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1 OXFORD CAMBRIDGE AND RSA EXAMINATIONS Advanced Subsidiary General Certificate of Education Advanced General Certificate of Education MATHEMATICS 4737 Decision Mathematics MARK SCHEME Specimen Paper MAXIMUM MARK 7 This mark scheme consists of 5 printed pages and 3 blank pages. OCR 004 Registered Charity Number: [Turn over

2 1 (i) For attempt at the bipartite graph For correct graph B1 3 For the correct incomplete matching (ii) Alternating paths are: S C=P D and T A=L B=M F or S C=P B=M D and T A=L B=P D=M F For attempt at an alternating path or S C=P B=M F and T A=L B=P D For one correct path or T A=L B=M D and S C=P D=M F For the second path correcct or T A=L B=M F and S C=P D B1 4 For correct matching (iii) A B C D E F For appropriate zeros and ones (e.g.) to L correspond with minimum cost matching M For a correct table O P S T (i) Adding a dummy column gives Reducing: then B1 For a dummy column (equal entries, 6 ) For reducing rows For reducing columns Four lines are needed to cover zeros For covering zeros in the reduced matrix Augmenting: or For correct augmentation process Hence Adam Glasgow and Betty Swansea For these two allocations correct and either Clive Manchester, Dave London or Clive Manchester, Eleanor London or Dave London, Eleanor Manchester 7 For any one of the correct possibilities 0 0 (ii) Without Betty, reduced matrix is For new reduced matrix 0 1 Hence either Adam Glasgow, Clive Swansea, Dave London, Eleanor Manchester or Adam Glasgow, Clive Manchester, Dave Swansea, Eleanor London For either of the correct new possibilities

3 3 (i) 1 0 0 5 5 1 0 7 7 0 0 min(1, 5) = 5 0 7 1 min(7, 7) = 7 1 0 0 1 min(8, 7) = 7 min(10, ) = 0 min(, 5) = 5 min(14, ) = 0 min(, 7) = 7 1 min(7, ) = 7 min(8, ) = 8 8 For dealing with route min

Maximum number of crates is 8 B1 7 For correct number (ii) New maximin values are 15, 7,, 1,,, For appropriate re-calculation Hence new route is (0; 0) (1; 0) (; 0) (3; 0) For correct new route New

3 3 3 (i) min(1, 5) = min(7, 7) = min(8, 7) = 7 min(10, ) = 0 min(, 5) = 5 min(14, ) = 0 min(, 7) = 7 1 min(7, ) = 7 min(8, ) = 8 8 For dealing with route min column For at least 6 minima correct For dealing with maximin column For Stage 1 section of table all correct For completely correct table Route is (0; 0) (1; ) (; ) (3; 0) B1 For correct route Maximum number of crates is 8 B1 7 For correct number (ii) New maximin values are 15, 7,, 1,,, For appropriate re-calculation Hence new route is (0; 0) (1; 0) (; 0) (3; 0) For correct new route New maximum number of crates is 3 For correct number 10 4 (i) B1 For correct arcs and activities (activity on arc network or equivalent with activity at node) For correct process for forward pass For correct process for reverse pass For all early and late times correct Minimum completion time is 8 hours For correct minimum time stated Critical activities are A, B, C, F B1 For correct critical activities Start telephoning at am B1t 7 For stating the appropriate time of day (ii) For resource histogram with axes labelled For correct heights 3, 3, 5, 8, 8, 8, 5, 1 Maximum number of people needed is 8 B1 3 For correct number stated (iii) Time H A A A D D D 11 1 A A A D D D 1 1 B B D D D 1 C C C C C 3 C C C C C For substantially correct attempt 3 4 C C C C C For a correct schedule 4 5 C C C C C 5 6 E E E 6 7 F Start telephoning at am B1 3 For correct time stated 13 [Turn over

4 4 5 (i) Capacity is = 80 litres/sec For correct use of zero from AB For correct value 80 (ii) Maximum flow is 80 For relevant use of max flow/min cut So flow of = 300 is not possible For completely correct proof (iii) For correct method for excess and backflow For all initial excess capacities correct 3 For all initial backflows correct (iv) Augment by 70 along SBCDT (e.g.) B1 For identifying a correct augmentation New excesses and backflows are as shown above For modifying excesses and backflows Now augment by 50 along SACDT (e.g.) For continuing the process as far as possible Final excesses and backflows are as shown above 4 For a completely correct solution (v) B1t For showing the augmented flow correctly The value of the augmented flow is 80 litres/sec, For comparing with results from (i) or (ii) and so is the maximum possible 3 For a completely correct explanation (Or equivalent explanation based on the disconnectedness of S and T in (iii)) 14

5 5 6 (i) X Y Z row min A B 4 3 C col max Play-safe for Rose is B B1 For correct statement Play-safe for computer is Y B1 For correct statement Not stable as 3+ 0 B1 3 For use of max row min and min col max (ii) Row C is dominated by row B B1 For correct statement or explanation X Y Z A B1 For new matrix B 4 3 (iii) Expected pay-off with X is 1 a+ 4(1 a) = 4 3a B1 For showing given answer correctly and with Y is 3a+ 3(1 a) = 3 B1 For correct value 3 and with Z is 4a+ (1 a) = + a B1 For correct expression + a B1t For correct diagram Required value of a is 0.4 B1t For correct value Score of 1 or : play A; score of 3, 4 or 5: play B; Throw die again if it shows a six B1t 6 For a correct decision rule (iv) X Y Z A B1 For correct pay-off matrix for the computer B 4 3 Add 4 to each value B1 For a correct explanation (v) z = 0.6 B1 For the correct value of z p= 1. P =.8 B1 For the correct values of both p and P The computer should choose X with probability 0.4 and Z with probability 0.6 B1 For a correct description of the strategy On average the computer will lose no more than.8 points per game B1 4 For a correct interpretation of P 17

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physicsandmathstutor.com ADVANCED GCE MATHEMATICS 4737 Decision Mathematics 2 Candidates answer on the Answer Booklet OCR Supplied Materials: 8 page Answer Booklet Graph paper Insert for Questions 5 and 6 (inserted) List of Formulae