Computational modelling techniques Exercise set 3 Solutions

Transcription

1 Computational modelling techniques Exercise set 3 Solutions 1. For the following two data sets, construct a divided difference table. What conclusions can you make about the data? Would you use a low-order polynomial as an empirical model? If so, what order? (i) x y x y In this case a 3 rd order polynomial is suitable Data y = 1x 3 + 2x 2 + 3x Data Poly. (Data) (ii) x y ,316 x y 1 4, ,5 15, ,25 1, , , , , , , ,25 4, , ,675 2, , , A lower degree polynomial is not suitable in this case.

2 2. In the following data, X is the Fahrenheit temperature and Y is the number of times a cricket chirps in 1 minute. Is there a trend in the data? Are any of the data points outliers? Construct a divided difference table. Is smoothing with a low-order polynomial appropriate? If yes, choose an appropriate polynomial and fit using the least-squares criterion of best fit. Plot the fitted model against the data. X Y y = 3,703x - 130, Series1 Linear (Series1) The data looks linear. No outliers. X Y , , ,5 3 1,5 4, ,25 0, , , ,875-0,375 0, , , , , ,2631 0, , , , , , , , , ,01223 The second row of divided differences contains small values and also there is a change in sign (both positive and negative values, interspersed). Thus, a linear model is suitable.

3 3. Same as above, for the following data representing the population of the United States from 1790 to Year Observed population ,929, ,308, ,240, ,638, ,866, ,069, ,192, ,443, ,558, ,156, ,948, ,995, ,972, ,711, ,755, ,669, ,697, ,323, ,2, ,505, ,709, ,416,000 If we plot the data we notice no significant outlier. Furthermore, the data seems to follow an increasing polynomial trajectory, possibly quadratic Population Population

4 Year Population Ddiff 1 Ddiff 2 Ddiff 3 Ddiff 4 Ddiff 5 Ddiff , , , , ,9125-0, ,5 3, , , , , , , ,4667-0, , ,5 37,0125 1, , ,167-37,1167-1, , ,5 9, , , , , , , ,333-32,6792-0, , , , , , ,83-70,7417-2, , , ,6625 3, , ,1667-4, , ,17-57,5792 0, , , , , , ,64-82,02-19,3047-1, , , ,564 50, , , If we aim for divided differences that are small in absolute value, the appropropriate degree for the polynomial can be taken to be 4. Let us fit a 4-degree polynomial. f(x) = ax 4 + bx 3 + cx 2 + dx + e Minimize the sum of squares of absolute deviations: S = y i f(x i ) 2 = y i (ax 4 i + bx 3 i + cx 2 i + dx i + e) 2 The partial derivatives of all the coefficients a, b, c, d, e should be 0. a = 2 y i (ax 4 i + bx 3 i + cx 2 4 i + dx i + e) x i = 0 b = 2 y i (ax 4 i + bx 3 i + cx 2 3 i + dx i + e) x i = 0 c = 2 y i (ax 4 i + bx 3 i + cx 2 2 i + dx i + e) x i = 0 d = 2 y i (ax 4 i + bx 3 i + cx 2 i + dx i + e) x i = 0 e = 2 y i (ax 4 i + bx 3 i + cx 2 i + dx i + e) = 0

5 We get the system of equations: A b C d e free 3,77384E+27 1,97617E+24 1,03594E+ 5,43644E+17 2,85609E+14 3,01582E+22 1,97617E+24 1,03594E+ 5,43644E+17 2,85609E+14 1,503E+11 1,54442E+19 1,03594E+ 5,43644E+17 2,85609E+14 1,503E ,91291E+15 5,43644E+17 2,85609E+14 1,503E ,05622E+12 2,85609E+14 1,503E f(x) x x 3 + (15. 8E + 5)x 2 + (19. 9E + 8)x + (9. 4E + 11) Population y = 0,0745x 4-558,73x 3 + 2E+06x 2-2E+09x + 9E Population Poly. (Population) Alternatively, you may find this polynomial by using a spreadsheet and adding a polynomial trend curve to the plot. If you fit a quadratic polynomial (either as above or with a trend curve) you will notice that the two models are very similar. To understand why, notice that already the third divided differences have alternating signs and their values are relatively small (with respect to the original data). 4. In a town there are two supermarkets only, namely Welcome and Park. A marketing research indicated that a consumer of Welcome may switch to Park in his/her next shopping with a probability of α=0.3, while a consumer of Park may switch to Welcome in his/her next shopping with a probability of β=0.4. (i) Write a model describing the probability that a Welcome consumer will still be a Welcome consumer after n shopping.

6 w 0 =1, p 0 =0 w n =(1-α)w n-1 +βp n-1 ; p n =αw n-1 +(1-β)p n-1 An equivalent way of formulating it is (w n,p n )=(w n-1,p n-1 )P, where P = 1 α β α 1 β (ii) What is the market share of the two shops in the long run? Let w be the market share of Welcome at time t and p the market share of Park at time t. (1 α)w + βp = w αw + (1 β)p = p p + w = 1 w = β α + β = 0.57 p = α α + β = 0.43 (iii) If at n = 0 a consumer belongs to Park, what happens on his/her fourth shopping? 4 P = = (0 1)P 4 = ( ) With probability he becomes a Welcome consumer and with probability he stays a Park consumer. 5. Consider a gambler in a series of games. In each game he either loses 1 or he wins 1. He starts with M Euros and will continue to play until he reaches N Euros (N>M), or he loses all his sum of money. (i) Write a DTMC modelling his sum of money after each game. Let P ij be the probability of going from state i to state j. p if j = i + 1, 0 < i < N 1 p if j = i 1, 0 < i < N P ij = 1 if i = j {0, N} 0 otherwise The probability matrix is: p 0 p P = 1 p 0 p

7 (ii) What is the probability of having M+2 Euros after 2 games? How about after 3 games? After 2 games, the player can have M+2 EUR only if he wins in both games: P(M+2) = p*p = p 2. After 3 games, the only possible cases are: 3 wins (M+3); 2 wins one loss (M+1), 2 losses, one win (M-1) and 3 losses (M-3). Thus, the probability of having M+2 EUR after 3 games is 0. (iii) What is the probability of having M Euros after 4 games? In order to have M EUR after 4 games, the player should have won twice and lost twice (WWLL, WLWL, WLLW, LLWW, LWLW, LWWL). The probability is 6p 2 (1-p) 2. (iv) If M=5 and N=10, what is the probability of losing all money in at most 7 steps? How about in exactly 7 steps? To lose in at most 7 steps, the player should lose in 1, 2, 3, 4, 5, 6 or 7 steps. At each step he loses 1EUR, so he cannot lose all money in less than 5 steps. In 5 steps he loses all money with probability (1-p) 5. He cannot lose in 6 steps because he should win at least once, and then he has 1EUR left to play one more round. There are 5 ways of losing all money in exactly 7 steps: WLLLLLL, LWLLLLL, LLWLLLL, LLLWLLL, LLLLWLL, giving probability 5p(1-p) 6. Probability of losing in at most 7 steps is (1-p) 5 +5p(1-p) Two alternative designs are submitted for a landing module to enable the transport of astronauts to the surface of Mars. The mission is to land safely on Mars, collect several hundred pounds of samples from the planet s surface, and then return to the shuttle in its orbit around Mars. The alternative designs are displayed together with their reliabilities in the following two figures. Which design would you recommend to NASA? What assumptions are required? Are the assumptions reasonable? Alternative 1

8 Alternative 2 r 1 = 0.993*( *0.005)*( *(1-0.99*0.999))*0.98*(1-(1-0.99*0.99)*(1-0.99*0.99)) r 1 = 0.993* * *0.98* = r 2 = 0.998*( *0.005)*(1-0.01*0.001*0.002)*0.98*( ) r 2 = 0.998* * *0.98* = The reliability of the second design is higher than the reliability of the first design. Also, for landing and rockets the second alternative has more parallel options, so it is less likely to fail. The assumption encoded in the design is that the mission is successful whenever power, communication, landing, storage and rockets work properly (at least one of the corresponding components when more are connected in parallel, more complex relation for landing in first design). The less reasonable assumption is that the design has no impact on the parameters of the actual mission: the minimum number of rockets required for propulsion probably has an impact on the actual force that can be generated. Note that the second design has better power supply. Even if we replace it with one of the same quality (reliability 0.993) the second design is still better, but the difference is reduced: r 2 = Write a DTMC for the craps game, described in the following: a. Two dice are rolled b. If the outcome is 7 or 11: win c. If the outcome is 2,3, or 12: loss d. Any other outcome: keep rolling until either the outcome is the same as in the first roll, or the outcome is 7 i. Eventual outcome is 7: loss ii. Eventual outcome is the same as in the first roll: win We make the assumption that what matters is the sum of the two values on the dice (so, if a player has (2,2) in the first round and (3,1) in the next one, he wins). The following table lists all possible combinations from rolling two dice, grouped by their sum, and the resulting probability of each sum.

9 Sum Combinations Probability 2 {(1,1)} 1/36 3 {(1,2), (2,1)} 2/36 4 {(1,3), (2,2), (3,1)} 3/36 5 {(1,4), (2,3), (3,2), (4,1)} 4/36 6 {(1,5), (2,4), (3,3), (4,2), (5,1)} 5/36 7 {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} 6/36 8 {(2,6), (3,5), (4,4), (5,3), (6,2)} 5/36 9 {(3,6), (4,5), (5,4), (6,3)} 4/36 10 {(4,6), (5,5), (6,4)} 3/36 11 {(5,6), (6,5)} 2/36 12 {(6,6)} 1/36 Next, to construct the DTMC, we follow the rules of the game. Essentially each different sum corresponds to one state in the game, with the following exceptions: - Sums 7 and 11 are merged into the win state - Sums 2, 3 and 12 are merge into the loose state - The initial start state does not correspond to an actual sum The probabilities shown on the edges come from the probability of reaching the next state (= probability of corresponding sum, or sum of probabilities if the transition is to a state corresponding to more sums). The resulting model is shown below.