Determining the Matrix Representation - Assignment Example

Title: Assessment 2 Name: Institutional Affiliation: Tutor: Date: QUESTION 1 In solving this equation, it is important to first consider that the equation has both the first and second differentials. Therefore, the equation represented will take the form: Y’’-9y’+3=0 From this equation, we can then apply the Laplace transformation leading to the final equation taking the form of (s²y-2s-1) -9(sy-2s) + 3y=0 From here, we collect the like terms together, which in this case will be the values of y. We bring all values of Y forming: Y (s²-9s+3) Completing the equation we obtain Y (s²-9s+3) = 2s+19 Making Y the subject if the formula, we get, Y =    2s+19/ (s²-9s+3) In order to find Y, we shall have to find the inverse of the Laplace transformation 2s + 19                A            B            C                                      =                +             +                                                             s²-9s+3               s          s-3        s + 2 We obtain the values for A B and C as follows: A(s-3)(S+2) + Bs(S+2) + Cs(S-3) = 2s +19 For s=0 -6A=19 A = 19/6 For s=3 15B=25 B=5/3 For s=-2 10c=15 C= 1.5 This will give our final solution as 19/6 + (5/3) e^3t + (1.5) e^-2t (ii) -2T’’ = 10-4T It is necessary to rearrange this question in order to come up with a proper equation that can be solved using the Laplace transformation. As such, the equation shall take the form -2T’’ +4T -10 =0 As such, it is now possible to open the components of the equation -2[S²y+2S+1] + 4[2Sy+1] – 10y =0 Bringing the like terms together (in this case, the value Y), we have, Y [-2S²+8S- 10] = 2S+6 It is equally important to note that the value S squared represents the characteristic equation of this differential equation. After getting the above equation, we go a step further and divide the equation on the left had side with that on the right hand side, therefore making Y the subject of the resulting equation. This will be as shown below: Y = 2S +6/ (-2S +8S -10) Simplifying the denominator we get: Y = 2S +6/ (S +4S -5) Simplifying the denominator further, we get Y= 2S + 6 / [S-1] [S+5] Assigning the values A and B on the two equations in the denominator, we have: 2S + 6 / [S-1] [S+5] = A/(S-1) + B/ (S+5) A(S+5) + B(S-1) = 2S+6 Let S be 1 6A = 8 A= 4/3 Let S be -5 -6B=-4 B = -2/3 This gives us the final equation as 4/3[S-1] + -2/3[S+5] This gives the value of y at Y = (4/3) e^t – (2/3) e^-5t QUESTION 2. From this question, we have to find the probabilities of the states for the next 3 hours. Here, I shall consider deriving a transitional matrix highlighting the probabilities of the outcomes within the given 45 minute period. The result is a 4x4 matrix as illustrated below: 0.5 0.5 0 0 0.1 0.3 0.6 0 0 0.5 0.4 0.1 0.1 0 0.8 0.1 Following the above matrix, it is possible to come up with the computation for the 3 hour period considering that the 3 hours is 4x45 minutes. The resulting computation will give us a P⁴ and a 4x4 matrix as shown below: 0.14 0.401 0.42 0.039 0.088 0.3996 0.4596 0.0528 0.0814 0.3895 0.4766 0.0525 0.0834 0.4025 0.459 0.551 Therefore, according to the requirements of the question, obtaining the grade most likely to be after the 3 hour exam, the criteria will be to look out for the resulting probability obtained in the row number 3. This gives us the approximate value of 0.459. This means that after successfully completing the 3 hour exam, you will pass with a possible grade of 0.459 b.) In solving the second part, the key guiding factor is the fact that they are descried as steady states. The states will only be considered to be steady states if they are time independent, that is they do not rely on the times 1,2 3, and 4. As such, in the computation, the states are steady. All states in the simple chain communicate and the distribution is stationary. Further, it is true that vP⁴=v following from the fact that (vP)P = v. We go further and determine the stationary distributions of the transitional matrix by multiplying it with a (1x4)π matrix. The resulting equation will take the form: (π 1, π2, π3, π4) X 0.5 0.5 0 0 0.1 0.3 0.6 0 0 0.5 0.4 0.1 0.1 0 0.8 0.1 An important point to note when carrying out the computations with phi (π) is that the resulting summations of all phis above, that is π1 + π2 + π3 +π4 = 1. This therefore means that to find the steady state results for a group of 14 students, we shall multiply the resulting phis with a factor of 14. This results with the respective probabilities that fall in the four categories: fail, acceptable, standard, genius. From the above matrix, we shall have the value of π as: Π1 (0.5 + 0.1 + 0.1) Π2 (0.5 + 0.3 + 0.5) Π3 (0.6 + 0.4 + 0.8) Π4 (0.1 + 0.1) Solving the above calculations from the matrix form, we find the values of Π1Π2, Π3, and Π4 to be (0.127, 0.3968, 0.4286, 0.047) respectively. This means that with the consideration of 14 in a class, the stationary state results for a class with 14 pupils will be: 2 students with fails, 5 with acceptable grades, 6 with standard and 1 with genius. The values are truncated since we are referring to the number of students. For example, I avoided using values such 0.6 students (for the genius category) QUESTION 3. Resource Tables (X) Chairs (Y) Constraints Finishing 3 2 90 Carpentry 5 3 230 Profit per unit 85 55 Objective function: P = 85X +55Y Finishing Constraint 3X +2Y ≤ 90 Carpentry Constraint 5X + 3Y≤230 Conditions for the functions to hold, X, Y ≥0 From here, we include the slack variables, therefore making the inequalities to be full equations. The introduction of slack variables is an important step as it helps in analyzing the objective functions. Therefore, we shall have: 3X +2Y +s1 =90 5X + 3Y + s2 =230 Further, since the unused hours will obviously not result to any profit, the objective function can also be incorporated with the slack variables but now there will be zero coefficients. As such, the equation will be in the form: P= 85X + 55Y +0s1 + 0s2 P - 85X - 55Y - 0s1 - 0s2 = 0 From here, we write the three equations linearly and seek to find the value for which P will be maximum 5X + 3Y +s1 +0 s2 =230 3X +2Y +0s1 + s2 =90 P – 85X – 55Y -0s1 -0s2 = 0 We design a table that will represent this information as follows: BASIC VARIABLE X Y S1 S2 p Value S1 5 3 1 0 0 230 S2 3 2 0 1 0 90 P -85 -55 0 0 1 0 From the values about in s1 and s2, the general assumption at this stage is that all the hours available in remain unutilized and therefore the slack variables still maintain the highest values. Moreover, the variables s1 and s2 now represent the initial solution mix. As such, the table draws up the initial solution for X=0, Y=O, s1=230, s2=90 and P=0 Further, we select the pivot column from the table above. Ideally, the pivot column is identified by looking out for the column with the highest negative between X and Y. In our case, the column with the highest negative is column X. We also look out for the pivot row. The pivot row is identified by the fact that it is the row with the smallest result after dividing each value in the pivot row with the results shown in the value. In our case, the two values are 230/5 and 90/3. These two give 46 and 30 respectively. We shall therefore choose 30, which is represented by the row s2. BASIC VARIABLE X Y S1 S2 p Value S1 5 3 1 0 0 230 X 1 2/3 0 1/3 0 30 P -85 -55 0 0 1 0 Dividing the row s2 by 3, we obtain the values shown above. Further, to make the values in the pivotal column to equal to zero, we come up with equations linking row s2 with the other two rows. Each equation is then multiplied all through the remaining rows with the effect of making it possible to reduce the values in the pivotal column to zero. As such, the equation for the row s1 will be -5R1+ R2. Similarly, the equation for the row p is 85.R1+R3 BASIC VARIABLE X Y S1 S2 p Value S1 0 -1/3 -1 -5/3 0 80 X 1 2/3 0 1/3 0 30 P 0 5/3 0 28.33 1 2400 This therefore means that if 30 tables are manufactured, then the number of carpentry hours used are reduced by a total of 150 hours that is a product of 30 by 5. Similarly, the values changes from 230 to 80. What this intimates is that the profit is increased by a large margin of 2400 pounds as the value is increased from zero to 2400. BASIC VARIABLE X Y S1 S2 p Value S1 0 -1/3 -1 -5/3 0 80 X 1 2/3 0 1/3 0 30 P 0 5/3 0 28.33 1 2400 The last row in this table contains no negative numbers. What this means is that we have obtained the maximum value of P. Therefore, in the event that 80 chairs are made, then the value of the tables consequently changes to 30. The optimal solution for this problem is therefore X= 30 Y=80 Profit = 2400 S1=0 S2=0 QUESTION 4. In determining the matrix representation of the above problem, the key consideration is that there are 3 countries. Therefore, in the matrix, we have to show the movement within the three states A, B and C. This therefore results in a 3x3 matrix with each row in the matrix showing the movement. The matrix will hence take the form [1 0 0] [0.1 0.6 0.3] [0.1 0.4 0.5] In determining when the plane will take up permanent residence, the key factor to watch out for is the number of successive steps that being in a certain state will outnumber the possibility of being in the other states. As such, the probability of being in a particular state at a defined time t=1 will be the row number 2 in the matrix representation. In this case, we have our introductory/ initial position set out as [1 0 0]. The transition matrix is representative of the probability of moving from through the transition matrix. This will require us to develop a probability distribution showing the probability of being in any states at the designated time t=1 Considering that we need to start from any of the other two states, we shall have our theta for the value a = 0, b =10(0.1 +0.32) and c= 10[0.03 +0.4(0.1 + 0.4)] Calculating the total sum of the three we obtain 2.3+3.3+0= 5.6 Therefore the plane will take up permanent residency after 5.6 weeks. QUESTION 5. The stochastic transitional matrix is representative of the movement throughout the designated time steps (days). From the diagram, all the three states are communicating with each other apart from state C which does not communicate with the state A. Therefore, a transitional probability matrix will form a 3x3 matrix with the above formation [0.4 0.5 0.1] [0.5 0.4 0.1] [0 0.05 0.95] The probability of being in state C after s two time steps entails finding the probability of moving to state C either from states A or B. therefore, all the possibilities involved in moving from states A or B will be calculated and summed up in order to come up with a conclusive probability of such an outcome: P(0.5) x P(0.1) = 0.05 This is the probability of moving from A to B then to C P(0.3) x P(0.1) = 0.03 This represents the probability of moving from B to A then to C P(0.05) x P(0.1) = 0.005 This is the probability of moving from C to B and back to C Therefore, to find the probability of being in state C after two time steps will be the sum total of the 3 probabilities. This will give us 0.085 as the answer. (iii) In finding the steady state probabilities, we need to develop the probabilities of being in a particular state as the time interval gets bigger. As such, the steady state probability provides a projection of the long run probabilities of being in each and every state within a markov chain. From our transition matrix, we shall form the equation below. [0.4 0.5 0.1] [p,p,p] [0.5 0.4 0.1] = [p,p,p] [0 0.05 0.95] Kindly note that the matrix [p,p,p] has been multiplied with the entire 3x3 matrix above. 0.4p1+0.5p2+0.1p2 = p1 0.5p1+0.4p2+0.1p3 = p2 0.05p2+0.95p3 =p3 -0.6p1+0.5p2+0.1p3 =0 …………….i 0.5p1-0.6p2+0.1p3 =0 ………………ii 0.05p2-0.95p3 = 0 ……………..iii From here, we can then solve for the corresponding values of p1, p2 and p3 respectively. Making p2 the subject in equation iii, we get p2=19p3 Going ahead to equation 2, we substitute p2 with p3 0.5p1-11.4p3+0.1p3=0 Making p1 the subject, we get, 0.5p1=11.3p3, therefore, p1=22.6p3 Since p1+p2+p3=1, we can find the values of the three steady states as follows: 22.6p3+19p3+p3=1 P3=0.02347 Therefore p2 = 19p3= 0.446 And p1 = 22.6p3 = 0.5304 The questions asks the steady state probability of being in state A which is given by P1 in our solution above. Therefore, Steady state probability of being in state A = 0.5304 Steady state probability of being in state B = 0.446 Steady state probability of being in state C = 0.0235 Read More

From the above matrix, we shall have the value of π as: Π1 (0.5 + 0.1 + 0.1) Π2 (0.5 + 0.3 + 0.5) Π3 (0.6 + 0.4 + 0.8) Π4 (0.1 + 0.1) Solving the above calculations from the matrix form, we find the values of Π1Π2, Π3, and Π4 to be (0.127, 0.3968, 0.4286, 0.047) respectively. This means that with the consideration of 14 in a class, the stationary state results for a class with 14 pupils will be: 2 students with fails, 5 with acceptable grades, 6 with standard and 1 with genius. The values are truncated since we are referring to the number of students.

For example, I avoided using values such 0.6 students (for the genius category) QUESTION 3. Resource Tables (X) Chairs (Y) Constraints Finishing 3 2 90 Carpentry 5 3 230 Profit per unit 85 55 Objective function: P = 85X +55Y Finishing Constraint 3X +2Y ≤ 90 Carpentry Constraint 5X + 3Y≤230 Conditions for the functions to hold, X, Y ≥0 From here, we include the slack variables, therefore making the inequalities to be full equations. The introduction of slack variables is an important step as it helps in analyzing the objective functions.

Therefore, we shall have: 3X +2Y +s1 =90 5X + 3Y + s2 =230 Further, since the unused hours will obviously not result to any profit, the objective function can also be incorporated with the slack variables but now there will be zero coefficients. As such, the equation will be in the form: P= 85X + 55Y +0s1 + 0s2 P - 85X - 55Y - 0s1 - 0s2 = 0 From here, we write the three equations linearly and seek to find the value for which P will be maximum 5X + 3Y +s1 +0 s2 =230 3X +2Y +0s1 + s2 =90 P – 85X – 55Y -0s1 -0s2 = 0 We design a table that will represent this information as follows: BASIC VARIABLE X Y S1 S2 p Value S1 5 3 1 0 0 230 S2 3 2 0 1 0 90 P -85 -55 0 0 1 0 From the values about in s1 and s2, the general assumption at this stage is that all the hours available in remain unutilized and therefore the slack variables still maintain the highest values.

Moreover, the variables s1 and s2 now represent the initial solution mix. As such, the table draws up the initial solution for X=0, Y=O, s1=230, s2=90 and P=0 Further, we select the pivot column from the table above. Ideally, the pivot column is identified by looking out for the column with the highest negative between X and Y. In our case, the column with the highest negative is column X. We also look out for the pivot row. The pivot row is identified by the fact that it is the row with the smallest result after dividing each value in the pivot row with the results shown in the value.

In our case, the two values are 230/5 and 90/3. These two give 46 and 30 respectively. We shall therefore choose 30, which is represented by the row s2. BASIC VARIABLE X Y S1 S2 p Value S1 5 3 1 0 0 230 X 1 2/3 0 1/3 0 30 P -85 -55 0 0 1 0 Dividing the row s2 by 3, we obtain the values shown above. Further, to make the values in the pivotal column to equal to zero, we come up with equations linking row s2 with the other two rows. Each equation is then multiplied all through the remaining rows with the effect of making it possible to reduce the values in the pivotal column to zero.

As such, the equation for the row s1 will be -5R1+ R2. Similarly, the equation for the row p is 85.R1+R3 BASIC VARIABLE X Y S1 S2 p Value S1 0 -1/3 -1 -5/3 0 80 X 1 2/3 0 1/3 0 30 P 0 5/3 0 28.33 1 2400 This therefore means that if 30 tables are manufactured, then the number of carpentry hours used are reduced by a total of 150 hours that is a product of 30 by 5. Similarly, the values changes from 230 to 80. What this intimates is that the profit is increased by a large margin of 2400 pounds as the value is increased from zero to 2400.

BASIC VARIABLE X Y S1 S2 p Value S1 0 -1/3 -1 -5/3 0 80 X 1 2/3 0 1/3 0 30 P 0 5/3 0 28.33 1 2400 The last row in this table contains no negative numbers. What this means is that we have obtained the maximum value of P. Therefore, in the event that 80 chairs are made, then the value of the tables consequently changes to 30.

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