ERR method to select best of three projects

ERR (external rate of return) assumes net receipts are reinvested outside the project at the external rate ε. Receipts are compounded to the end of the life at ε, expenditures (negative flows) are discounted to year 0 at ε, and ERR i' solves

Σ E_t(P/F, ε, t)(F/P, i', N) = Σ R_t(F/P, ε, N - t)

Here ε = 20%, MARR = 10%, N = 6 (cash flows in Rs.).

Step 1: ERR of each project

Project A:

• Receipts compounded to year 6 at ε: FW(R) = 26,200(2.4883) + 29,000(2.0736) + 30,200(1.7280) + 31,000(1.4400) + 31,000(1.2000) + 26,000(1.0000) = 285,354
• Expenditures discounted to year 0 at ε: PW(E) = 64,000(1.0000) = 64,000
• ERR: 64,000(1 + i')^6 = 285,354 → (1 + i') = (4.4587)^(1/6) → ERR = 28.29%

Project B:

• Receipts compounded to year 6 at ε: FW(R) = 39,200(2.0736) + 38,000(1.7280) + 38,000(1.4400) + 38,000(1.2000) + 38,000(1.0000) = 285,269
• Expenditures discounted to year 0 at ε: PW(E) = 68,000(1.0000) + 4,000(0.8333) = 71,333
• ERR: 71,333(1 + i')^6 = 285,269 → (1 + i') = (3.9991)^(1/6) → ERR = 25.99%

Project C:

• Receipts compounded to year 6 at ε: FW(R) = 20,500(2.4883) + 40,600(2.0736) + 40,000(1.7280) + 39,000(1.4400) + 39,000(1.2000) + 32,400(1.0000) = 339,679
• Expenditures discounted to year 0 at ε: PW(E) = 75,500(1.0000) = 75,500
• ERR: 75,500(1 + i')^6 = 339,679 → (1 + i') = (4.4991)^(1/6) → ERR = 28.49%

All three are acceptable, so an incremental comparison is needed (order of investment A 64,000 < B 68,000 < C 75,500). Incremental cash flows are found and ERR is applied on them in the same way.

Step 2: incremental ERR

Increment A → B: [-4000, -30200, 10200, 7800, 7000, 7000, 12000] ERR(B - A) = 14.32% > 10%, so B is better than A.

Increment B → C: [-7500, 24500, 1400, 2000, 1000, 1000, -5600] ERR(C - B) = 39.79% > 10%, so C is better than B.

(Check: A → C gives 27.10% > 10%.)

Answer: Select Project C. C has the highest ERR of its own (28.49%), and each extra investment, from A to B and then from B to C, earns more than the MARR of 10%.