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Question
Prepare the worksheets yourself for examples used in sections 3.1, 3.2, 3.3 and 3.4 respectively. Add two new more records in each worksheet (with your own assumed values) and verify whether the computations are correct.
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Solution
Constants Setup: H3 = 28 (Days in Month), H4 = 35% (DA Rate), H5 = 40% (HRA Rate), H6 = 12% (PF Rate).
Data and Computations Table:
| Emp ID | Name | Basic Pay | Attendance (Days) | Basic Earned (Col F) | DA (Col G) | HRA (Col H) | PF (Col N) | Net Salary |
| 101 | Amit | ₹ 40,000 | 28 | ₹ 40,000 | ₹ 14,000 | ₹ 16,000 | ₹ 4,800 | ₹ 65,200 |
| 102 | Priya | ₹ 50,000 | 21 | ₹ 37,500 | ₹ 13,125 | ₹ 15,000 | ₹ 4,500 | ₹ 61,125 |
| 103* | Raj (New) | ₹ 30,000 | 28 | ₹ 30,000 | ₹ 10,500 | ₹ 12,000 | ₹ 3,600 | ₹ 48,900 |
| 104* | Neha (New) | ₹ 60,000 | 14 | ₹ 30,000 | ₹ 10,500 | ₹ 12,000 | ₹ 3,600 | ₹ 48,900 |
Basic Earned: =(C4/H3) * D4 → (40,000/28) × 28 = 40,000
DA (Dearness Allowance): =F4 * $H$4 → 40,000 × 35% = 14,000
HRA (House Rent Allowance): =F4 * $H$5 → 40,000 × 40% = 16,000
PF (Provident Fund): =F4 * $H$6 → 40,000 × 12% = 4,800
Net Salary: =(F4 + G4 + H4) - N4 → (40,000 + 14,000 + 16,000) - 4,800 = 65,200
Verification: For the new record Neha, her 14 days of attendance correctly cut her Basic Earned in half (₹ 30,000), and all dynamic downstream headers adapt perfectly.
Calculates asset depreciation evenly across its lifespan using the SLN function.
Constants Setup: C3 = Cost (₹1,000,000), D3 = Salvage Value (₹100,000), E3 = Useful Life (10 Years).
| Year | Opening Book Value | Depreciation Formula (Col K) | Yearly Depreciation Expense | Accumulated Depreciation | Closing Book Value |
| 1 | ₹ 1,000,000 | =SLN(C3, D3, E3) | ₹ 90,000 | ₹ 90,000 | ₹ 910,000 |
| 2 | ₹ 910,000 | =SLN(C3, D3, E3) | ₹ 90,000 | ₹ 180,000 | ₹ 820,000 |
| 3* | ₹ 820,000 | =SLN(C3, D3, E3) | ₹ 90,000 | ₹ 270,000 | ₹ 730,000 |
| 4* | ₹ 730,000 | =SLN(C3, D3, E3) | ₹ 90,000 | ₹ 360,000 | ₹ 640,000 |
SLN = `("Cost" − "Salvage")/"Life"`
= `(1,000,000 − 100,000)/10`
= ₹ 90,000 per year
The newly appended Years 3 and 4 track a linear drop of exactly ₹90,000 each. The locked variables block cell reference shifts when formulas are dragged down.
Constants Setup: F3 = Cost (₹500,000), G3 = Salvage Value (₹50,000), H3 = Useful Life (5 Years).
| Year | Opening Book Value | Book Value DB Formula Structure (Col G) | Yearly Depreciation Expense | Accumulated Depreciation | Closing Book Value |
| 1 | ₹ 500,000 | =DB(F3, G3, H3, 1) | ₹ 185,500 | ₹ 185,500 | ₹ 314,500 |
| 2 | ₹ 314,500 | =DB(F3, G3, H3, 2) | ₹ 116,680 | ₹ 302,180 | ₹ 197,820 |
| 3* | ₹ 197,820 | =DB(F3, G3, H3, 3) | ₹ 73,391 | ₹ 375,571 | ₹ 124,429 |
| 4* | ₹ 124,429 | =DB(F3, G3, H3, 4) | ₹ 46,163 | ₹ 421,734 | ₹ 78,266 |
Fixed Depreciation Rate: Excel’s internal DB logic evaluates the rate using:
`1 - ("Salvage"/"Cost")^((1/"Life"))`
= `1 - ((50,000)/(5000,000))^((1/5))`
= 37.1%
Year 3 Computation: ₹ 197,820 (Opening) × 37.1%
= ₹ 73,391
Year 4 Computation: ₹ 124,429 (Opening) × 37.1%
= ₹ 46,163
The dynamic computations for newly added periods 3 and 4 show standard declining balance behaviour where total depreciation decreases sequentially over time.
