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Edexcel A-level Further Maths revision notes

Dynamic programming

Section 4
Year 2
Year 2: this is content the exam board adds beyond the AS subject content, for the full A-level.
1 specification point

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 9FM0 section 4

Checked against Edexcel 9FM0 section 4. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

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In the exam: Formulae booklet provided · calculator allowed in every paper

Year 2
Year 2: this is content the exam board adds beyond the AS subject content, for the full A-level.

Principles of dynamic programming. Bellman's principle of optimality. Stage variables and state variables. Use of tabulation to solve maximum, minimum, minimax or maximin problems.

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Explanation

  • Dynamic programming solves a multi-stage decision problem by working backwards from the finish. Bellman's principle of optimality states that any part of an optimal path is itself optimal: if the best route from SS to TT passes through a vertex VV, then the part of it from VV to TT must be the best route from VV to TT.
  • That is what makes the backward calculation valid, because the best value from each vertex can be computed once and then reused. Two kinds of variable organise the work.
  • The stage variable says how far through the process a decision is being made, for example which column of the network or which month of a production plan. The state variable says which situation the process is in at that stage, for example which vertex of that column or how much stock is held.
  • The tabulation lists, for every stage and state, the possible actions, the value of each action and the best of them, and the optimal value is read from the final row. The same table solves four kinds of problem: maximum and minimum total, where actions are added to the value already found;
  • minimax, where the value of a route is its largest arc and the aim is to make that as small as possible; and maximin, where the value is its smallest arc and the aim is to make that as large as possible.
Worked example

A network has arcs SA=12SA=12, SB=9SB=9, AC=7AC=7, AD=15AD=15, BC=11BC=11, BD=6BD=6, CT=10CT=10, DT=8DT=8. Find the route of maximum total length from SS to TT.

  1. 1.Stage 1 (states CC, DD): the only choice is to go to TT, so the values are C:10C: 10 and D:8D: 8.
  2. 2.Stage 2, state AA: AC+10=17AC+10=17 or AD+8=23AD+8=23, so the best is 2323 using ADAD.
  3. 3.Stage 2, state BB: BC+10=21BC+10=21 or BD+8=14BD+8=14, so the best is 2121 using BCBC.
  4. 4.Stage 3, state SS: SA+23=35SA+23=35 or SB+21=30SB+21=30, so the best is 3535 using SASA.
  5. 5.Tracing the recorded choices forwards gives S→A→D→TS\to A\to D\to T.

Answer: The maximum route is S A D TS\,A\,D\,T of length 3535.

Common mistakes

  • Don't fall into the trap of working forwards and choosing the best arc at each step, which is a greedy method and can miss the optimum.
  • Don't fall into the trap of adding arc lengths in a minimax or maximin problem instead of taking a maximum or minimum.
  • Don't fall into the trap of recording only the values and not which action achieved them, so the optimal route cannot be traced back.

Exam tip

Set the table out with columns for stage, state, action, value and the best value; the mark scheme awards method marks for the table's structure as well as for the numbers.

Tier 1 · Easy

ORIGINAL

1.

State Bellman's principle of optimality, and explain what is meant by a stage variable and by a state variable in a dynamic programming problem.

(4)

(Total for Question 1 is 4 marks)

Tier 2 · Standard

ORIGINAL

1.

A network has arcs SA=4SA=4, SB=7SB=7, SC=5SC=5, AD=9AD=9, AE=3AE=3, BD=6BD=6, BE=8BE=8, CD=2CD=2, CE=11CE=11, DT=5DT=5, ET=4ET=4. Use dynamic programming to find the route of maximum total length from SS to TT, setting out your table by stage and state.

(6)

(Total for Question 1 is 6 marks)

Tier 3 · Hard

ORIGINAL

1.

A network has arcs SA=12SA=12, SB=9SB=9, AC=7AC=7, AD=15AD=15, BC=11BC=11, BD=6BD=6, CT=10CT=10, DT=8DT=8. Find both the minimax route and the maximin route from SS to TT, giving the value of each.

(8)

(Total for Question 1 is 8 marks)

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