BUS 1700 Managing Operations

EOQ and Safety Stock Practice Problems

SOLUTIONS

Problem 1

A local distributor for a national tire company expects to sell 9,600 steel belted radial

tires of a certain size and tread design next year.

Demand is stable, annual holding cost is

$16 per tire, and ordering cost is $75.

a) What is the EOQ?

The parameters are D = 9600/year, K = $75/order, h = $16/year.

Calculate the

EOQ by substituting the values in the formula as follows:

∗ = √ 2 ℎ = √ 2 × 75 × 9600 16 = 300

b) How many times per year does the store order?

=

∗ = 9600 300 = 32

c) What is the total annual cost if the EOQ is ordered?

= ℎ ∗ 2 +

∗ = 16 300 2 + 75 9600 300 = $4,800

Problem 2

Daily ice cream demand at I-Scream parlor is normally distributed with a mean of 400

quarts and a standard deviation of 120.

I-Scream has its ice cream supplied by a

wholesaler who charges $8 per quart.

The wholesaler has a five-day delivery lead time,

and charges a $90 flat fee for each delivery.

I-Scream estimates that the inventory related

holding cost (i.e., cost of capital and physical holding cost, etc.) to be 25% of the dollar

value of inventory per year. I-Scream wants 90% availability.

Assume 360 days in the

year.

(z = 1.28 for A = 90%.)

[Hint: The inventory holding cost, h, can be calculated by multiplying the cost per quart

and the holding cost rate (i.e., 25%).]

The parameters are as follows: annual demand is D = 400 × 360/year, the fixed ordering

cost is K = $90/order, and the annual holding cost is h = $8/quart × 25%/year =

$2/quart/year.

a) What is the EOQ, Q*?

∗ = √ 2 ℎ = √ 2 × 90 × 400 × 360 2 = 3,600

b) Find the optimal reorder point, ROP.

The optimal reorder point can be calculated as follows

=

×

+ √2 ×

= 400 × 5 + 1.28√1202 × 5 ≈ 2,343

c) To reduce consumer complaints due to occasional stock-outs, I-Scream is

considering a new strategy that will require 99% ice cream availability. Quantify

the impact of this strategy on Q* and ROP. (z=2.33 for A=99%.)

The optimal order quantity will remain unchanged. The new reorder point is

=

×

+ √2 ×

= 400 × 5 + 2.33√1202 × 5 ≈ 2,625

The new strategy will increase the optimal reorder point by 2,625/2,343 – 1 =

12%.

Problem 3

A distributor sells 100 cases of a sports drink each week.

The distributor operates 50

weeks per year.

It takes the supplier two weeks to deliver the sports drink to the

distributor at a cost of $8 per case.

The inventory related holding cost (capital, insurance,

etc.) for the distributor is 25% of the dollar value of inventory per year.

Each order

placed with the supplier costs the distributor $20.

This cost includes labor, forms, etc.

[Hint: The inventory holding cost, h, can be calculated by multiplying the cost per case

and the holding cost rate (i.e., 25%).]

a) What order quantity minimizes the distributor’s total inventory holding and

ordering cost?

Remember that the demand (D) and inventory holding cost (h) should have the

same units.

Annual demand is 100×50 = 5,000 cases/year.

Hence, the input

parameters to calculate the EOQ quantity are as follows: D = 5,000 cases/year, K

= $20/order, h = $8/case × 25%/year = $2/case/year.

Using the EOQ formula, we

find

∗ = √ 2 ℎ = √

2 × 20 × 5,000 2

≈ 316

b) What is the optimal reorder point for the distributor?

Since the distributor has to wait for two weeks to receive the shipment from the

supplier, it has to ensure that there is enough inventory to cover the demand over

the lead-time. That is,

=

×

= 100 cases/week × 2 weeks = 200 cases.

c) Assuming the distributor selects the order quantity specified in part (a), how many

orders are placed per year?

Number of orders placed per year =

∗ = 100×50 316 = 15.82

d) Assuming the distributor selects the order quantity specified in part (a), what is

the distributor’s total annual inventory holding and ordering cost?

The total annual cost for the distributor is the sum of the annual inventory

holding, and ordering costs.

Total annual cost = ℎ ∗ 2 +

∗ = 0.25 × 8 × 316 2 + 20 × 100×50 316 ≈ $633

After some time, the distributor realizes that they sometimes run out of stock.

They hire

a consultant to figure out the reason.

The consultant points out that the distributor is not

taking into account the weekly fluctuations in demand and that the reorder point should

be higher.

Analyzing past data, the consultant finds that weekly demand is normally

distributed with a mean of 100 cases and standard deviation of 20 cases.

e) What is the optimal reorder point that would allow the distributor to achieve 99%

availability?

(The z-score corresponding to 99% availability is z = 2.33.)

Weekly demand is normally distributed with mean μ = 100 and σ = 20. The lead- time LT = 2 weeks.

The reorder point, which would allow the distributor to

achieve 99% availability, is

=

×

+ √ × 2 = 100 × 2 + 2.33√2 × 202 ≈ 266.

Problem 4

Dull Computers purchases a special printed circuit board (PCB) from a supplier for use in

a new laptop.

It will take Dull Computers 1 week to transmit an order to the supplier.

After receiving the order, the supplier manufactures PCBs in 2 weeks.

Shipping PCBs to

Dull takes an additional 1 week.

Dull estimates that the average demand and its standard deviation are 10,000 units per

week and 3,000 units per week, respectively. Dull’s management would like the in-stock

rate (or availability) to be at least 98%.

a) What is the minimum safety stock that achieves 98% availability? (z=2.05 for A=0.98.)

LT = order transmission + manufacturing + shipping = 1 + 2 + 1 = 4 weeks

= √2 ×

= 2.05√3,0002 × 4 = 12,300

b) Dull’s management believes that the required level of safety stock is too high, and

seeks the help of a consulting group (BS and Associates) to make some changes in

the supply chain.

BS and Associates propose building a new information system that

can transmit orders in real-time.

Thus, the project would reduce the information

transmission time from 1 week to 0 weeks.

Calculate the % decrease in the safety

stock after the implementation of this project.

LT = order transmission + manufacturing + shipping = 0 + 2 + 1 = 3 weeks

= √2 ×

= 2.05√3,0002 × 3 = 10,652

ℎ

12,300 − 10,652 12,300 = 13.40%

Problem 5

A local company purchases paint for use in its production process.

Monthly demand is

constant at 675 gallons.

A fixed ordering cost of $25 is incurred by the company each

time an order is placed.

The paint currently costs $22 per gallon, and the manufacturer

charges an additional shipping fee of $3/gallon.

The cost of carrying the paint in

inventory is 12.5% per year.

What is the optimal order quantity?

D = 675/month = 8,100/year, K = $25/order, h = $25/gallon × 12.5%/year

∗ = √ 2

ℎ = √

2 × 25 × 8,100 25 × 0.125

= 360

Note that we added the shipping cost of $3/gallon to the purchase cost of $22/gallon

to find the total cost of acquiring one gallon.

Problem 6

Adding lead times greater than zero to the EOQ model:

a. Increases the EOQ

b. Decreases the EOQ

c. Increases the ROP

d. Decreases the ROP

e. None of the above

Answer c.

The basic EOQ model assumes that lead times are zero.

We can simply place a new

order when the inventory level drops to zero.

So, ROP = 0.

When lead times are greater

than zero, ROP equals demand during the lead time multiplied by the lead time.

Since

both demand during the lead time and the lead time are now greater than zero, ROP is

now greater than zero.

So, ROP increases from zero to a number greater than zero.

51作业君版权所有