91Ó°ÊÓ

Inventory is often a company's most valuable asset, second only to its human resources, and thus organisations must have enough inventory on hand to satisfy the demands of their consumers. Companies that do not use inventory management tactics such as these will likely store too much inventoryduring periods of low demand while simultaneously storing too little inventory during periods of strong demand. Either issue results in squandered chances.

Given,

Annual demand (D) = 13,000 units

Weekly demand (d) = 250 units

Standard deviation(S) = 40 units

Ordering cost per unit (O) = $ 100

Lead time (L) = 4 weeks

Annual inventory carrying cost (C) = $0.65 per unit

Service probability = 98%

$$\begin{gathered} \mathrm{EOQ}=\sqrt{\frac{2×\mathrm{A}×\mathrm{O}}{\mathrm{H}}} \\ \mathrm{Where}, \\ \mathrm{A}=\mathrm{Annual}\mathrm{Demand} \\ \mathrm{O}=\mathrm{Ordering}\mathrm{cost}\mathrm{per}\mathrm{unit} \\ \mathrm{H}=\mathrm{Carrying}\mathrm{cost}\mathrm{per}\mathrm{unit} \end{gathered}$$

$$\begin{gathered} \mathrm{EOQ}=\sqrt{\frac{2×\mathrm{A}×\mathrm{O}}{\mathrm{C}}} \\ =\sqrt{\frac{2×13,000×100}{0.65}} \\ =2000\mathrm{Units}. \end{gathered}$$

So, the optimal order quantity is 2000 units.

Given,

Lead time (L) = 4 weeks

Standard deviation(S) = 40 units

$$\begin{gathered} {\mathrm{σ}}_{{}_{\mathrm{L}}}=\sqrt{\mathrm{L}+{\left(\mathrm{S}\right)}^2} \\ =\sqrt{4+{\left(40\right)}^2} \\ =\sqrt{6400} \\ =80\mathrm{Units} \end{gathered}$$

So, the standard deviation during the lead time is 80 units.

Given,

Lead time (L) = 4 weeks

Standard deviation = 80 units

Service probability of 98% ,the value of z is 2.05

Weekly demand (d) = 250 units

$$\begin{gathered} \mathrm{Reorder}\mathrm{Point}=\left(\mathrm{d}×\mathrm{L}\right)+\left(\mathrm{z}×{\mathrm{σ}}_{\mathrm{L}}\right) \\ =\left(250×4\right)+\left(2.05×80\right) \\ =1,000+164 \\ =1,164\mathrm{units} \end{gathered}$$

So,the reorder point is 1,164 units.

Given,

Safety stock = 64 units

Standard deviation = 80 units

$$\begin{gathered} \mathrm{z}=\frac{\mathrm{safety}\mathrm{stock}}{{\mathrm{σ}}_{\mathrm{L}}} \\ =\frac{64}{80} \\ =0.80 \end{gathered}$$

So, for the value of z is 0.80, the service probability will be 79%