Forecasting the Adoption of Ebooks

Case 1 Ebook

Date: 3/9/2013

Case 1: Forecasting the Adoption of E-Books

Product | p | q | Market Structure (wa=0.3) | Product Characteristics (wb=0.7) | Weighted Numerical Score |

CD player | 0.055 | 0.378 | 5 | 5 | 5/20.4=0.245 |

Cable television service | 0.100 | 0.060 | 7 | 3 | 4.2/20.4=0.206 |

Home Personal Computer | 0.121 | 0.281 | 4 | 6 | 5.4/20.4=0.265 |

Cellular telephone | 0.008 | 0.421 | 3 | 7 | 5.8/20.4=0.284 |

Weighted Average for e-book | 0.0684 | 0.2990 | | | |

Question 1

The size of the market for e-books in the long-run:

293.7 million x 46.7% x 8% = 10.973 million

(Total US population x percentage of US population reading literature x Percentage of traditional paper books purchased online in 2003)

It will take 10 years to reach 95% penetration of the potential market.

Question 2

I predict that the sale of the e-book when it first became available is 0.751milion (pm) (Assumption: each innovator buy one book.)

Question 3

The long-run total adoption of e-books would be 10.973 million. (Total US population x percentage of US population reading literature x Percentage of traditional paper books purchased online in 2003)

Question 4

I do expect the market for e-books tend to be guided by imitators rather than innovators. According to the table above, I found that the weighted average for e-book of q (0.0684) is larger than p (0.2990).

In addition, the word of mouth takes an important role in the adoption of e-books. The information about the e-book from the innovators contribute much to the adoption of e-books since people tend to hesitate to change their old reading habit.

Question 5

S(t)=[p+(q/m)N(t-1)][m-N(t-1)]

Year 1: [0.0684+(0.2990/10.973) X (0)][10.973-0] = 0.751 million

Year 2: [0.0684+(0.2990/10.973) X (0.751)][10.973-0.751] = 0.908 million

Year 3: [0.0684+(0.2990/10.973) X (1.659)][10.973-1.659] = 1.059 million

Year 4: [0.0684+(0.2990/10.973) X (2.717)][10.973-2.717] = 1.176 million

Year 5:...