OCR D2 (Decision Mathematics 2) 2011 January

1 Four friends, Amir (A), Bex (B), Cerys (C) and Duncan (D), are visiting a bird sanctuary. They have decided that they each will sponsor a different bird. The sanctuary is looking for sponsors for a kite \(( K )\), a lark \(( L )\), a moorhen \(( M )\), a nightjar \(( N )\), and an owl \(( O )\). Amir wants to sponsor the kite, the nightjar or the owl; Bex wants to sponsor the lark, the moorhen or the owl; Cerys wants to sponsor the kite, the lark or the owl; and Duncan wants to sponsor either the lark or the owl.

1. Draw a bipartite graph to show which friend wants to sponsor which birds. Amir chooses to sponsor the kite and Bex chooses the lark. Cerys then chooses the owl and Duncan is left with no bird that he wants.
2. Write down the shortest possible alternating path starting from the nightjar, and hence write down one way in which all four friends could have chosen birds that they wanted to sponsor.
3. List a way in which all four friends could have chosen birds they wanted to sponsor, with the owl not being chosen.

2 Amir, Bex, Cerys and Duncan all have birthdays in January. To save money they have decided that they will each buy a present for just one of the others, so that each person buys one present and receives one present. Four slips of paper with their names on are put into a hat and each person chooses one of them. They do not tell the others whose name they have chosen and, fortunately, nobody chooses their own name. The table shows the cost, in \(\pounds\), of the present that each person would buy for each of the others. As it happens, the names are chosen in such a way that the total cost of the presents is minimised.
Assign the cost \(\pounds 25\) to each of the missing entries in the table and then apply the Hungarian algorithm, reducing rows first, to find which name each person chose.

3 The table lists the duration, immediate predecessors and number of workers required for each activity in a project.

1. Represent the project by an activity network, using activity on arc. You should make your diagram quite large so that there is room for working.
2. Carry out a forward pass and a backward pass through the activity network, showing the early event times and late event times clearly at the vertices of your network. State the minimum project completion time and list the critical activities.
3. Draw a resource histogram to show the number of workers required each hour when each activity begins at its earliest possible start time.
4. Show how it is possible for the project to be completed in the minimum project completion time when only six workers are available.

4 Answer parts (v) and (vi) of this question on the insert provided. The diagram represents a system of pipes through which fluid can flow. The weights on the arcs show the lower and upper capacities of the pipes in litres per second.
\includegraphics[max width=\textwidth, alt={}, center]{33995efa-7ede-4e83-89d3-7b6c8be8d955-4_703_789_479_678}

1. Which vertex is the source and which vertex is the sink?
2. Cut \(\alpha\) partitions the vertices into the sets \(\{ A , B , C \} , \{ D , E , F , G , H , I \}\). Calculate the capacity of cut \(\alpha\).
3. Explain why partitioning the vertices into sets \(\{ A , D , G \} , \{ B , C , E , F , H , I \}\) does not give a cut.
4. (a) How many litres per second must flow along arc \(D G\) ?
   (b) Explain why the arc \(A D\) must be at its upper capacity. Hence find the flow in \(\operatorname { arc } B A\).
   (c) Explain why at least 7 litres per second must flow along arc \(B C\).
5. Use the diagrams in the insert to show a minimum feasible flow and a maximum feasible flow. The upper capacity of \(B C\) is now increased from 8 to 18 .
6. (a) Use the diagram in the insert to show a flow of 19 litres per second.
   (b) List the saturated arcs when 19 litres per second flows through the network. Hence, or otherwise, find a cut of capacity 19 .
7. Explain how your answers to part (vi) show that 19 litres per second is the maximum flow.

5 A card game between two players consists of several rounds. In each round the players both choose a card from those in their hand; they then show these cards to each other and exchange tokens. The number of tokens that the second player gives to the first player depends on the colour of the first player's card and the design on the second player's card. The table shows the number of tokens that the first player receives for each combination of colour and design. A negative entry means that the first player gives tokens to the second, zero means that no tokens are exchanged. Let the stages be \(0,1,2,3,4,5\). Stage 0 represents arriving at the sanctuary entrance. Stage 1 represents visiting the first bird, stage 2 the second bird, and so on, with stage 5 representing leaving the sanctuary. Let the states be \(0,1,2,3,4\) representing the entrance/exit, kite, lark, moorhen and nightjar respectively.

1. Calculate how many minutes it takes to travel the route $$( 0 ; 0 ) - ( 1 ; 1 ) - ( 2 ; 2 ) - ( 3 ; 3 ) - ( 4 ; 4 ) - ( 5 ; 0 ) .$$ The friends then realise that if they try to find the quickest route using dynamic programming with this (stage; state) formulation, they will get the route \(( 0 ; 0 ) - ( 1 ; 1 ) - ( 2 ; 2 ) - ( 3 ; 3 ) - ( 4 ; 1 ) - ( 5 ; 0 )\), or this in reverse, taking 27 minutes.
2. Explain why the route \(( 0 ; 0 ) - ( 1 ; 1 ) - ( 2 ; 2 ) - ( 3 ; 3 ) - ( 4 ; 1 ) - ( 5 ; 0 )\) is not a solution to the friends' problem. Instead, the friends set up a dynamic programming tabulation with stages and states as described above, except that now the states also show, in brackets, any birds that have already been visited. So, for example, state \(1 ( 234 )\) means that they are currently visiting the kite and have already visited the other three birds in some order. The partially completed dynamic programming tabulation is shown opposite.
3. For the last completed row, i.e. stage 2, state 1(3), action 4(13), explain where the value 18 and the value 6 in the working column come from.
4. Complete the table in the insert and hence find the order in which the birds should be visited to give a quickest route and find the corresponding minimum journey time.

   Stage	State	Action	Working	Suboptimal minimum
   \multirow{4}{*}{4}	1(234)	0	10	10
   	2(134)	0	14	14
   	3(124)	0	12	12
   	4(123)	0	17	17
   \multirow{12}{*}{3}	1(23)	4(123)	\(17 + 6 = 23\)	23
   	1(24)	3(124)	\(12 + 2 = 14\)	14
   	1(34)	2(134)	\(14 + 3 = 17\)	17
   	2(13)	4(123)	\(17 + 4 = 21\)	21
   	2(14)	3(124)	\(12 + 2 = 14\)	14
   	2(34)	1(234)	\(10 + 3 = 13\)	13
   	3(12)	4(123)	\(17 + 3 = 20\)	20
   	3(14)	2(134)	\(14 + 2 = 16\)	16
   	3(24)	1(234)	\(10 + 2 = 12\)	12
   	4(12)	3(124)	\(12 + 3 = 15\)	15
   	4(13)	2(134)	\(14 + 4 = 18\)	18
   	4(23)	1(234)	\(10 + 6 = 16\)	16
   \multirow{12}{*}{2}	1(2)	3(12) 4(12)	\(20 + 2 = 22\)	21
   	1(3)	2(13) 4(13)	\(21 + 3 = 24 18 + 6 = 24\)	24
   	1(4)
   	2(1)
   	2(3)
   	2(4)
   	3(1)
   	3(2)
   	3(4)
   	4(1)
   	4(2)
   	4(3)
   \multirow{4}{*}{1}	1
   	2
   	3
   	4
   0	0	1 2 3 4

6 Answer this question on the insert provided. Four friends have decided to sponsor four birds at a bird sanctuary. They want to construct a route through the bird sanctuary, starting and ending at the entrance/exit, that enables them to visit the four birds in the shortest possible time. The table below shows the times, in minutes, that it takes to get between the different birds and the entrance/exit. The friends will spend the same amount of time with each bird, so this does not need to be included in the calculation. Let the stages be \(0,1,2,3,4,5\). Stage 0 represents arriving at the sanctuary entrance. Stage 1 represents visiting the first bird, stage 2 the second bird, and so on, with stage 5 representing leaving the sanctuary. Let the states be \(0,1,2,3,4\) representing the entrance/exit, kite, lark, moorhen and nightjar respectively.

1. Calculate how many minutes it takes to travel the route $$( 0 ; 0 ) - ( 1 ; 1 ) - ( 2 ; 2 ) - ( 3 ; 3 ) - ( 4 ; 4 ) - ( 5 ; 0 ) .$$ The friends then realise that if they try to find the quickest route using dynamic programming with this (stage; state) formulation, they will get the route \(( 0 ; 0 ) - ( 1 ; 1 ) - ( 2 ; 2 ) - ( 3 ; 3 ) - ( 4 ; 1 ) - ( 5 ; 0 )\), or this in reverse, taking 27 minutes.
2. Explain why the route \(( 0 ; 0 ) - ( 1 ; 1 ) - ( 2 ; 2 ) - ( 3 ; 3 ) - ( 4 ; 1 ) - ( 5 ; 0 )\) is not a solution to the friends' problem. Instead, the friends set up a dynamic programming tabulation with stages and states as described above, except that now the states also show, in brackets, any birds that have already been visited. So, for example, state \(1 ( 234 )\) means that they are currently visiting the kite and have already visited the other three birds in some order. The partially completed dynamic programming tabulation is shown opposite.
3. For the last completed row, i.e. stage 2, state 1(3), action 4(13), explain where the value 18 and the value 6 in the working column come from.
4. Complete the table in the insert and hence find the order in which the birds should be visited to give a quickest route and find the corresponding minimum journey time.

   Stage	State	Action	Working	Suboptimal minimum
   \multirow{4}{*}{4}	1(234)	0	10	10
   	2(134)	0	14	14
   	3(124)	0	12	12
   	4(123)	0	17	17
   \multirow{12}{*}{3}	1(23)	4(123)	\(17 + 6 = 23\)	23
   	1(24)	3(124)	\(12 + 2 = 14\)	14
   	1(34)	2(134)	\(14 + 3 = 17\)	17
   	2(13)	4(123)	\(17 + 4 = 21\)	21
   	2(14)	3(124)	\(12 + 2 = 14\)	14
   	2(34)	1(234)	\(10 + 3 = 13\)	13
   	3(12)	4(123)	\(17 + 3 = 20\)	20
   	3(14)	2(134)	\(14 + 2 = 16\)	16
   	3(24)	1(234)	\(10 + 2 = 12\)	12
   	4(12)	3(124)	\(12 + 3 = 15\)	15
   	4(13)	2(134)	\(14 + 4 = 18\)	18
   	4(23)	1(234)	\(10 + 6 = 16\)	16
   \multirow{12}{*}{2}	1(2)	3(12) 4(12)	\(20 + 2 = 22\)	21
   	1(3)	2(13) 4(13)	\(21 + 3 = 24 18 + 6 = 24\)	24
   	1(4)
   	2(1)
   	2(3)
   	2(4)
   	3(1)
   	3(2)
   	3(4)
   	4(1)
   	4(2)
   	4(3)
   \multirow{4}{*}{1}	1
   	2
   	3
   	4
   0	0	1 2 3 4