---
title: "Forecast sales for year 2010"
subject: "Engineering Economics"
chapter: 1
key: "semester_6/engineering_economics:forecast-sales-2010"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/forecast-sales-2010
site: BCT Engineering
---

# Forecast sales for year 2010

Engineering Economics · Chapter 1: Introduction

- Asked in: 2065 Kartik, 10 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/forecast-sales-2010
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-1

## Question

Forecast the sales for year 2010.
| Year | 2000 | 2001 | 2002 | 2003 | 2004 | 2005 | 2006 | 2007 |
|---|---|---|---|---|---|---|---|---|
| Sales Rs. '000 | 416 | 287 | 307 | 268 | 378 | 523 | 457 | 587 |

## Short answer

Using a least-squares straight-line trend Y = 402.875 + 16.696X (X = 2×(year − 2003.5)), the forecast sales for 2010 is 619.9, approximately Rs 620 thousand.

## Answer

#### Method
Fit a straight-line trend Y = a + bX by least squares. There are 8 years (even), so take the mid-point 2003.5 as origin and use X = 2 × (year − 2003.5), giving X = −7, −5, −3, −1, 1, 3, 5, 7 (ΣX = 0). Then a = ΣY/n and b = ΣXY/ΣX².

| Year | X | Y (Rs '000) | XY | X² |
|---|---|---|---|---|
| 2000 | −7 | 416 | −2,912 | 49 |
| 2001 | −5 | 287 | −1,435 | 25 |
| 2002 | −3 | 307 | −921 | 9 |
| 2003 | −1 | 268 | −268 | 1 |
| 2004 | 1 | 378 | 378 | 1 |
| 2005 | 3 | 523 | 1,569 | 9 |
| 2006 | 5 | 457 | 2,285 | 25 |
| 2007 | 7 | 587 | 4,109 | 49 |
| Total | 0 | 3,223 | 2,805 | 168 |

a = 3,223 / 8 = 402.875
b = 2,805 / 168 = 16.696 (per 2 years, so about 8.35 per year)

**Trend equation: Y = 402.875 + 16.696 X**, with X = 2 × (year − 2003.5)

#### Forecast for 2010
X = 2 × (2010 − 2003.5) = 13
Y = 402.875 + 16.696 × 13 = 402.875 + 217.05 = **619.9**

**Answer: Forecast sales for 2010 = about Rs 620 thousand.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With 8 years, why is the origin taken at 2003.5 with X = 2 x (year - 2003.5)?
   A. To make SigmaX = 0 with integer X values for an even number of years
   B. Because 2003.5 is the year of highest sales
   C. To make the slope equal to 1
   D. Because the trend must start in an odd year
   Answer: A. Using the midpoint makes SigmaX = 0 and X = -7...7, which simplifies a = SigmaY/n and b = SigmaXY/SigmaX^2. (Key point: Coding the origin)

2. With SigmaY = 3,223, SigmaXY = 2,805, SigmaX^2 = 168, n = 8, what are a and b?
   A. a = 3,223, b = 2,805
   B. a = 402.875, b = 16.696
   C. a = 402.875, b = 350.6
   D. a = 16.696, b = 402.875
   Answer: B. a = SigmaY/n = 402.875 and b = SigmaXY/SigmaX^2 = 16.696. (Key point: Trend coefficients)

3. What value of X is used for 2010?
   A. 6.5
   B. 7
   C. 13
   D. 15
   Answer: C. X = 2 x (2010 - 2003.5) = 13. (Key point: X for 2010)

4. What is the forecast for 2010?
   A. About Rs 403 thousand
   B. About Rs 217 thousand
   C. About Rs 1,400 thousand
   D. About Rs 620 thousand
   Answer: D. Y = 402.875 + 16.696 x 13 = 619.9. (Key point: Forecast)

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