Case Northwestern Memorial Hospital: Minimum par level $=$ Average demand $\backslash$times Average lead time $\backslash$times $(1+$ Safety stock factor$).,$ or $($D $\backslash$times L $\backslash$times $(1+$ S$).).$

Consider two scenarios: $(1)$ variable lead time $($L$)$ with three$-$day average, and $(2)$ fixed lead time of one day. Assume fixed demand $($D$)$ for both cases:

$$ When L averages three $($and varies as much as $5$ days$)$ days, nervous customers want more safety stock. IF$,$ S $=0\mathrm{.}3(30$ per cent more than normally required$).$ Minimum par $=$ D $\backslash$times $3\backslash$times $1\mathrm{.}3=3\mathrm{.}9$D$.$

$$ When L is almost always one day $($L averages one day$),$ customers feel comfortable with less safety stock. IF$,$ S $=0\mathrm{.}1.$ Minimum par $=$ D $\backslash$times $1\backslash$times $1\mathrm{.}1=1\mathrm{.}1$D

What would the theoretical reduction in minimum par level$($in percentage$)$ when lead time decreases from three day to one day as stated in two scenarios above $($Use L$,$D$,$S mentioned in the scenario$$s for computation$)?$

$(1\mathrm{.}1/5)*100\%$

$(2\mathrm{.}8/3\mathrm{.}9)*100\%$

$(1\mathrm{.}1/3\mathrm{.}9)*100\%$

$(3\mathrm{.}9/2\mathrm{.}8)*100\%$