1. Example of a Decision Tree Problem: The Payoff Table
The management also estimates the profits when choosing from the three alternatives (A, B, and C) under the differing probable levels of demand. These costs, in thousands of dollars are presented in the table below: 21

2. Example of a Decision Tree Problem: Step 1
Example of a Decision Tree Problem: Step 1. We start by drawing the three decisions A B C 22

3. Example of Decision Tree Problem: Step 2
Example of Decision Tree Problem: Step 2. Add our possible states of nature, probabilities, and payoffs A B C
High demand (.4)
Medium demand (.5)
Low demand (.1)
$90k
$50k
$10k
$200k
$25k
-$120k
$60k
$40k
$20k 23

4. Example of Decision Tree Problem: Step 3
Example of Decision Tree Problem: Step 3. Determine the expected value of each decision
High demand (.4)
Medium demand (.5)
Low demand (.1) A
$90k
$50k
$10k
EVA=.4(90)+.5(50)+.1(10)=$62k
$62k 24

5. Example of Decision Tree Problem: Step 4. Make decision
High demand (.4)
Medium demand (.5)
Low demand (.1) A B C
$90k
$50k
$10k
$200k
$25k
-$120k
$60k
$40k
$20k
$62k
$80.5k
$46k
Alternative B generates the greatest expected profit, so our choice is B or to construct a new facility. 25

6. Location Factor Rating Example 1: we are considering two different cities Richmond, Birmingham for the location of a medium-sized Red Bakery Firm. The bakery will produce an assortment of bakery goods on site and will sell directly to retail customers as well as whole sale do grocery stores, restaurants, etc. The factors shown in Table-1 have been evaluated for two cites.
Good Excellent

7. The total score can be computed for each site
The total score can be computed for each site. This is done by first converting the rating for each non-cost factor to a numerical score. The conversion for the example is shown in Table-2 using a 10-point scale .

8. The location with the highest total score is then the best choice
The location with the highest total score is then the best choice. The total scores are as follows: S1 =15(8)+5(6)+5(10)+5(2)+10(8)+60(6) S2 =15(10)+5(4)+5(8)+5(6)+10(6)+60(10) S1 =650 S2 =900 This scoring system, therefore, indicates that alternative 2, Birmingham, is preferred.

9. Location Factor Rating Example 2:
Key Scores
Success (out of 100) Weighted Scores
Factor Weight France Denmark France Denmark
Labor availability and attitude (.25)(70) = 17.5 (.25)(60) = 15.0
People-to- car ratio (.05)(50) = 2.5 (.05)(60) = 3.0
Per capita income (.10)(85) = 8.5 (.10)(80) = 8.0
Tax structure (.39)(75) = 29.3 (.39)(70) = 27.3
Education and health (.21)(60) = 12.6 (.21)(70) = 14.7
Totals
Table 8.4
© 2011 Pearson Education, Inc. publishing as Prentice Hall

10. Location Factor Rating Example 3:
Labor pool and climate
Proximity to suppliers
Wage rates
Community environment
Proximity to customers
Shipping modes
Airport service
LOCATION FACTOR
.30
.20
.15
.10
.05
WEIGHT 80
100 60 75 65 85 50
Site 1 91 95 90 92
Site 2 72
Site 3
SCORES (0 TO 100)
Weighted Score for “Labor pool and climate” for
Site 1 = (0.30)(80) = 24
Copyright 2011 John Wiley & Sons, Inc., R.Taylor

11. Location Factor Rating
24.00
20.00
9.00
11.25
6.50
4.25
2.50
77.50
Site 1
19.50
18.20
14.25
12.00
4.60
3.25
80.80
Site 2
27.00
15.00
10.80
9.50
4.50
82.05
Site 3
WEIGHTED SCORES
Site 3 has the highest factor rating
Copyright 2011 John Wiley & Sons, Inc.,R.Taylor

12. Locational Break-even analysis:
Example 1: Potential locations at Albany, Baker and Casper have the cost structures shown in Table for a product expected to sell for $130. a)Find the most economical location for an expected to sell volume of units per year. b)What is the expected profit if the site selected in (a) is used ? c)For what output range is each location best? .

13. a)
A: TC:$ $75(6.000) =$
B: TC:$ $50(6.000) =$ *
C: TC:$ $25(6.000) =$
Therefore the most economical location is B
b) Expected profit (using B )
P=$130 (6.000)-$ =$ /Yr
c) From the graph, for quantities between 0 and 2000 A is best, between B is best and C is best for greater than 8000 units.

14. Locational Break-Even Analysis Example 2
Three locations:
Selling price = $120
Expected volume = 2,000 units
Akron $30,000 $75 $180,000
Bowling Green $60,000 $45 $150,000
Chicago $110,000 $25 $160,000
Fixed Variable Total
City Cost Cost Cost
Total Cost = Fixed Cost + (Variable Cost x Volume)
© 2011 Pearson Education, Inc. publishing as Prentice Hall

15. Locational Break-Even Analysis Example–
$180,000 –
$160,000 –
$150,000 –
$130,000 –
$110,000 –
$80,000 –
$60,000 –
$30,000 –
$10,000 –
Annual cost
| | | | | | |
,000 1,500 2,000 2,500 3,000
Volume
Bowling Green cost curve
Akron cost curve
Chicago cost curve
Akron lowest cost
Chicago lowest cost
Bowling Green lowest cost
Figure 8.2
© 2011 Pearson Education, Inc. publishing as Prentice Hall

16. North-West&VAM
Warehouse Factory Supply Demand Solving: North-west: $1015 VAM : $779

17. LAYOUT - Line Balancing
a. CT= production time per day/output per day
/stages/station
/ stages/station
b. Stages,worker ,workstation number= Sum of task times/ CT

18. d. Sum of task times/ (CT x number of stations,workers etc)
d. Sum of task times/ number of stations,workers etc

19. /