0.8() (0.) + 0.8(0.8) ] [ ] 0. = ] = ] [ ] 0 0. = 0.8 ] = [ ] [ ] (c) Is P regular? Is Q regular? Why or why not? P is not regular because P is an absorbing stochastic matrix. Q is regular because Q has all positive entries. Problem 3 ( points total) The victims of a disease are classified into three states: cured, dead, and sick. Once cured, a person is permanently cured because they become immune. Each year 70% of the sick are cured, 0% remain sick, and 0% die. We model this as an absorbing stochastic process where the regular time interval is one year and the states are Cured, Dead, and Sick.

5 (a) Make a transition diagram for this Markov process. Indicate the absorbing states. The absorbing states are Cured and Dead, which in our diagram are the states with no strictly outbound arrows. (b) Make the absorbing stochastic matrix A = [ ] I S 0 R for this Markov process, putting the columns and rows indexing the absorbing states first. Write down the sub-matrices S and R explicitly. so that and Cured Dead Sick Cured Dead 0 0. Sick [ ] I S A = = R S = [ ] 0.7, R = [0.] 0. (c) Use the fundamental matrix F = (I R) to estimate how long someone who is sick will stay sick. (The ijth entry of F is the expected number of times the process will be in nonabsorbing state i if if starts in nonabsorbing state j. )

6 The fundamental matrix F = (I R) = ( 0.) = /(0.8) = 5/4. The rows and columns of F are indexed by the nonabsorbing states, but there is only one nonbsorbing state here, which is Sick. So, the single entry of the matrix F is telling us how long someone stays in nonabsorbing state sick once they enter non-absorbing state sick, which is 5/4 =.5 years. (d) In the long term, what is the probability that a sick person is cured, and what is the probability that a sick person dies? In other words, what will be the long-term distribution between the absorbing states of patients who contract the disease? Hint: the formula for the stable matrix of an absorbing stochastic process is [ ] I S(I R) 0 0 The hint gave us the formula again, so we should use it: First ] [5 ] ] S(I R) = = 4 [ [ and the stable matrix is 7 [ ] 0 I S(I R 8 ) = So in the long term, 7 8 of the patients who enter state Sick will end up in absorbing state Cured, and 8 of the patients who enter state Sick will end up in absorbing state Dead. So the long term prognosis of someone who becomes sick with the disease is that they have a 7 8 chance of surviving. Problem 4 Form the initial simplex tableau for the following LP problem: MInimize 4x+8y subject to the constraints x + 5y 4x y 6 x + y 30 x 0 y 0 Label the columns of the tableau with the variables they represent. Indicate which variables are in Group I and Group II. Stop when you have set up the initial simplex tableau. Do not finish solving the problem.

7 To use the simplex tableau, we must convert the objective function to a maximization problem and convert two of the constraints to standard form of [linear polynomial] [number]. So our new problem is Maximize 4x 8y subject to the constraints We obtain the initial tableau x 5y 4x y 6 x y 30 x 0 y 0 x y u v w M u v w M The Group I variables are x and y. The Group II variables are u, v, w and M (Note that I labeled the rows of the tableau with the Group II variables). Problem 5( points total) A nutritionist working for NASA must minimize the weight of food carried on a shuttle run while meeting minimum nutritional requirements. She has two types of foods to choose from: A and B. Each tube of food A contains 4 units of protein, units of carbohydrates, units of fat, and weighs pounds. Each tube of food B contains 3 units of protein, 6 units of carbohydrates, unit of fat, and weighs pounds. The minimum requirements are 40 units of protein, 50 units of carbohydrates, and 4 units of fat. (a) (9 points) Let x denote the number of tubes of food A brought on board the shuttle, and let y be the number of tubes of food B brought on board. Give the set of inequalities that x and y must satisfy. First we find the inequality constraint based on protein requirements. The total protein on board will be ( number of tubes of food A) ( units protein per tube of A )+ ( number of tubes of food B) ( units protein per tube of B ) = (x) (4) + (y)(3) = 4x + 3y This total protein must be at least 40 units, so we have the constraint 4x+3y 40. Next the total carbohydrates on board will be ( number of tubes of food A) ( units carbohydrate per tube of A )+

8 ( number of tubes of food B) ( units carbohydrate per tube of B ) = (x) () + (y)(6) = x + 6y There must be at least 50 units of carbohydrates, so we have the inequality x + 6y 50 as a constraint. Similarly we obtain an inequality constraint x + y 4 for the requirements for fat. It does not make sense to allow negative amounts of tubes, so we also have the inequalitites x 0 and y 0 in our set. To summarize, the family of inequalities x and y must satisfy is 4x + 3y 40 x + 6y 50 x + y 4 x 0 y 0 (b) (3 points) Find the objective function: Express the total weight of food as a function of x and y that the nutritionist would like to minimize. The objective function is the total weight of the food brought on board. This is ( number of tubes of food A) ( weight in pounds of each tube of A )+ ( number of tubes of food B) ( weight in pounds of each tube of B ) = (x)() + (y)() = x + y Stop: don t solve the LP problem! Problem 6(0 points) Maximize x + 3y + 5z subject to the constraints x + z 0 3y + z 4 x 0, y 0, z 0 State the values of x, y, z for which the objective function is maximized. Since there are 3 variables in the problem we must use the simplex method. There are two constraints besides the non-negativity constraints, so we need two slack variables u, v 0 for these, and then the variable M will represent the value of the objective we wish to make as large as possible. We convert the maximization problem into the following system of equations: This system yields the initial tableau: x +z +u = 0 3y +z +v = 4 x 3y 5z +M = 0

9 x y z u v M u v M We are in the feasible set because there are no negative numbers in the upper right box of the tableau: our goal is therefore to get rid of all the negative numbers in the bottom left box of the tableau by pivoting. We choose the column with the most negative entry in the bottom left box as our pivot column: Column 3 has -5. To choose the pivot entry, we compare the positive ratios 0/ and 4/: 0/ is smaller, so we pivot about the in Column 3. R x y z u v M R R R 3 + 5R x y z u v M z v M We still have a negative number in the lower left box, so we are not yet at a maximum. We pivot in Column which has the -3 in the lower left box; to choose the pivot entry, there is only one positive ratio, 9/3, to consider. So we pivot about the 3 in column : 3 R x y z u v M R 3 +3R x y z u v M z y M There are no negative numbers in the lower left box of the tableau, so this tableau gives a maximum for the objective. The Group I variables are x, u, v and we set them to zero: the Group II variables are y = 9 3, z = 5, and M = 44. So, the objective function x + 3y + 5z achieves its maximum value at the point x = 0, y = 9 3 and z = 5.