Mark Scheme (Results) June GCE Decision D2 (6690) Paper 1

Transcription

1 Mark (Results) June 20 GCE Decision D2 (6690) Paper

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3 EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is The Edexcel Mathematics mark schemes use the following types of marks: M marks: method marks are awarded for knowing a method and attempting to apply it, unless otherwise indicated. A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned. B marks are unconditional accuracy marks (independent of M marks) should not be subdivided. 3. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes and can be used if you are using the annotation facility on epen. bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark

4 June 20 Decision Mathematics D Mark Question. (a) A B C D E F A B C D E F B3, 2,, 0 (3) A C B E F D A M A = 07 A (3) (c) Delete A 9 M B E RMST weight = 6 Lower bound = = 9 km A M A (4) 0

5 (a)b 2B 3B M A 2A (c)m A 2M 2A Notes: One double entry correct Two double entries correct Three double entries correct NN route, each letter appearing once, condone lack of return vertex CAO CAO Finding my RMST accept 6 for both marks Either or 6 seen Adding on two least arcs, accept and 9 or AC and AB 9 CAO 2. (a) Adds a column of four zeros and 0. B () (c) Shadow costs A B C D 0 x M -9 2 x x - 0 A x x 6 M x x A A B C D 20 θ θ θ 7 θ θ 2 θ M Entering cell: C Exiting cell: 3C Aft A B C D DM A (4) (4) 9 2

6 (a) B: cao Notes: (c) M: Finding all 8 shadow costs A: cao 2M: Finding missing four improvement indices no extra zeros 2A: cao M: A valid route, their most negative II chosen, only one empty square used, θ s balance. Aft: consistent; their entering and exiting cells stated clearly 2DM: An improved solution 7 entries only, (so must now be using one of my negative indices as the entering cell). Must ft from their valid route. 2A: cao 3. (a) P 7x+ z+ 4s = 320 M A b.v x y z r s t Value r y t P (2) b.v. x y z r s t Value Row ops. 3 r R + 2 R 3 2M 2Aft y R 2 2 R 3 x R 3 2 M 2A P R 4 + 7R 3 3A (c) P = 376 x = 8 y = z = 0 r = 4 s = 0 t = 0 M Aft A (3) 0 (5) 3

7 (a) (c) Notes: M: One equal sign, P and 320 present A: cao M: correct pivot located, attempt to divide row. If choosing negative pivot M0M0 in A: pivot row correct including change of b.v. 2M: (ft) Correct row operations used at least once or stated correctly. 2Aft: Looking at non zero-and-one columns, one column ft correct 3A: cao. M: At least 4 values stated. Reading off bottom row, or negative values get M0. Aft: Their four basic variables correct ft from their table. 2A: cao 4

8 4. (a) S plays S plays 2 S plays 3 L plays -4 - L plays L plays Row 3 dominates row so row may be deleted. S plays S plays 2 S plays 3 L plays M L plays Let Laura play 2 with probability p and 3 with probability (- p) If Sam plays : Laura s gain is 3p 3(- p) = 3 + 6p If Sam plays 2: Laura s gain is p + 0 (- p) = p If Sam plays 3: Laura s gain is 2p + 2 (- p) = 2 4p M A (3) B2,ft,0 (2) (c) 3+ 6p = p M 7p = 3 A 3 p = 7 Laura should play row : never, row 2: 3 of the time and 7 Aft row 3: 4 of the time 7 and the value of the game is 3 to her. A 7 (4) 9 5

9 (a) (c) Notes: M: Matrix reduced correctly. Could be implicit from equations. 2M: Setting up three probability equations, implicit definition of p. A: CAO Bft: At least two lines correct, accept p> or p<0 here. Must both be function of p. 2B: 3 lines cao, 0 p, scale clear (or line = ), condone lack of labels. Rulers used. 3M: Finding their correct optimal point, must have three lines, and setting up an equation to find 0 p. A: CAO 2Aft: All three options listed must ft from their p, check page, no negatives. 3A: CAO 6

10 5. (a) a = b = 5 c = 3 Flow = 49 B, B B, B (4) M A (c) e.g. SBEHT - 7 together with M A either SBEHDAFGT 2 or SBCEHDAFGT - 2 A2,,0 (d) 58 B (e) e.g. (2) (4) () M A (2) (f) Max flow = min cut Cut through HT, HG, GF, FT Value 58 M A (2) 5 7

11 (a) B: a = cao 2B: b = 5 cao 3B: c = 3 cao 4B: 49 cao Notes: (c) (d) M: Two numbers on each arc A: cao M: One valid flow augmenting route found and value stated. A: Flow increased by at least 2 2A: A second correct flow 3A: Flow increased by 9 and no more B: cao (e) M: Consistent flow pattern > 5 A: cao (f) M: Must have attempted (e), S to T, and made an attempt at a cut. A: cut correct may be drawn. Refer to max flow-min cut theorem three words out of four. 8

12 6. (a) Task A Task B Task C Worker P Worker Q Worker R Let x ij Where ij j if worker does the task = 0 otherwise x indicates worker i being assigned to task j, i { P, Q, R} { A, B, C}, B B Minimise 27x + 3x + 25x + 26x + 30x + 34x + 35x + 29x + 32x B B PA PB PC QA QB QC RA RB RC Subject to: xpa + xpb + xpc = x + x + x = QA QB QC xra + xrb + xrc = x + x + x = PA QA RA x + x + x = PB QB RB x + x + x = PC QC RC M A A (7) Since we need to maximise first subtract all entries from some n 4 M Task A Task B Task C Worker P Worker Q 9 5 A Worker R (2) 9 9

13 (a) Notes: B: defining variables 2B: defining variables 3B: minimise 4B: cao M: At least 3 equations, coefficients of. Accept inequalities here should be precisely 9 variables. A: cao 3 equations correct accept slack variables if defined 2A: cao 6 equations correct accept slack variables if defined 7. M: subtracting from some n 4 condone up to two errors A: correct (a) Stage State Action Dest. Value 0 H H- London London 36 5 = 3* I I London London 38 4 = 34* F FH H = 54 FI I = 56* G GH H = 53 M A GI I = 55* (2) 2 C CF F = 92* 2M 2A CG G = 92* D DF F = 9 DG G = 93* E EF F = 9* 3A EG G = 90 (3) 3 A AC C = 09 3M 4Aft AD D = * AE E = * B BC C = 05 BD D = 06* 5Aft BE E = 05 (3) 4 London London A A -5 + = 06* 4M 6Aft London B B = 03 Optimal expected income is Aft (3) Optimal schedules are: London A D G I London (or v.v.) Bft London A E F I London (or v.v) B (2) 3 0

14 (a) Notes: Throughout section (a): Condone lack of destination column and/or reversed stage numbers throughout. Only penalise incorrect result in Value ie ignore working values. Penalise absence of state or action column with first two A marks earned only Penalise empty/errors in stage column with first A mark earned only. M: First stage completed. A: CAO Penalise * errors only once in the question on the first occurrence 2M: Second stage completed. Penalise reversed states here and at end. Bod if something in each cell. 2A: Any 2 states correct. (Penalise * errors only once in the question). 3A: All 3 states correct. (Penalise * errors only once in the question). 3M: 3 rd stage completed. Bod if something in each cell. 4Aft: A or B state correct. (Penalise * errors only once in the question). 5Aft: A and B states correct. (Penalise * errors only once in the question). 4M: 4 th stage completed. Bod if something in each cell. 6Aft: Final, state correct. (Penalise * errors only once in the question). 7Aft: CAO Bft: route correct, consistent with their working penalise reversed states again here. Condone absence of London 2B: both routes cao. London to London.

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