Solved Problems

Solved Problem 6-1 Patterson Electronics supplies microcomputer circuitry to a company that incorporates microprocessors into refrigerators and other home appliances. One of the components has an annual demand of 250 units, and this is constant throughout the year. The carrying cost is estimated to be $1 per unit per year, and the ordering cost is $20 per order.
1. To minimize cost, how many units should be ordered each time an order is placed?
2. How many orders per year are needed with the optimal policy?
3. What is the average inventory if costs are minimized?
4. Suppose the ordering cost is not $20 and Patterson has been ordering 150 units each time an order is placed. For this order policy to be optimal, what would the ordering cost have to be?
Solution
1. The EOQ assumptions are met, so the optimal order quantity is
  
  $EOQ = Q^{*} = \sqrt{\frac{2 D C_{o}}{C_{h}}} = \sqrt{\frac{2 (250) 20}{1}} = 100 units$ $EOQ = Q^{*} = \sqrt{\frac{2 D C_{o}}{C_{h}}} = \sqrt{\frac{2 (250) 20}{1}} = 100 units$
2. Number of orders per year $= \frac{D}{Q} = \frac{250}{100} = 2.5$ $= \frac{D}{Q} = \frac{250}{100} = 2.5$ orders per year
  
  Note that this would mean that in one year the company places three orders and in the next year it needs to place only two orders, since some inventory would be carried over from the previous year. It averages 2.5 orders per year.
3. Average inventory $= \frac{Q}{2} = \frac{100}{2} = 50$ $= \frac{Q}{2} = \frac{100}{2} = 50$ units
4. Given an annual demand of 250, a carrying cost of $1, and an order quantity of 150, Patterson Electronics must determine what the ordering cost would have to be for the order policy of 150 units to be optimal. To find the answer to this problem, we must solve the traditional EOQ equation for the ordering cost. As you can see in the calculations that follow, an ordering cost of $45 is needed for the order quantity of 150 units to be optimal.
  
  $\begin{array}{l} Q & = & \sqrt{\frac{2 D C_{o}}{C_{h}}} \\ C_{o} & = & Q^{2} \frac{C_{h}}{2 D} \\ = & \frac{{(150)}^{2} (1)}{2 (250)} \\ = & \frac{22, 500}{500} = $ 45 \end{array}$ $\begin{array}{l} Q & = & \sqrt{\frac{2 D C_{o}}{C_{h}}} \\ C_{o} & = & Q^{2} \frac{C_{h}}{2 D} \\ = & \frac{{(150)}^{2} (1)}{2 (250)} \\ = & \frac{22, 500}{500} = $ 45 \end{array}$
Solved Problem 6-2 Flemming Accessories produces paper slicers used in offices and in art stores. The minislicer has been one of its most popular items: annual demand is 6,750 units and is constant throughout the year. Kristen Flemming, owner of the firm, produces the minislicers in batches. On average, Kristen can manufacture 125 minislicers per day. Demand for these slicers during the production process is 30 per day. The setup cost for the equipment necessary to produce the minislicers is $150. The carrying cost is $1 per minislicer per year. How many minislicers should Kristen manufacture in each batch?

Solution

The data for Flemming Accessories are summarized as follows:

$\begin{array}{l} D & = & 6, 750 units \\ C_{s} & = & $ 150 \\ C_{h} & = & $ 1 \\ d & = & 30 units \\ p & = & 125 units \end{array}$ $\begin{array}{l} D & = & 6, 750 units \\ C_{s} & = & $ 150 \\ C_{h} & = & $ 1 \\ d & = & 30 units \\ p & = & 125 units \end{array}$

This is a production run problem that involves a daily production rate and a daily demand rate. The appropriate calculations are shown here:

$\begin{array}{l} Q^{*} & = & \sqrt{\frac{2 D C_{o}}{C_{h} (1 - d / p)}} \\ = & \sqrt{\frac{2 (6, 750) (150)}{1 (1 - 30 / 125)}} \\ = & 1, 632 \end{array}$ $\begin{array}{l} Q^{*} & = & \sqrt{\frac{2 D C_{o}}{C_{h} (1 - d / p)}} \\ = & \sqrt{\frac{2 (6, 750) (150)}{1 (1 - 30 / 125)}} \\ = & 1, 632 \end{array}$
Solved Problem 6-3 Dorsey Distributors has an annual demand for 1400 units of a metal detector. The cost of a typical detector to Dorsey is $400. The carrying cost is estimated to be 20% of the unit cost, and the ordering cost is $25 per order. If Dorsey orders in quantities of 300 or more, it can get a 5% discount on the cost of the detectors. Should Dorsey take the quantity discount? Assume the demand is constant.

Solution

The solution to any quantity discount model involves determining the total cost of each alternative after quantities have been computed and adjusted for the original problem and every discount. We start the analysis with no discount:

$\begin{array}{l} EOQ (no discount) & = & \sqrt{\frac{2 (1, 400) (25)}{0.2 (400)}} \\ = & 29.6 units \\ Total cost (no discount) & = & Material cost + Ordering cost + Carrying cost \\ = & $ 400 (1, 400) + \frac{1, 400 ($ 25)}{29.6} + \frac{29.6 ($ 400) (0.2)}{2} \\ = & $ 560, 000 + $ 1, 183 + $ 1, 183 = $ 562, 366 \end{array}$ $\begin{array}{l} EOQ (no discount) & = & \sqrt{\frac{2 (1, 400) (25)}{0.2 (400)}} \\ = & 29.6 units \\ Total cost (no discount) & = & Material cost + Ordering cost + Carrying cost \\ = & $ 400 (1, 400) + \frac{1, 400 ($ 25)}{29.6} + \frac{29.6 ($ 400) (0.2)}{2} \\ = & $ 560, 000 + $ 1, 183 + $ 1, 183 = $ 562, 366 \end{array}$

The next step is to compute the total cost for the discount:

$\begin{array}{l} EOQ (with discount) & = & \sqrt{\frac{2 (1, 400) (25)}{0.2 ($ 380)}} \\ = & 30.3 units \\ Q (adjusted) & = & 300 units \end{array}$ $\begin{array}{l} EOQ (with discount) & = & \sqrt{\frac{2 (1, 400) (25)}{0.2 ($ 380)}} \\ = & 30.3 units \\ Q (adjusted) & = & 300 units \end{array}$

Because this last economic order quantity is below the discounted price, we must adjust the order quantity to 300 units. The next step is to compute total cost:

$\begin{array}{l} Total cost (with discount) & = & Material cost + Ordering cost + Carrying cost \\ = & $ 380 (1, 400) + \frac{1, 400 (25)}{300} + \frac{300 ($ 380) (0.2)}{2} \\ = & $ 532, 000 + $ 117 + $ 11, 400 = $ 543, 517 \end{array}$ $\begin{array}{l} Total cost (with discount) & = & Material cost + Ordering cost + Carrying cost \\ = & $ 380 (1, 400) + \frac{1, 400 (25)}{300} + \frac{300 ($ 380) (0.2)}{2} \\ = & $ 532, 000 + $ 117 + $ 11, 400 = $ 543, 517 \end{array}$

The optimal strategy is to order 300 units at a total cost of $543,517.
Solved Problem 6-4 The F. W. Harris Company sells an industrial cleaner to a large number of manufacturing plants in the Houston area. An analysis of the demand and costs has resulted in a policy of ordering 300 units of this product every time an order is placed. The demand is constant, at 25 units per day. In an agreement with the supplier, F. W. Harris is willing to accept a lead time of 20 days, since the supplier has provided an excellent price. What is the reorder point? How many units are actually in inventory when an order should be placed?

Solution

The reorder point is

$ROP = d \times L = 25 (20) = 500 units$ $ROP = d \times L = 25 (20) = 500 units$

This means that an order should be placed when the inventory position is 500. Since the ROP is greater than the order quantity, $Q = 300,$ $Q = 300,$ an order must have been placed already but not yet delivered. So the inventory position must be

$\begin{array}{l} Inventory position & = & (Inventory on hand) + (Inventory on order) \\ = & 500 = 200 + 300 \end{array}$ $\begin{array}{l} Inventory position & = & (Inventory on hand) + (Inventory on order) \\ = & 500 = 200 + 300 \end{array}$

There would be 200 units on hand and an order of 300 units in transit.
Solved Problem 6-5 The B. N. Thayer and D. N. Thaht Computer Company sells a desktop computer that is popular among gaming enthusiasts. In the past few months, demand has been relatively consistent, although it does fluctuate from day to day. The company orders the computer cases from a supplier. It places an order for 5,000 cases at the appropriate time to avoid stockouts. The demand during the lead time is normally distributed, with a mean of 1,000 units and a standard deviation of 200 units. The holding cost per unit per year is estimated to be $4. How much safety stock should the company carry to maintain a 96% service level? What is the reorder point? What will the total annual holding cost be if this policy is followed?

Solution

Using the table for the normal distribution, the Z value for a 96% service level is about 1.75. The standard deviation is 200. The safety stock is calculated as

$SS = Z σ_{dLT} = 1.75 (200) = 375 units$ $SS = Z σ_{dLT} = 1.75 (200) = 375 units$

For a normal distribution with a mean of 1,000, the reorder point is

$\begin{array}{l} ROP & = & (Average demand during lead time) + SS \\ = & 1, 000 + 350 = 1, 350 units \end{array}$ $\begin{array}{l} ROP & = & (Average demand during lead time) + SS \\ = & 1, 000 + 350 = 1, 350 units \end{array}$

The total annual holding cost is

$THC = \frac{Q}{2} C_{h} + (SS) C_{h} = \frac{5, 000}{2} 4 + (375) 4 = $ 11, 500$ $THC = \frac{Q}{2} C_{h} + (SS) C_{h} = \frac{5, 000}{2} 4 + (375) 4 = $ 11, 500$

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