DETERMINISTIC MODELS FOR
INVENTORY CONTROL
•
Basically, Deterministic models keeps the condition that
all variables take values which are known exactly.
However, Deterministic models do remove some of the
assumptions made in the classical analysis or EOQ.
1.
Model for discounted unit cost: Many suppliers quote
lower prices for larger orders, so a procedure is
described for finding the overall lower cost.
Production Order Quantity (Finite Replenishment rate):
This is typical of a finished goods store at the end of a
production line. Here units arrive at a finite rate and
stock accumulates during a production run. Remove the
assumption of instantaneous replenishment used for
EOQ and calculate new optimal batch sizes.
2.
• Quantity Discount in Unit Cost
Remove EOQ’s assumption that all costs are
fixed (constant values which never change)
Address problems where costs vary with the
quantity ordered
Two main approaches: Valid Total Cost
Curve & Rising Delivery Cost
• Valid Total Cost Curve:
In practice, normally a sliding scale of unit costs is
applied, so that every unit bought has a different
price
Frequently the unit cost decreases in steps with
supplier offering a reduced price on all units if more
than a certain number are bought
There are two main characteristics of valid cost
curve:
1. The valid total cost curve always rises to the
left of a valid minimum
2. There are two possible positions for the overall
minimum cost, either at valid minimum or cost
breakpoint
1. Valid Minimum:
The minimum point on the cost curve is
within the range of valid order quantities for
this particular unit cost
2. Invalid Minimum:
The minimum point on the cost curve falls
outside the valid order range for this
particular unit cost
•
1.
2.
3.
4.
5.
6.
7.
Procedure for finding the lowest Total
Cost with varying unit cost
Take the next lowest unit cost curve
Calculate the minimum point, Q= 2DS/ic
Is this point valid??
If not, calculate cost at breakpoint to the left of
valid range
If yes, calculate the cost of the valid minimum
Compare the costs of all the points considered
& select lowest
Finish
Quantity Discounts
Price per unit decreases as
order quantity increases
CoD CcQ
TC =
+
Q
2
+ PD
where
P = per unit price of the item
D = annual demand
ORDER SIZE
PRICE
0 - 99
100 – 199
200+
$10
8 (d1)
6 (d2)
Quantity Discount
Model
TC = ($10 )
TC (d1 = $8 )
Inventory cost ($)
TC (d2 = $6 )
Carrying cost
Ordering cost
Q(d1 ) = 100 Qopt
Q(d2 ) = 200
Quantity Discount:
Example
QUANTITY
PRICE
1 - 49
50 - 89
90+
$1,400
1,100
900
Co = $2,500
Cc = $190 per computer
D = 200
Q = 2DS/ic = (2)(2500)(200)/(190) = 72.5 pcs
So, calculating the cost for Q=72.5:
TC
= P x D + (Co x D)/Q + (Cc x Q)/2
= (1100)(200) + (2500x 200)/72.5+ (190x72.5)/2)
= $233784 a year (point A)
So, calculating the cost for Q=90:
TC
= P x D + (Co x D)/Q + (Cc x Q)/2
= (900)(200) + (2500x 200)/90+ (190x90)/2)
= $194105 a year (point B)
• Exercise:
Annual demand for an item is 2000 units, each
order costs $10 to place and annual holding cost
is 40% of unit cost. The unit cost depends on the
quantity ordered as follows:
 Unit cost is $1 for order quantities less than 500
 $0.80 for quantities between 500 and 999 each
 $0.60 for quantities of 1000 or more
What is the optimal ordering policy??
Listing the variables:
D = 2000 units a year
S = $10 per order
i = 40% of unit cost a year
C = ?? (depend on order quantities)
•
Notice that:
–
Total cost is:
TC = c x D + (S x D)/Q + (ic x Q)/2
Follow the Procedure:
Taking the lowest cost curve:
c = $0.60, valid for Q of 1000 or more
Q = (2)(D)(S)/(ic) = (2)(2000)(10)/(0.4)(0.60) = 408.2
This is an invalid minimum as Q is not greater than 1000
So, calculating the cost of the breakpoint:
TC = c x D + (S x D)/Q + (ic x Q)/2
= (0.6)(2000) + (10 x 2000)/1000 + (0.4 x 0.6 x 1000)/2
= $1340 a year (point A)
• Taking the next lowest cost curve:
c = $0.80, valid for Q between 500 and 999
Q = 2DS/ic = (2)(2000)(10)/(0.4)(0.80)
= 353.6
This is an invalid minimum as Q is not between 500 and 999
So, calculating the cost of the break point at the lower end:
TC
= c x D + (S x D)/Q + (ic x Q)/2
= (0.8)(2000) + (10 x 2000)/500 + (0.4x0.8x500)/2
= $1720 a year (point B)
• Taking the next lowest cost curve:
c = $1.00, valid for Q less than 500
Q = 2DS/ic = (2)(2000)(10)/(0.4)(1.00)
= 316.2
This is valid minimum as Q is less than 500
Then, calculating the cost at this valid minimum:
TC = c x D + (S x D)/Q + (ic x Q)/2
=(1.00)(2000)+(10 x 2000)/316.2+ (0.4x1.00x316.2)/2
= $2126.49 a year (Point C)
• How to Choose the best ordering policy?
Compare the results at point A, B & C:
1. Point A: Q=1000, cost= $1340 a year
2. Point B: Q=500, cost= $1720 a year
3. Point C: Q=316.2, cost= $2126.49 a year
Therefore, the best policy is to order batches
of 1000 units, placing order every six months
with total annual costs of $1340
• Exercise 2:
A company works for 50 weeks a year during which
demand for a product is constant at 10 units a week.
The cost of placing order, including delivery charges
is estimated to be $150. The company aims for 20%
holding cost on assets employed. The supplier of the
item quotes a basic price of $250 a unit with discount
of 10% on orders of 50 units or more, 15% on orders
of 150 units or more and 20% on orders of 500 units
or more. What is the optimal order quantity for the
item??
Listing the variables:
D = 10 x 50 = 500 units a year
S = $150 per order
i = 20% x c (unit cost)
Follow the Procedure:
Taking the lowest cost curve:
c = $200, valid for Q of 500 or more
Q = (2)(D)(S)/(ic) = (2)(500)(150)/(0.2)(200) = 61.2
This is an invalid minimum as Q is not greater than 500
So, calculating the cost of the breakpoint to the left, Q=500:
TC = c x D + (S x D)/Q + (ic x Q)/2
= (200)(500) + (150 x 500)/500 + (0.2 x 200 x 500)/2
= $110,150 a year (point A)
Taking the next lowest cost curve:
c = $212.50, valid for Q between 150 and 500
Q = 2DS/ic = (2)(500)(150)/(0.2)(212.50)
= 59.4
This is an invalid minimum as Q is not between 150 and 500
So, calculating the cost of the break point to the left, Q=150:
TC = c x D + (S x D)/Q + (ic x Q)/2
= (212.50)(500) + (150 x 500)/150 + (0.2x212.50x150)/2
= $109,938 a year (point B)
Taking the next lowest cost curve:
c = $225, valid for Q between 50 and 150
Q = 2DS/ic = (2)(500)(150)/(0.2)(225)
= 57.7
This is a valid minimum as Q is between 50 and 150
So, calculating the cost at this valid minimum:
TC = c x D + (S x D)/Q + (ic x Q)/2
= (225)(500) + (150 x 500)/57.7 + (0.2x225x57.7)/2
= $115,098 a year (Point C)
Compare the results at point A, B & C:
1. Point A: Q=500, cost= $110,150 a year
2. Point B: Q=150, cost= $109,938 a year
3. Point C: Q=57.7, cost= $115,098 a year
So, Choose to order in batches of 150 units with annual
cost of $110,000
• Rising Delivery Cost
Extended version of discounted price analysis. The
only difference is that the valid cost curve is reversed,
so we look for optimal order quantities to the left of a
valid minimum and calculate the cost of break points
at the upper limit of valid ranges. Address changes in
the reorder cost due to the amount of delivery
Example:
Pasir ent., uses 4 tonnes of fine industrial sand every day.
This sand costs $20 a tonne to buy, and $1.90 a tonne to
store for a day. Deliveries are made by modified lorries
which carry up to 15 tonnes, and each delivery of a load
or part load costs $200. Find the cheapest way to ensure
continuous supplies of sand??
• Solution:
Notice that unit cost is constant
Reorder cost is $200 for each lorry load
Orders up to 15 tones need one lorry with a reorder
cost of $200, orders between 15 and 30 tones need
two lorries with reorder cost of $400, orders between
30 and 45 tones need three lorries with reorder costs
of $600, and so on.
Other variables:
D
= 4 tones a day
c
= $20 a tone
H
= $1.90 a tone a day
•
Taking the Lowest Cost Curve:
S = $200, valid for Q less than 15 tonnes
Q = √(2)(D)(S)/H
= √(2)(200)(4)/1.90
= 29.0 tonnes
This is invalid minimum as Q is not less than 15 tonnes
Calculating the cost at the break point:
TC = c x D + (S x D)/Q + (H x Q)/2
= (20)(4) + (200x4)/15 + (1.90x15)/2
= $147.58 (Point A)
Taking the next lowest cost curve:
S = $400, valid for Q between 15 and 30 tonnes
Q = √(2)(D)(S)/H
= √(2)(400)(4)/1.90
= 41.0 tonnes
This is invalid minimum as Q is not between 15 and 30 tonnes
Calculating the cost at the break point (to the right of the range):
TC = c x D + (S x D)/Q + (H x Q)/2
= (20)(4) + (400x4)/30 + (1.90x30)/2
= $161.83 (Point B)
Taking the next lowest cost curve:
S = $600, valid for Q between 30 and 45 tonnes
Q = √(2)(D)(S)/H
= √(2)(600)(4)/1.90
= 50.3 tonnes
This is invalid minimum as Q is not between 30 and 45 tonnes
Calculating the cost at the break point (to the right of the range):
TC = c x D + (S x D)/Q + (H x Q)/2
= (20)(4) + (600x4)/45 + (1.90x45)/2
= $176.08 (Point C)
Taking the next lowest cost curve:
S = $800, valid for Q between 45 and 60 tonnes
Q = √(2)(D)(S)/H
= √(2)(800)(4)/1.90
= 58.0 tonnes
This is a valid minimum as Q is between 45 and 60 tonnes
Calculating the cost at the valid minimum:
TC = c x D + √(2)(S)(I)(A)
= (20)(4) + (2)(800)(1.90)(4)
= $190.27 a day (Point D)
Compare the results at point A, B,C & D:
1. Point A: Q=15 tonnes
2. Point B: Q=30 tonnes
3. Point C: Q=45 tonnes
4. Point D: Q=58 tonnes
cost= $147.58
cost= $161.83
cost= $176.08
cost= $190.27
So, the best choice is to order 15 tonnes at a time,
with deliveries needed every 15/4 = 3.75 days
• Production Order Quantity (Finite
Replenishment Rates)
Used when the rate of production is greater than
demand, thus goods will accumulate at a finite
rate while the line is operating ( P – D)
Concern with the stock of finished goods at the
end of a production line
Main Assumptions:




A single item is considered
Demand is known, constant and continuous
All costs are known exactly and do not vary
No shortages are allowed
Replenishment of stock often occurs at a finite
rate rather than instantaneously
• The purpose of this analysis is to find the optimal
batch size (optimal order quantity) when the
production rate is greater than demand
• Other significant objectives are finding total cost,
production time, cycle time and the perfect time
to place an order (ROL)
• Modified Formula EOQ:
Q = 2DS/ic x P/P – D
Point A=(P – D) X PT
Production Quantity
Model
• An inventory system in which an order is
received gradually, as inventory is
simultaneously being depleted
• AKA non-instantaneous receipt model
– assumption that Q is received all at once is relaxed
• p - daily rate at which an order is received over
time, a.k.a. production rate
• d - daily rate at which inventory is demanded
Production Order Quantity Model
• Answers how much to order and when
to order
• Allows partial receipt of material
– Other EOQ assumptions apply
• Suited for production environment
– Material produced, used immediately
– Provides production lot size
• Lower holding cost than EOQ model
POQ Model Inventory Levels
Inventory Level
Production portion of cycle
Demand portion of cycle with no
supply
Supply
Begins
Supply
Ends
Time
Production Quantity Model
(cont.)
Inventory
level
Q(1-d/p)
Maximum
inventory
level
Q
(1-d/p)
2
Average
inventory
level
0
Order
receipt period
Begin
End
order order
receipt receipt
Time
POQ Model Equations
p = production rate
d = demand rate
Maximum inventory level = Q - Q d
p
=Q1- d
p
2CoD
Q
d
Average inventory level =
12
p
CoD CcQ
d
TC = Q + 2 1 - p
Qopt =
d
Cc 1 p
Production Quantity Model:
Example
Cc = $0.75 per yard
Co = $150
d = 10,000/311 = 32.2 yards per day
2CoD
Qopt =
TC =
2(150)(10,000)
Cc 1 - d
p
Co D
+
Q
D = 10,000 yards
p = 150 yards per day
CcQ
2
=
32.2
0.75 1 150
1-
d
p
= 2,256.8 yards
= $1,329
Q
2,256.8
Production run/time = p =
= 15.05 days per order
150
Production Quantity Model:
Example (cont.)
Number of production runs =
10,000
D
=
2,256.8
Q
Maximum inventory level = Q x 1 -
d
p
= 1,772 yards
Copyright 2006 John
Wiley & Sons, Inc.
12-30
= 4.43 runs/year
= 2,256.8 X
1-
32.2
150
• Example:
Demand for an item is constant at 1800 units a year. The
item can be made at a constant rate of 3500 units a year.
Unit cost is $50, batch set-up cost is $650 and holding cost is
30% of value a year. What is the optimal batch size for the
item? If production set-up time is 2 weeks, when should this
be started?
SOLUTION:
Listing the variables we know:
D = 1800 units a year
P = 3500 units a year
c
= $50 a unit
S = $650 a batch
i
= 0.3 x $50 = $15 unit a year
• Find the optimal batch order (optimal order quantity):
Q = 2DS/ic x P/P - D
= (2)(650)(1800)/15
x 3500/3500-1800
= 566.7
 Production time, PT, is:
PT = Q/P
= 566.7/3500 = 0.16 years
= 8.4 weeks
 Cycle Time, CL, is:
CL = 2S/ic x D
x P/P – D
= (2)(650)/(15)(1800) x
= 0.31 years/16.4 weeks
3500/3500-1800
• Variable Cost, VC:
VC
x
P – D/P
=
2 x S x ic x D
=
(2)(650)(15)(1800) x
(3500 – 1800)/3500
= $4129 a year
• Total Cost :
TC = c x D + VC
= 50 x 1800 x 4129
= $94,129 a year
• If it takes 2 weeks to set up production (lead time), then the time
to start production (replenishment) can be found from the
calculation of the Reorder level.
ROL = LT x D = 2 x (1800/52) = 70 (round up)
• Then, the best policy is to start making a batch of 567 units
whenever stocks fall to 70 units.
Reasons for Variability in Production
Most variability is caused by waste or
by poor management. Specific
causes include:
 employees, machines, and suppliers produce
units that do not conform to standards, are late
or are not the proper quantity
 inaccurate engineering drawings or
specifications
 production personnel try to produce before
drawings or specifications are complete
 customer demands are unknown