Chapter 17, Problem 4RP

Explanation of Solution

a.

New copies of the book sold by the publisher each year:

Let,

$\begin{array}{l}\text{NB: New book}\hfill \\ \text{B1: Once-used book}\hfill \\ \text{B2: Twice-used book}\hfill \\ \text{B3: Thrice-used book}\hfill \\ \text{NSB: Books not sold back}\hfill \end{array}$   

The transition probability matrix “P” for the given problem is shown below:

	NB	B1	B2	B3	NSB
NB	0	0.9	0	0	0.1
B1	0	0	0.8	0	0.2
B2	0	0	0	0.6	0.4
B3	0	0	0	0	1

For an organization in which each book is classified into one of “s” groups (here s= 4).

Let,

$\begin{array}{l}{\text{P}}_{\text{ij}}\text{ - Denotes the fraction of books which begin a time period in group i begin the next time period}\hfill \\ \text{in group j}\text{.}\hfill \\ {\text{H}}_{\text{i}}\text{ - Denotes the number of group i books hired at the beginning of each period}\text{.}\hfill \\ {\text{N}}_{\text{i}}\text{ - Denotes the limiting number of group i books}\text{.}\hfill \end{array}$

To determine ${\text{N}}_{\text{i}}$, use the steady-state condition:

$\text{Number of books entering the state i during each period = Number of books leaving the state i during each period}$

Here,

$\text{Amount entering group i= Amount leaving group i}$

This implies,

${\text{H}}_{\text{i}}\text{+}{\displaystyle \sum _{k\ne i}{N}_k{p}_{ki}}={\text{N}}_{\text{i}}\text{+}{\displaystyle \sum _{k\ne i}{p}_{ik}(i=1,\mathrm{...},s)}$

Let steady state census for ${\text{NB;B1;B2;B3 be N}}_{\text{1}}{\text{;N}}_{\text{2}}{\text{;N}}_{\text{3}}{\text{;N}}_{\text{4}}$

Explanation of Solution

b.

Average profit per book:

The bookstore’s profit on each type of book is as given below:

$\begin{array}{l}\text{New book: \$6}\hfill \\ \text{Once-used book: \$3}\hfill \\ \text{Twice-used book: \$2}\hfill \\ \text{Thrice-used book: \$1}\hfill \end{array}$

If the steady-state census is the representative of the bookstore’s sales, the total profit on books is as follows:

$\begin{array}{l}\begin{array}{l}\text{Total profit =(6}\times {\text{N}}_{\text{1}}\text{)+}\left(\text{3}\times {\text{N}}_{\text{2}}\text{'}\right)+\left(\text{2}\times {\text{N}}_{\text{3}}\right)\text{+}\left(\text{1}\times {\text{N}}_{\text{4}}\right)\hfill \\ \text{ = }\left(\text{6}\times \text{1}\right)\hfill \end{array}\end{array}$