--- title: "Find P using gradient formula" subject: "Engineering Economics" chapter: 2 key: "semester_6/engineering_economics:find-p-gradient-10-12-14" source: https://www.bctengineer.com/q/semester_6/engineering_economics/find-p-gradient-10-12-14 site: BCT Engineering --- # Find P using gradient formula Engineering Economics · Chapter 2: Interest and Time Value of Money - Asked in: 2070 Magh, 4 marks - Video: [Arithmetic Gradient - Engineering Economics Lightboard](https://www.youtube.com/watch?v=ykQ7M9UNuok) — Engineering Economics Guy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/engineering_economics/find-p-gradient-10-12-14 - Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-2 ## Question Find the value of P if i = 10%. Use gradient formula also. [Figure: cash flow diagram, years 0 to 6; receipts 10 at year 0, 12 at year 1, 14 at year 2, then 16 at years 3, 4, 5 and 6; P = ? as a downward arrow at year 2] ## Short answer Break the series into uniform and gradient parts, move all to year 2 at 10% interest using P/F, P/A, and P/G factors. The equivalent present worth is P = 90.02. ## Answer #### Assumption P is the single equivalent amount at **year 2** (where the arrow is drawn) of all the receipts: 10 at year 0, 12 at year 1, 14 at year 2, and 16 at years 3, 4, 5 and 6. i = 10%. ```text 16 16 16 16 12 14 ^ ^ ^ ^ 10 ^ ^ | | | | ^ | | | | | | -----+----+----+----+----+----+----+---> year 0 1 2 3 4 5 6 | v P = ? ``` #### Splitting the series Receipts = uniform 10 (years 0-6) + gradient 0, 2, 4, 6 (years 0-3, G = 2) + uniform 6 (years 4-6). #### Value at year 2 of each part 1. Uniform 10 at years 0, 1, 2: 10 (F/A, 10%, 3) = 10 × 3.310 = 33.100 2. Uniform 10 at years 3 to 6 (4 payments): 10 (P/A, 10%, 4) = 10 × 3.1699 = 31.699 3. Gradient 0, 2, 4, 6 at years 0 to 3. Its (P/G, 10%, 4) value (4.3781) falls one period before the first term, i.e. at year −1, so 2 (P/G, 10%, 4)(F/P, 10%, 3) = 2 × 4.3781 × 1.3310 = 11.655 4. Uniform 6 at years 4, 5, 6: 6 (P/A, 10%, 3) = 6 × 2.4869 = 14.921 at year 3, so at year 2: 14.921 (P/F, 10%, 1) = 14.921 × 0.9091 = 13.565 P = 33.100 + 31.699 + 11.655 + 13.565 = **90.02** Check by direct compounding to year 2: 10(1.21) + 12(1.1) + 14 + 16(0.9091 + 0.8264 + 0.7513 + 0.6830) = 12.1 + 13.2 + 14 + 50.72 = 90.02. **Answer: P = 90.02 (in the units of the figure), equivalent to the whole series at year 2.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. At which year is P located? A. Year 0 B. Year 2, where the arrow is drawn C. Year 6 D. Year -1 Answer: B. P is the equivalent at year 2. (Key point: Where is P) 2. Gradient 0,2,4,6 at years 0-3 with (P/G,10%,4)=4.3781: where does that value fall? A. At year 2 B. At year 0 C. One period before the first term, year -1, so it is multiplied by (F/P,10%,3) D. At year 3 Answer: C. P/G places the value one period before the gradient start. (Key point: Gradient value) 3. Why is the 10 uniform split into two parts? A. Because 10 changes each year B. One via (P/G) C. Not needed D. Years 0-2 via (F/A,10%,3) and years 3-6 via (P/A,10%,4) Answer: D. Both bring to year 2 from either side. (Key point: Uniform 10) 4. What is P? A. 90.02 B. 33.1 C. 80.5 D. 101.2 Answer: A. Components 33.100 + 31.699 + 11.655 + 13.565 = 90.02. 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