• Focus on minimizing costs
– EOQ
– Linear Programming
• Two types of inventory costs (IC):
– Order/Setup Costs (OCs), and
– Carrying Costs (CCs)
IC = OC + CC
• We will use following variables:
TC - Total Inventory Costs
P
- Cost to place one purchase order
Q
- Quantity - Units ordered in a single order; or
D
- Demand - number of inventory needed in year
C
- Carrying Cost for 1 unit for entire year
• Economic Order Quantity (EOQ)
model determines:
– Optimal amount of inventory to
produce/purchase at given time
• Discussion applicable to
production runs and orders
• OCs:
– Purchasing inventory Æ all costs involved
in placing order
– Manufacturing inventory Æ cost of setting
up production line
• CCs:
– All costs involved in holding inventory
• E.g., interest on inventory cost,
insurance, breakage and storage cost
• Remember that ICs are
divided into 2 parts:
– OCs
– CCs
• 1st Æ Focus on OCs
1
• OCs:
– cost to place 1 order (P),
multiplied by
– # of orders placed
• # of orders placed:
# of Orders =
• All OCs in a year:
– (Cost to Place an Order) x (# of orders placed):
P(
D
)
Q
Annual Demand For Units
D
=
Size of Order
Q
• Annual CCs:
(Annual Carrying Cost Per Unit) x (# of Units in Inventory)
• Is # of units in Inventory = units in order (Q)?
• NO! We buy units, sell them, buy some more:
• Average = ½ of amount of order: _Q_
2
• Carrying costs:
– C times the number of you units that
your have on hand at any given
time.
– On average, your firm has Q/2 units
on hand at any given time.
C(
_Q_
2
TC: OCs + CCs
)
_Q_
_D_
P(
Q
) + C(
2
)
2
• To get optimal amount to
order to minimize ICs:
– Take 1st derivative of
equation with respect to Q
– Set derivative to 0
– Solve for Q
• Rewrite Inventory Cost equation to
make it easier to use Power Rule for
taking derivatives:
TC =
PDQ-1 + ½CQ1
∂/∂Q TC =
-PDQ-2 + ½C
0 =
-PDQ-2 + ½C
-½C =
-PDQ-2
-½CQ2 =
Q2
-PD
=
_2PD_
C
_______
Q
=
/ _2PD_
√
• EOQ Formula gives optimal amount of units to
order
• Lead Time is time it takes to obtain new inventory
once order placed
– E.g., Lead Time is 2 days if:
• It takes 2 days for inventory to arrive once
ordered
• Assuming inventory sold/used at
uniform rate, Rate of Usage is:
Rate of Usage =
___Annual Demand For Units___
Number of Working Days in Year
• Safety Stock is extra inventory carried as buffer against
fluctuation in demand
• Reorder Point formula can be modified to reflect Safety
Stock:
(“EOQ Formula”)
C
• Reorder Point is point in time where existing
inventory is just sufficient to cover demand until
new inventory arrives
• Reorder Point can be calculated as follows:
Reorder Point = Rate of Usage x Lead Time
• E.g., we sell 10 units a day & it takes 2 days to
receive new inventory once ordered
– Reorder Point is 20 units
• Firm should place order once inventory
level drops to 20 units
• E.g., Assume
– Annual demand for units is 5,000 units (D)
– Annual Carrying Cost of 1 unit is $10 per year (C)
– Setup Cost is $1,000 (P)
– Co. scheduled 4 equal production runs for
upcoming year
– Co. has 250 business days/year
– Sales occur uniformly
– Production takes 1 day
Reorder Pt. = (Rate of Usage x Lead Time) + Sfty. Stk.
3
• Using EOQ formula Æ optimal run is 1,000
units:
_________
EOQ = √2(P)(D)/(C)
= 1,000 units
________________
EOQ = √[2($1,000)(5,000)]/$10 = 1,000 units
• Co. saves following by using production runs of 1,000
units
– rather than the 1250-unit run (5,000/4) currently
planned:
Inventory Costs =
P(
_D_
Q
)+(
_CQ_
2
)
Cost @ 1250-Unit Run =
[$1000 (5000/1250) + ½ (1250 x 10)]
$10,250
Cost @ 1000-Unit Run =
[$1000 (5000/1000) + ½ (1000 x 10)]
-10,000
Inventory Cost Savings =
• How do you:
– Maximize Sales Revenue/CM, or
– Minimize costs
– When dealing with constraints
• We will discuss Graphical approach
$250
• Since Co. has 250 business days Æ
– Co. sells 20 units/day (5,000/250).
• If Co. needs 1 day to produce units Æ
– Co. should begin production run once
inventory drops to 20 units
– If Co wants Safety Stock Æ
• Add Safety Stock to 20-unit Reorder
Point
Constrained Optimization
• We want to either:
– Maximize profits/CM, or
– Minimize costs
• This is easy, you make either infinity units to maximize
profits/CM or zero units to minimize costs
• Problem have a constraint.
– Limited Capacity Æ Max profits/CM
– Need to achieve goals Æ Min costs
Graphical Approach
• 1st state your goal
• E.g.,
– Maximize CM / Minimize cost
– Mathematical formula for item being maximized
or minimized
• Called Objective Function (OB)
• Variables in OB are your alternative courses of
action Æ What you can change
– E.g., # of each type of product that you will
produce
4
• Assume:
– Co can make 2 types of bolts:
• Bolt A or
• Bolt B
– Bolt A has CMU =10¢
– Bolt B has CMU =12¢
– Co wants to know # of each type of bolt to produce to
maximize CM:
– OB:
Maximize .10A + .12B
– Where
• “A” Æ# of A Bolts produced
• “B” Æ# of B Bolts produced
• Without more information, Co.
would produce ∞ # of every bolt
• With this production, Co. would
have ∞ CM
• In reality, there is limit to # of bolts
that Co. can produce
• Assume each bolt must pass through following
machines, & time required on each machine
differs, as shown below:
Machine I
Machine II
Machine III
A Bolt
.1 min
.1 min
.1 min
B Bolt
.1 min
.4 min
.5 min
• Add constraints to OB:
Description
OB:
Constraints:
• In 1 day, there are 240, 720, &160 minutes
available, on Machine I, Machine II and
Machine III, respectively
• How many of each type of bolts should Co.
produce in 1 day?
•
•
•
subject to:
Only 240 mins available on Machine I.
.1A + .1B ≤ 240
Only 720 mins available on Machine II.
.1A + .4B ≤ 720
Only 160 mins available on Machine III.
.1A + .5B ≤ 160
Can't manufacture negative # of bolts.
Next, graph production area that meets all of
constraints
– Feasible Region
Graph consists of Cartesian axis with the OB
Variables as x-axis & y-axis
E.g., Last constraint alone means you are dealing
with top right quarter of Cartesian axis:
Problem
Max .10(A) + .12(B)
A,B≥0
• Now, graph other constraints, which are
inequalities
– either ≥ or ≤
• Area covered by such an inequality consists of
1 line that divides Cartesian plane & 1 side of
that line.
• Formula of line is inequality formula with =
substituted for ≤ or ≥
• To graph inequality Æ graph line that divides
the Cartesian plane & chose side of line that
satisfies inequality
29
5
•
To graph 1st constraint (.1A+.1B ≤ 240)
– Graph line (.1A+.1B = 240)
– Easiest way to graph line is identify points where line crosses
each axis
• We know that B=0 on any point on A-axis
• Therefore, point where line crosses A-axis has 0 as B
coordinate.
• So, we replace B with 0 in equation & solve for A:
• Where does line cross B-axis?
• We know that A=0 at any point on B-axis
• Therefore, point where line crosses B-axis has 0 as A
coordinate
• So, we replace A with 0 in equation and solve for B:
.1A + .1B =
240
240
.1(0) + .1B =
240
.1A + .1(0) =
240
.1B =
240
.1A =
240
B =
240/.1
B =
2400
.1A + .1B =
A =
240/.1
A =
2400
• So, line crosses A-axis at (2400, 0)
• So, line crosses B-axis at (0, 2400)
32
• With these 2 points, you can now graph
the line:
• Now, we need to pick side of line that satisfies
inequality
• Easiest way to do this is to test 1point on 1 side
of line
– You test point by substituting coordinates
into inequality formula, & check if coordinates
produce a value that satisfies inequality
• If point tested satisfies inequality, then every
point on same side of line will satisfy inequality
• Easiest point to test is origin (0,0) because you
are dealing with zeros as variables
• So, plug origin into inequality & see if inequality
is true:
.1A + .1B ≤ 240 Inequality
.1(0) + .1(0) ≤ 240 Test the Origin
0 ≤ 240 True Statement
• So, Feasible Region for 1st constraint
includes side of line that contains origin
• Feasible Region that satisfies 1st & last
constraint consists of:
• To graph 2nd constraint (.1A+.4B ≤ 720)
– Graph line, .1A+.4B = 720
– We know that line crosses A-axis at (7200,
0):
.1A + .4B =
720
.1A + .4(0) =
720
.1A =
720
A =
A =
720/.1
7200
6
• Line crosses B-axis at (0,1800):
.1A + .4B =
720
.1(0) + .4B =
720
.4B =
720
B =
720/.4
B =
1800
• Feasible Region for 2nd constraint
includes side of line that contains origin
• Feasible Region that satisfies 1st, 2nd &
last constraints consists of following:
• By testing origin, we see that side that contains
origin satisfies inequality:
.1A + .4B ≤ 720 Inequality
.1(0) + .4(0) ≤ 720 Test the Origin
0 ≤ 720 True Statement
• To graph 3rd constraint (.1A+.5B ≤ 160)
– Graph line, .1A+.5B = 160
– Line crosses A-axis at (1600, 0):
.1A + .5B =
160
.1A + .5(0) =
160
.1A =
• Line crosses B-axis at (0, 320):
.1A + .5B = 160
.1(0) + .5B = 160
.5B = 160
B = 160/.5
B = 320
160
A =
160/.1
A =
1600
• By testing origin, we see side that
contains origin satisfies inequality:
.1A + .5B ≤ 160 Inequality
.1(0) + .5(0) ≤ 160 Test the Origin
0 ≤ 160 True Statement
7
• Feasible Region for 3rd constraint
includes side of line that contains
origin
• Feasible Region that satisfies all of
constraints is:
• Once, you have identified Feasible Region that
satisfies all constraints
– You have to decide which points within Feasible
Region maximize CM
– Corners points of Feasible Region are most
extreme points
• Therefore, corner points represent production
levels that produce most extreme CM
– E.g., highest & lowest CM
• The fact that Feasible Region no
longer touches lines for 1st & 2nd
constraints tells you that 1st & 2nd
constraints are not “binding”
• Time on Machines I and II could be
unlimited and it would not affect our
production possibilities
• To find highest CM
– Plug each corner points into original OB
[.10(A)+.12(B)]
– Check to see which point produces highest CM:
Corner
(1600 , 0)
(0 , 320)
(0 , 0)
.10(A)+.12(B)
.1 (1600) + .12 (0) =
CM
160 + 0 =
$160.00
.1 (0) + .12 (320) = 0 + 38.4 =
38.40
.1 (0) + .12 (0) =
0+0 =
0
• Co. generates highest CM by producing 1600 A
Bolts and 0 B Bolts
THE END
© Dr. Michael Constas 2013
8