Determining Optimum Order Quantity, Stock Out Probability, And Process Capability

Model and Calculation for Optimum Order Quantity of Teddy Bower Boots

1 – (Order quantity) Teddy Bower Boots

Question 1.1

According to the single-period stock model, by Stevenson and Hojati, (2007)

(Cs) Under-stocking cost = estimated demand or shortage cost = difference between the price of sales per pair and the cost per pair

Thus, Cs = 54 – 40= $14 per pair

Overstocking cost (Co).On the other hand = estimated demand = cost per pair- the salvage value per pair

Co = 40-27 = $13

Question 1.2

Service level also called the critical ratio = Cs / (Cs + Co) =14/ (14+13)

Service level = 0.5185

It is important for the Company to determine the z-score for the demand normal distribution that produces a certain probability which is equal to the service level of 0.5185

Therefore, 51.85% of the area under the normal curve must be to the right side of the maximum stocking level.  Using the standard normal table, with an area of about 0.5185, the z-score will be 0.0464

Maximum quantity stock = µ + z = 400 + (0.0464*150) = 407 pairs

The Teddy Bower Boots should order 407 pairs of boots from the suppliers because this quantity is the optimum standard that shall maximize the expected profit.  

Question 1.3

The case assuming Teddy Bower will order 450 pairs, and for that 450 pairs, there will be excess after the season end. But they anticipate the discount rate of that excess will not be 50% as it was in the original case. So what is it? We can get it after knowing the new discounted price.

Z-score at 450 pairs of boots

Inventory = z + µ = Z*150 + 400 = 450

150z=50

By making z the subject of the formula by dividing both sides by 150 we have the z-score = 0.3333

Hence, the new discount rate is 33.33 % and is less than, the earlier one, 50%

2 – (Stock out Probability) Aircraft Manufacturing Plant

1^st question

We know that

When a bin consisting 250 objects is consumed, the order for top-up for the succeeding bin is in course and it takes around 3 hours for it to reach.

Henceforth, a stock-out situation will occur the moment 250 items are consumed in 3 hours. Since, standard deviation (µ) of exhaustion every hour is 30,

  Standard deviation of exhaustion for the period of the lead time of 3 hours

= µ of hourly exhaustion x Lead time

Calculating Stock Out Probability for Aircraft Manufacturing Plant

= 30x Square root (3) = 51.96

Placing in terms of value Z score,

Systematic exhaustion in 3 hours + Z= 250

=3 x 60 + 51.96*Z = 250

Otherwise, 51.96*Z +180 = 250

= Z = 70/51.96 = 1.347∞1.35

Using the Standard Normal distribution table, Probability corresponding to Z-score will be 0.9115

And so probability of no stock out is 0.9115

= 1 – 0.9115=0.08851

THUS, THE PROBABILITY OF FASTENER STOCKOUT = 0.08851

2^nd Question

The beset probability of stockout is 1 % = 0.01

Therefore, the probability of no stockout = 1 – 0.01 = 0.99

Using the formula, NORMSINV, the comparable Z value for 0.99:

That is, Z is equals NORMSINV (0.99) = 2.326

Therefore, the smallest bin size (in regards to the number of fasteners) = Normal demand in 3 hours + Z * standard deviation of demand in 3 hours.

= 180 + 51.96.Z

= 300.85 is equivalent to 301 

THE SMALLEST NUMBER OF FASTENERS A FULL BIN SHOULD CONSIST IS 301

3– (Process Capability) Laboratory Testing

Question 3.1

Mean (μ) = 36.5 minutes

The Standard deviation (σ) = 9 minutes

Therefore, the Upper Specification Limit (USL) = 40 minutes

And the Upper Capability Measure (UCM) = (40-36.5)/ (3*9) = 0.130

Reasoning: Standard deviation of the process is higher than the mean, and so the spread of the process =3*9 (27) is larger than the spread of the portrayal limit (=40-36.5 = 3.5). Consequently, being the proportion of the spread of the portrayal limit to the spread of this process shows that the UCM of the process is essentially trivial.

Value of Z equals (40-36.5)/9 = 0.39

Question 3.2

The probability that the turnaround time of the laboratory testing process is longer than its upper specification limit equals to 1 – 0.65 = 0.35

Double of UCM = 0.1296*2 = 0.240

In order to twofold the UCM, the standard deviation = 4.5 = (40-36.5)/0.2593/3

Reasoning: Continuing with the previous reasoning, it is now clear that capability measure in our scenario is the ratio of the spread of the portrayal limit to the spread of the entire process mean. In an attempt to double the capability portion, the range of the average has to be reduced partly by half.

Probability/ 2 = 0.35/2 = 0.175

Corresponding Z value = NORMSINV (1-0.175) = 0.937

Standard deviation = (40-36.5)/0.937 = 3.745

Reasoning: The probability of turnaround time of this process being greater than object time (higher portrayal frontier) is directly proportional to the discrepancy between the real time and the process mean in the numerator alongside the spread of the course in the denominator. So, in an attempt to diminish the probability of turnaround time being bigger than object time, whichever the standard deviation or spread of the process mean should be condensed or reduced (Vespignani, 2012).

References

Stevenson, W.J., and Hojati, M., 2007. Operations management (Vol. 8). Boston: McGraw-Hill/Irwin.

Vespignani, A., 2012. Modeling dynamical processes in complex socio-technical systems. Nature physics, 8(1), p.32.