GATE 2015 CS (CS3) – Question 17
For function points: external inputs I=30, outputs O=60, enquiries E=23, files F=8, external interfaces N=2. Their weights are 4,5,4,10,7. Of fourteen adjustment factors, four are not applicable, four have value 3, and the remaining factors have value 4. Find the adjusted function-point metric.
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Correct answer: 612 to 613
Explanation
The unadjusted count is 30×4+60×5+23×4+8×10+2×7=606. Of fourteen factors, four contribute zero, four contribute twelve, and six contribute twenty-four, giving total influence thirty-six. The standard adjustment is 0.65+0.01×36=1.01. Adjusted points are 606×1.01=612.06, within the 612–613 published range.