If demand for an inventory item is normally distributed with a mean of 150 units per day and standard deviation of 2 units/day, lead time is normally distributed with an average of 4 days and standard deviation of 0.5, and service level is 95% (z=1.65) , at what point should it be reordered?

128

328

548

748

850

ROP (xy) = 724 units to be recorded

Step-by-step explanation:

Given:-

- Demand for an inventory item(x) is normally distributed with :

                       mean ux = 150 units/day

                       standard deviation sx = 2 units/day

- lead time(y) is normally distributed with:

                       mean uy = 4 days

                       standard deviation sy = 0.5 days

- Service level is 95% (zs=1.65):

Find:-

at what point should it be reordered?

Solution:-

- We will deonte (xy) the number of units to be serviced to be normally distributed with mean and standard deviation:

                       uxy = ux*uy = 150*4 = 600 units

                       sxy = √[(uy*sx^2) + ux^2*sy^2 ]

                              =√[(4*2^2) + 150^2*0.5^2 ]

                              = 75.10659

- At the service level of 95% the confidence interval would be:

                       ROP (xy) = uxy + zs*sxy )

                       ROP (xy) = 600 + 1.65*75.10659

                       ROP (xy) = 724 units to be recorded      

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.