QUESTION ONE Let 7C = Total Cost MC = Marginal Cost AC = Average Cost

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Rudolf Waters
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1 ANSWER QUESTION ONE Let 7C = Total Cost MC = Marginal Cost AC = Average Cost Q = Number of units AC = 7C MC = Q d7c d7c 7C Q Derivation of average cost with respect to quantity is different from marginal cost. Thus, the analysis is wrong. (b) (i) 7C = MC Q 8Q 11 7C = Q 3 Q 11Q f 7C = 14Q 11Q Q (ii) AR = Q 7R 7R = AR Q = 7R = Q 8Q 8QQ (iii) Profit = Total revenue Total cost Π = 7R 7C = Q 8Q - 1 Π = Q 6Q 11Q Q 14Q 11Q 1 (iv) For maximum or minimum, dπ point) dπ Q = -Q 1Q 11 Q 1Q + 11 = = 1 π Second order condition; When Q = 11, d π d Q = 11 or Q = 1 = -Q + 1 = - (11) + 1 = - 1 < = > (Max. turning

2 When Q = 1, d π = -(1) + 1 = 1 > = > (min. turning point) Level of output that maximizes profit in 11 units. (v) MR = d TR = 16Q QUESTION TWO (i) (ii) (iii) (iv) Transition matrix is one whose elements are probabilities that a process will change from one state to another state in a defined period of time. Initial probability vector is a vector that gives the initial (starting) probability of each state. When initial probability vector is multiplied by the transition matrix we get future or predicted probability vector. Equilibrium state in the longterm or steady state of a Markov process. Provided the assumptions of Markov process persist, the system finally reaches an equilibrium called steady or long term states i.e. a state where no further net changes occur. At equilibrium the following holds. (Equilibrium state vector)(transition matrix) = (Equilibrium state vector) An absorbing state is a state which cannot be left once it has been entered. (b) (i) Transition matrix, A B C D M A.95 B. C D (ii) Number of employees in each class after one year: A 6 B 9 C 15 D A 75 B 93 C D Number of employees in each class after two years:

3 A B 9.6 C D

4 Class Number of Employees A 9 B 93 C 18 D 189 Total 5 QUESTION THREE (a) Two events are said to be statistically independent if the occurrence of one event does not influence the occurrence of the other event. (b) Men that had minor accident =. x 5 = 1 Men that had safety instructions given they had minor accident =.3 x 1 = 3 Men that had no safety instructions given they had minor accident = 1 3 = 7 Overall No. of men that had safety instructions =. x 5 = 1 Overall No. of men that had no safety instructions =.8 x 5 = 4 SI NSI TOTAL A NA Total 1 4 Let A = Minor Accident NA = No minor accident SI = Had safety instructions NSI = Had no safety instructions 393 P = P NA SI 1 (i) NA NSI (ii).97 (c) (i) λ x λ.4.4 P(X = X) = P(X = ) = =.673 x!! (ii) P(x ) = P(x =, 1 or ) = P(x = ) + P(x = 1) + P(x = ) = + +! 1!!

5 QUESTION FOUR = =.99 (a) (i) Statistical hypothesis is a statement about a population which may be true or not true. (ii) A test of hypothesis is a decision rule based on a random sample of a population. The test help us to decide on whether to reject or fail to reject the null hypothesis. (i) Type I error is the error that is committed when we reject a true hypothesis. (ii) Type II error is the error that is committed when we accept a false hypothesis. (iii) Level of significance refers to the maximum allowance probability of type I error. (b) Is hypothesis H :µ 1 = µ H 1 :µ 1 µ Level of significance - Let α = 5%.5 n 1 +n = = 18 d.f t critical =.1 Decision rule: reject H if t calculated < -.1 or When t-calculated >.1 Combined variance of the two samples, S = n S1 n S n1 n S = X t calculated = 1 X S n1 n t calculated = 1.8 = = Since t calculated (1.8) < t critical (.1), we fail to reject H and conclude that the mean standard hours for employees in the factories is the same. (ii) Conditions of the test are:

6 - The two sampled population is normally distributed. - The two populations are statistically independent. - The standard deviations of the two populations are approximately equal. Interpretation of the outcome: - The two populations are the same. QUESTION FIVE (a) = a + b1 X y 1 + b X + b 3 X 3 = X1 +.75X y +.43X 3 S 1 =.19 S =.34 S 3 =.18 = 5% =.5 t =.9 d.f = 4 4 = Confidence interval = Point estimate t standard error (i) t 1 = b 1.1 = S % confidence interval = = < 1< (ii) t =.59 95% confidence interval.34 =.75 ±.9.34 =.75 ± < < (iii) t 3 = % confidence interval =.43 ± ± < 3 <.86

7 Assumptions made: - Sum of errors is zero - Errors are normally distributed - Errors are independent - Regression coefficients are small - Sample size is small and hence t-statistic is used. - Predictor variables are independent. The assumptions are reasonable because confidence interval estimate is symmetrical about the mean. (c) Regression coefficient.43 gives the strongest evidence of being statistically discernible because it has the biggest t ratio. (d) X 1 should be dropped because it is not statistically significant Its t ratio <.9 1is therefore not necessary. QUESTION SIX (a) (i) Feasible solution is one that satisfies all the constraints in the problem. (ii) Transportation problem is a special case of linear programming that is concerned with the transportation or allocation of goods from various sources to various destinations. The purpose of transportation problem is to schedule the conveyance of goods between sources and destinations in such a way that costs are minimized and contributions are maximized. (iii) Assignment problem is a special case of a transportation problem that is concerned with the determination of the most economical (optimal) way of allocating tasks to facilities. Assignment problem assumes that a task can only be assigned one facility and one facility to one task. The number of tasks must be equal to the number of facilities. (b) Introduce a dummy sales person E, to make numbers of rows equal to number of columns A B C D E

8 Reduce each row by the largest figure in the row and ignore the resulting negative sign because it is a maximization problem A B C D E Minimum No. of lines = 4 < 5 = 7 solution is not optimal A B C D E Minimum number of lines = 5 = No. of rows on columns Solution is optimal. Optimal solution Sales person District to be assigned Ratings A 1 9 B 3 96 C 5 93 D 94 E 4 Maximum total ratings QUESTION SEVEN (a) (i) (ii) Dominance is a situation in which a certain strategy is better than the other(s). In a game situation a strategy is said to dominate the other(s) if it is superior. Saddle point is one that has the smallest value in its rows and the largest value in its column. The solution or value of pure strategy game is given by the saddle point of the payoff matrix.

9 (iii) (iv) When a game has no saddle point, players result to mixed strategy game. A mixed strategy arises when a player decides to adopt one option part of the time and the other option(s) the rest of the time. Value of the game is the expected payoff to the winner when they play their best strategies. It can also be defined as the average payoff to a winner over a long series of plays. (b) (i) Player X Y 1 Y Y 3 Y 4 Min X X Max Max. min Min Max It is not possible to determine the value of the game using maximum and minimax because there is no saddle point. (ii) 1 + X = 1 X = 1 = X 1 Y 1 + Y + Y 3 + Y 4 = 1 1 y, Use B s pure strategies to get A s expected payoffs. 1. V 1 = X 1 + 4X = X 1 + 4(1-X 1 ) = 4 X 1 X 1 V 1 4. V = X 1 + 3X = X 1 + 3(1 X 1 ) = 3 X 1 X 1 3 V 3 3. V 3 = 3X 1 + X = 3X 1 + (1 X 1 ) = + X 1 X 1 1 V V 4 = - X 1 + 6X = - X 1 + 6(1-X 1 ) = 6 7X 1 X 1 6/7 1

10 V 4 6-1

11 Plot the four lines as a function of X ½ 1 R 3 Feasible region Maximum of the minimum occurs at X 1 = ½ This is the intersection of lines, 3 and 4. Optimal solution X 1 = ½, X = ½, V = ½ A to play strategy 1, 5% of his time and 5% of the time to play strategy ; after which he will win ½ points. QUESTION EIGHT (a) Attributes that makes beta distribution chosen in PERT analysis are: - It has the best fit to uncertain project completion times. - It is uni-modal - It has finite limits - It is very similar to normal distribution except that it has a slight skew. - It help us to adequately estimate expected project completion time and its variance. - It can help us to measure the uncertainty in the estimation. 1 B A E D F C 3 G H

12 (ii) Critical path is B E G - H Shortest project duration = 14 weeks. (iii) Time float : TF = LFT EST D TF (A) = 6 4 = TF (B) = 7 7 = TF (C) = = 7 TF (D) = = TF (E) = TF (F) = 11 7 = TF (G) = TF (H) = H 4 G 3 F 3 3 E 6 D 4 C B A 4

13 Time (wk) Workers Reschedule F by wks Reschedule C by 7 wks (iv) Comment - Minimum number of workers needed throughout the duration is 6 workers. - C and F are pushed to the end of their slack times to ensure that there is a smooth engagement of the resources.

56:171 Operations Research Final Exam December 12, 1994

56:171 Operations Research Final Exam December 12, 1994 Write your name on the first page, and initial the other pages. The response "NOTA " = "None of the above" Answer both parts A & B, and five sections