Chapter 4.17, Problem 4P

Explanation of Solution

Finding optimal solution for the problem:

In the problem, it is given that the James Breed bakes two types of cakes, cheesecakes and Black forest cakes. At most 65 cakes can be baked during any month. Also, it takes 50cents to hold cheesecake and 40cents to hold a black forest cake. The costs per cake and the demand for cakes which must be met on time are given.

The tableau is given below

Item	Month 1	Month 2	Month 3
	Demand	Cost/cake(S)	Demand	Cost/Case(S)	Demand	Cost/Cake(S)
Cheesecake	40	3.00	30	3.40	20	3.80
Black forest	20	2.50	30	2.80	10	3.40

Let,

${\text{x}}_{\text{tc}}$=Number of cheesecakes produced during month $\text{t}$

${\text{x}}_{\text{tb}}$=Number of Black forest cakes produced during the month $\text{t}$

${\text{i}}_{\text{tc}}$=Number of cheesecakes in the inventory at the end of month $\text{t}$

${\text{i}}_{\text{tb}}$=Number of Black forest cakes in the inventory at the end of month $\text{t}$. Here, $\text{t=1,2,3}$

Thus the objective function is,

Minimize,

$\begin{array}{c}{\text{Z=3x}}_{\text{1c}}\text{+3}{\text{.4x}}_{\text{2c}}\text{+3}{\text{.8x}}_{\text{3c}}\text{+2}{\text{.5x}}_{\text{1b}}\text{+2}{\text{.8x}}_{\text{2b}}\text{+3}{\text{.4x}}_{\text{3b}}\\ \text{+0}{\text{.5(i}}_{\text{1c}}{\text{+i}}_{\text{2c}}{\text{+i}}_{\text{3c}}\text{)+0}{\text{.4(i}}_{\text{1b}}{\text{+i}}_{\text{2b}}{\text{+i}}_{\text{3b}}\text{)}\end{array}$

The problem is subjected to the constraints that during any month, he can bake at most 65 cakes.

Thus,

$\begin{array}{c}{\text{x}}_{\text{1c}}+{\text{x}}_{\text{1b}}\le 65\\ {\text{x}}_{\text{2c}}+{\text{x}}_{\text{2b}}\le 65\\ {\text{x}}_{\text{3c}}+{\text{x}}_{\text{3b}}\le 65\end{array}$

Also, it is given than the amount of cakes in the inventory is equal to the difference of cakes previously in the inventory plus the total production and the demand.

Thus, the constraints become,

$\begin{array}{c}{\text{i}}_{\text{1c}}={\text{x}}_{\text{1c}}-40\\ {\text{i}}_{\text{2c}}={\text{i}}_{\text{1c}}+{\text{x}}_{\text{2c}}-30\\ {\text{i}}_{\text{3c}}={\text{i}}_{\text{2c}}+{\text{x}}_{\text{3c}}-20\\ {\text{i}}_{\text{1b}}={\text{x}}_{\text{1b}}-20\end{array}$

$\begin{array}{c}{\text{i}}_{\text{2b}}={\text{i}}_{\text{1b}}+{\text{x}}_{\text{2b}}-30\\ {\text{i}}_{\text{3b}}={\text{i}}_{\text{2b}}+{\text{x}}_{\text{3b}}-10\end{array}$

In the above, all the variables are greater than 0.

Therefore, the LP can be formulated as,

Minimize,

$\begin{array}{c}{\text{Z=3x}}_{\text{1c}}\text{+3}{\text{.4x}}_{\text{2c}}\text{+3}{\text{.8x}}_{\text{3c}}\text{+2}{\text{.5x}}_{\text{1b}}\text{+2}{}_{}\end{array}$