HW2

HW 2

[Name of the Writer]

[Name of the Institution]

HW 2

Question no 7.9

Graph.

The entire cost of units, with similar crossover point.

4 x = 15000 + 1.82 x

2.18 x = 15000

x = 15000/2.18

x=6.881

Question 8.5(PART A)

For Maitland

Space

Space= Weight x score value

Space= 0.30 x 60

Space= 18

Costs

=0.25 x 40

=10

Traffic Density

=0.20 x 50

= 10

N. Income

=0.15 x 50

= 7.5

Zoning Laws

=0.10 x 80

= 8

For Baptist Church

Space

=0.30 x 70

=21

Costs

= 0.25 x 80

= 20

Traffic Density

= 0.25 x 80

= 20

Neighborhood Income

= 0.15 x 70

= 10.5

Zoning Laws

= 0.10 x 20

= 2

North side Mall

Space

=0.30 x 80

=24

Costs

= 0.25 x 30

= 12

Traffic Density

= 0.25 x 60

= 6

Neighborhood Income

= 0.15 x 40

= 6

Zoning Laws

= 0.10 x 90

= 9

If we look closely at the results the total scores are

Maitland= 53.5

Baptist Church= 69.5

North side Mall = 58.5

The highest Score is 69.5 so Baptist point is the possible and likely option.

PART B

If the factors space and traffic density are reversed than the than the table will have following values

Factor

Maitland

Baptist Church

North side Mall

Space

12

14

16

Costs

10

20

7.5

Traffic Density

15

24

18

Neighborhood Income

7.5

10.5

6

Zoning Laws

8

2

9

52.5

70.5

56.5

By viewing above Baptist Church is still the best option.

Question 8.23

PART A

x-cordinate of the centre of gravity Cx is represented as:

Cx=idix×QiiQi

y-cordinate of the centre of gravity Cy is represented as:

Cy=idiy×QiiQi

Using the equations for the x and y coordinates, we get our value for Cx as,

Cx=25×2,000+25×5,000+55×10,000+50×7,000+80×10,000+70×20,000+90×14,0002,000+5,000+10,000+7,000+10,000+20,000+14,000

Cx=4,535,00068,000

Cx=66.69

We can calculate Cy as,

Cy=45×2,000+25×5,000+55×10,000+45×10,000+20×7,000+50×10,000+20×20,000+25×14,0002,000+5,000+10,000+7,000+10,000+20,000+14,000

Cy=2,055,00068,000

Cy=30.22

Hence for the center of gravity, we get our coordinates as 66.69 and 30.22.

PART B

If the 103 prjected by 20 %

105 projected by 20%

103=10,000 x 1.2

=12,000

105= 10,000 x 1.2

=12,000

Cx=25×2,000+25×5,000+55×12,000+50×7,000+80×12,000+70×20,000+90×14,0002,000+5,000+12,000+7,000+12,000+20,000+14,000

Cx=4,805,00072,000

Cx=66.73

For Cy

Cy=45×2,000+25×5,000+55×12,000+45×12,000+20×7,000+50×12,000+20×20,000+25×14,0002,000+5,000+12,000+7,000+12,000+20,000+14,000

Cy=2,355,00072,000

Cy=32.70

Question 9.4

Based on the information and tables it is said that designed could be as follow.

Materials and Welding

Drills and Grinders

Room 1= Benders

Room 2 = Materials

Room 3=Welding

Room 4=Drills

Room 5= Grinders

Room 6= Lathes

The length in feet would be 60 feet and width would be 40 feet wide.

Question 12.1

By solving the table of L.houts Plastic Charlotte Inventory Levels, we’ll get the following table.

L. Houts Plastic Charlotte Inventory Levels

Item Code no.

Average Inventory

Value

($/UNIT)

Annual Dollar($)

Percentage of Annual Dollar Volume($)

Class

Total Percentage of Annual Dollar Volume($)

1289

400

3.75

1500

44.49%

Class A

80.08%

2347

300

4

1200

35.59%

Class A

80.08%

2349

120

2.5

300

8.90%

Class B

17.13%

2363

75

1.5

112.5

3.34%

Class B

17.13%

2394

60

1.75

105

3.11%

Class B

17.13%

2395

30

2

60

1.78%

Class B

17.13%

7844

12

2.05

24.6

0.73%

Class C

2.79%

6782

20

1.15

23

0.68%

Class C

2.79%

9111

6

3

18

0.53%

Class C

2.79%

8210

8

1.8

14.4

0.43%

Class C

2.79%

8310

7

2

14

0.42%

Class C

2.79%

Total Dollar Volume

3371.5

100.00 %

100.00%

Question 12.5

DATA

Demand Rate, D = 19500

Setup/order cost, S = 25

Holding cost, H=4 $

Unit Price=?

Part a)

EOQ = SQRT ((2* Demand * Order Cost)/ Holding Cost)

EOQ= SQRT ((2*19500*25)/4)

=493.71 Units

=494 Units

Part b)

Annual holding costs for the workbooks = (Quantity of items) * (Holding Cost)/ 2

= 494 * 4/2

= 987.42 $

Part c)

Annual ordering costs for the Workbooks = (Demand* Order Cost)/ Q

Annual ordering costs for the Workbooks = (19500 * 25)/493.71

= 987.42 $

Question no 7.12

In order to solve the problem and make decision in choosing in software, projected volume

is considered. The volume of 80 is above the crossover point so Keith must rent the HP

software.

Question no 12.12

Part a

EOQ = 2DS/H

EOQ = 2(6,000)(30)/10

EOQ = 36,000

EOQ= 189,74 unit

Part b

Average inventory = Q/2 = 94.87 unit

Part c

Optimal number of orders per year= D/Q

=6,000/189.74

=31.62

Part d

Optimal number of days between any two order= working days per year/ number of orders

= 250 days/ 31.62

= 7.91

Part e

Annual cost of ordering and holding cost = (D/Q) S + (Q/2) H

= Working days per year/ number of orders

= 250 days / 31.62

= 7.91

Part f

Total Cost = $ 1897.36 + 6,000 (100) = $ 601,897.36

Question 14.3

Part A

Part B

Question no 14.17

The lot for lot solution for a data is given below.

Period

1

2

3

4

5

6

7

8

9

10

11

12

Gross Requirement

30

20

40

10

40

60

20

80

10

90

70

10

Beginning Inventory

50

20

0

0

0

0

0

0

0

0

0

0

Net Requirement

0

0

40

10

40

60

20

80

10

90

70

10

Time phase Net Requirement

40

10

40

60

20

80

10

90

70

10

0

Planned Order Release

40

10

40

60

20

80

10

90

70

10

0

Planned Deliveries

40

10

40

60

20

80

10

90

70

10

Ending Inventory

20

0

0

0

0

0

0

0

0

0

0

0

The gross cost would last for 10 weeks, so

Holding Cost = 20 * 2.5 = 50

Total Cost = 10*175 + 50 = $1800