Chapter 13, Problem 21P

Demand for walnut fudge ice cream at the Sweet Cream Dairy can be approximated by a normal distribution with a mean of 21 gallons per week and a standard deviation of 3.5 gallons per week. The new manager desires a service level of 90 percent. Lead time is two days, and the dairy is open seven days a week. (Hint: Work in terms of weeks.)

a. If an ROP model is used, what ROP would be consistent with the desired service level? How many days of supply are on hand at the ROP, assuming average demand?

b. If a fixed-interval model is used instead of an ROP model, what order size would be needed for the 90 percent service level with an order interval of 10 days and a supply of 8 gallons on hand at the order time? What is the probability of experiencing a stockout before this order arrives?

c. Suppose the manager is using the ROP model described in part a. One day after placing an order with the supplier, the manager receives a call from the supplier that the order will be delayed because of problems at the supplier's plant. The supplier promises to have the order there in two days. After hanging up, the manager checks the supply of walnut fudge ice cream and finds that 2 gallons have been sold since the order was placed. Assuming the supplier's promise is valid, what is the probability that the dairy will run out of this flavor before the shipment arrives?

Summary Introduction

To determine: The reorder point and days of supply on hand.

Introduction: Reorder point is the point at which the firm has to replenish its inventory in order to avoid shortages or any lag in production.

Answer to Problem 21P

The reorder point and days of supply on hand is 9 gallons and 3days.

Explanation of Solution

Given information:

$\begin{array}{c}\text{Demand}\left(\text{Mean}\right)\text{,}\overline{\text{d}}=21\text{gallons/week}\\ \text{Standard}{\text{deviation,σ}}_{\text{d}}=3.5\text{gallons/week}\\ \text{LT}=\text{2days}\\ \text{Service level}=90\text{\%}\end{array}$

Formula:

$\text{ROP}=\overline{\text{d}}\left(\text{LT}\right)+\text{z}\left({\text{σ}}_{\text{d}}\right)\sqrt{\text{LT}}$

Calculation of Reorder point and number of days of supply:

Using z-factor table, the z value corresponds to 0.90. The closest value of probability is 0.8997 which corresponds to z = 1.28.

$\begin{array}{c}\text{ROP}=21\left(\frac{2}{7}\right)+1.28\left(3.5\right)\sqrt{\frac{2}{7}}\\ =9\text{gallons}\text{(rounded)}\end{array}$ (1)

Reorder point is calculated by multiplying 21 with $\left(\frac{2}{7}\right)$ and adding the resultant with the product of $1.28\left(3.5\right)\sqrt{\frac{2}{7}}$, which yields 9 gallons.

$\begin{array}{c}\text{Days}\text{of}\text{supply}=\frac{9}{\frac{21}{7}}\\ =\frac{9}{3}\\ =3\text{days}\text{of}\text{supply}\text{on}\text{hand}\text{at}\text{ROP}\end{array}$

Number of days of supply is calculated by dividing 9 with the ratio of 21 and 7 which yields 3 days of supply on hand at ROP.

Hence, the reorder point and days of supply on hand is 9 gallons and 3days.

Summary Introduction

To determine: The probability of experiencing stock out before the order arrives.

Introduction: Safety stock refers to the quantity of additional or extra stock which are maintained to mitigate the risks and uncertainty which arises due to fluctuation in supply and demand.

Answer to Problem 21P

The probability of experiencing stock out before the order arrives is 14.23%.

Explanation of Solution

Given information:

$\begin{array}{c}\text{Demand}\left(\text{Mean}\right)\text{,}\overline{\text{d}}=21\text{gallons/week}\\ \text{Standard}{\text{deviation,σ}}_{\text{d}}=3.5\text{gallons/week}\\ \text{LT}=\text{2days}\\ \text{Service level}=90\text{\%}\end{array}$

$\begin{array}{c}\text{Order}\text{interval}=10\text{days}\\ \text{Supply}\text{on}\text{hand}\text{at}\text{order}\text{time}=8\text{gallons}\end{array}$

Formula:

$\text{ROP}=\overline{\text{d}}\left(\text{LT}\right)+\text{z}\left({\text{σ}}_{\text{d}}\right)\sqrt{\text{LT}}$

Calculation to determine the probability of stock out before arriving of the order:

$\begin{array}{c}\text{ROP}=21\left(\frac{2}{7}\right)+\text{z}\left(3.5\right)\sqrt{\frac{2}{7}}\\ 8=6+1.871\text{z}\\ \text{z}=1.07\end{array}$

Using z-factor table, the lead service level is 0.8577.

$\begin{array}{c}\text{Risk of stock out before the order arrives }=\text{1}-\text{0}\text{.8577}\\ =\text{0}\text{.1423}\\ =\text{14}\text{.23\%}\end{array}$

 The risk of stock out is calculated by finding the z-factor from using the ROP of 8, mean demand of 21 and standard deviation of 3.5 and lead time of $\left(\frac{2}{7}\right)$. Substituting the values in the formula and solving for z, the z value obtained is 1.07.

For z=1.07, the corresponding value from z-factor table is 0.8577. The probability to find the risk of stock out is difference between 1 and 0.8577 which gives 14.23%.

Hence, the probability of experiencing stock out before the order arrives is 14.23%.

Summary Introduction

To determine: The probability of stock out situation.

Introduction: Safety stock refers to the quantity of additional or extra stock which are maintained to mitigate the risks and uncertainty which arises due to fluctuation in supply and demand.

Answer to Problem 21P

The probability of experiencing stock out is 29.81%.

Explanation of Solution

Given information:

$\begin{array}{c}\text{Demand}\left(\text{Mean}\right)\text{,}\overline{\text{d}}=21\text{gallons/week}\\ \text{Standard}{\text{deviation,σ}}_{\text{d}}=3.5\text{gallons/week}\\ \text{LT}=\text{2days}\\ \text{Service level}=90\text{\%}\end{array}$

Calculation to determine the probability of stock out situation:

$\begin{array}{c}\text{ROP}=9\text{gallons}\\ \text{Stock}\text{after}\text{one}\text{day}=9-2\\ =7\text{gallons}\end{array}$

The ROP is 9 gallons (refer equation (1)) and 2 gallons had been sold after placing order. Stock after day 1 of placing order is the difference between 9 and 2 which is 7 gallons.

$\begin{array}{c}\text{ROP}=21\left(\frac{2}{7}\right)+\text{z}\left(3.5\right)\sqrt{\frac{2}{7}}\\ 7=6+1.871\text{z}\\ \text{z}=0.53\end{array}$

Using z-factor table, the lead service level is 0.7019.

$\begin{array}{c}\text{Risk of stock out before the order arrives }=\text{1}-\text{0}\text{.7019}\\ =\text{0}\text{.2981}\\ =\text{29}\text{.81\%}\end{array}$

The risk of stock out is calculated by finding the z-factor from using the ROP of 7, mean demand of 21 and standard deviation of 3.5 and lead time of $\left(\frac{2}{7}\right)$. Substituting the values in the formula and solving for z, the z value obtained is 0.53.

For z = 0.53, the corresponding value from z-factor table is 0.7019. The probability to find the risk of stock out is difference between 1 and 0.7019 which gives 29.81%.

Hence, the probability of experiencing stock out is 29.81%.