Dynamic programming Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Dynamic programming questions. See exactly how to solve problems on dynamic-programming, bellman, stage-state-action, minimisation.

dynamic-programmingbellmanstage-state-actionminimisationmaximisationmaximin-minimax
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
A route must be chosen through a staged network so as to keep the total cost as low as possible. The states are arranged in stages as follows: stage 00 contains AA; stage 11 contains BB, CC, DD; stage 22 contains EE, FF; stage 33 contains GG. Each arc joins a state to a state in the next stage, and the arc values are AB=7AB=7, AC=33AC=33, AD=3AD=3, BE=46BE=46, BF=27BF=27, CE=37CE=37, CF=24CF=24, DE=41DE=41, DF=26DF=26, EG=18EG=18, FG=16FG=16. Using Bellman's principle of optimality and working backwards through the stages, find the minimum total cost of a route from AA to GG.

Worked solution

  1. Identify the stages, states and arcs of the network

    stages 0 to 3,  source A,  sink G\text{stages }0\text{ to }3,\;\text{source }A,\;\text{sink }G

    Dynamic programming labels each state by the stage it belongs to; a route uses exactly one arc per stage from the source to the sink.

  2. Work backwards from the final stage, giving each state its optimal value

    g(A)=45,  g(B)=43,  g(C)=40,  g(D)=42,  g(E)=18,  g(F)=16g(A)=45,\;g(B)=43,\;g(C)=40,\;g(D)=42,\;g(E)=18,\;g(F)=16

    Starting at the stage just before the sink and moving back one stage at a time, each state is given its best value using the values already found at the next stage.

  3. Trace the optimal route forwards from the source

    ADFG(3+26+16=45)A\to D\to F\to G\quad\left(3+26+16=45\right)

    Following the optimal choice recorded at each state gives the route, whose arc values add up to the optimal total.

  4. State the minimum total cost

    minimum total cost=45\text{minimum total cost}=45

    The least possible total cost of a route from AA to GG is 4545.

Answer
4545
Question 2
2 markseasy
A piece of equipment is to be kept in service over several stages. In this staged network each arc value is the net cost, in suitable units, of a keep-or-replace decision. The states are arranged in stages as follows: stage 00 contains AA; stage 11 contains BB, CC, DD; stage 22 contains EE, FF; stage 33 contains GG. Each arc joins a state to a state in the next stage, and the arc values are AB=19AB=19, AC=12AC=12, AD=32AD=32, BE=43BE=43, BF=14BF=14, CE=29CE=29, CF=31CF=31, DE=24DE=24, DF=38DF=38, EG=11EG=11, FG=37FG=37. Using Bellman's principle of optimality and working backwards through the stages, find the minimum total cost of a route from AA to GG.

Worked solution

  1. Identify the stages, states and arcs of the network

    stages 0 to 3,  source A,  sink G\text{stages }0\text{ to }3,\;\text{source }A,\;\text{sink }G

    Dynamic programming labels each state by the stage it belongs to; a route uses exactly one arc per stage from the source to the sink.

  2. Work backwards from the final stage, giving each state its optimal value

    g(A)=52,  g(B)=51,  g(C)=40,  g(D)=35,  g(E)=11,  g(F)=37g(A)=52,\;g(B)=51,\;g(C)=40,\;g(D)=35,\;g(E)=11,\;g(F)=37

    Starting at the stage just before the sink and moving back one stage at a time, each state is given its best value using the values already found at the next stage.

  3. Trace the optimal route forwards from the source

    ACEG(12+29+11=52)A\to C\to E\to G\quad\left(12+29+11=52\right)

    Following the optimal choice recorded at each state gives the route, whose arc values add up to the optimal total.

  4. State the minimum total cost

    minimum total cost=52\text{minimum total cost}=52

    The least possible total cost of a route from AA to GG is 5252.

Answer
5252
Question 3
2 markseasy
A factory schedules its production over several stages. Each arc value is the cost, in suitable units, of meeting demand by the corresponding production decision. The states are arranged in stages as follows: stage 00 contains AA; stage 11 contains BB, CC, DD; stage 22 contains EE, FF; stage 33 contains GG. Each arc joins a state to a state in the next stage, and the arc values are AB=29AB=29, AC=21AC=21, AD=7AD=7, BE=46BE=46, BF=16BF=16, CE=48CE=48, CF=12CF=12, DE=18DE=18, DF=40DF=40, EG=11EG=11, FG=5FG=5. Using Bellman's principle of optimality and working backwards through the stages, find the minimum total cost of a route from AA to GG.

Worked solution

  1. Identify the stages, states and arcs of the network

    stages 0 to 3,  source A,  sink G\text{stages }0\text{ to }3,\;\text{source }A,\;\text{sink }G

    Dynamic programming labels each state by the stage it belongs to; a route uses exactly one arc per stage from the source to the sink.

  2. Work backwards from the final stage, giving each state its optimal value

    g(A)=36,  g(B)=21,  g(C)=17,  g(D)=29,  g(E)=11,  g(F)=5g(A)=36,\;g(B)=21,\;g(C)=17,\;g(D)=29,\;g(E)=11,\;g(F)=5

    Starting at the stage just before the sink and moving back one stage at a time, each state is given its best value using the values already found at the next stage.

  3. Trace the optimal route forwards from the source

    ADEG(7+18+11=36)A\to D\to E\to G\quad\left(7+18+11=36\right)

    Following the optimal choice recorded at each state gives the route, whose arc values add up to the optimal total.

  4. State the minimum total cost

    minimum total cost=36\text{minimum total cost}=36

    The least possible total cost of a route from AA to GG is 3636.

Answer
3636
Question 4
2 markseasy
Stock is to be managed over several stages. Each arc value is the combined ordering and holding cost of the corresponding decision. The states are arranged in stages as follows: stage 00 contains AA; stage 11 contains BB, CC, DD; stage 22 contains EE, FF; stage 33 contains GG. Each arc joins a state to a state in the next stage, and the arc values are AB=25AB=25, AC=43AC=43, AD=41AD=41, BE=13BE=13, BF=3BF=3, CE=5CE=5, CF=45CF=45, DE=34DE=34, DF=12DF=12, EG=4EG=4, FG=46FG=46. Using Bellman's principle of optimality and working backwards through the stages, find the minimum total cost of a route from AA to GG.

Worked solution

  1. Identify the stages, states and arcs of the network

    stages 0 to 3,  source A,  sink G\text{stages }0\text{ to }3,\;\text{source }A,\;\text{sink }G

    Dynamic programming labels each state by the stage it belongs to; a route uses exactly one arc per stage from the source to the sink.

  2. Work backwards from the final stage, giving each state its optimal value

    g(A)=42,  g(B)=17,  g(C)=9,  g(D)=38,  g(E)=4,  g(F)=46g(A)=42,\;g(B)=17,\;g(C)=9,\;g(D)=38,\;g(E)=4,\;g(F)=46

    Starting at the stage just before the sink and moving back one stage at a time, each state is given its best value using the values already found at the next stage.

  3. Trace the optimal route forwards from the source

    ABEG(25+13+4=42)A\to B\to E\to G\quad\left(25+13+4=42\right)

    Following the optimal choice recorded at each state gives the route, whose arc values add up to the optimal total.

  4. State the minimum total cost

    minimum total cost=42\text{minimum total cost}=42

    The least possible total cost of a route from AA to GG is 4242.

Answer
4242
Question 5
2 markseasy
A project is to be staffed over several stages. Each arc value is the cost, in suitable units, of the corresponding hiring-and-firing decision. The states are arranged in stages as follows: stage 00 contains AA; stage 11 contains BB, CC, DD; stage 22 contains EE, FF; stage 33 contains GG. Each arc joins a state to a state in the next stage, and the arc values are AB=4AB=4, AC=23AC=23, AD=2AD=2, BE=21BE=21, BF=38BF=38, CE=13CE=13, CF=15CF=15, DE=41DE=41, DF=9DF=9, EG=35EG=35, FG=46FG=46. Using Bellman's principle of optimality and working backwards through the stages, find the minimum total cost of a route from AA to GG.

Worked solution

  1. Identify the stages, states and arcs of the network

    stages 0 to 3,  source A,  sink G\text{stages }0\text{ to }3,\;\text{source }A,\;\text{sink }G

    Dynamic programming labels each state by the stage it belongs to; a route uses exactly one arc per stage from the source to the sink.

  2. Work backwards from the final stage, giving each state its optimal value

    g(A)=57,  g(B)=56,  g(C)=48,  g(D)=55,  g(E)=35,  g(F)=46g(A)=57,\;g(B)=56,\;g(C)=48,\;g(D)=55,\;g(E)=35,\;g(F)=46

    Starting at the stage just before the sink and moving back one stage at a time, each state is given its best value using the values already found at the next stage.

  3. Trace the optimal route forwards from the source

    ADFG(2+9+46=57)A\to D\to F\to G\quad\left(2+9+46=57\right)

    Following the optimal choice recorded at each state gives the route, whose arc values add up to the optimal total.

  4. State the minimum total cost

    minimum total cost=57\text{minimum total cost}=57

    The least possible total cost of a route from AA to GG is 5757.

Answer
5757

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