Respuesta :
Solution :
Given data:
Annual demand, D = 180,000 chairs
Ordering cost, F = $ 150 per order
Annual holding cost per unit, C = $25
Lead time of order, L = 5 days
Standard deviation of order during lead time = 30
a). The optimal order quantity
[tex]$=\sqrt{\frac{2FD}{C}}$[/tex]
[tex]$=\sqrt{\frac{2\times 150 \times 180,000 }{25}}$[/tex]
= 1469.69
= 1470 (rounding off)
b). The Z value of the customer service of 90%,
i.e., the probability of 0.90 as per normal distribution table = 1.29
∴ Safety stock = Z value x standard deviation of order during lead time
= 1.29 x 30
= 38.7
= 39 (rounding off)
c). The reorder point
[tex]= \text{Average demand per day} x \text{Demand lead time (day) + Safety stock}[/tex]
[tex]$=\frac{\text{annual demand}}{\text{300 days}} \times \text{ Demand Lead time (days) + Safety stock}$[/tex]
[tex]$=\frac{180,000}{300} \times 5 + 39$[/tex]
= 3039
d). The optimal annual total inventory cost
[tex]$\text{= Annual ordering cost + Annual Inventory carrying cost}$[/tex]
[tex]$\text{= Number of orders} \times \text{Ordering cost + Average inventory} \times }$[/tex] [tex]$\text{Inventory holding cost per unit per year}$[/tex]
[tex]$=\frac{\text{annual demand}}{\text{optimum order quantity}} \times \text{ordering cost+}\frac{\text{optimum ordering cost}}{2}\times C$[/tex]
[tex]$=\frac{180,000}{1470} \times 150 + \frac{1470}{2} \times 25$[/tex]
= 18367.34 + 18375
= $ 36,742.34
