Question

In a furniture manufacturing facility, the daily demand to quick-dry glue is normally distributed with a
constant average of 5 kg. and constant standard deviation 1. The manufacturer procures the glue from a
glue supplier. Both facilities work on a 24/7 basis. The fixed ordering cost is $60 per order. While keeping
the glue in inventory for one hour, maintaining the adhesiveness of one kilogram costs $0.01. Since it is
an important component of the furniture products, a shortage in glue is not a desirable situation.
(a) In the beginning, it was told that the lead time of the glue supplier is constant and 4 days from order to
delivery. Suppose the probability of glue shortage is allowed to be 0.005 at maximum. What should be
the safety stock level and what is the optimal inventory policy? Carry out the whole analysis
and show all of your calculations.
(b) Later, it is realized that there were some issues with the delivery reliability of the glue supplier. The
lead time was actually normally distributed with a constant average of 4 days and constant variance
0.20. Suppose the maximum probability of glue shortage is reduced to 0.001. Assuming independence
between the demand and the lead time, what should be the safety stock level and what is the optimal
inventory policy? Carry out the whole analysis and show all of your calculations.

Answer 1

The daily average demand is given as 5Kg.Therefore,

Mean of glue required daily(µ) = 5 Kg

Std. deviation of glue required(σ) = 1 Kg

(a)

As given in question, the maximum glue shortage allowed is 0.005.

Therefore the confidence interval is 99.5%.[ 1-0.005= 0.995 or 99.5%]

The z value for 99.5% confidence interval is 2.807  

Let the maximum daily usage of glue be x.

Then using the formulae: (x-µ)/σ = z-value

We put the values to find x: (x-5)/1= 2.807

x=7.807 Kg

Safety stock= (Maximum Daily usage*Maximum lead time)-(Average Daily usage*Average lead time)

Safety Stock= (7.807*4)-(5*4)= 11.228 Kg

Note:- In part (a) maximum lead time is same as average lead time.

Optimum Inventory policy:

Economic Order Quantity

EOQ= (2AO/H)1/2

Where A= Average daily usage in Kg= 5

O= Ordering cost= 60$

H= Holding cost/ Kg daily= 24*0.01= 0.24 $/Kg per day

EOQ= (2*5*60/.24)1/2 = 50 Kg

Re-order point= (Lead time* Average daily usage) + Safety stock

Re-order point= (4*5) + 11.228= 31.228 Kg.

(b)

As next part, the maximum glue shortage allowed is 0.001.

Therefore the confidence interval is 99.9%.[ 1-0.001= 0.999 or 99.9%]

The z value for 99.5% confidence interval is 3.291  

Let the maximum daily usage of glue be x1.

Then using the formulae: (x1-µ)/σ = z-value

We put the values to find x1: (x1-5)/1= 3.291

x1=8.291 Kg

Similarly we calculate maximum lead time.

Let the maximum lead time be x2

µ (Mean Time)= 4 days

σ ( Standard deviation)= .2

The confidence interval is 99.9%. The z value for 99.5% confidence interval is 3.291

using the formulae: (x2-µ)/σ = z-value (Where x= maximum lead time)

We put the values to find x: (x2-4)/.2= 3.291

x2= 4.6582

Safety stock= (Maximum Daily usage*Maximum lead time)-(Average Daily usage*Average lead time)

New Safety Stock= (8.291*4.6582)-(5*4)= 18.6211 Kg

Optimum Inventory policy:

EOQ= (2*5*60/.24)1/2 = 50 Kg [ It remains same]

Re-order point= (Lead time* Average daily usage) + Safety stock

Re-order point= (4*5) + 18.6211= 38.6211 Kg.