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Unit 4 Review — General Maths

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★ General Maths Unit 4 — End of Unit Review

Loans, Investments and Annuities • Graphs and Networks • Networks and Decision Mathematics

This review covers all three topics of Unit 4: compound interest and reducing-balance loans, annuities (future value and present value), and comparing financial options; graph theory, Euler and Hamiltonian paths, Dijkstra’s shortest path, and minimum spanning trees; and critical path analysis, network flow, the Hungarian assignment algorithm, and two-person zero-sum games. Allow approximately 90–120 minutes for this review.

Loans, Investments and Annuities

  1. Q1 — Compound interest recurrence

    Fluency

    A credit card charges interest at 1.5% per month on any unpaid balance. The current balance is $2500 and no repayments are made.

    (a) Write the recurrence relation for the balance after n months.

    (b) Calculate the balance after 3 months.

    (c) How much interest has accumulated after 3 months?

    (a) V0 = 2500; Vn+1 = 1.015 × Vn

    (b) V3 = 2500 × (1.015)3 = 2500 × 1.04568 = $2614.20

    (c) Interest = $2614.20 − $2500.00 = $114.20

  2. Q2 — Reducing balance loan

    Fluency

    A personal loan of $8000 charges 2% interest per month. Monthly repayments are $250.

    (a) Write the recurrence relation for the outstanding balance.

    (b) Calculate the balance after the first two repayments.

    (c) Is the loan balance reducing? Explain why or why not.

    (a) V0 = 8000; Vn+1 = 1.02 × Vn − 250

    (b)
    V1 = 1.02 × 8000 − 250 = 8160 − 250 = $7910.00
    V2 = 1.02 × 7910 − 250 = 8068.20 − 250 = $7818.20

    (c) Yes, the balance is reducing. Month 1 interest = $8000 × 0.02 = $160. The $250 repayment exceeds the $160 interest, so $90 reduces the principal each month. The balance will be fully paid off in time.

  3. Q3 — Loan repayment table

    Understanding

    A loan of $3000 is taken at 3% per month with monthly repayments of $500.

    (a) Complete a repayment table for the first 4 months (showing: Month, Opening Balance, Interest, Repayment, Closing Balance).

    (b) In which month is the loan fully repaid?

    (a) Repayment table:

    MonthOpeningInterest (3%)RepaymentClosing
    1$3000.00$90.00$500.00$2590.00
    2$2590.00$77.70$500.00$2167.70
    3$2167.70$65.03$500.00$1732.73
    4$1732.73$51.98$500.00$1284.71

    (b) Continuing:
    Month 5: 1284.71 × 1.03 = 1323.25 − 500 = $823.25
    Month 6: 823.25 × 1.03 = 847.95 − 500 = $347.95
    Month 7: 347.95 × 1.03 = 358.39 − 358.39 = $0 (final payment ≈ $358.39)
    The loan is fully repaid in Month 7.

  4. Q4 — Future value annuity

    Understanding

    Priya deposits $300 per month into a savings account earning 0.5% per month.

    (a) Write a recurrence relation for the balance after n months (starting from V0 = 0).

    (b) Calculate the balance after the first 3 months using the recurrence relation.

    (c) Use the future value formula to find the balance after 24 months.

    (a) V0 = 0; Vn+1 = 1.005 × Vn + 300

    (b)
    V1 = 1.005 × 0 + 300 = $300.00
    V2 = 1.005 × 300 + 300 = 301.50 + 300 = $601.50
    V3 = 1.005 × 601.50 + 300 = 604.51 + 300 = $904.51

    (c) FV = D × ((1+r)n−1) / r = 300 × ((1.005)24−1) / 0.005
    (1.005)24 = 1.12716
    FV = 300 × 0.12716 / 0.005 = 300 × 25.432 = $7629.60

  5. Q5 — Present value annuity (drawdown)

    Understanding

    At retirement, Lucas has $200 000 in a superannuation account earning 0.4% per month. He withdraws $1200 per month.

    (a) Write the recurrence relation.

    (b) Calculate the balance after the first 2 withdrawals.

    (c) Is the fund growing, stable, or reducing? Explain.

    (a) V0 = 200 000; Vn+1 = 1.004 × Vn − 1200

    (b)
    V1 = 1.004 × 200 000 − 1200 = 200 800 − 1200 = $199 600
    V2 = 1.004 × 199 600 − 1200 = 200 398.40 − 1200 = $199 198.40

    (c) Reducing. Monthly interest = $200 000 × 0.004 = $800. Since the $1200 withdrawal exceeds the $800 interest earned, $400 of principal is consumed each month. The fund will eventually be exhausted.

  6. Q6 — Comparing financial options

    Understanding

    Mia needs to borrow $15 000 over 3 years (36 months). Two options are available:

    • Option A (flat rate): 8% p.a. simple interest on the original principal.
    • Option B (reducing balance): 12% p.a. (1% per month). Monthly repayment = $498.

    (a) For Option A, find the total interest and monthly repayment.

    (b) For Option B, find the total repaid and total interest.

    (c) Which option costs Mia less in total interest? Explain why the lower stated rate (Option A) does not necessarily mean a lower cost.

    (a) Option A (flat rate):
    Interest = $15 000 × 0.08 × 3 = $3600
    Total repaid = $15 000 + $3600 = $18 600
    Monthly repayment = $18 600 ÷ 36 = $516.67/month

    (b) Option B (reducing balance):
    Total repaid = $498 × 36 = $17 928
    Total interest = $17 928 − $15 000 = $2928
    Monthly repayment = $498/month

    (c) Option B costs less ($2928 vs $3600 in interest), despite having a higher nominal rate (12% vs 8%). Flat rate interest is always charged on the full original principal — even as the debt reduces. Reducing balance interest is charged only on the outstanding balance, which falls each month, so much less interest accumulates overall.

  7. Q7 — Saving for a goal

    Problem Solving

    Amara wants to save $20 000 for a house deposit. She has $5000 already saved and plans to add $400 per month into an account earning 0.5% per month.

    (a) Write the recurrence relation (V0 = 5000).

    (b) After 12 months she has approximately $10 028. After 24 months she has approximately $15 362. After 36 months she has approximately $21 015. In which year does she reach her $20 000 goal?

    (c) If the interest rate drops to 0.3% per month (all else unchanged), write the new recurrence relation. Would she reach $20 000 faster or slower? Explain without calculating.

    (a) V0 = 5000; Vn+1 = 1.005 × Vn + 400

    (b) After 36 months (year 3): balance ≈ $21 015 > $20 000. After 24 months: balance ≈ $15 362 < $20 000. The goal is reached during Year 3 (between month 25 and month 36).

    (c) New recurrence: Vn+1 = 1.003 × Vn + 400. She would reach the goal slower. A lower interest rate means each month’s balance earns less, so the fund grows at a reduced pace. Since $400 monthly contributions remain unchanged, it will simply take more months to accumulate the same $20 000.

Graphs and Networks

  1. Q8 — Degree sequences and Handshaking Lemma

    Fluency

    A graph has vertices A, B, C, D, E, F with edges: AB, AC, AD, BC, BE, CF, DE, EF.

    (a) Find the degree of each vertex.

    (b) Verify the Handshaking Lemma.

    (c) How many vertices have odd degree? What does this imply for Euler trails and circuits?

    (a) deg(A)=3 (AB,AC,AD)  •  deg(B)=3 (AB,BC,BE)  •  deg(C)=3 (AC,BC,CF)
    deg(D)=2 (AD,DE)  •  deg(E)=3 (BE,DE,EF)  •  deg(F)=2 (CF,EF)

    (b) Sum of degrees = 3+3+3+2+3+2 = 16 = 2 × 8 edges ✓

    (c) Odd-degree vertices: A, B, C, E — there are 4 odd-degree vertices. This means no Euler trail and no Euler circuit exist (Euler trail requires exactly 2 odd-degree vertices; Euler circuit requires 0).

  2. Q9 — Euler and Hamiltonian paths

    Fluency

    A graph has 6 vertices with degree sequence: 2, 2, 3, 3, 4, 4.

    (a) Does an Euler circuit exist? An Euler trail?

    (b) Explain why we cannot determine from the degree sequence alone whether a Hamiltonian path exists.

    (a) Odd-degree vertices: the two vertices of degree 3 — exactly 2 odd-degree vertices.
    No Euler circuit (not all even). Euler trail exists, starting and ending at the two degree-3 vertices.

    (b) A Hamiltonian path visits every vertex exactly once (not every edge). Its existence depends on the specific connections between vertices (the graph’s structure), not just the degrees. Two graphs can have identical degree sequences but one may have a Hamiltonian path and the other not. There is no simple degree-based rule for Hamiltonian paths.

  3. Q10 — Dijkstra’s shortest path algorithm

    Understanding

    Find the shortest path from A to F in the weighted graph with edges: AB=4, AC=2, BC=3, BD=5, CD=4, CE=6, DE=2, DF=7, EF=3.

    Dijkstra’s algorithm:

    Start: A=0. Tentative: B→4, C→2.

    Visit C(2): update B→min(4, 2+3)=4 (no change), D→2+4=6, E→2+6=8.

    Visit B(4): update D→min(6, 4+5)=6 (no change).

    Visit D(6): update E→min(8, 6+2)=8 (no change), F→6+7=13.

    Visit E(8): update F→min(13, 8+3)=11.

    Visit F(11). Done.

    Shortest distance A to F = 11
    Trace back: F(11) ← E+3; E(8) ← D+2; D(6) ← C+4; C(2) ← A+2.
    Shortest path: A → C → D → E → F = 2+4+2+3 = 11

  4. Q11 — Minimum spanning tree

    Understanding

    Use Kruskal’s algorithm to find the minimum spanning tree for the graph with vertices A, B, C, D, E and edges: CE=2, AB=3, BC=4, AC=5, BD=6, CD=7, AD=8, DE=9.

    Kruskal’s algorithm (add edges in order of increasing weight, skipping any that create a cycle):

    CE=2 ✓ — adds {C,E}

    AB=3 ✓ — adds {A,B}

    BC=4 ✓ — connects {A,B} to {C,E}: component {A,B,C,E}

    AC=5 ✕ — A and C already connected (skip)

    BD=6 ✓ — connects D to {A,B,C,E}: component {A,B,C,D,E}

    All 5 vertices connected with 4 edges. Done.

    MST edges: CE=2, AB=3, BC=4, BD=6. Total weight = 15.

  5. Q12 — Road network planning

    Problem Solving

    A council needs to connect 5 towns (A–E) with sealed roads. Possible roads and costs (km): AB=8, AC=6, AD=10, AE=12, BC=4, BD=9, BE=7, CD=5, CE=11, DE=3.

    (a) Find the minimum spanning tree (minimum total road length to connect all towns) using Kruskal’s algorithm. Show all steps.

    (b) A direct road DE (3 km) is completed. State the total minimum network distance and list which roads are built.

    (a) Kruskal’s algorithm:
    Sorted edges: DE=3, BC=4, CD=5, AC=6, BE=7, AB=8, BD=9, AD=10, CE=11, AE=12

    DE=3 ✓ — {D,E}
    BC=4 ✓ — {B,C}
    CD=5 ✓ — connects {B,C} to {D,E}: {B,C,D,E}
    AC=6 ✓ — connects A to {B,C,D,E}: {A,B,C,D,E}. Done (4 edges, 5 vertices).

    MST: DE=3, BC=4, CD=5, AC=6. Total = 18 km.

    (b) DE is already in the MST. The minimum network is identical: roads built are DE (3 km), BC (4 km), CD (5 km), AC (6 km). Total minimum distance = 18 km.

Networks and Decision Mathematics

  1. Q13 — Critical path analysis

    Understanding

    A project has activities: A(3 days), B(5 days)→A, C(4 days)→A, D(6 days)→B and C, E(2 days)→D.

    (a) Perform a forward scan to find the EST of each activity and the minimum project duration.

    (b) Perform a backward scan to find the LST of each activity.

    (c) Find the float of each activity and identify the critical path.

    (a) Forward scan:
    EST(A)=0  •  EST(B)=3  •  EST(C)=3
    EST(D) = max(3+5, 3+4) = max(8, 7) = 8
    EST(E) = 8+6 = 14
    Minimum project duration = 14+2 = 16 days

    (b) Backward scan:
    LST(E) = 14  •  LST(D) = 14−6 = 8  •  LST(B) = 8−5 = 3  •  LST(C) = 8−4 = 4
    LST(A) = min(3, 4)−3 = 3−3 = 0

    (c) Floats: A=0−0=0 • B=3−3=0 • C=4−3=1 • D=8−8=0 • E=14−14=0
    Critical path: A → B → D → E (all float = 0), duration = 3+5+6+2 = 16 days ✓
    Activity C has 1 day of float — it can be delayed by 1 day without affecting the project.

  2. Q14 — Network flow and minimum cut

    Understanding

    A flow network has the following edge capacities:

    • S→A = 7, S→B = 5
    • A→C = 4, A→T = 3
    • B→C = 5
    • C→T = 9

    (a) Verify flow conservation: if flow on S→A = 7 (using all capacity), A→C = 4, A→T = 3, S→B = 5 (all), B→C = 5, C→T = 9, show that conservation holds at nodes A, B, and C.

    (b) What is the maximum flow into T?

    (c) Identify the minimum cut and verify it equals the maximum flow.

    (a) Flow conservation:
    Node A: in = S→A = 7; out = A→C + A→T = 4+3 = 7 ✓
    Node B: in = S→B = 5; out = B→C = 5 ✓
    Node C: in = A→C + B→C = 4+5 = 9; out = C→T = 9 ✓

    (b) Flow into T = A→T + C→T = 3 + 9 = 12 units (= flow out of S = 7+5 = 12 ✓)

    (c) Minimum cut: consider {S→A = 7, S→B = 5} (the source-side cut): capacity = 7+5 = 12 = maximum flow ✓

  3. Q15 — Hungarian algorithm

    Understanding

    Three workers (A, B, C) are to be assigned to three tasks (1, 2, 3). Costs ($):

    Task 1Task 2Task 3
    A937
    B582
    C461
    Apply the Hungarian algorithm to find the optimal assignment and minimum total cost.

    Row reduction (mins: A=3, B=2, C=1):
    A:[6,0,4]   B:[3,6,0]   C:[3,5,0]

    Column reduction (col mins: T1=3, T2=0, T3=0; subtract 3 from col 1):
    A:[3,0,4]   B:[0,6,0]   C:[0,5,0]

    Zeros: (A,T2), (B,T1), (B,T3), (C,T1), (C,T3)
    Independent zeros: A→T2, B→T1, C→T3 — all different rows and columns. Done.

    Optimal assignment: A→Task 2, B→Task 1, C→Task 3
    Costs from original: A→T2=3, B→T1=5, C→T3=1
    Minimum total cost = 3+5+1 = $9

  4. Q16 — Two-person zero-sum game

    Understanding

    A payoff matrix for a two-player game (payoffs to row player A) is:

    B1B2
    A141
    A225

    (a) Check for a saddle point.

    (b) If no saddle point exists, find the optimal mixed strategy for each player and the value of the game.

    (a) Row minima: A1=1, A2=2. Maximin=2 (A2). Column maxima: B1=4, B2=5. Minimax=4 (B1).
    2 ≠ 4 → No saddle point.

    (b) a=4, b=1, c=2, d=5. D = (4+5)−(1+2) = 9−3 = 6.
    Row player: p = (5−2)/6 = 3/6 = 1/2 (A plays A1 with prob 1/2, A2 with prob 1/2)
    Column player: q = (5−1)/6 = 4/6 = 2/3 (B plays B1 with prob 2/3, B2 with prob 1/3)
    Value: V = (4×5−1×2)/6 = (20−2)/6 = 18/6 = 3
    A gains 3 units on average per play.

  5. Q17 — Multi-concept applied problem

    Problem Solving

    A logistics company is planning a new delivery operation involving three decisions:

    (a) Project scheduling: The setup has activities: P(2 days), Q(4 days)→P, R(3 days)→P, S(5 days)→Q and R, T(1 day)→S. Find the critical path and total project duration. Which activity has the most float?

    (b) Route capacity: Goods flow from depot S to distribution centre T through a network: S→A=10, S→B=8, A→C=6, A→D=5, B→C=4, B→D=7, C→T=9, D→T=8. Find the maximum daily goods flow using augmenting paths.

    (c) Staff assignment: Three drivers (X, Y, Z) are assigned to three routes (1, 2, 3) with delivery times (hours): X:[3,5,4], Y:[2,7,3], Z:[6,4,5]. Find the assignment that minimises total delivery time.

    (a) Critical path:
    Forward: EST(P)=0, EST(Q)=2, EST(R)=2, EST(S)=max(2+4,2+3)=max(6,5)=6, EST(T)=6+5=11. Duration=11+1=12 days.
    Backward: LST(T)=11, LST(S)=6, LST(Q)=6−4=2, LST(R)=6−3=3, LST(P)=min(2,3)−2=0.
    Floats: P=0, Q=0, R=3−2=1, S=0, T=0.
    Critical path: P→Q→S→T (duration 2+4+5+1=12 days). Activity R has the most float (1 day).

    (b) Max flow:
    S→A→C→T: min(10,6,9)=6. Remaining: S→A=4, A→C=0, C→T=3.
    S→A→D→T: min(4,5,8)=4. Remaining: S→A=0, A→D=1, D→T=4.
    S→B→D→T: min(8,7,4)=4. Remaining: S→B=4, B→D=3, D→T=0.
    S→B→C→T: min(4,4,3)=3. Remaining: S→B=1, B→C=1, C→T=0.
    No more augmenting paths. Maximum flow = 6+4+4+3 = 17 units/day.
    Verify: min cut {C→T=9, D→T=8} = 17 ✓

    (c) Assignment:
    Row reduction (mins: X=3, Y=2, Z=4): X:[0,2,1] Y:[0,5,1] Z:[2,0,1].
    Column mins: C1=0, C2=0, C3=1. Subtract 1 from C3: X:[0,2,0] Y:[0,5,0] Z:[2,0,0].
    Zeros: (X,R1), (X,R3), (Y,R1), (Y,R3), (Z,R2), (Z,R3).
    Independent zeros: X→R1, Z→R2, Y→R3. All different rows and columns. Done.
    Optimal assignment: X→Route 1 (3 hrs), Y→Route 3 (3 hrs), Z→Route 2 (4 hrs).
    Minimum total delivery time = 3+3+4 = 10 hours.