economic production quantity (epq).pptx
DESCRIPTION
financeoperations managementTRANSCRIPT
Economic Production Quantity (EPQ)
• Production done in batches or lots
production capacity > usage or demand ratefor a part for the part
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Assumptions of EPQsimilar to EOQexcept orders are received
incrementally during production
Economic Production Quantity (EPQ)
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Economic Production Quantity (EPQ)
3
dpp
QI
p
Q
d
Q
SQ
DH
I
max
max
;lengthRun ;length Cycle
2Cost
Setup
Annual
Cost
Holding
Annual
TC
dp
p
H
DSQ
20TC = Total annual cost
Q = Order quantity (units)H = Annual holding cost
per unitD = Annual DemandS = Ordering (or setup) cost per order
Q0 = Optimal run or order quantityp = Production rated = Usage or demand rateImax = Maximum inventory level
EPQ ExampleHoldit Inc. produces reusable shopping bags. Demand is 20,000 bags per day, 5 days per week, 50 weeks per year. Production is 50,000 per day. The setup cost is $200 and the annual holding cost rate is $.55 per bag. Calculate the EPQ, the total cost, the cycle length and optimal production run length.
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H = $0.55 per bag S = $200 D = 20,000 bags x 50 wks x 5 days
d = 20,000 bags per day p = 50,000 bags per day
dp
p
H
DSQ
20
850,772050
50
55.
)200)(000,000,5(20
GG
GQ
EPQ Example
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H = $0.55 per bag d = 20,000 bags per day p = 50,000 bags per day
S = $200 D = 20,000 bags x 50 wks x 5 days
dpp
QIS
Q
DH
I
max
max 2
TC
bags 46,710 30000000,50
850,77max I
$25,690 002850,77
5)55(.
2
710,46TC
million
Holdit Inc. produces reusable shopping bags. Demand is 20,000 bags per day, 5 days per wk, 50 wks per yr. Production is 50,000 per day. Setup cost is $200 and annual holding cost rate is $.55 per bag. Calculate total cost.
EPQ ExampleHoldit Inc. produces reusable shopping bags. Demand is 20,000 bags per day, 5 days per week, 50 weeks per year. Production is 50,000 per day. The setup cost is $200 and the annual holding cost rate is $.55 per bag. Calculate cycle length and optimal production run length.
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H = $0.55 per bag S = $200 D = 20,000 bags x 50 wks x 5 days
d = 20,000 bags per day p = 50,000 bags per day
p
Q
d
Q lengthRun ;length Cycle
days 3.89every 000,20
850,77length Cycle
orderper days 56.1000,50
850,77lengthRun
EOQ with Quantity Discounts• Price reductions are often offered as incentive to buy
larger quantities• Weigh benefits of reduced purchase price against
increased holding cost
R = per unit price of the itemD = annual demand
Annualholdingcost
PurchasingcostTC = +
Q2
H DQ
STC = +
+Annualorderingcost
RD +
Total Cost with Purchase Cost
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Cost
EOQ
TC with PD
TC without PD
PD
0 Quantity
Adding Purchasing costdoesn’t change EOQ
Total Cost with Quantity Discounts
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Best Purchase Quantity Procedure
begin with the lowest unit price
compute the EOQ for each price range
stop when find a feasible EOQ
Is EOQ for the lowest unit price
feasible?Yes:
it is the optimal order
quantity
No: compare total cost at
all break
quantities
larger than
feasible EOQ
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The quantity that yields the lowest total cost is optimum
Carrying Cost are Constant
• There will be a single minimum point.• All curves will have their minimum point
at the same quantity.
1. Compute the minimum point.
2. Only one of the unit prices will have the minimum point in its feasible range since ranges do not overlap. Identify the range.
a) If the feasible minimum point is on the lowest price range, that is the optimal order quantity
b) If on any other range, compute the total cost for the minimum point and for the price breaks of all lower unit costs.
Example: Fixed Carrying CostThe maintenance department of a large hospital uses
about 816 cases of liquid cleanser annually. Ordering costs are $12, carrying costs are $4 per case a year, and the new price schedule indicates that orders of less than 50 cases will cost $20 per case, 50 to 79 case will cost $18 per case, 80 to 99 cases will cost $17 per case, and larger orders will cost $16 per case. Determine the optimal order quantity and the total cost.
Order Quantity(units) Price/unit($)
0 to 2,499 $1.20
2,500 to 3,999 1.00
4,000 or more .98
D = 816 S = $12 H = $4 case per year
1. Compute the common minimum
cases 7069.97 = 4
2(816)(12) =
H
2DS = QO
2.
TC (70) = (70/2)4 + (816/70)12 + 18(816) = $14 968 Since there are lower cost ranges, each must be checked against the minimum cost generated by 70 cases at $18.
TC (80) = (80/2)4 + (816/80)12 + 17(816) = $14 154
TC (100) = (100/2)4 + (816/100)2 + 16(816) = $13 354
Example: Quantity DiscountsBelow is a quantity discount schedule for an item with
an annual demand of 10,000 units that a company orders regularly at an ordering cost of $4. The annual holding cost is 2% of the purchase price per year. Determine the optimal order quantity.
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Order Quantity(units) Price/unit($)0 to 2,499 $1.202,500 to 3,999 1.004,000 or more .98
D = 10,000 units S = $4
units 2,000 = 0.02(1.00)
4)2(10,000)( =
H
2DS = QO
units 2,020 = 0.02(0.98)
4)2(10,000)( =
H
2DS = QO
H = .02R R = $1.20, 1.00, 0.98
Interval from 0 to 2499, the Qo value is feasible
Interval from 2500-3999, Qo value is NOT feasible
Interval from 4000 & up, Qo value is NOT feasible
Example: Quantity Discounts
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Order Quantity Price/unit($)0 to 2,499 $1.202,500 to 3,999 1.004,000 or more .98
units 1,826 = 0.02(1.20)
4)2(10,000)( =
H
2DS = QO
Quantity Discount Models
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2500 4000
An
nu
al cost
0 Quantity
EOQs (not feasible)
1st break quantity
2nd break quantity
1st range total cost
curve
2nd range total cost curve
3rd range total cost curve
EOQ
TC(0-2499) = (1826/2)(0.02*1.20) + (10000/1826)*4+(10000*1.20) = $12,043.82
TC(2500-3999)= $10,041
TC(4000&more)= $9,949.20
Therefore the optimal order quantity is 4000 units
Example: Quantity Discounts
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Q2H D
QSTC = + RD +
Concept CheckWhich of the following is FALSE about EOQ?
A. It determines how many to order.B. The EOQ always results in the lowest total
cost.C. The model minimizes total cost by
balancing carrying and order costs.D. The model is robust and works even if all
assumptions are not exact.
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Concept CheckWhich is NOT a difference between EOQ and
EPQ?
A. A different formula is used.B. EPQ is used mainly for producing batches,
and EOQ is for receiving orders.C. Quantity is received gradually in EPQ.D. Demand can be variable for EPQ but not for
EOQ.
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Concept CheckWhich is NOT an assumption of both EOQ and EPQ?
A. Demand is known with certainty and is constant over time
B. No shortages are allowedC. Order quantity is received all at onceD. Lead time for the receipt of orders is constant
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What’s next?• EOQ models give HOW MANY to order• Now look at WHEN to order
– Reorder Point (ROP)
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d = Demand rate (units per day or week)LT = Lead time (in days or weeks)Note: Demand and lead time must have the same time units.
ROP = d LT
Example: ROPAnnual Demand = 1,000 units
Days per year = 365
Lead time = 7 days
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units/day 2.74 = days/year 365
units/year 1,000 = d
units 20or 19.18=(7days)units/day 2.74=L =ROP d
When inventory level reaches 20 units, place the next order.
Fixed Order Quantity/Reorder Point Model
Safety Stock1. Variability of
demand and lead time 2. Service Level2a. Lead time service level
2b. Annual service level
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Reorder Point = Expected demand + Safety Stock (ROP) during lead time
When to Reorder with EOQ Ordering
• Reorder Point – When inventory level drops to this amount, the item is reordered.
• Safety Stock - Stock that is held in excess of expected demand due to variability of demand and/or lead time.
• Service Level – Probability demand will not exceed supply.– Lead time service level: probability that demand will not exceed
supply during lead time. – Annual service level: percentage of annual demand filled.
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Determinants of the Reorder PointRate of
demandLead time
Demand and/or lead
time variability
Stockout risk (safety stock)
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ROP Expected demandSafety stockduring lead time
Safety Stock
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LT Time
Expected demandduring lead time
Maximum probable demandduring lead time
ROP
Qu
an
tity
Safety stock
Safety stock reduces risk ofstockout during lead time
Reorder Point
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z = Safety factor; number of standard deviations above expected demanddLT = The standard deviation of demand during lead time
Safety Stock = z.dLT
The ROP based on a normalDistribution of lead time demand
Demand During Lead Time
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ROP with Lead Time Service Level• variable demand during a lead time
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ROP = expected demand during lead time + safety stock
z = Safety factor; number of standard deviations above expected demanddLT = The standard deviation of demand during lead time
ROP = + z.dLT
ROP with Lead Time Service Level• variable demand and constant lead time
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ROP = (average demand x lead time) + z x st. dev. of demand
during lead time(demand and lead time measures in same time units)
sd = standard deviation of demand per day
sdLT = sd LT
ROP with Lead Time Service Level• both demand and lead time are variable
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ROP = (avg. demand x avg. lead time) + z x st. dev. of demand
in lead time(demand and lead time measures in same time units)
sd= standard deviation of demand per day
sLT= standard deviation of lead time
sdLT = (average lead time x sd2)
+ (average daily demand) 2sLT2
Example 1: ROP with Lead Time Service Level
Calculate the ROP required to achieve a 95% service level for a product with average demand of 350 units per week and a standard deviation of demand during lead time of 10. Lead time averages one week.
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From Table 12-3 (p434), z for 95% = 1.65
ROP = 350 + ZsdLT
= 350 + 1.65 (10)
= 350 + 16.5 = 366.5 ≈ 367
A new order should be placed when inventory level reaches 367 units.
Example 2: ROP with Lead Time Service Level
Calculate the ROP and amount of safety stock required to achieve a 90% service level for a product with variable demand that averages 15 units per day with a standard deviation of 5. Lead time is consistently 2 days.
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From Table 12-3 (p434), z for 90% = 1.28
ROP = (15 units x 2 days) + ZsdLT
= 30 + 1.28 ( 2) (5)
= 30 + 8.96 = 38.96 ≈ 39
Safety stock is about 9 units and a new order should be placed when inventory level reaches 39 units.
Example 3: ROP with Lead Time Service Level
Calculate the ROP for a product that has an average demand of 150 units per day and a standard deviation of 16. Lead time averages 5 days, with a standard deviation of 2. The company wants no more than 5% stockouts.
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service level = 1 – 5% = 95%From Table 12-3 (p434), z for 95% = 1.65
Place a new order when inventory level reaches 1004 units
ROP = (150 units x 5 days) + 1.65sdlt
= (150 x 5) + 1.65 (5 days x 162) + (1502 x 12)
= 750 + 1.65 (154) = 1,004 units
ROP Using Annual Service Level1. Calculate
2. Use a table to find the z value associated with E(z)
3. Use the z value in the appropriate ROP formula,
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dLT
annualSLQzE
)1(
)(
dLTzROP timeleadduringdemandexpected
SLannual = annual service level E(z) = standardized expected number of units short during an order cycle.
Min/Max model• similar to fixed order-quantity/reorder point
(ROP) model• difference:
– if at order time, Q on hand < min (ROP), then order quantity = max – Q on hand(max EOQ + ROP)
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Inventory Models• EOQ/ROP model
– Order size constant, time between orders changes
• Fixed Order Interval/Order up to Level Model– orders placed at fixed time intervals– determine how much to order to bring inventory level up
to a predetermined point (order up to level)– used widely for retail– consider expected demand during lead time, safety
stock, and amount on hand– demand or lead time can be variable
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Comparing Inventory Models
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EOQ/ROP
Fixed Interval/
Order up to
Disadvantagesrequires a larger safety stockincreases carrying costcosts of periodic reviews
Fixed Order Interval: Benefits and Disadvantages
• Benefits– grouping items from same supplier
can reduce ordering/shipping costs– practical when inventories
cannot be closely monitored
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Fixed Order Interval/Order up to Level Model
Determining the order interval
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OI = order interval (in fraction of a year)S = fixed ordering cost per purchase orders = variable ordering cost per SKU included in the order (line item)
(assume s is the same for every SKU)
n = n number of SKUs purchased from the supplierRj = unit cost of SKUj , j = 1, …, ni = annual holding cost rateDj = annual demand of SKUj , j = 1, …., n
Total Annual Inventory Cost:
TC =
Optimal Order Interval:
OInsSiR
OIDj
j 1)(.
2
.
jj RDi
nsSOI
)(2*
Fixed Order Interval/Order up to Level Model
Determining the Order up to Level
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LTOIzLTOId d
Stock
Safety
interval
protection during
demand Expected
I
handon Amount IQ
max
max
= Average daily or weekly or monthly demand OI = Order interval (length of time between ordersLT = Lead time in days or weeks or monthsz = Safety factor; # of standard deviations above expected demandd = Standard deviation of daily or weekly or monthly demand
= 20 (30 + 10) + (2.32) (4) 30 + 10 = 800 + 2.32 (25.298) = 858.7 or 859 units stock up to level
Example: Fixed Order Interval ModelAverage daily demand for a product is 20 units, with a standard deviation of 4 units. The order interval is 30 days, and lead time is 10 days. Desired service level is 99%. If there are currently 200 units on hand, how many should be ordered?
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LTOIzLTOId d maxI
maxI
Amount to order = 859 – 200 = 659 units
Coordinated Periodic Review Model
• determines an order interval (OI) and order up to level for reviewing every stock keeping unit (SKU)– calculate a multiple (mi ) of OI for each SKUi
– use this to determine the optimal OI for each SKUi
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To use:Compare on hand inventory of each SKU to its ROP
(forecast demand for next OI + lead time + safety stock)if on hand is less: order a quantity that brings the on
hand level to SKU’s order up-to levelthe order up-to level is enough for the next OI + LT.
Single Period Model• Single period model
– model for ordering of perishables and other items with limited useful lives
• Shortage cost Cs
– generally the unrealized profits per unit– Revenue per unit – Cost per unit
• Excess cost Ce
– cost per unit - salvage per unit for items left over at the end of a period
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GOAL = find order quantity (stock level) that minimizes total excess and shortage costs.
Single Period Model• Continuous stocking levels
– Identifies optimal stocking levels– Optimal stocking level balances unit
shortage and excess cost
• Discrete stocking levels– Desired service level is equaled or
exceeded• Compare service level to cumulative probability of
demand48
Optimal Stocking Level
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Service Level
So
Quantity
Ce
Cs
Balance point
Service level (SL) =Cs
Cs + CeCs = Shortage cost per unitCe = Excess cost per unit
So = Optimum stocking level (i.e., order quantity
Example 1: Single Period Model
• Ce = $0.20 per unit• Cs = $0.60 per unit• Service level = Cs/(Cs+Ce) = .6/(.6+.2)• Service level = .75
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Service Level = 75%
Quantity
Ce
Cs
Stockout risk = 1.00 – 0.75 = 0.25
Example 2: Single Period Model
The Poisson table (App. B, Table C) Cs is unknown Ce = $500
provides these values for a mean of 2.0: Number of FailuresCumulative Probability0 .1351 .4062 .6773 .8574 .9475 .983
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s.406, so .406($500 )$500s
ss
CC C
C
Optimum stock level = 2, then SL between
Cs = $343.17
The range of shortage cost is $343.17 to $1,047.99.
.677, so .677($500 )$500s
s ss
CC C
C
Cs = $1,047.99.
A company usually carries 2 units of a spare part that costs $500 and has no salvage value. Part failures can be modeled by a Poisson distribution with a mean of 2 failures during the useful life of the equipment. Estimate the range of shortage cost for which stocking 2 units of this spare part is optimal.
Review: Inventory Models• EOQ models used to determine order size
– Simple model for many types of inventory– Trade-off between carrying and ordering costs– Quantity discount model adds purchasing costs
and compares total cost for various order sizes (that is still a feasible EOQ)
• EPQ models used to determine production lot size– Used when producing and depleting items at
same time– Trade-off between carrying and setup costs– Consider production and usage rate
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Review: Inventory Models• ROP (reorder point)
– Determines at what quantity (when) to re-order – Consider expected demand during lead time and safety stock – Trade-off cost of carrying safety stock & risk of stockout
• Fixed Order Interval Model– Used when orders placed at fixed time intervals – determine
how much to order– Used widely for retail– Consider expected demand during lead time, safety stock,
and amount on hand– Demand or lead time can be variable – Need more safety stock, but not continuous monitoring
• Single-Period Model– Determines at what quantity (when) to re-order – Used when can’t carry goods to next period (e.g.. perishables)– Trade-off cost of shortages & of excess (wasted) inventory
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Define the term inventory and list the major reasons for holding inventories.
Discuss the objectives of inventory management. List the main requirements for effective inventory
management. Describe the A-B-C approach and perform it.Describe Basic Inventory Control SystemsBe able to describe and solve problems using:
EOQ, EPQ, ROP, Fixed Order Interval Model, Single Period Model.
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