Assignment; Basic Inventory Model
Problem 1: A toy manufacturer uses approximately 32000
silicon chips annually. The Chips are used at a steady rate
during the 240 days a year that the plant operates. Annual
holding cost is 60 cents per chip, and ordering cost is $24.
Determine
a) How much should we order each time to minimize our
total cost
b) How many times should we order
c) what is the length of an order cycle
d) What is the total cost
What is the Optimal Order Quantity
2 DS
EOQ 
H
D = 32000, H = .6, S = 24
2(32000)( 24)
EOQ 
 1600
.6
How many times should we order
Annual demand for a product is 32000
D = 32000
Economic Order Quantity is 1600
EOQ = 1600
Each time we order EOQ
How many times should we order ?
D/EOQ
32000/1600 = 20
what is the length of an order cycle
working days = 240/year
32000 is required for 240 days
1600 is enough for how many days?
(1600/32000)(240) = 12 days
What is the Optimal Total Cost
The total cost of any policy is computed as
TC  (Q / 2) H  ( D / Q ) S
The economic order quantity is 1600
TC  .6(1600 / 2)  24(32000 / 1600)
TC  480  480
TC  960
This is the total cost of the optimal policy
Assignment; Problem 2
Victor sells a line of upscale evening dresses in his boutique.
He charges $300 per dress, and sales average 30 dresses per
week. Currently, Vector orders 10 week supply at a time from
the manufacturer. He pays $150 per dress, and it takes two
weeks to receive each delivery. Victor estimates his
administrative cost of placing each order at $225. His
inventory carriyng cost including cost of capital, storage, and
obsolescence is 20% of the purchasing cost. Assume 52 weeks
per year.
a) Compute Vector’s total annual cost of inventory system
(carrying plus ordering but excluding purchasing) under
the current ordering policy?
b) Without any EOQ computation, is this the optimal policy?
Why?
Assignment; Problem 2
c) Compute Vector’s total annual cost of inventory system
(carrying plus ordering but excluding purchasing) under the
optimal ordering policy?
d) Compute EOQ and total cost of the system
e) What is the ordering interval under optimal ordering
policy?
f) When do you order?
g) What is average inventory and inventory turns under the
original policy and under the optimal ordering policy?
Inventory turn = Demand divided by average inventory.
Average inventory = Max Inventory divided by 2.
Average inventory is the same as cycle inventory.
h) Compute the flow time under the two policies.
Current Policy
Problem 6.3
flow unit = one dress
flow rate d = 30 units/wk
costost C = $150/unit
fixed order cost S = $225
H = 20% of unit cost.
lead time L = 2 weeks
ten weeks supply
Q = 10(30) = 300 units.
52 weeks per year
a) Annual demand = 30(52) = 1560
Number of orders/yr = D/Q = 1560/300
= 5.2
(D/Q) S = 5.2(225) = 1,170/yr.
Average inventory = Q/2 = 300/2 = 150
H = 0.2(150) = 30
Annual holding cost = H (Q/2) = 30(150) = 4,500 /yr.
b) Without any computation, is this the optimal policy?
Why?
Without any computation, is EOQ larger than 300 or
smaller
Why
c) Total annual costs = 1170+4500 = 5670
Optimal Policy
d) Compute EOQ
Q* = EOQ =
2 DS
2(1560)225
= 153 units.

H
30
His annual cost will be
4,589
e) What is the ordering interval under optimal ordering
policy?
We order D/Q = 1560/153 = 10.2 times
A year is 52 weeks.
Therefore we order every 52/10.2 = 5.1 weeks
Optimal Policy
f) When do you order (re-order point) ?
An order for 153 units two weeks before he expects to run out.
That is, whenever current inventory drops to
30 units/wk * 2 wks = 60 units
which is the re-order point.
Whenever inventory reaches 60 we order 153.
This process is repeated 10.2 times a year. Every 5.1 weeks.
Inventory Turns
g) What is average inventory and inventory turns under the
original policy and under the optimal ordering policy?
Inventory turns = yearly demand / Average inventory
Average inventory = cycle inventory = I = Q/2
Current policy inventory turns = D/(Q/2)= 1560/(300/2)
Current policy Inventory turns = 10.4 times per year.
Optimal policy inventory turns = D/(Q/2)= 1560/(153/2)
Optimal policy inventory turns = = 20.4
turns roughly double
Flow time
h) Compute the flow time under the two policies.
Average inventory = cycle inventory = I = Q/2
Current average inventory = 300/2 = 150
Throughput?
R?
R= D
R= 30 /week
Current flow time
RT= I
30T= 150  T= 5 weeks
Optimal average inventory = 153/2) = 76.5
Optimal flow time RT=I  30T= 76.5
T = 2.55 weeks
Did we really need this computations?
Assignment; Problem 3
Complete Computer (CC) is a retailer of computer
equipment in Minneapolis with four retail outlets.
Currently each outlet manages its ordering
independently. Demand at each retail outlet averages
4,000 per week. Each unit of product costs $200, and CC
has a holding cost of 20% per annum. The fixed cost of
each order (administrative plus transportation) is $900.
Assume 50 weeks per year. The holding cost will be the
same in both decentralized and centralized ordering
systems. The ordering cost in the centralized ordering is
twice of the decentralized ordering system.
Assignment; Problem 3
Decentralized ordering: If each outlet orders individually.
Centralized ordering: If all outlets order together as a single
order.
a) Compute EOQ in decentralized ordering
b) Compute the cycle inventory for one outlet and for all
outlets.
c) Compute EOQ in the centralized ordering
d) Compute the cycle inventory for all outlets and for one
outlet
e) Compute the total holding cost + ordering cost (not
including purchasing cost) for the decentralized policy
f) Compute the total holding cost plus ordering cost for the
centralized policy
Decentralized Policy
Four outlets
Each outlet demand
D = 4000(50) = 200,000
S= 900
C = 200
H = .2(200) = 40
If all outlets order together
in a centralized ordering,
then S= 1800
2DS
Q
H
2(200000)( 900)

40
=3000
With a cycle inventory of 1500 units for each outlet.
The total cycle inventory across all four outlets equals 6000.
With centralization of purchasing the fixed order cost is S =
$1800.
2(4  200000)( 2  900)
EOQ 
40
=8485
and a cycle inventory of 4242.5.
Total holding & ordering costs under the two policies
TC  S ( D / Q)  H (Q / 2)
Decentralized
TC  900(200000 / 3000)  40(3000 / 2)
TC  60000  60000  120000
Decentralized: TC for all 4 warehouses = 4(120000)=480000
Centralized
TC  1800(800000 / 8485)  40(8485 / 2)
TC  169711  169700  339411
339411 compared to 480000 about 30% improvement