--- title: "Deposit now to draw growing monthly amount" subject: "Engineering Economics" chapter: 2 key: "semester_6/engineering_economics:deposit-now-growing-withdrawal" source: https://www.bctengineer.com/q/semester_6/engineering_economics/deposit-now-growing-withdrawal site: BCT Engineering --- # Deposit now to draw growing monthly amount Engineering Economics · Chapter 2: Interest and Time Value of Money - Asked in: 2076 Shrawan, 3 marks - Video: [How To Calculate The Present Value of an Annuity](https://www.youtube.com/watch?v=RU-osjAs6hE) — The Organic Chemistry Tutor - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/engineering_economics/deposit-now-growing-withdrawal - Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-2 ## Question How many rupees should you deposit now so that you will be able to draw Rs 5000 at the end of this month which increases by 2 percent per month for 10 years. Bank interest rate is 12% per year. ## Short answer With nominal 12% compounded monthly, i=1%. Use geometric gradient present worth: P = 5,000 × [1 - (1.02/1.01)^120] / (0.01 - 0.02) = Rs 11,30,894. ## Answer #### Data and assumption First withdrawal A₁ = Rs 5,000 at the end of month 1, growing by g = 2% per month for n = 120 months. The interest of 12% per year is taken as nominal, compounded monthly: i = 1% per month. #### Formula (geometric gradient present worth) For i ≠ g: P = A₁ [1 − ((1 + g)/(1 + i))ⁿ] / (i − g) ```text A1=5000 5100 5202 ... (x1.02 each month) ^ ^ ^ ^ --+-------+-----+--- ... ---+--> month 0 1 2 3 120 P = ? ``` #### Calculation (1 + g)/(1 + i) = 1.02/1.01 = 1.009901 ((1.02)/(1.01))¹²⁰ = 3.26179 i − g = 0.01 − 0.02 = −0.01 P = 5,000 × (1 − 3.26179) / (−0.01) = 5,000 × 226.179 = **Rs 1,130,894** **Answer: Deposit now about Rs 11.31 lakh.** (If 12% were an effective annual rate, i would be 0.9489% per month and the answer would be a little larger.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which formula applies to withdrawals growing 2% per month? A. P = A(P/A,i,n) B. Geometric gradient P = A1[1 - ((1+g)/(1+i))^n]/(i - g) C. P = G(P/G,i,n) D. P = A/i Answer: B. Percent growth each period is a geometric series. (Key point: Formula type) 2. What monthly rate is used? A. 12% B. 2% C. 1% (12% nominal compounded monthly) D. 0.9489% Answer: C. i = 12/12 = 1%, with n = 120. (Key point: Monthly rate) 3. Since g (2%) exceeds i (1%), what happens in the formula? A. The formula fails B. P is negative C. The ratio is below 1 D. i - g is negative and the ratio^n exceeds 1, so the result is still positive Answer: D. Both numerator and denominator are negative: 5,000 x 226.18 = 1,130,894. (Key point: Case i < g) ## Related in Engineering Economics - [Define equivalent worth and ROR; deposit to draw monthly](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aequivalent-worth-ror-draw-monthly-10000) — Engineering Economics, chapter 2 - [Monthly deposits with quarterly compounding, future amount](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Adeposit-quarterly-monthly-five-years) — Engineering Economics, chapter 2 - [Capitalized worth and infinite-period deposit](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Acapitalized-worth-monthly-yearly-four) — Engineering Economics, chapter 2 - [Mutual fund from growing salary, 15 years](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Amutual-fund-salary-growth-15yr) — Engineering Economics, chapter 2 - [Compounding semi-annually and monthly, 10% nominal](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Adeposit-10000-semiannual-monthly) — Engineering Economics, chapter 2 - [Car loan interest rate and lump sum at year 3](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Acar-loan-installments-lump-sum-year3) — Engineering Economics, chapter 2 ## Related in other subjects - [Best installed capacity at 10% interest](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Abest-installed-capacity-10pct) — Hydro Power, chapter 3 - [Capitalised value from Rs 20,000 monthly rent](https://www.bctengineer.com/question.md?key=semester_8%2Fconstruction_management%3Avalue-building-20000-rent-2078) — Construction Management, chapter 13 - [Symbol rate, entropy and information rate (6 symbols per 10 μs)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Asymbol-rate-six-symbols-10us) — Communication Systems, chapter 6 - [Rate analysis: M20 (1:1½:3) RCC per 10 m³](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Arate-m20-rcc-10m3) — Estimating and Costing, chapter 4