Inventory 2

MODEL 2: EOQ without Instantaneous (noninstantaneous) Receipt (Fixed-Order Quantity Model with Usage)

This model sometimes called Production Order Quantity Model. In the basic model we have assumed that the entire inventory, which is ordered, arrives simultaneously. In many cases, however, this is not a valid assumption, because the vendor delivers the order in partial shipment or portions ever a period of time.

In such cases inventory is being used while new inventory is still being received and the inventory does not build up immediately to its maximum point. Instead, it builds up gradually when inventory is received faster than it is being used; then it declines to its lowest level as incoming shipments stop and the use of inventory continues. This version of the EOQ model is known as “Noninstantaneous Receipt Model” also referred to as the “Gradual Usage Model ” and “Production Lot Size Model”. In this, noninstantaneous receipt model, the order quantity is received gradually over time, and the inventory level is depleted at the same time it is being replenished. This concept is illustrated in Fig.3.

We can apply the above stated to an optimum production lot size, where the finished goods are being sold while each lot is produced. In this case, the inventory of finished goods doesn’t build up immediately to its maximum point, as in case of Model1. Instead, it builds up gradually as goods are produced faster than they are being sold; then it declines to its lowest point as production of a particular batch ceases although sales continue (see fig 3).

Inventory level

Optimal order/lot size (units)

Inventory rises as incoming shipments (or production) Production Exceeds usage/salesRate

Maximum inventory level (Imax), (Incoming Shipments or production stops)X or

X0 orQ0 Inventory declines as use/sales

continue

New receipts/production begin

t1 t2 TimeSupply begins supply ends

t0

Fig. 3 : Inventory with receipt over time.

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% into (R-C)/Rinventory

Let R = Daily receipt/production rate in units c = use rate in units daily

EOQ Model OLS Model

X0 = Q0 =

X0 (MU) = Q0 (MU)=

Imax = Xo(1 – c/R) Imax = Qo(1 – c/R)

t1 = X0 / R t1 = Qo/R

t2 = Imax/C t2 = Imax/C

t0 = t1 + t2 = X0/C t0 = t1 + t2 = Qo/C

Ke = Ke =

Example 32:

İzmir Wine Co. bottles 5000 cases of a particular rose wine annually. The set-up cost per run is 90 MU. Factory cost is 5 MU per case. Carrying costs on finished goods inventory is 20 %. Production rate is 100 cases per day and sales amount to 14cases/day.

How many cases should be bottled per production run?

Solution :

Q0 = = = 1023 Cases/Run

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Example 33 :

A plastics moulding firm produces and uses 24 000 Teflon bearing inserts annually. The cost of setting up for production is 70 MU and the weekly rate is 1000 units. If the production cost is 5.50 MU per unit and the annual carrying cost is 1.50 MU per unit, how many units should the firm produce during each production run? What is the maximum inventory that the firm will stock?Solution :

The demand and the production rates must be in the same units, so we arbitrarily put both into annual terms, assuming a 52-week year.

Q0 = = = 2040 Inserts/Run

t1 = Q0 / R = 2040 / 1000 = 2 weeks

so the firm will be producing inserts about every month.

Imax = Q0 (1 – c/R) = 2040(1-24000/52000) = 1098.5 Inserts

Example 34 :

The Ege Creamery Co. produces ice-cream bars for vending machines and has an annual demand for 72000 bars. The Co. has the capacity to produce 400 bars per day. It takes only a few minutes to adjust the production set-up cost estimated at 7.50 MU per set up for the bars, and the firm is reluctant to produce too many at one time because the storage cost (refrigeration) is relatively high at 1.50 MU/bar/year. The firm supplies vending machines with its “ Ege bars “ on 360 days of the year.

a) What is the most economical number of bars to produce during any one production run?

a) What is the optimal length of the production run in days?c) What is the total inventory cost?d) What is the maximum inventory level?

Solution :

a) Q0 = = = 1200 Bars/Run

b) t1 = Qo/R = 1200/400 = 3 Days to = Qo/C = 1200/72000 = 0.017 Year ≈ 6 Days

c) Ke= = = 900 MU

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d) Imax = Qo(1-c/R) = 1200(0.5) = 600 Bars

[ N0 = C/Q0 = 72000/1200 =60 runs /year ]

Example 35 :

A contractor has to supply 10 000 bearings per day to an automobile manufacturer. He finds that, when he starts a production run, he can produce 25000 bearings per day. The cost of holding a bearing in stock for one year is 0.02 MU and the set-up cost of a production is 18 MU.

How many times should the production runs occur?

Solution:

Q0 = = = 105 000 Bearings

No = C/Qo = (10000x365)/105000 = 34.76 ≈ 35 Runs

Example 36:

A firm has a yearly demand for 52000 units of a product, which it produces. The cost of setting up for production is 80 MU and weekly production rate is 1000 units. The carrying cost is 3.50 MU/unit/year.

How many units should the firm produce on each production run?

Solution :

Q0 = = = ∞

They should have an infinite (continuous) run.

Example 37 :

Mr. Demir wonders what effect an annual inventory costs will result if he allows his supplier to deliver the # 1713 valve orders gradually rather than all at once. The analyst analyses this request and develops these estimates:

C = 10000 valves/yearE = 0.40 MU/valve/yearB = 5.50 MU/orderc = 40 valves per day (10000 values/year)R = 120 valves per day.

a) Calculate the EOQ.b) Calculate the total inventory costs.

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c) Should the supplier be allowed to deliver valves gradually?

Solution :

a) The EOQ:XO = = = 624.26 valves

b) The new total inventory costs are:

Ke= = =171.27 MU

c) The EOQ and total inventory costs from example 23, when the valves were delivered all at once were

EOQ = 524.4 valves Ke = 209.76 MU

The estimated savings to the Co. for allowing its supplier to deliver the # 1713 valve gradually are

Savings = -171.27 + 209.76 = 38.50 MU /year

Result:

The Supplier should be allowed to deliver the # 1713 valves gradually.

Example 38 :

A large hotel serves banquets and several restaurants from central kitchen in which labour is shifted among various stations and jobs. Salad consumption (demand) is virtually constant and known to be 30 000 Salads/year. Salads can be produced at a rate of 45000/year. Salads cost 0.40 each and it costs 4MU to set up the salad line.

Carrying costs of salads, high because of spoilage, are estimated to be 90% of the cost of a salad. No stock outs are allowed. The hotel would like to establish an operating doctrine for salad operation.

Solution:

First we can set the reorder point at R0 = 0, because labour can be shifted to the salad operation instantaneously, and the production rate is greater than the demand rate.

Finding Q0;

Q0 = = = 1414 Salads

Example 39 :

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Azim, Corporation is currently purchasing sales units faster than it is able to sell them as a result; management has requested a re-evaluation of its ordering policy. An analysis of the current sales market reveals the following; anticipated annual demand 10000 units, daily sales 300 units, daily arrivals of units 400 units. The relevant cost data, obtained from the sales office is: order cost 15 MU/order, Unit cost 1.50 MU, carrying cost 20% at unit cost. Determine the order policy that Azim should use.

Solution:

X0 = = = 2000 units

AZIM should initiate an order policy requiring lot shipments of size 2000 units.

Ke = = = 150 MU

Total Cost = Ka + Ke = 10000 ´ 1.50 + 150 = 15150 MU

to = (Xo / C) 365 = (2000 / 10000)´365 = 73 days

N0 = C/Xo = 10000/2000 = 5 orders / year

Example 40:

ABC-Lite., Inc., a manufacturing concern produces fixtures for use in one of its assemblies. The department producing the fixtures is capable and producing 6250 units/week. However, total current demand is only 5000 fixtures / week. The firm operates 52 weeks/year and has developed the following pertinent cost data:

unit cost of manufactured fixtures 7.50 MU; set-up cost 104 MU; holding cost/unit/year 0.10 MU;

To maintain a smooth flow of materials, the production department requires a 4-week lead-time on production requests. Using this information, determine the following:

a) Optimal lot size (optimal production quantity),b) Total annual inventory cost,c) The reorder point.

Solution :

a) Q0 = = = 52000 fixtures

Thus the optimal production lot size (OLS) is 52 000 fixtures per run.

b) Ke= = 1 030 MU

The total annual inventory cost, given a optimal lot size of 52000 fixtures is 1030 MU/year.

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c) R0 = clt + tlt

= 5000 ´ 4 = 20 000 fixtures

The reorder point is 20000 fixtures.

Note: This policy requires 5 production runs per year.

N0 = C/Q0 = (5000 x 52)/520 000 = 5 runs

Example 41:

A toy manufacturer uses 48.000 Rubber wheels per year for its popular dumb truck series. The firm makes its own wheels, which it can produce at a rate of 800 per day. The day trunks are assembled uniformly over the entire year. Carrying cost is 1.00 MU/wheel/year. Set up cost for a production run of wheels is 45 MU. The firm operates 240 days per year. Determine each of the following

a) Optimal run size,b) Minimum total annual cost of carrying and set up.c) Cycle timed) Run timee) Imax, i.e. maximum inventory level.

Solution:

a) Q0= = = 2400 wheelsb) Ko = = = 1800 MUc) to = Qo/C = 2400/ (4800/240) = 12 days

Thus a run of wheels will be made every 12 days.d) t1 = Qo/R = 2400/800 = 3 days

Thus each run will require 3 days to complete.e) Imax = Qo(1-c/R) = 2400(1- 200/800) =1800 wheels

Example 42:

Assume that The Gul Carpet Store in Izmir has its own manufacturing facility in which it produces Demir Carpet. We further assume that the ordering cost, “s” is the cost of setting up the production process to make Demir carpet. Recall that E=0.75/meter and C= 10000 m/year.

The manufacturing facility operator the same days the store is open (i.e. 311 days) and produces 150 m of the carpet per days. Determine the optimal order size, total inventory cost, the length of the time to receive an order, the number of orders per year, and the maximum inventory level.

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Solution

s = 150MU/orderE=0.75 per unitC=10000m c=10000/311 =32.2m/day

Qo= = = 2 256.8 m

This value is substituted into the following formula to determine total minimum inventory cost:

Ke= = = 1 329 MU

or

Ke= (C/Q0).s + (Q0/2)E(1-c/R) = (10000/2256.8)150 + (2 256.8/2)(0.75)(1-32.2/150) = 1 329 MU

The length of time to receive an order for this type of manufacturing operation is commonly called the length of the production run. It is computed as follows

t1 = Q0/R = 2 256.8/150 = 15.05 days/order

The number of orders per year is actually the number of production runs that will be made

No = = 4.43 runs/year

The maximum inventory level is.

Imax= Qo(1- c/R) = 2256.8 (1-32.2/150) = 1 772 m

Example 43:

The manager of a bottling plant which bottles soft drinks needs to decide how long a

`run` of each type of drink to ask the lines to process. Demand for each type of drink is

reasonably constant at 80 000 per month (a month has 160 production hours). The bottling

lines fill at a rate of 3000 bottles per hour but take an hour to change over between different

drinks. The cost of each changeover (cost of labor and lost production capacity) has been

calculated at 100 MU/hour. Stock-holding costs are counted at 0.1 MU/bottle-month.

a. How many bottles the companies produce an each run?

b. The staffs who operate the lines have devised a method of reducing the changeover

time from 1 hour to 30 minutes. How would that change the ELS?

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Solution:

a. ELS= √2Cs/[E(1-c/R)] = √[2*80000*100]/[0.1(1-500/3000)] = 13 856 bottles

b. New s = 50

New X0 =√(2*50*80000)/[0.1(1-50/3000)] = 9 798 bottles

Example 44:

Jantsan Co. makes and sells specialty hubcaps for the retail automobile after market.

Jantsan`s forecast for its hubcap is 1000 units next year, with an average daily demand of 4

units/ However, the production process is most efficient at 8 units per day (so the company

produces 8 per day but uses only 4 per day). Given the following values, solve for the

optimum number of units per order.

Solution:

Annual demand = C= 1000 units

Setup cost = s= 10 MU

Holding cost = E= 0.50 MU/unit-year

Daily production rate = R= 8 units/day

Daily demand rate = 4 units/day

Q0 = √2Cs/[E(1-c/s)] = √(2*100*10)/[0.5(1-4/8)] = 282.8 = ~ 283 hubcaps.

Example 45:

Flemming Accessories producers paper slicers used in offices and art stores. The

minislicer has been one of its most popular items; annual demand is 6750 units, and is

constant throughout the year. Minislicers are produced in batches. An average the firm can

manufacture 125 minislicers/day. Demand for these slicers during the production process is

30 day. The set-up cost for the equipment necassary to produce the minislicers is 150 MU.

Carrying costs are 1 MU/minislicers per yaer. How many minislicers should Flemming

manufacture in each batch?

Solution:

Q0 = √2Cs/[E(1-c/s)] = √(2*6750*150)/[1(1-30/125)] = 1 632 minislicers/run

Example 46:

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Jan Getry is the owner of a small company that produces electric scissors use to cut

fabric. The annual demand is for 8000 scissors and Jan produces the scissors in batches. On

the average, Jan produces 150 scissors per day and during the production process, demand for

scissors has been about 40 scissors per day. The cost to set up the production process is 100

MU and it costs Jan 0.30 MU to carry one pair of scissors for one year. How many scissors

should Jan produce in each batch?

Solution:

Q0 = √2Cs/[E(1-c/R)] = √(2*8000*100)/[0.3(1 -40/150)] = 2 697 scissors /run

Quantity Discounts

Quantity discounts are price reductions for large orders offered to customers to induce them to buy in large quantities. If quantity discounts are offered, the customers must weigh the potential benefits of reduced purchase price and fewer orders that will result from buying in large quantities against the increase in carrying costs caused by higher average inventories. The buyer’s goal with the quantity that will minimize total cost, where the total cost is the sum of carrying cost, ordering cost, ordering cost and purchasing.

K = Carrying Cost + Ordering Cost + Purchasing Cost = (X/2)E + (C/X)B + Cp

In the basic EOQ model, determination of the order size does not involve the purchasing cost. The rationale for not including unit price is that under the assumption of no quantity discounts, price per unit is the same for all order sizes.

When quantity discounts are offered, there is a separate U-shaped total-cost-curve for each unit price. Each curve has a minimum; those points are not necessarily feasible. The objective of the quantity discount model is to identify an order quantity that will represent the lowest total cost for the entire set of curves.

There are two general cases of the model.

i. Carrying costs are constant (e.g. 2 MU/unit)ii. Carrying costs are stated as a percentage of purchase price ( e/g 20 % of unit

price)

When carrying costs are constant, there will be single EOQ, which is the same for all the cost curves.

Ka

Kb

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Kc

Ea,b,c

Fig. 4: When carrying costs are constant, all curves have the same EOQ.

When carrying costs are specified as a percentage of unit price, each curve will have a different EOQ.

Fig. 5: When carrying costs are giving as a percentage of unit prices.

The procedure for determining the overall EOQ differs slightly depending on which of these two cases is relevant.

A- When carrying costs that are constant, the procedure is as follows:

1.Compute the common EOQ.

2. Only one of the unit prices will have the EOQ in its feasible range since the range do not overlap/ Identify that range.

a) If the feasible EOQ is on the lowest price range, that is the optimal order quantity.

b) If the EOQ is in any other range, compute the total cost for the EOQ and for the price breaks of all lower unit costs. Compare the total cost is the optimal order quantity.

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EOQ aEOQ bEOQ c

KaKb

Kc

CCa

CCb

carrying cost

Example 47:

The maintenance department of a large hospital uses about 816 cases of liquid cleanser annually. Ordering cost is 12MU, carrying costs are 4MU/cases, and the price schedule is as follows:

Range Price1 to 49 20 MU 50 to 79 18 80 to 99 17100 or more 16

Determine the optimal order quantity and the total cost.

Solution:

1. Compute the common EOQ = Xo = = = 70 cases

2. The 70 cases can be bought at 18 MU/case, since 70 falls in the range of 50-79 cases. The total cost to purchase 816 cases/ a year at a rate of 70 cases/order, will be:

K70 = (C/X) B + (X/2)E + Cp = (816/70)12 + (70/12)4 + 81618 = 14 968 MU

Since the cost ranges exist, each must be checked against the minimum cost generated by 70 cases at 18 MU each. In order to buy at 17MU/case, at least 80 cases must be purchased. The total cost at 80 cases will be

K80 = (816/80)12 + (80/2)4 + 81617 = 14 154 MU

To obtain a cost of 16 MU / case, at least 100 cases per order are required, and the total cost will be

K100 = (816/100)12 + (100/2)4 + 81616 = 13 354 MU

Therefore, since 100 cases per order yields the lowest total cost, 100 cases is the overall optimal order quantity.

B. When carrying costs are expressed as a percentage of price, determine the best purchase quantity with the following procedure:

1. Beginning with the lowest price, compute the EOQ for each price range until a feasible EOQ is found (i.e. until an EOQ is found that falls in the quantity range for its price)

2. If the EOQ for the lowest price is feasible, it is the optimal order quantity. If the EOQ is not the lowest price range, compare the total cost at the price break for all lower

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prices with the total cost of the largest feasible EOQ. The quantity that yields the lowest total cost is the optimum.

Example 48: (A Quantity Discount with Constant Carrying Cost)

Comptech Computers wants to reduce a large stock of microcomputers it is discontinuing. It has offered the Eastern Mediterranean University Bookstore at Northern Cyprus a quantity discount pricing schedule if they will purchase the microcomputers in volume, as follows.

Quantity Price 1-49.1.1 1400MU50-89.1.1 110090+ 900

The annual carrying cost for the University for a microcomputer is 190 MU, the ordering cost is 2500 MU and the annual demand for this particular model is estimated to be 200 units. The University Bookstore units to determine if it should take advantage of this discount an order the basic EOQ order size.

Solution:

First determine the EOQ and total cost with the basic model..

B=2500MUE=190MUC=200 units /year

X0 = = = 72.5 = 73 Computers

Although we will use Xo=72.5 in the subsequent computation, realistically the order size would be 73 computers. This order size is eligible for the firs discount of 1100 MU. Therefore this price is used to compute total cost as follows:

TC=K =(C/X)B + (X/2)E +Cp = (2500/2)200+(72.5/2)190 +1100(200)= 233784 MU

Since there is a discount for a large order size than 50, this total cost of 233784 MU units must be compared with total cost with an order size of 50 and a price of 900 MU, as follows.

K= (2500/2)200 + (90/2)190 + 900(200) = 194105 MU

Since this total cost is lower (194105MU<233784MU), the maximum discount price should be taken and 90 units should be ordered.

Example 49:

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A manufacturing firm has been offered a particular item it uses according to the following discount pricing schedule provided by the supplier.

Quantity Price 1-199 65MU200-599 59600+ 56

The manufacturing company uses 700 of the items annually, the annually carrying cost is 14MU/unit, and the ordering cost is 275MU.Determine the amount the firm should order.

Solution:

First, determine the optimal (economic) order size and the total cost with basic EOQ model.

X0 = = = 165.83 unitsKe= = = 2 321.64 MU

Next compare the order size with the second-level quantity discount with an order size of 200 and a discount price of 59MU.

K= (275/200)700+ (14/2)20 +59(700)= 43 662.50 MU

This discount results in a lower cost.

Finally, compare the current discounted order size with the fixed-price discount for X=600

K= (700/200)275 + (600/2)14 + 56(700)= 43 720.83 MU

Since this total cost is higher, the economic order is 200 with a total cost of 43662.50MU

Example 50:

Surge Electric uses 4000 toggle switches a year. Switches are priced as follows:

Range Unit Price E= Zp1-499 0.90 MU500-999 0.851000+ 0.82

It costs approximately 18 MU to prepare an order and receive it, and carrying costs are 18% of the purchase price per unit on an annual basis. Determine the optimal order quantity and the annual cost.

Solution:

Range Unit Price E = Zp1-499 0.90 MU 0.18(0.90)=0.1620

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500-999 0.85 0.18(0.85)=0.15301000+ 0.82 0.18(0.82)=0.1476

a) Find the EOQ for each price, starting with the lowest price, until a feasible EOQ is located,EOQ0.82 = = = 988 switches

Since 988 switches will cost 0.85 MU each rather than 0.82 each, 988 is not feasible EOQ. Next try 0.85 per unit.

EOQ0.85 = = 970 switches

This is feasible; 970 fall in the 0.85 MU ranges of 500-999.

b) K970 = (4000/970) 18 + (970/2) 0.153 + 0.85(4000) = 3 548 MU

K1000 = (4000/1000) 18 + (1000/2) 0.1476 + 0.82(4000) = 3 426 MU Thus the minimum-cost order size is 1000 switches.

Example 51: (Quantity Discount with Carrying cost as a % of price.)

The Izmir Saglik Hospital uses disposable surgical packages for many routine operations rather than sterilising and packaging the necessary bandages and instrument. It uses approximately 100 Surgical Packages unit each month. Elective surgery is scheduled for times when the schedule for other surgery is low, so that demand for Surgical Packages is fairly uniform. Each Surgical Packages costs 35MU in quantities of less than 75, and 32,50 if purchased in quantities of 75 or more. The hospital estimates that it costs 8MU to process and receive an order. The cost of holding (carrying) inventory is estimated to the 12% of the purchase per year. Determine the economic order quantity and the total cost.

Solution:

Range Unit Price E=Z.p__________ 1-75 35MU 35(0.12) = 4.275+ 32.50MU 32.5(0.12) = 3.9

a) Find the EOQ for the each price, starting with the lowest price, until a feasible EOQ is located.

X32.50 = = = 70 units

This quantity is not feasible. The hospital would not receive the lowest price if it purchased 70 packages each time. Next try 35 MU per unit.

X35 = = = 67.61 = 68 units

This is a feasible order quantity, but another step will determine of it is the best. The hospital will achieve a sizeable saving on the item cost if it purchases 75 units in each order.

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Lets use Total Cost equation to evaluate this possibility and find the total annual cost if 68 is used as the order quantity.

K68 = + Cp = + 1200(35) = 42 283.97 MU

If the hospital buys the Surgical Packages in quantities of 75 the total annual cost will be

K75 = (75/2)3.90 + (1200/75)8 + 1200(32.5) = 39 274.25 MU

The proper order quantity is 75 surg packs.

Example 52:

A small manufacturing firm uses roughly 3400 kg. of chemical dye a year.Currently the firm purchases 300 kgs per order and pays 3MU/kg. The supplier has just announced that orders of 1000kgs. or more will be filled at a price of 2 MU/kg.

The manufacturing firm incurs a cost of 100 MU each time it submits an order and assign an annual holding cost of 17% of the purchase price per kg.

a. Determine the order size that will minimise the total cost.

b. If the supplier offered the discount at 1500 kg instead of 1000 kg’s, what order size would minimise total cost?

Solution:

a) C = 3400kg/yearp = 2

E = Zp = 2(0.17) = 0.34

X 2MU/kg = = = 1 414 kgs

Since the quantity is feasible at 2MU/kg it is the optimum.

b) When the discount is offered at 1500 kgs, the EOQ for the 2 MU/kg range is no longer feasible. Consequently it becomes necessary to compute the EOQ for 3 MU/kg and compare the total cost for that size with the total cost using the price break quantity (i.e 1500).

X 3MU/kg = = = 1 155 kgs

K 1155 = (3400/1155)100 + (1155/2)(0.17)(3) + 3400(3) = 10 789 MU K 1500 = (3400/1500)100 + (1500/2)(0.17)(2) + 3400(2) = 7 282 MU

Hence, because it would result in a lower total cost, 1500 is the optimal order size.

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How to Determine the Optimum Safety Stock Level when Out-of-Stock Costs are Known

How much should we buy when it is time to replenish the inventory? This question was answered through Model 1 and 2. Here we want to introduce quantitative techniques that will help us to answer the question: “ When should we replenish the inventory?” This when-to-order point is called the “Reorder Point”.

Lead Time

If you call for home delivery of pizza and it takes 2 hrs for it to arrive, then 2 hrs. is the lead-time. If you order is going to be delivered 30 calendar days after, then 30 days is the lead-time. Usage of an item during lead-time is known as lead-time demand.

Fig. 6: Inventory level with constant demand and constant lead-time.

The assumptions

1) Constant demand2) Constant lead-time

are hardly ever true.

Planned demand of an item can be affected by

-Unexpected market acceptance. -By the weatheror - By a strike.

The lead-time varies too

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inventorylevel

EOQ

time

leadtime

order received

order placed

reorder level

reorder

point

orderplaced

orderreceived

stockout

inve

ntor

y le

vel

late arrival

normal leadtime

- A supplier may run into problems (strikes, floods, break downs) or

- The transportation company may experience delays.

Variations in the lead-time or in demand often cause stock-outs, the condition that exists when the inventory on the hand is not sufficient to cover needs.

Note that inventory level does not increase to original points, since unfilled orders have to be filled. (We are assuming here that the customer waits for delivery and does not cancel the order)

a) Inventory level with constant demand and excessively long lead time.

b) Inventory level with excessive demand and constant lead-time.Fig.7: Inventory levels.

Stock-outs are undesirable because they can be quiet expensive. Lost sales and disgruntled customers are examples of external costs. Idle machines and employee ill will are

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stockout

order received

normalleadtime

reorder pointlevel

order placed

inve

ntor

y le

vel

examples of internal costs. Management desire to avoid stock-outs leads to further consideration of when to order and reorder.

Safety Stock

Safety-stocks constitute one of the major means of dealing with the uncertainties associated with variations in demand and lead-time. Safety-stocks are amounts of inventory held in excess of regular usage quantities in order to provide specified levels of protection against stock-out. Safety-stocks are extra inventory held as a badge, or protection against the possibility of a stock-out.

It is obvious that a safety stock has two effects on a firm’s costs. It will decrease the cost of stock-outs, but increase carrying costs.

The cost of stock-out multiplied by the number of stock-outs prevented by the safety stock gives the cost reduction figure. Note that this cost addition is continuing -even permanent- in nature because the safety stock is always a part of total inventory. Note that also because the safety stock does not often define in quantity, we do not divide it by 2 to get average inventory.

Example 53:

A nationwide trucking firm has an average demand of 10 new tires per week and receives deliveries from a Izmit tire company about 20 business days (5 days/week) after placing an order. If the firm seeks to maintain a safety stock of 15 tires, what is the order point?

Solution:

Order point (OP) = clt +SS

= (10 tires/week)(4 weeks) + 15 tires = 55 tiresExample 54:

A producer of Gül Shampoo uses 400 litres per week of a chemical, which is ordered in EOQ of 5000 litres at a quantity discount cost of 3.75 MU/litre. The procurement lead-time is maintained. The storage cost is 0.01 MU/litre-week.

Find a) The maximum inventory on hand.b) The average inventory maintained.

c) The order point in units.

Solution:

a) Maximum Inventory = I max = Safety Stock + EOQ

I max = 200 +5000 = 5200 litres

b) Average Inventory = Iave= (Imax + Imin)/2 = (5200 + 200)/2 = 2 700 liters

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c) Order point = OP = c lt + SS

= 400 x 2 + 200 = 1 000 litres

The optimum safety stock to carry is determined in the light of two goals, which are somewhat hostile to each other:

1) to minimise the cost of stock-outs while also2) to minimise carrying costs on the safety stock.

The decision of how much safety stock to carry is not an easy one. Every approach to this problem has its own limitations. To determine an approach level of safety stock, we will use the probability approach, the most satisfactory approach.

-We shall assume a constant lead-time;-We shall assume that each lot is delivered all at one time.

Under these assumptions, a stock-out can be caused only by an increase in demand after the reorder point has been reached. If the increase had occurred before the reorder point was reached, a purchase order would have been placed at the moment the inventory.

Example 55:

The cost of being out of stock for a particular item = 50 MU/unit Cu ; cost of carrying one item in safety stock = 10 MU / unit.

EOQ = 3600 units average daily usage = 50 items/day. No= 5 ordersLead-time = 6 days Inventory record card is given in the following table. How much safety stock the firm

should carry?

Use during reorder Period, units

No. of times this quantity was used

Use probability

150 3 3/100 =0.03200 4 4/100=0.04250 6 6/100=0.06300 68 68/100=0.68350 9 9/100=0.09400 7 7/100=0.07450 3 3/100=0.03

100 1.00

Solution:

Order point = Lead time x Usage/day= 6 days x 50 units/day = 300 units

Ordering point is 300 units. The company will be safe 81% of the time (0.68 + 0.06 0.04 +0.03), but it will be out of stock 19% of the time (0.09 +0.07 +0.03) . We are concerned

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over this figure of 19 %. To reduce or avoid this shortage, we could carry some safety stock and pick the one, which yields the lowest total for

1. Cost of stock-outs plus2. Carrying costs on the safety stock.

Thus we would consider carrying a safety stock of:

i. 50 Units

This would cover a usage of 350 during the reorder point. We would be out of stock only when usage is 400 or 450 units. (0.07 +0.03 = 0.10 of the time)

ii. 100 Units

This would cover a usage of 350 or 400 units during the reorder point. We would be out of stock only when usage is 450 units. This would be 0.03 of the time.

iii . 150 Units

This would cover a usage of 350,400 or 450 during the reorder point. We would never run out of stock with this amount of safety stock. If the optimal order number is 5 orders per year, the firm, therefore, be in danger of running out of item 5 times during the year. EOQ thus affects the reorder point.

The cost of being out of stock for the four courses of action- No safety stock- 50 units- 100 units- 150 units are shown in the table.

Safety stock

Prob. Of being out

No. short Expected annual

cost shortage

Total annual

cost

Annual carrying

cost

Total cost/year(stock-out +

carrying)0 0.09 when

we use is 350

50 50x0.09x50MUx15=1125

0.07 when we use is 400

100 100x0.07x50MUx5=1750

0.03 when we use is 450

150 150x0.03x50 MUx5=1125

4000 0 4000MU

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50 0.07 when we use is 400

50 50x0.07x50MUx5=875

0.03 when we use is 450

100 100x0.03x50MUx5=750

1625 50x10MU=500MU

2125MU

100 0.03 when we use is 450

50 50x0.03x50MUx5=375

100x10MU=1000

1375MU

150 0 0 150x10MU=1500

1500MU

The appropriate safety stock is 100 units. Adoption of the safety stock policy would change the reorder point. If 100 motors are held as safety stock, then the reorder point is determined as follows.

Reorder point = average daily use x lead time + Safety Stock = 50 x 6 + 100 = 400 units

Example 56:

Demir Sales has used a fixed-quantity model as a basis for establishing its inventory policy for the past three years. During the time the sales pattern has been fluctuating. In such a way that demand remains uncertain. For the past year, Demir Sales manager has kept a record of actual demand during the reorder period. The results are shown in Table below.

Use during reorder

period (units)

Frequency of use

Relative frequency of

use

Probability of use

Cumulative probability

totals50 4 4/100 0.04 0.04100 9 9/100 0.09 0.13150 12 12/100 0.12 0.25200 45 49/100 0.49 0.74250 12 12/100 0.12 0.86300 8 8/100 0.08 0.94350 6 6/100 0.06 1.00

100 100/100 1.00

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Current reorder point for Demir Sales is 200 units, a policy, which involves no stock-out 74 % of the time. If 50 units to 250 units increase the reorder point then the management will be the safe 26 % of the time.

Suppose Demir Sales wants to know the total annual cost of various safety stock possibilities. The cost of running out of stock is set at 10 MU/unit. Calculate these potential costs when the reorder point is 200 units and the company orders 6 times per year [Shortage cost is 10 MU/unit]

Solution:

Reorder pt. (preset) = 200 units

Usage above reorder level = 250,300,350

Usage probabilities = 0.12,0.08,0.06

Shortage cost = 10 MU/unit.

Orders per year = 6

Safety stock

Usage Units short

Probability of

shortage

Annual Cost

(10Mux6xunits short)

Expected Annual

Cost

Total Expected Annual

Cost

0 250 50 0.12 3000 360300 100 0.08 6000 480350 150 0.06 9000 540 1380

50 300 50 0.08 3000 240350 100 0.06 6000 360 600

100 350 50 0.06 3000 180 180150 350 0 0 0 0 0

Safety stock of o units results in a total expected annual cost of 1380 MU, and a safety stock of 50 units yields cost of 600 MU and 100 units safety stocks has total cost of 180 MU. A safety stock of 150 units yields MU. If the management of Demir Sales were to institute a safety stock policy that would minimize the total expected annual cost, the policy would require a safety stock of 150 units.

Service Level

As told before when variability is present in demand or lead-time, the possibility that actual demand will exceed expected demand is created. Safety stock will reduce the risk of running out of inventory (a stock-out) during lead-time. The reorder point then increases by the amount of safety stock:

Order point (OP) = clt + SS

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Stock-out protection is needed only during lead-time.Because it costs a lot of money to hold safety stock, a manager must carefully weight the cost of carrying safety stock-out against the reduction in stock-out risk it provides, since the service level increases as the risk of stock-out decreases. Order cycle service level can be defined as

" The probability that demand will not exceed supply during lead time"(i.e that the amount of stock on hand will be sufficient to meet demand)

Hence, Service level of 95% implies a probability of 95% that demand will not exceed supply during lead-time.

An equivalent statement that demand will be satisfied in 95% of such instances does

not mean that 95% of demand will be satisfied. The risk of stock-out is the complement of service level; stock-out risk is 5%.

Service models can be used when demand variability is present. The first model can be used if an estimate of expected demand during lead-time and its standard deviation are available.

Fig. 8: The OP based on a normal distribution of lead-time demand.

The formula is

OP = Expected Demand during lead time + Z s Clt

The models generally assume that any variability in demand rate or lead-time can be adequately described by a normal distribution.

The value of “Z” depends on the stock-out risk that the manager is willing to accept generally, the smaller the risk the manager is willing to accept, the greater the value of Z. The value of Z is obtained from normal distribution table.

Example 57:

Suppose that the manager of a construction supply house determined from the historical records that the lead-time demand for sand average 50 tons. In addition, suppose the

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Service level

expected demand OP

(probability of nostockout)

risk of stockout

Q

ZOSAFETY STOCKS z scale

manager determined that demand during lead-time could be described by a normal distribution that has a mean of 50 tons and a standard deviation of 50 tons.

Answer these questions, assuming that the manager is willing to accept a stock-out risk of no more than 3%.

a. What value of "Z " is appropriate?b. How much safety stock should be held?c. What reorder point should be used?

Solution:

Expected lead time demand = 50 tonss clt = 5 tonsRisk = 3%

a. From the normal-distribution table, using a service level of 1 - 0.03 = 0.97 , you obtain a value of Z = + 1.88

b. Safety Stock = Z s clt = 1.88(5) = 9.40 tonsc. OP = c lt + Z s clt = 50 + 9.40 = 59.40 tons When data on lead-time demand are not readily available the above formula cannot be

used. Nevertheless, data are generally available on daily or weekly demand, and on the length of lead-time. Using that data, a manager can determine whether demand and /or lead time variable and if variability exists in one or both, the related standard deviation(s). For those situations the following formulas can be used.

* If only demand is variable, then “sclt = sc “ and the reorder point is;

OP = c LT where c = average daily or weekly demand LT= lead time in days or weeks.

* If only lead-time is variable, then sclt = cslt , the order point is; OP = cLT + Zcslt

* If both demand and lead-time are variable, then

OP = cLT + Z

Example 58:

A restaurant use an average of 50 jars a special sauce each week. Weekly usage of sauce has a standard deviation of 3 jars. The manager is willing to accept no more than a 10% risk of stock out during lead-time, which is two weeks. Assume the distribution of usage is normal.

a. Which are the above formulas is appropriate for this situation? Why?b. Determine the value of zc. Determine the reader point.

Solution:

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SL=90%

3600 5136

c = 50 jars/weeksc = 3 jars/weekLT = 2 weeksRisk = 10%, Service Level = 90%

a) Because only demand is variable (i.e. has a std. dev.). Formula cLT + Z sc is appropriate.

b) SL = 90% Z = +1.28

c) OP = cLT + Z sc = 50 2 + 1.28 3 = 100 + 5.43 = 105.43 jars

Example 59: (OP for variable demand and constant lead-time)

The housekeeping department of a motel uses approximately 400 washcloths per day. The actual amount tends to vary with the number of guest or any given night. Usage can be approximated by a normal distribution that has a mean of 400 a std. deviation of 9 washcloths per day. A linen supply company delivery towels and washes cloths with a lead-time of 3 days. If the motel policy is to maintain a stock-out risk of 2%, what is the minimum number of washcloths that must be on hand at reorder time, and how much of that amount can be considered safety stock?

Solution:

c = 400 washcloths/day Z for 98% is about + 2.055LT = 3 days OP = cLT + Z sc = 400(3) + 2.055 9

sc = 9 washcloths/day = 1200 + 32.03 = 1232 washcloths

Risk = 2% ® SL = 98%

Safety Stock = 32 wash cloths

Example 60: (OP for constant demand and variable lead time)

The motel uses approximately 600 bars of soap each day and this tens to vary more than a few bars either way. Lead-time for soap delivery is normally distributed with a mean of 6 days and a standard deviation of 2 days. A service level of 90% is required. Find the reorder point.

Solution:

c = 600 bars/day SL = 90% Z = +1.28LT = 6 days slt = 2 days

OP = c LT + Z c (slt)

= 600(6)+1.28(600)(2)

= 5136 bars of soap

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Example 61: (OP for variable demand rate and variable lead time)

The motel replaces broken glasses at a rate of 25 per day. In the past, this quantity has

tended to vary normally and have a standard deviation of 3 glasses/day. Glasses are ordered

from İzmir supplier. Lead time is normally distributed with an average of 10 days and a

standard deviation of 2 days. What reorder point should be used to achieve level of 95%?

Solution:

c = 25 glasses/day LT = 10 days sc = 3 glasses/day slt = 2 days

SL =95% z = 1.65

OP = cLT + Z = 25(10) + 1.65

= 334 glasses

MODEL III : Inventory Management with Planned Stock-outs

(Back-order Model)

Up to this we have been concerned with methods which prevent stock-outs. However,

in certain situations management may find it desirable from a cost point of view to not only

allow stock-out but to plan for them. It is quite common, for example, not to find the sofa you

want in the fabric you want at your local furniture store. The shopkeeper will, however, order

exactly what you want if you will wait for delivery.

Back-orders

The specific type of stock-out we are concerned with here is called a “back-order”.

When we speak of an item being back-ordered, we imply that:

1. The customer placed an order.

2. The supplier was out of stock in that item.

3. The customer does not withdraw the order.

4. The customer waits until the next shipment arrives.

5. The supplier fills the customer’s order when the next shipment arrives

If however, the customer will withdraw the order when the item is found to be out of

stock, the back-order model is not appropriate. In back-order model, we will have the

additional “cost of back-ordering”. Back-ordering cost is composed of two different costs.

1. Any cost of handling the backorder (special handling, follow-up, labour)

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2. Whatever loss of customer goodwill occurs as a result of having to backorder an item.

Management scientists have determined that the optimum units per economic lot size ( X0)

and number of units back-ordered (S) (owed to customers when a new economic lot arrives)

and also Imax, i.e. maximum inventory level possible when economic lot arrives.

Fig. 9: Inventory behaviour in back-order situation

Formulas:

Economic Order Quantity : X0 = Stock-out Quantity : S = X0 (E/(E+d)) or S = Ke/d

: S = Total Inventory Costs : Ke = {(X0-S)2/2X0}E +(C/X0)B +(S2/2X0)d

: Ke = ` Stock-out time : tS = S/c (c = daily demand rate

Time inventory available : t1 = (X0 - S)/c

Complete Inventory Cycle Time: T0 = X0/c

Maximum Inventory : Imax= Xo(d/(E+d))

: Imax=

Example 62:

Mr. Ahmet Doğru is the Renault dealer in İzmir. After having explained the back

order model to him, he believes that his situation is one for which the assumptions of this

model hold true. Mr. Doğru has come up with these estimates:

C (annual demand) = 400 units

E (annual carrying cost for inventory expressed as MU/unit/year) = 800 MU

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Time

inve

ntor

yle

vel

x0s=no. of unitsbackordered

Imax

T k t1 ts

+-

complete inventorycycle

time during whichinventory is available

time during whichthere is

a stockout

S

B (cost per order) = 100 MU

d (cost to maintain 1 unit on back order status for 1 year) = 150 MU

a) Complete the optimum XO and S.

b) Complete the total annual cost of inventory system.

Solution:

a) X0 = =

= 25 cars

S = 25.166 (800/(800+150)) = 21.19 = 21 cars as back-order

Mr. Doğru would plan to have 21 cars back-ordered at the time each shipment of 25 cars,

the economic lot, arrived.

c) Total annual inventory cost with XO = 25 cars and back-orders = 21 cars

Ke = = = 3 178.88 MU

MODEL IV. Inventory Management With Planned Stock-outs (No Back-orders)

In manufacturing organisations, stock-outs can halt production, idling expensive labour and facilities, other costs of being out of stock include the cost of expediting replacement inventory the cost of a loss of sales and a loss of goodwill and other intangible costs.

The stock-out costs could be included in the EOQ and this would increase the order quantity, which also decreases the frequency of exposure to the risk (cost) of the stock-out.

Let d = cost of stock-out

EOQ (With stock-out) = XSO =

We can use carrying costs to imply something about “d”

Implied d/order = (Carrying cost/year)/(orders/year) d/order = [(X0/2)E]/N0

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We assume that the firm is willing to carry inventory up to the point where this carrying cost is just offset by the stock-out cost, i.e. same cost /order can be used as estimate of “d”.

Example 63:

A glassware manufacturer with an annual demand of 500 unit has ordering costs of 45 MU/order and carrying costs of 15MU/m –year

a) Compute the implied stock-out cost.b) Compute revised EOQ with stock out costs.

Solution:

a. XO= = = 55 units

Orders/yr. = C/X0 = 500/55 = 9 orders

Implied SO cost/order = d = [(X0/2)E]/N0 = [(55/2)15]/9 = 46 MU/order

b. EOQ (with SO) = = = 78 units

Example 64:

A firm has an annual demand of 1000 units, ordering cost of 10MU/order and carrying cost of 10MU/unit–year. Stock-out costs are estimated to be about 40MU each time when the firm has an exposure to stock-out. How much safety stock is justified by the carrying costs?

Solution: XO= = = 100 units

Orders /yr = C/X0 = 1000/100 = 10 orders/year

Stock-out costs = 40MU 10 orders = 400MU/year

At carrying cost of 10 MU/unit–year, the 400MU will fund (400 MU/year)/(10 MU/unit-year) = 40 units of safety stocks

SOLVED PROBLEMS

Example 1:

A contractor has to supply 10.000 bearings per day to an automobile manufacture. He finds that, when he starts a production run, he can produce 25.000 bearings per day. The cost

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of holding a bearing in stock for one year is 0.02 MU and the set up cost of a production run is 18.00 MU. How frequently should production runs be made?

Solution:

We assume that run sizes are constant, that a new run will be started whenever the inventory is zero and that the sole reason for producing for inventory is to obtain lower production costs.

Q0 = = =105 000 bearings

t0 = Q0/C = (105 000 bearings/day) / (10 000 bearings/day) = 10.5 days

Example 2:

A subcontractor undertakes to supply diesel engines to a truck manufacturer at the rate of 25 per day. There is a clause in the contract penalising him 10 MU /engine /day late for missing the scheduled delivery date. The finds that the cost of holding a completed engine in stock is 16 MU/month. His production process is such that each month (30 days) he starts batches of engines throw the shops, and all these engines are available for delivery any time after the end of the month

What should his inventory level be at the beginning of each month? (i.e. immediately after taking into stock the engines made in the previous month and then shipping engines to fill unsatisfied demand from the previous month.)

Solution:

In this problem we are concerned with balancing the costs of holding inventory against the costs of delayed deliveries to customer. Since the subcontractor has already decided to produce a batch of engines every month, he has fixed the size of each batch at 25 x 30 = 750 engines.

Imax = XO(d/(E+d)) = 30(25) [10/((16/10)+10)] = 712 engines

Example 3:

A contractor undertakes to supply diesel engines to a truck manufacturer at a rate of 25 per day. He finds that the cost of holding a completed engine in stock is16 MU per month, and there is a clause in the contract penalising him 10 MU/engine /day late for missing the scheduled delivery date. Production of engines is in batches, and each time anew batch is started there are set-up costs of 10.000 MU. How frequently should batches be started and what should be the initial inventory level at the time each batch is completed?

Solution:

E = 16/30 MU/day d = 10 MU/day s = 10 000 MU c = 25 engines/day

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XO = =

= 968.2458 (1.02632) = 993.73 Engines

S = Xo[E/(E+d)] = 993.73 [16/30/(16/30 + 10)] = 50.31 Engines

Imax = XO [d/(E+d)] = 993.73 [ 10/(16.30 + 10)] = 943.42 Engines

t0 = Xo/C = 993.73/25 = 39.75 = 40 days

It would be better to start a new batch approximately every 40 days. Thus the cost of inventory will be

Ke= = = 503.1MU

or Ke = Sd. = 50.31 ´ 10 = 503.1 MU

Example 4:

ABC-Lite, a manufacturing concern produces electric appliances. The company expects next years sales to be 180 000 units. Each run requires an outlay of 100 MU for machine set-up and each unit carried inventory costs 9 MU.

It is estimated that the cost of permitting a back order is 16 MU/ Unit/yr. Each back-order is filled as soon as the production run is completed.

Determine: a) The optimal size for each production run,

b) The maximum level of inventory that the manufacturer can expect to have on hand,

c) The time between runs.

Solution:

a) Qo = = = 2 500 units

b) Imax= Q0[d/(E+d)] = 2 500 [ 16/(9+16)] =2 500 (0.64) 1 600 units

c) to = Q0/C = 2 500/180 000 = 0.014 year = 5 days

Example 5:

An Operations Management of Nationwide Car Pentals must decide on the number

of vehcles of a certain model to alocate to his agency in the Güzelyurt area on a one-time

basis. The cars are obtained from an auto leasing firm at a cost of 20 MU/day. Naionwide

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rents the cars to its customers for 30 MU/day. If a car is not used, the auto leasing firm will

give nationwide an 8 MU rebate. Records of past demand have yielded the empirical

probability distribution shown. How many units of this model should Nationwide stock if it

seeks to balance the costs of overstocking and understocking?

Demand (# of cars) Prob. of demand P(C ) Cumulative prob. Of demand P(C )6 (or less) 0.00 1

7 0.03 18 0.07 0.979 0.15 0.90

10 0.20 0.7511 0.23 0.5512 0.15 0.3213 0.12 0.1714 0.05 0.05

15 (or more) 0.00 0.00

Solution:

If the Nationwide overstocks the loss/unit for every excess unit at the end of the period

will be

K0 = Cost/unit – SV/unit = 20 MU/unit – 8 MU/unit = 12 MU/unit

If the Nationwide under stocks, the opportunity cost for every unit Nationwide could

gain, but did not stock will be

Ku = Price/unit - Cost/unit = 30 MU/unit – 20 MU/unit = 10 MU/unit

Then the critical probability is

P(c ) = K0/ K0 + Ku = 12 / 12+ 10 = 0.545

We chose to stock 11 cars because our calculated P(C ) value of 0.545 is closer o

cumulative probability of 0.55. We would be inclined to stock less rather than more because

the cost of overstocking (12 MU) is >the cost of under stocking (10 MU)

SOR = 0.545

Because the SOR is the complement of service level

SL = 100% - SOR% SL= 1 – 0.545 = 0.455

Service level would be 45.5%

Example 6:

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DEMIR SPORTSWEAR can purchase a special shipment of wool caps for 1.85 MU

each. The caps can be sold for 4.95 MU during the fall season, but any that have not been sold

by NEW YEAR will be reduced to 1.50 MU. The following probability distribution has been

estimated for demand for these caps prior to New Year. Caps must be purchased in dozens.

Assume that any caps unsold by new year can be sold at the lower price. How many caps

should DEMIR purchase?

Demand (Dozens)

Prob. That demand will equal this amount

Prob. That demand will equal or exceed this level

4 0.05 1

5 0.15 0.95

6 0.25 0.80

7 0.25 0.55

8 0.15 0.30

9 0.10 0.15

10 0.05 0.05

Solution:

If the store overstocks, the loss per unit for every excess unit at the end of the year

will be

K0 = Cost/unit – SV/unit = 1.85 MU/unit – 1.50 MU/unit = 0.35 MU/unit

If the store understocks, the opportunity cost for every unit the store could sell but did

not stock will be

Ku = Price/unit - Cost/unit = 4.95 MU/unit – 1.85 MU/unit = 3.10 MU/unit

The critical probability

P(c ) = K0/ K0 + Ku = 0.35 / 0.35 + 3.10 = 0.101

DEMIR SPORTSWEAR must purchase 9 dozens because 9 dozens of caps has a cumulative

probability of 0.15 which is >critical probability (0.101)

Example 7:

Hot cup cafe is faced with the decision of how many cups of coffee to prepare before

a football game at the local stadium. Each cup of coffee costs 0.03 MU per cup and sells for

0.12 MU per cup. Past records indicate that 38000 cups are enough to prevent any shortage

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and this is the number prepared before each game. Unsold coffee is disposed of at a total loss.

The following data summarizes the sales history

a. How many cups of coffee should be prepared prior to each game?

b. What is the long-run expected loss under the current policy?

Demand Frequency27000 728000 1229000 2030000 2531000 1532000 1033000 534000 535000 1

Solution:

K0 = Cost/unit – SV/unit = 0.12 MU/unit – 0.03 MU/unit = 0.09 MU/unit

Ku = Price/unit - Cost/unit = 0.03 MU/unit – 0.00 MU/unit = 0.03 MU/unit

P(c ) = K0/ K0 + Ku = 0.03 / 0.03 + 0.09 = 0.25

Demand Relative probability

Prob. (at least “X” units demanded)

27000 0.07 128000 0.12 0.9329000 0.20 0.8130000 0.25 0.6131000 0.15 0.3632000 0.10 0.2133000 0.05 0.1134000 0.05 0.0635000 0.01 0.0136000 0.00 0.00

The critical fractile is 0.25. This value is bracketed by 0.36 and 0.21, these values

correspond to inventory levels of 31000 and 32000 respectively. Select as the optimal

inventory level of two levels. The optimal inventory level is 31000 units.

b) To determine the long-run expected loss under the current policy, the difference between

this cost and the cost of the optimal policy is the long-run expected loss.

* The cost of the current policy = 38000*0.03 = 1140 MU

* The cost of the optimal policy = 31000*0.03 = 930 MU

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Thus the long-run expected loss = 1140 – 930 = 210 MU.

Example 8:

The City Chronicle has a daily newspaper demand that varies from between 20000

and 24000 copies per day. The paper costs 0.08 MU per issue to produce and generates

revenue of 0.20 MU per issue. Unsold papers have no value.

a. What is the optimal level of papers to stock?

b. What service level would that optimal level corresponds with?

Solution:

Ku.P( C)* = Ko[I - P( C)]

P( C)* = Ko/Ko+Ku

Where,

K0 = Cost/unit – SV/unit= 0.08 – 0 = 0.08

Ku = Price/unit - Cost/unit = 0.20 – 0.08 = 0.12

P( C)* = Ko/Ko+Ku = 0.08 /0.08 + 0.12 = 0.40

The 0.40 is the equating probability that demand will be exceeded and represents a

stock out risk established on the basis of the basis of costs given.

Thus the corresponding SERVİCE LEVEL is

Service level = 1 – Stock out risk

SL = 1-0.40 = 0.60

Iopt = C min + % SL (inventory)

Iopt = 20000 + 0.6 (24000 – 20000) = 22400 units.

Example 9:

BMW Auto sales is offering a special car attachment at the unheard-of-price 100

MU/unit. The attachment costs MNW 70 MU. Unsold units can be salvaged for 30 MU/unit.

Management has projected the following weekly demand pattern:

Weekly demand (units) Probability of demand35 0.10

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36 0.1537 0.2538 0.2539 0.1540 0.10

41+ 01.00

a. Using marginal analysis, determine the otimal stock level

b. Suppose that restocking is a continual process. If the unit is not sold in one period, it is

held over the next period. However, there is an additional cost of 15 MU for handling

and storage. What is the optimal stock level under these conditions? (Use marginal

analysis and assume that any unsold unit is held over for one period only)

Solution:

a. Selling price/unit = 100 MU

Cost/unit = 70MU

Salvage value /unit = 30 MU

K0 = Cost/unit – SV/unit= 70 – 30 = 40 MU (marginal loss)

Ku = Price/unit - Cost/unit = 100 – 70 = 30 MU (marginal profit)

P( C)* = Ko/Ko+Ku =40 / 40+30 = ~0.57 (critical ratio)

Weekly demand (units) Probability of demand Cum.Prob.35 0.10 1.0036 0.15 0.9037 0.25 0.7538 0.25 0.5039 0.15 0.2540 0.10 0.10

41+ 0 01.00

The optimal inventory level equals to 37 units.

b. K0 = Cost/unit – SV/unit=85– 30 = 55 MU (marginal loss)

Ku = Price/unit - Cost/unit = 100 – 70 = 30 MU (marginal profit)

P( C)* = Ko/Ko+Ku =55 / 55+30 = ~0.65 (critical ratio)

So, the optimal inventory level equals to 37 units.

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Example 10:

The ABC Co. is planning to stock a new product. The Co. Has developed the following

information:

Annual usage = 5400 units

Cost of the product = 365 MU/unit

Ordering cost = 55 MU/order

Carrying cost = 28% /year of inventory value held.

a Determine the optimal number of units per order

b. Find the optimal number of orders/year

c. Find the annual total inventory cost

Solution:

a. X0 = √2CB/zp = √2*5400*55/365*0.28 = ~76 units/order

b. N0 = C/X0 =5400 / 76 = ~71 orders/year

c. Ke= √2CBzp = √2*5400*55*365*0.28 = 7791.46 MU/year.

Example 11:

Holding costs are 35 MU/unit/year. The ordering cost is 120 MU/order and sales are

relatively constant at 400 month.

a. What is the optimal order quantity?

b. What is the annual total inventory cost?

Solution:

a. X0 = √2CB/E= √2*(400*12)*120/35 = 181.42 units/order

b. Ke= √2CBE= √2*(400*12)*120*35 = 6349.80 MU/year

Example 12:

Azim furniture company handles several lines of furniture, one of which is the popular

Layback Model TT chair. The manager, Mr. Farmerson, has decided to determine by use of

the EOQ model the best quantity to obtain in each order. Mr. Farmerson has determined from

past invoices that he has sold about 200 chair during each of the past five years at a fairly

uniform rate and he expects to continue at that rate. He has estimated that preparation of an

order and other variable costs associated with each order are about 10 MU, and it costs him

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about 1.5 % per month (or 18% per year) to hold items in stock. His cost for the chair is 87

MU.

a. How many layback chairs should be ordered each time?

b. How many orders would there be?

c. Determine the approximate lenght of a supply order in days?

d. Calculate the minimum total inventory cost

e. Show and verify that the annual holding cost is equal to the annual ordering cost (due

to rounding, show these costs are approximately equal)

Solution:

a. X0 = √2CB/zp = √2*200*10 /0.18*87 = 15.98 =~16 chairs/order

b. N0 = C/X0 =200 /16= 12.5 orders/year

c. t0 = X0/C * 365 = 16 /200 *365 = 29.2 days

d. Ke= √2CBzp = √2*200*10*0.18*87 = 250.28 MU/year

e. N0 *B = 12.5 * 10 = 125 MU

x/2*zp = 15.98/2 * 0.18*87 = 125.1 MU

Example 13:

A. Leyla Tas has determined that the annual demand for #6 screws is 100000 screws.

Leyla, estimates that it costs 10 MU every time when an order is placed. This cost includes

wages, the cost of the forms used in placing the order and so on. Furthermore, she estimates

that the cost of carrying are screw in inventory for a year is one-half of 0.01 MU. Assume that

the demand is constant throughout the year.

a. How many #6 screws should Leyla order at a time to minimize total inventory cost?

b. How many orders per year would be placed? What would the annual ordering cost be?

c. What would the average inventory be? What would the annual holding cost be?

Solution:

a. E= 0.01/2 = 0.005

X0 = √2CB/E= √2*100000*10 / 0.005 = 20000 screws/order

b. N0 = C/X0 = 100000/20000= 5 orders/year

Total ordering cost = N*B = 5*10 = 50 MU/year

c. Average inventory = x/2 = 20000/2 = 10000 units

Total holding cost = x/2*E= 10000 * 0.005 = 50 MU/year

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B. It takes approximately 8 working days for an order of #6 screws to arrive once the

order has been placed. The demand is fairly constant, and on the average the store sells 500

screws each day. What is the ROP?

Solution:

ROP = Ro= use rate * lead time = c*tlt = 500 screws/day * 8 days = 4 000 screws.

C. The manager believes that Leyla places too many orders for screws /year. He

believes that an order should be placed only twice/year. If Leyla follows her manager`s

policy, how much more would this cost every year over the ordering policy that she

developed, if only two orders were placed each year, what effect would this have on the ROP?

Solution:

Twice a year = Ke= NB + X/2*E = 2*10 + 50000/2*0.005 = 20 + 125 = 145

5 times a year = Ke= √2CBE = √2*100000*10*0.005 = 100 MU

Extra cost for manager’s offer = 45 MU

No effect on ROP.

P R O B L E M S

1. The Izmir Petrol Co. uses 600 quarts of oil per month. The unit cost of a quart of oil to the co. is 0.40 MU; the cost of placing an order is 10 MU.Carrying charges are set at 22% of unit cost. For all practical purposes demand is continuous and uniform.

a) Determine the optimum order size, (EOQ)b) What is the total annual cost of optimum order policy, Ke?c) If lead-time is one month, what is the re-order pt.?

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d) What is the annual cost of the total system? (This includes cost to meet demand)

2. A local maintenance firm uses 500 screwdrivers per month. The unit cost at a screwdriver is 0.50 MU. Ordering cost is 8 MU/order. Lead-time for processing an order is one month. The firm estimates that inventory holding charges a 0.20 MU per 1.00 MU valuation per unit – year. For purposes of maintaining an emergency stock, management insists that the stock level never fall below 50 units.

a) Determine the economic order quantity, (EOQ) b) Determine the optimum time between orders. c) What is the reorder point? d) What is the total annual cost of this company’s system?

3. Product X is a standard item in Azim’s inventory. Final assembly is performed on an assembly line, which is in operation every day. One of the components of product X, component HD08 is produced in another department at the rate of 100 units per day. The assembly line uses HD08 at the rate of 40 per day. The firm operates 250 days each year. Set up costs total 50 MU and the average annual holding cost is 0.50 MU/unit-year. Each component HD08 costs 7 MU and requires a lead-time of 7 days. Using this data, determine the following;

a) The optimal production lot size , OLS, b) The reorder point, Xro

c) The total annual cost of the optimal lot-size policy, Ke, d) What is the total annual cost? (k)

4. Shoe Plaza produces one style at shoe for which demand has averaged 1500 units per month. Each pair of these shoes that’s produced costs shoe plaza 20 MU. Holding costs are set at 6% of cost per pair per year. Each set up requires an outlay of 400mu. Assuming no shortages are permitted, lead-time is two-months and “Replacement-Runs” is instantaneous, determine the following

a) The optimum production lot- sizeb) The total yearly cost to meet demand,c) The number of production runs per year,d) The time between orders,e) The reorder point.

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