CELE Engineering Mathematics — Engineering EconomyExam Answer Templates

Marks

1

Topic

Time Value of Money

Difficulty

easy

Template Id

T1

Examiner Tip

Examiners look for TWO ideas in one sentence: (1) present is worth more and (2) because it can earn interest. Both must appear to earn the mark.

Model Answer

The time value of money is the principle that a sum of money available today is worth more than the same sum available in the future, because money available today can earn interest over time.

Question Type

very_short_answer

Answer Structure

• One concise sentence: state that present money is worth more than future money due to its earning capacity [1 mark]

Scoring Breakdown

Common Mark Deductions

• Vague answer such as 'money has value' without mentioning time or interest
• Confusing time value with inflation without mentioning interest earning

Key Phrases To Include

• worth more today
• earn interest
• earning capacity
• future equivalent

Marks

1

Topic

Nominal vs Effective Interest Rate

Difficulty

easy

Template Id

T2

Examiner Tip

The '− 1' is the most commonly dropped part. Write the full formula on one line and underline or circle the '− 1' to remind yourself it must be there.

Model Answer

i_eff = (1 + r/m)^m − 1 where r = nominal annual rate, m = number of compounding periods per year.

Question Type

very_short_answer

Answer Structure

• Write the formula: i_eff = (1 + r/m)^m − 1 [1 mark]
• Define variables (optional but good practice at 1-mark level)

Scoring Breakdown

Common Mark Deductions

• Writing (1 + r)^m instead of (1 + r/m)^m — dividing by m is essential
• Forgetting the '− 1' at the end
• Using i_eff = r/m (this gives only the per-period rate, not the effective annual rate)

Key Phrases To Include

• i_eff
• (1 + r/m)^m
• minus 1
• nominal rate
• compounding periods

Marks

2

Topic

Simple and Compound Interest

Difficulty

easy

Template Id

T3

Examiner Tip

Write the keyword 'principal only' for simple interest and 'interest on interest' for compound interest — these phrases signal conceptual understanding and are specifically rewarded.

Model Answer

Simple interest is computed only on the original principal, so the interest earned each period is constant. Formula: F = P(1 + in) Compound interest is computed on the accumulated balance (principal plus previously earned interest), so interest earns interest each period. Formula: F = P(1 + i)^n where P = present worth, i = interest rate per period, n = number of periods, F = future worth.

Question Type

short_answer

Answer Structure

• Define simple interest and state its formula: F = P(1 + in) [1 mark]
• Define compound interest and state its formula: F = P(1+i)^n [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing F = P + Pin (equivalent but missing the factored form examiners prefer)
• Using 'n' as years without clarifying it is periods (loses precision)
• Swapping the formulas between simple and compound

Key Phrases To Include

• principal only
• F = P(1 + in)
• accumulated balance
• interest on interest
• F = P(1+i)^n

Marks

2

Topic

Compound Interest

Difficulty

easy

Template Id

T4

Examiner Tip

Write (1.12)^8 = 2.4760 explicitly before multiplying. This shows you computed the compound factor and earns the substitution mark even if your final multiplication is slightly off.

Model Answer

Given: P = ₱50,000 i = 12% per year = 0.12 n = 8 years Required: Future worth F Solution: F = P(1 + i)^n F = 50,000(1 + 0.12)^8 F = 50,000(1.12)^8 F = 50,000 × 2.47596 F = ₱123,798 Answer: The future worth is ₱123,798.

Question Type

numerical

Answer Structure

• Given block — list P, i, n with units [0.5 mark]
• Write correct formula: F = P(1+i)^n [0.5 mark]
• Correct substitution and computation of (1.12)^8 [0.5 mark]
• Correct final answer with peso sign [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using simple interest formula F = P(1+in) instead of compound — most common error
• Not converting percentage: using i = 12 instead of i = 0.12
• Omitting the peso sign in the final answer

Key Phrases To Include

• F = P(1+i)^n
• (1.12)^8 = 2.4760
• ₱123,798

Marks

2

Topic

Nominal vs Effective Interest Rate

Difficulty

medium

Template Id

T5

Examiner Tip

Always write the intermediate step r/m = 0.18/12 = 0.015. This step alone earns a mark and prevents you from using the wrong per-period rate.

Model Answer

Given: Nominal rate r = 18% per year = 0.18 Compounding frequency m = 12 (monthly) Required: Effective annual rate i_eff Solution: i_eff = (1 + r/m)^m − 1 i_eff = (1 + 0.18/12)^12 − 1 i_eff = (1 + 0.015)^12 − 1 i_eff = (1.015)^12 − 1 i_eff = 1.19562 − 1 i_eff = 0.19562 Answer: The effective annual interest rate is 19.56%.

Question Type

numerical

Answer Structure

• Given: r = 0.18, m = 12 [0.5 mark]
• Correct formula: i_eff = (1 + r/m)^m − 1 [0.5 mark]
• Substitution: (1 + 0.015)^12 [0.5 mark]
• Final answer: 19.56% [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Reporting i_eff = 1.5% (the monthly rate, not the effective annual rate)
• Writing (1 + 0.18)^12 without dividing by m
• Forgetting to subtract 1 at the end
• Leaving the answer as a decimal (0.1956) without converting to percentage

Key Phrases To Include

• i_eff = (1 + r/m)^m − 1
• r/m = 0.015
• m = 12
• 19.56%

Marks

3

Topic

Present Worth — Compound Interest

Difficulty

easy

Template Id

T6

Examiner Tip

Write both forms: P = F(1+i)^(−n) and P = F/(1+i)^n — this demonstrates full understanding and often earns the formula mark twice in partial-credit schemes.

Model Answer

Given: F = ₱50,000 i = 12% per year = 0.12 n = 8 years Required: Present worth P Solution: Using the present worth formula: P = F(1 + i)^(−n) P = 50,000(1 + 0.12)^(−8) P = 50,000(1.12)^(−8) Compute (1.12)^8 = 2.47596 Therefore (1.12)^(−8) = 1/2.47596 = 0.40388 P = 50,000 × 0.40388 P = ₱20,194 Answer: The present worth is ₱20,194. Verification (optional): F = 20,194(1.12)^8 = 20,194 × 2.47596 ≈ ₱50,000 ✓

Question Type

numerical

Answer Structure

• Given block: F, i, n correctly identified [0.5 mark]
• Correct formula stated: P = F(1+i)^(−n) [1 mark]
• Correct compound factor computed: (1.12)^(−8) = 0.40388 [0.5 mark]
• Correct multiplication and boxed final answer with unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using F = P(1+i)^n but forgetting to isolate P — solving for F instead
• Confusing the present-worth factor with the compound-amount factor
• Arithmetic error in computing (1.12)^8 — use logarithms or sequential multiplication

Key Phrases To Include

• P = F(1+i)^(−n)
• present worth factor
• (1.12)^8 = 2.47596
• ₱20,194

Marks

3

Topic

Annuities — Future Worth and Present Worth

Difficulty

medium

Template Id

T7

Examiner Tip

Note that P = F(1+i)^(−n) is a useful cross-check: P = 18,340 × (1.08)^(−6) = 18,340/1.58687 ≈ ₱11,557. Writing this verification line earns goodwill from examiners and confirms your answer.

Model Answer

Given: A = ₱2,500 per year (end-of-year, ordinary annuity) i = 8% per year = 0.08 n = 6 years Required: (a) Future worth F; (b) Present worth P Solution: (a) Future Worth: F = A × [(1+i)^n − 1] / i F = 2,500 × [(1.08)^6 − 1] / 0.08 (1.08)^6 = 1.58687 (1.08)^6 − 1 = 0.58687 0.58687 / 0.08 = 7.3359 ← Future Worth Factor (F/A) F = 2,500 × 7.3359 F = ₱18,340 (b) Present Worth: P = A × [(1+i)^n − 1] / [i(1+i)^n] P = 2,500 × [0.58687] / [0.08 × 1.58687] P = 2,500 × 0.58687 / 0.12695 P = 2,500 × 4.6229 P = ₱11,557 Answer: (a) F = ₱18,340 (b) P = ₱11,557

Question Type

numerical

Answer Structure

• Given block with A, i, n [0.5 mark]
• Correct future-worth annuity formula and computation [1 mark]
• Correct present-worth annuity formula and computation [1 mark]
• Both answers correctly boxed with units [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using F = A(1+i)^n instead of the annuity series formula — treating it as a single lump sum
• Forgetting the (1+i)^n denominator in the present-worth formula
• Applying annuity-due (beginning-of-period) adjustment when not required
• Using n = 5 instead of n = 6 by miscounting periods

Key Phrases To Include

• ordinary annuity
• end-of-period
• F = A[(1+i)^n−1]/i
• P = A[(1+i)^n−1]/[i(1+i)^n]
• (1.08)^6 = 1.58687

Marks

2

Topic

Perpetuity

Difficulty

easy

Template Id

T8

Examiner Tip

The word 'indefinitely,' 'forever,' or 'in perpetuity' in a problem is your trigger to use P = A/i. Mention the word 'perpetuity' in your answer — it signals to the examiner that you recognized the problem type correctly.

Model Answer

Given: A = ₱15,000 per year (perpetual/infinite series) i = 10% per year = 0.10 Required: Present worth P of a perpetuity Solution: For a perpetuity (infinite annuity): P = A / i P = 15,000 / 0.10 P = ₱150,000 Answer: A deposit of ₱150,000 must be made today to fund the scholarship indefinitely.

Question Type

numerical

Answer Structure

• Identify as a perpetuity (infinite series) [0.5 mark]
• State formula P = A/i [0.5 mark]
• Correct substitution and answer [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using the finite annuity formula with a very large n instead of P = A/i
• Dividing by 10 instead of 0.10 — giving ₱1,500 (100× error)
• Not recognizing 'indefinitely' as the keyword for perpetuity

Key Phrases To Include

• perpetuity
• infinite series
• P = A/i
• ₱150,000

Marks

3

Topic

Straight-Line Depreciation

Difficulty

easy

Template Id

T9

Examiner Tip

After computing d, verify: d × n + S = C → 20,000 × 9 + 20,000 = 180,000 + 20,000 = 200,000 ✓. This is a fast, one-line verification that proves your depreciation is correct — write it explicitly.

Model Answer

Given: First cost C = ₱200,000 Salvage value S = ₱20,000 Useful life n = 9 years Required: (a) Annual depreciation d; (b) Book value BV₄ Solution: (a) Straight-line annual depreciation: d = (C − S) / n d = (200,000 − 20,000) / 9 d = 180,000 / 9 d = ₱20,000 per year (b) Book value at end of year 4: BV₄ = C − (d × t) BV₄ = 200,000 − (20,000 × 4) BV₄ = 200,000 − 80,000 BV₄ = ₱120,000 Answer: (a) Annual depreciation = ₱20,000/year (b) Book value after 4 years = ₱120,000

Question Type

numerical

Answer Structure

• Given block: C, S, n [0.5 mark]
• Correct SL formula and annual depreciation d = ₱20,000 [1 mark]
• Correct book value formula BV_t = C − dt and BV₄ = ₱120,000 [1 mark]
• Correct units and boxed answers [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using C alone without subtracting salvage S: d = 200,000/9 — the most common SL error
• Using BV₄ = C − 4d correctly but with wrong d because of previous error (error carried forward — examiner may still award the BV mark if method is correct)
• Confusing the book value at END of year 4 with the beginning of year 4 (off by one period)

Key Phrases To Include

• d = (C − S)/n
• straight-line method
• BV_t = C − dt
• ₱20,000 per year
• ₱120,000

Marks

3

Topic

Sum-of-Years-Digits Depreciation

Difficulty

medium

Template Id

T10

Examiner Tip

Note that SYD accelerates depreciation — year 1 always has the LARGEST charge. As a quick check, the year-1 SYD fraction must be the largest (8/36 here). If your year-1 fraction is smaller than year-2's, you have the formula inverted.

Model Answer

Given: C = ₱250,000 S = ₱25,000 n = 8 years Depreciable amount = C − S = 250,000 − 25,000 = ₱225,000 Required: SYD depreciation for year 1 (d₁) and year 2 (d₂) Solution: Sum of years digits: SYD = n(n+1)/2 = 8(9)/2 = 36 SYD depreciation formula: dₜ = [(n − t + 1)/SYD] × (C − S) Year 1 (t = 1): d₁ = [(8 − 1 + 1)/36] × 225,000 d₁ = [8/36] × 225,000 d₁ = 0.22222 × 225,000 d₁ = ₱50,000 Year 2 (t = 2): d₂ = [(8 − 2 + 1)/36] × 225,000 d₂ = [7/36] × 225,000 d₂ = 0.19444 × 225,000 d₂ = ₱43,750 Answer: Year 1 depreciation = ₱50,000 Year 2 depreciation = ₱43,750

Question Type

numerical

Answer Structure

• Compute depreciable amount C − S = ₱225,000 [0.5 mark]
• Compute SYD = n(n+1)/2 = 36 [0.5 mark]
• Apply dₜ = [(n−t+1)/SYD] × (C−S) for year 1: ₱50,000 [1 mark]
• Apply formula for year 2: ₱43,750 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using t from the END of life rather than from the beginning: using (t/SYD) instead of [(n−t+1)/SYD]
• Forgetting to subtract salvage: using C instead of (C−S) = 225,000
• Computing SYD as 1+2+...+8 = 36 correctly but then reversing the fraction direction

Key Phrases To Include

• SYD = n(n+1)/2
• dₜ = [(n−t+1)/SYD] × (C−S)
• SYD = 36
• ₱50,000
• ₱43,750
• accelerated depreciation

Marks

3

Topic

Annuities — Period Conversion

Difficulty

medium

Template Id

T11

Examiner Tip

Period consistency is the single most tested concept in board-exam annuity problems. Write the conversion step explicitly: 'Convert to monthly: i = 6%/12 = 0.5%; n = 3 × 12 = 36.' This earns a mark even before you touch the formula.

Model Answer

Given: A = ₱2,000 per month Nominal rate r = 6% per year compounded monthly Per-period rate i = 6%/12 = 0.5% per month = 0.005 n = 3 years × 12 months/year = 36 periods Required: Future worth F Solution: F = A × [(1+i)^n − 1] / i F = 2,000 × [(1.005)^36 − 1] / 0.005 Compute (1.005)^36: (1.005)^36 = 1.19668 (1.005)^36 − 1 = 0.19668 0.19668 / 0.005 = 39.336 ← Series compound-amount factor F = 2,000 × 39.336 F = ₱78,672 Answer: The future worth of the monthly deposits is ₱78,672.

Question Type

numerical

Answer Structure

• Convert to per-period rate: i = 0.5%/month = 0.005 [0.5 mark]
• Convert to total periods: n = 36 [0.5 mark]
• State and apply correct annuity formula [1 mark]
• Correct computation and final answer [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using i = 6% annual and n = 3 years without converting to monthly — most common and costliest error
• Using n = 3 (years) with i = 0.5% (monthly) — mixed periods
• Using the present-worth formula instead of future-worth formula

Key Phrases To Include

• i = 6%/12 = 0.5% per month
• n = 36 periods
• F = A[(1+i)^n−1]/i
• (1.005)^36 = 1.19668
• ₱78,672

Marks

5

Topic

Economic Comparison — Present Worth Method

Difficulty

hard

Template Id

T12

Examiner Tip

Always state the LCM in the first line of your solution. Board examiners immediately check whether you recognized the need for a common study period. Draw a simple cash-flow timeline below your Given block — mark the capital outlays with downward arrows and the AOC with downward arrows. This visual earns no extra mark but prevents you from missing a repeat purchase.

Model Answer

Given: Machine A: P_A = ₱180,000; AOC_A = ₱25,000/yr; life = 6 years; S = 0 Machine B: P_B = ₱120,000; AOC_B = ₱35,000/yr; life = 4 years; S = 0 i = 10% per year Required: Compare total present worth over LCM period; recommend the better machine. Solution Step 1 — Determine LCM study period: LCM(6, 4) = 12 years Machine A: 2 cycles (purchases at year 0 and year 6) Machine B: 3 cycles (purchases at year 0, year 4, year 8) Solution Step 2 — Present Worth of Machine A over 12 years: PW_A = 180,000 + 180,000(P/F, 10%, 6) + 25,000(P/A, 10%, 12) (P/F, 10%, 6) = (1.10)^(−6) = 0.56447 (P/A, 10%, 12) = [(1.10)^12 − 1] / [0.10 × (1.10)^12] = [3.13843 − 1] / [0.10 × 3.13843] = 2.13843 / 0.31384 = 6.8137 PW_A = 180,000 + 180,000(0.56447) + 25,000(6.8137) PW_A = 180,000 + 101,605 + 170,343 PW_A = ₱451,948 Solution Step 3 — Present Worth of Machine B over 12 years: PW_B = 120,000 + 120,000(P/F,10%,4) + 120,000(P/F,10%,8) + 35,000(P/A,10%,12) (P/F, 10%, 4) = (1.10)^(−4) = 0.68301 (P/F, 10%, 8) = (1.10)^(−8) = 0.46651 PW_B = 120,000 + 120,000(0.68301) + 120,000(0.46651) + 35,000(6.8137) PW_B = 120,000 + 81,961 + 55,981 + 238,480 PW_B = ₱496,422 Solution Step 4 — Decision: PW_A = ₱451,948 < PW_B = ₱496,422 Answer: Machine A is more economical because it has the lower total present worth of costs (₱451,948 vs ₱496,422) over the 12-year LCM study period.

Question Type

numerical

Answer Structure

• Identify LCM study period = 12 years [0.5 mark]
• Identify repeat purchase years for each machine [0.5 mark]
• Compute (P/F) and (P/A) factors correctly [1 mark]
• Correct PW_A = ₱451,948 [1 mark]
• Correct PW_B = ₱496,422 [1 mark]
• Correct decision statement with comparison [1 mark]

Scoring Breakdown

Common Mark Deductions

• Comparing over different time horizons (6 years vs 4 years) without LCM — invalid comparison
• Forgetting the repeat capital cost at year 6 for Machine A or years 4 and 8 for Machine B
• Adding the annual operating costs as simple sums instead of discounting using (P/A)
• Selecting the machine with lower first cost (₱120,000) without completing the present worth analysis
• No decision statement at the end — a solution without a conclusion loses the decision mark

Key Phrases To Include

• LCM study period
• repeatability assumption
• LCM(6,4) = 12 years
• PW_A < PW_B
• select Machine A
• (P/F, i, n)
• (P/A, i, n)

Marks

2

Topic

Benefit-Cost Analysis

Difficulty

easy

Template Id

T13

Examiner Tip

Always end with an explicit decision: 'Since B/C = ___ > 1.0, the project IS justified' or '< 1.0, the project is NOT justified.' The comparison to 1.0 is the decision criterion and must appear in your answer.

Model Answer

Given: PW of Benefits = ₱800,000 PW of Costs = ₱550,000 Required: Benefit-cost ratio (B/C) and justification decision Solution: B/C = PW of Benefits / PW of Costs B/C = 800,000 / 550,000 B/C = 1.455 Decision: Since B/C = 1.455 > 1.0, the project is economically justified — benefits exceed costs. Answer: B/C ratio = 1.455; the project is economically justified.

Question Type

numerical

Answer Structure

• State formula: B/C = PW Benefits / PW Costs [0.5 mark]
• Compute B/C = 1.455 [0.5 mark]
• Decision: B/C > 1 → justified [1 mark]

Scoring Breakdown

Common Mark Deductions

• Inverting the ratio: C/B instead of B/C
• Stating B/C > 1 is acceptable without identifying the computed value
• Failing to give a clear decision statement

Key Phrases To Include

• B/C = PW Benefits / PW Costs
• B/C = 1.455
• B/C > 1
• economically justified

Marks

3

Topic

Break-Even Analysis

Difficulty

medium

Template Id

T14

Examiner Tip

Always verify by substituting Q back: TC_X = 80,000 + 12(10,000) = ₱200,000 and TC_Y = 120,000 + 8(10,000) = ₱200,000. If they match, your break-even is correct. Write this one line — it demonstrates rigor and is expected at board-exam level.

Model Answer

Given: Method X: FC_X = ₱80,000/yr, VC_X = ₱12/unit Method Y: FC_Y = ₱120,000/yr, VC_Y = ₱8/unit Required: Break-even volume Q_BE; preferred method above and below Q_BE Solution: At break-even, total cost of Method X = total cost of Method Y: TC_X = TC_Y FC_X + VC_X × Q = FC_Y + VC_Y × Q 80,000 + 12Q = 120,000 + 8Q 12Q − 8Q = 120,000 − 80,000 4Q = 40,000 Q = 10,000 units/year Decision: Below Q_BE (Q < 10,000 units): Method X has lower total cost (lower fixed cost dominates) Above Q_BE (Q > 10,000 units): Method Y has lower total cost (lower variable cost dominates) At Q = 10,000 units: both methods have equal cost of: TC = 80,000 + 12(10,000) = ₱200,000 ✓ Answer: Break-even volume = 10,000 units/year. Method X is preferred for volumes below 10,000 units; Method Y is preferred above 10,000 units.

Question Type

numerical

Answer Structure

• Set up cost equations for both methods [0.5 mark]
• Equate TC_X = TC_Y and solve algebraically for Q [1 mark]
• Q_BE = 10,000 units [0.5 mark]
• Correct preference statements above and below Q_BE [1 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting fixed costs and only equating VC_X × Q = VC_Y × Q
• Computing Q correctly but reversing the preference decisions (saying Y is preferred at low volumes)
• Not verifying the answer by substituting Q back into both cost equations

Key Phrases To Include

• break-even volume
• TC_X = TC_Y
• FC + VC × Q
• 10,000 units
• lower fixed cost
• lower variable cost

Marks

5

Topic

Net Present Worth — Investment Decision

Difficulty

hard

Template Id

T15

Examiner Tip

A complete 5-mark NPW answer MUST contain: (1) the factor formula, (2) the computed factor value, (3) the NPW calculation, and (4) the decision statement. Missing the decision statement is the most common way a candidate loses the last mark on an otherwise perfect computation.

Model Answer

Given: Initial investment P₀ = ₱100,000 (cost, cash outflow at t = 0) Annual return A = ₱30,000/year (cash inflow at end of each year) n = 5 years MARR = 15% per year = 0.15 Required: Net Present Worth (NPW); accept/reject decision Solution Step 1 — Present Worth of Benefits (inflows): PW_benefits = A × (P/A, 15%, 5) (P/A, 15%, 5) = [(1.15)^5 − 1] / [0.15 × (1.15)^5] (1.15)^5 = 2.01136 (1.15)^5 − 1 = 1.01136 0.15 × 2.01136 = 0.30170 (P/A, 15%, 5) = 1.01136 / 0.30170 = 3.3522 PW_benefits = 30,000 × 3.3522 = ₱100,566 Solution Step 2 — Net Present Worth: NPW = PW_benefits − Initial Investment NPW = 100,566 − 100,000 NPW = ₱+566 Solution Step 3 — Decision: Since NPW = +₱566 > 0, the investment yields a return EXCEEDING the MARR of 15%. Answer: NPW = +₱566 > 0; the investment is ACCEPTABLE. (The project barely exceeds the MARR — the actual IRR is slightly above 15%.) Note: If NPW = 0, the investment earns exactly the MARR. If NPW < 0, reject.

Question Type

numerical

Answer Structure

• Given block with all four values clearly listed [0.5 mark]
• Identify and compute (P/A, 15%, 5) factor [1 mark]
• Compute PW of benefits = ₱100,566 [1 mark]
• Compute NPW = PW_benefits − initial cost [1 mark]
• Correct decision: NPW > 0 → accept [1 mark]
• Brief note on interpretation of NPW sign [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Simply summing the undiscounted cash flows: 5 × 30,000 = 150,000 − 100,000 = 50,000 — ignoring time value entirely
• Using the wrong factor: (F/A) instead of (P/A) for the annuity present worth
• Computing NPW but not providing a clear accept/reject decision with criterion
• Using i = 15 instead of i = 0.15 in the formula

Key Phrases To Include

• NPW = PW(benefits) − PW(costs)
• MARR
• (P/A, 15%, 5) = 3.3522
• NPW > 0
• acceptable
• exceeds MARR