Production and Operations Management (MAJOR 2)

Production and Operations Management (MAJOR 2)

نبحث عن أشخاص مهتمين باستمرارية الموقع ومايقدمه من مساعدة للطلبة... أرجو من المهتمين زيارة الموقع ورفع أي معلومات مهمة قد يرونها مفيدة لاصدقائهم في مواد أخرى. (التسجيل بسيط ولا يستغرق أكثر من دقيقتين)

محمد الجلال

(7) Chapter 12: Inventory Management

Types of Inventories:

-Finished Items

-Work in Process (WIP)

-Raw Materials

-Spare Parts (Replacement Parts)

-Goods-in-transit

Reasons to have inventory:

-To meet anticipated demand

-To smooth production requirement

-To decouple operations

-To help hedge against price increases

-To take advantage of quantity discounts

So, we need to have:

-A system to keep track of inventory

-A Reliable forecast of demand

-Knowledge of lead times

-Reasonable estimates of

Holding Costs (carrying costs)
Ordering Costs
Shortage Costs

-A Classification system

Inventory Counting Systems

Periodic System: Physical count of items made at periodic intervals.

Perpetual Inventory System: System that keeps track of removals from inventory continuously, thus monitoring current levels of each system.

Key Inventory Terms:

-Lead Time: Time interval between ordering and receiving the order.

-Holding (Carrying) Costs: Cost to carry an item in inventory for a length of time, usually a year.

-Ordering Costs: costs of ordering and receiving inventory.

-Shortage costs: Costs when demand exceeds supply.

Carrying Cost (Holding Cost): Rent, Heating, Lighting, Security, Obsolescence, Spoilage… costs that are carried while holding an inventory item.
Ordering Cost: Transportation, Faxes, Inspection, Receiving… Costs that are carried during the order stage of an item from the inventory.
Shortage Cost (Stock out cost): Loss of sales / Profit, loss in satisfaction, penalties… Costs that are usually undertaken when there is a shortage in inventory.

Note: Order Size IncreaseCarrying cost increaseOrdering Cost Decrease

Two Main Categories of Inventory Control Systems:

Continuous (Economic Order Quantity Models)

Fixed amount to order
Re-order at specific level reached
System is continuously monitored (computerized system to monitor the level)
(The Commonly Used)

Periodic (Fixes-Order-Interval Models)

Fixed time (period – ex. Every two weeks, etc.)
No monitoring, no records keeping (less direct control) (checked by the vendor instead)
Might be useful when inventory cannot be closely monitored
Each time an amount of order should be decided
Ex's of systems: University bookstore, Small retailer, Drug store (in small organizations)

Inventory Categories:

Periodic (4)
Continuous
EPQ(Economic Production Quantity)(3)
EOQ(Economic Order Quantity)

a)With Discount(2)

b)Without Discount (1)

Assumptions of EOQ Model:

-Only one product is involved

-Annual demand requirements known

-Demand Is evenly spread throughout the year

-Lead time does not vary

-Each order is received in a single delivery

The Inventory Cycle:

(1) EOQ without Discount Equation:

QOPT=

Total Cost = Annual Carrying Cost + Annual Ordering Cost

TC =

Open Page 554 or 562:

(Lookup the Examples)

Exercise: Solve Problem #3 (Ch 12) Page 594 Ed. 10(EOQ without Discount)

(1)

Answer:

Given:

D = 4860 bags

H = 75$

S = 10$

Equations:

EOQ =

Total Cost =

Average Quantity on Hand =

Number of Orders per Year =

Solution:

EOQ =
Average number of bags on hand= Q/2 = 36/2 = 18 bags
Number of Orders Per Year = 4860 / 36 = 135 order
Total Cost = (36/2)x75 + (4860/36)x10 = $2700
New EOQ =
New Total Cost = (37.75/2) x75 + (4860/37.75) x11 = $2831.78

Study Solved Problem #1 Page. 579 or 588

(2) EOQ withDiscount Equation:

Min. EOQ=

Optimal EOQ = (The Quantity with the least total cost)

Total Cost = Annual Carrying Cost + Annual Ordering Cost + Purchasing Cost

Total Cost =

Exercise: Problem #13 Page 596 (10th Ed)(EOQ with Discount)(2)

Answer:

Given:

-D: 18,000

-S: $96

-H: $0.60

1000 to 1999 / $1.25
2000 to 4999 / $1.20
5000 to 9999 / $1.15
10000 or more / $1.10

Requirements:

-A) Optimal Q

-B) Number of orders per year

Solution:

Min. EOQ =

Total Cost =

Number of Orders per Year =

A) Min. EOQ = Box

Total Cost2,000 = (2400/2)x0.60 + (18,000/2400)x96 + (1.20x18,000) = $23,040

Total Cost5,000= (5000/2)x0.60 + (18,000/5000)x96 + (1.15x18,000) = $22,525.6

Total Cost10,000 = (10,000/2)x0.60 + (18,000/10,000)x96 + (1.10x18,000) = $22,972.8

Then, the optimal order Quantity is 5,000 Box

B) D/QOptimal = 18,000/5,000 = 3.6 Orders

Economic Production Quantity (EPQ):

Check Book: Figure 12.6 – Page 564

(3) EPQ Equations:

Economic Run Qty Q0=

p = Production or Delivery Rate

u = Usage Rate (Daily) = D / "n" of operating days per year

Cycle Time =

Run Time =

Number of runs per year =

Imax = Maximum Inventory = ; I average =

Total CostMinimum = Carrying Cost + Setup Cost =

Build up Rate =

Note: The Run Time should ALWAYS be less than the Cycle Time

Ex, if we assume the run time to be 3 days, then cycle time should be greater than 3 days.

Exercise: Solve Problem #9 (Ch 12) Page 595 Ed. 10(EPQ)(3)

Answer:

Given:

D = 75,000 Hotdog/year

H = 0.45$ / Hotdog

S = 66$

P = 5,000 Hotdog/ day

u = D/n = 250 Hotdog Supplied to Restaurants / Day

Solution:

a)Economic Run Qty = = 4812.26 Hotdog

b)The number of runs per year = D/Q = 75,000 / 4812.26 = 15.58 runs

c)The length in days of a run (Run Time) = Q/P = 4812.26 / 5,000 = 0.96 Days

Periodic Control System: Fixed-Order-Interval Model(4)

-Orders are placed at fixed time intervals

-Order Quantity for next interval?

-Suppliers might encourage fixed intervals.

-May require only periodic checks of inventory levels.

-Risk of stock out.

-Fillrate – (Check Book on Definition of Filtrate)

Equations:

Q =

: Daily demand expected during protection interval.

OI: Order Interval (length of time between orders).

A: Amount on hand at re-order time.

Z: Number of Standard Deviations

: Standard Deviation of lead time demand.

Exercise(4):The corner drug store stocks a popular brand of sunscreen. The average demand for the sunscreen is 6 bottles per day. The producer checks the drugstore stock every 60 days. During one visit the drugstore had 8 bottles in stock. The lead time to receive an order is 5 days. Determine the order size for this order period that will enable the drugstore to maintain a 95% service level (z=1.65, = 1.2)

(4)

Answer:

Given:

: 6

LT: 5 Days

OI: 60

A: 8

Z: 1.65

: 1.2

Solution:

= = 397.96 items = 398 sunscreen bottles.

Exercise: Solve Problem #4 (Ch 12) Page 594 Ed. 10(EOQ without Discount)

(1)

Answer:

Given:

D = 40x260 = 10,400 box

H = $30

S = $60

Equations:

Solution:

EOQ =
Total Annual Cost = $
Yes…
The difference of 4 Boxes wouldn’t have any felt economic effect -.-

Exercise: Solve Problem #14 (Ch 12) Page 596 Ed. 10(EOQ with Discount)

(2)

Answer:

Given:

-D: 200x25 = 5,000 Stone

-S: $48

-H: $2 (Part A)

Less than 400 Stone / $10
400 to 599 Stone / $9
600 Stone or more / $8

Requirements:

-A) Optimal Q

Solution:

Min. EOQ =

Total Cost =

Number of Orders per Year =

A) Min. EOQ = Stone

Total Cost490= (490/2) x2 + (5,000/490)x48 + (9x5,000) = $45979.79

Total Cost600= (600/2)x2 + (5,000/600) x48 + (8x5,000) = $41,000

Then, the optimal order Quantity is 600 Stones

D/QOptimal = 5,000/600 = 8.33 Orders

Exercise: Solve Problem #11 (Ch 12) Page 595 Ed. 10(EPQ)

(3)

Answer:

Given:

D = 5x50x80 = 20,000 units/year

H = 10$ / Unit

S = 300$

P = 200 units/day

u = D/n = 80 units /day

Solution:

a)Economic Run Qty = = 1414 Units

b)Length of a production run = Q/P =1414/200 = 7 Days

c)Buildup Rate = P-U = 200 – 80 = 120 units/day

Exercise: Solve Problem #6 (Ch 12) Page 594 Ed. 10(EOQ without Discount)(1)

Answer:

Given:

D = 800 x12 = 9600 Crate

H = 10x0.35 = $3.5

S = 28$

Requirements:

-EOQ

-Total Cost

Solution:

EOQ = = 392 Crates

Total Cost =

(8) Chapter 14: MRP

Dependent & Independent Demand

Dependant Demand is certain

Independent Demand is uncertain

Bill-of-Materials: One of the three primary inputs of MRP; a listing of all of the raw materials, parts, subassemblies, and assemblies needed to produce one unit of a product.

Product Structure tree: A Visual depiction of the requirements in a bill of materials, where all components are listed by levels.

Now, MRP Tabulation Method:

-Gross Requirements

-Scheduled receipts

-On Hand

-Net Requirements

-Planned Order Receipts

-Planned Order Releases

EXCERCISES NEXT PAGE

Exercise: Solve Problem #4 (Ch 14) Page 682 Ed. 10(Part A Only)

Given:

-To Produce One E: 2x B (1xB = 4xJ, 2xF) & 3x J

-Then, to produce 1x E, we need 8xJ+4XF = 2x B + 3xJ = 1x E

Week / B.Inv / 1 / 2 / 3 / 4 / 5 / 6
Quantity / 80

E: (Lot-For-Lot) , LT= 2 Weeks

Gross Requirements / 80
Scheduled Receipts
Projected on Hand
Net Requirements / 80
Planned-Order Receipts / 80
Planned-Order Releases / 80

B: Lot-size: Multiple of 120 , LT= 2 Weeks, 60x On Hand

Gross Requirements / 160
Scheduled Receipts
Projected on Hand / 60 / 60 / 60 / 60 / 60 / 20 / 20
Net Requirements / 100
Planned-Order Receipts / 120
Planned-Order Releases / 120

J: Lot-size: Multiple of 30 , LT= 1 Week, 20x On Hand, Projected on Hand: 30, 30,30

Gross Requirements / 480 / 240
Scheduled Receipts / 30 / 30 / 30
Projected on Hand / 20 / 20 / 20 / 50 / 80 / 50 / 50
Net Requirements / 460 / 160
Planned-Order Receipts / 480 / 180
Planned-Order Releases / 480 / 180

Exercise: Solve Problem #5 (Ch 14) Page 682 Ed. 10(Part a,c,d)

Given:

-P = K + L + W

-K = 3G + 4H

-L = 2M + 2N

-W = 3Z

-On Hand: 20 L, 40 G, 200 H

-Scheduled Receipts: 10 K at week 3, 30 K at week 6, 200 W at week 3

-100P shipped week 6, 100P shipped week 7

-10% scrap allowance on each order of G

-H Min Order 200

Week / B.inv / 1 / 2 / 3 / 4 / 5 / 6 / 7 / 8
Quantity / 100 / 100

P: (Lot-For-Lot) , LT= 1 Weeks

Gross Requirements / 100 / 100
Scheduled Receipts
Projected on Hand
Net Requirements / 100 / 100
Planned-Order Receipts / 100 / 100
Planned-Order Releases / 100 / 100

K: Lot-For-Lot, LT= 2 Weeks

Gross Requirements / 100 / 100
Scheduled Receipts / 10 / 30
Projected on Hand / 10 / 10 / 10 / 30
Net Requirements / 90 / 70
Planned-Order Receipts / 90 / 70
Planned-Order Releases / 90 / 70

G: Lot-for-Lot , LT= 1 Week, *: 10% Scrap Allowance on G

Gross Requirements / 270 / 210
Scheduled Receipts
Projected on Hand / 40 / 40 / 40 / 40
Net Requirements / 230 / 210
Planned-Order Receipts / 253* / 231*
Planned-Order Releases / 253 / 231

H: Lot-for-Lot , LT= 1 Week, Min Order Size: 200

Gross Requirements / 360 / 280
Scheduled Receipts
Projected on Hand / 200 / 200 / 200 / 200 / 40
Net Requirements / 160 / 240
Planned-Order Receipts / 200 / 240
Planned-Order Releases / 200 / 240

Exercise: Solve Problem #7 (Ch 14) Page 683 Ed. 10

Given:

-80X needed week 6, 30X needed week 8

-Prepare MRP for D

-D: Batch 50 per order

-50 Dscheduled receiptsweeks1,3,5,7

-30 B on Hand, 20 D on Hand

-Lead Time = 1 week for 100 units, 2 weeks for 101 to 200 units, 3 weeks for 201-300

Week / B.inv / 1 / 2 / 3 / 4 / 5 / 6 / 7 / 8
Quantity / 80 / 30

X: Lot-For-Lot

Gross Requirements / 80 / 30
Scheduled Receipts
Projected on Hand
Net Requirements / 80 / 30
Planned-Order Receipts / 80 / 30
Planned-Order Releases / 80 / 30

B: Lot-For-Lot

Gross Requirements / 160 / 60
Scheduled Receipts
Projected on Hand / 30 / 30 / 20 / 30 / 30 / 30
Net Requirements / 130 / 60
Planned-Order Receipts / 130 / 60
Planned-Order Releases / 130 / 60

D: Lot size: 50

Gross Requirements / 260 / 240 / 120 / 90
Scheduled Receipts / 50 / 50 / 50 / 50
Projected on Hand / 20 / 70 / 70 / 120 / 10 / 60 / 20 / 50 / 10
Net Requirements / 140 / 180 / 100 / 40
Planned-Order Receipts / 150 / 200 / 100 / 50
Planned-Order Releases / 150 / 200 / 100 / 50

MRP in Services?

Service Application Such As:

-Professional Services

-Healthcare

-And Many other Services!

ERP

In Fact, ERP is just MRP II (Manufacturing resources planning).

MRP II evolved from MRP (Materials Requirements Planning)

Capacity Planning:

Required Capacity: Determining Capacity Requirements per Period…

Design/Available Capacity: Determining the available capacity by:

-Calculating total machine hours (considering all shifts) and total labor hours, and

-In parallel to look at layouts and details of processes to check:

Bottlenecks
Parallel Processes Pipelined
Periodic Shutdowns for Maintenance and So.

Exercise: Solve Problem #15 (Ch 14) Page 685 Ed. 10

Given:

Week / 1 / 2 / 3 / 4
Material / 40 / 80 / 60 / 70

Labor Standard Hours: 4 per ton

Machine Standard Hours: 3 per ton

Weekly Production Capacity: 300 for Labor, 200 for Machine

Answer: (Part A)

Week / 1 / 2 / 3 / 4
Material / 40 / 80 / 60 / 70
Labor Hours / 160 / 320 / 240 / 280
Machine Hours / 120 / 240 / 180 / 210

Utilization:

Week / 1 / 2 / 3 / 4
Labor / 53.33% / 106.67% / 80% / 93.33%
Machine / 60% / 120% / 90% / 105%