To determine the optimal number of production runs Gianna should have each year to minimize her total costs, we need to use the provided formula and the given data.
The relevant formula for determining the optimal order quantity [tex]\( q \)[/tex] in the inventory-problem is given by:
[tex]\[ q = \sqrt{\frac{2 \times \text{annual demand} \times \text{setup cost}}{k_1 + 2 \times k_2}} \][/tex]
Where:
- Annual demand ([tex]\( m \)[/tex]) = 144,000 cases
- Cost per year in electricity to store a case ([tex]\( k_1 \)[/tex]) = \[tex]$2
- Annual warehouse fees per case (\( k_2 \)) = \$[/tex]2
- Setup cost ([tex]\( \text{setup cost} \)[/tex]) = \$750
First, we need to compute the total storage cost per unit:
[tex]\[ \text{total storage cost per unit} = k_1 + 2 \times k_2 \][/tex]
[tex]\[ \text{total storage cost per unit} = 2 + 2 \times 2 \][/tex]
[tex]\[ \text{total storage cost per unit} = 2 + 4 \][/tex]
[tex]\[ \text{total storage cost per unit} = 6 \][/tex]
Now, substitute the values into the formula for [tex]\( q \)[/tex]:
[tex]\[ q = \sqrt{\frac{2 \times 144,000 \times 750}{6}} \][/tex]
[tex]\[ q = \sqrt{\frac{216,000,000}{6}} \][/tex]
[tex]\[ q = \sqrt{36,000,000} \][/tex]
[tex]\[ q = 6,000 \][/tex]
So, the optimal order quantity [tex]\( q \)[/tex] is 6,000 cases.
Next, to find the number of production runs needed each year, we use the formula:
[tex]\[ \text{production runs per year} = \frac{\text{annual demand}}{q} \][/tex]
[tex]\[ \text{production runs per year} = \frac{144,000}{6,000} \][/tex]
[tex]\[ \text{production runs per year} = 24 \][/tex]
Therefore, to minimize her total costs, Gianna should have 24 production runs each year.
