1

By: Prof. Y. Peter Chiu 12 / 2008

Adv-POM: Mid-Term #2Chaps.4,5,7

~ SOLUTIONs ~

2

#1-3 3,600(20%) 720

$40 ; 18.75% $7.50

$75

c i h

K

? / ? ** QT* 2 2(75)(720)

7.5120K

Qh

** 120

720

0.1667 2 QT years months

2.00 0.00 y

4 0.1667

0.3333

0.1667

2 0 0. 0

s

0

ear

0.3333 *

or

==>

* 720*0.00

*0.00

0

0

eff

eff

months years

T month T month

years T

yearR

R

#1

B:3,4,5

#2

B:3B:3

B:4B:4

3

#3

If Q=720/2=360 ( units purchased each time; twice a year)

*

( 180) 1 360 120

2 120 3601.6667G Q

G

**

*

*

( 120)

K (75)(720) (7.5)(120)

Q 2 120 2$900

G Q

h

G

Q

( 360) *

(0.6667)*($900)

$600

Addition Gal Costs Q G

B:5B:5

4

#4-5

' 19,2001 $1 1 0.75

76,800

1,600 (12) 19,200 (40)(8)(20)(12) 76,800 $4 0.25 $1 ; (50 150)(1.0) $200

2(200)(19,200)

0.753,200

h h

cih K

Q

#4

B:1,2B:1,2

B:1B:1

5

#5

1

1

3,200 0.1667 .

19,200

3,200 0.0417 .

76,800

of the working time does the company

produce these particula

0.0417 25% 0.1667

Qyears

Qyears

r items

B:2B:2

6

m&c

2*4*(30*12)0

(0.25)(1.8)

2*(6.25)*(15*12)0Q

(0.25)(3.6)

2*(2.4)*(10*12)0

(0.25)(2.0)

80

50

34

Bread

con

E Q

E

E Q

Solution #6

(E0 ) ($1.8)(80) ($3.6)(50) (2)(34) $392

Because Davis has $400 and it is enoughmoney to purchase using the EOQs.Therefore, optimal solutions will be the EOQs.

i iC Q

B:8B:8

7

C Q ($5.00)*Q if Q 0,3000 ，

$100; 20,000; 20%K i

0h (20%)($5.00) $1.00

*0

2(100)(20,000)Q 2,000

1.00

** 0 0

0 0 0*0

h QKG (Q ) C

Q 2

(20,000)(100) (1.00)(2,000) (20,000)(5.0)

2,000 2

1,000 100,000 1,000 $102,000

Solution #7 Supplier - A

B:6B:6

8

Source B

C(Q) = 5.10Q for Q 3000

C(Q) = (5.10)(3000) + 4.50(Q - 3000) for Q > 3000

= 15, 300 + 4.50Q -13,500

= 1,800 + 4.50Q

：

≤

j

1

2

5.10 Q 3000C(Q)

C 18004.50 Q 3000Q

Q

C $5.10 worse than Supplier A

1800C 4.50

Q

("X")

Solution #7 Supplier - B

B:6B:6

9

22 2

( C )QKG (Q) C

Q 2

1800(0.2)( 4.50)Q

(100) (20,000) 1,800 Q (20,000)( 4.50)

Q Q 2

38,000,000 90,180 0.45Q

Q

i

Solution #7 Supplier - BB:6B:6

10

'2 2

*2

' *2 2

*0 0

* *1 1 0 0

38,000,000G (Q) 0.45 0

Q

38,000,000 Q

0.4538,000,000

G (Q ) 90,180 (0.45)(9189)9,189

G (Q ) $102,000 versus $98,450

G (Q ) G (Q ), therefore,

9,189

$98,450

*2 2

*2

select G (Q ) $98,450 optimal

now buy Q 9,189 choose Supplier B

Solution #7 Supplier - B

B:6B:6

11

*

0

0

*

$0.50

($1.50 $0.50) $1.00

$1.00

$0.50 $1.00

0.6667

0.4

0.43

24,000

33

,000

25,290

u

u

u

Cumulative Distribution function of Demand

Crtical rate

Q

C

C

C

C C

Z

Q

Solution #8 : Newsboy model

B:7B:7

12

Week 1 2 3 4 5 6 7

Time-phased net reqts 290 180 150 320 280 210 170

S-M 620 0 0 600 0 380 0

Ending Inventory 330 150 0 280 0 170 0 930

# 9 (1)

(2) [360 (0.8)180] / 2

(3) [360 (0.8)(180) 2(0.8)(150)] / 3

(4) [360 144 240 3(

1:

0.8)(

$360

$252

$2

320)] / 4

48

$37 (8

(1)

)

4 : $

Start Week

Start Week

Stop

c

c

c

c

c

1 1 2 3

4 4 5

( )

(2) [360 (0.8)(280)] / 2

(3) [360 224 2(0.8)(210)] / 3

(1

360

$292

$306.7

4 : $360 ; $248) (2) = [360+(0.8

y 620

6

)(170)]/4

00

=Star

c

c

c c

Stop

r r r

y

t Week

Henr

cr

e

6 6 7

; 380y r r

Total Costs = 3*($320) + ($0.8)*(930) = $1,704 ( Silver-Meal)

B:11B:11

13

Starting in Week 1

Order Horizon

Total Holding cost

123

0$144 (0.8*180)$384 [0.8*180+2*(0.8)*150]

K=360 ∵ K is closer to period 3

∴1 1 2 3 620y r r r

#10B:12B:12

14

Starting in Week 4

Order Horizon

Total Holding cost

123

0$224 (0.8*280)$560 [224+2*(0.8)*210]

K=360

∵ K is closer to period 2

∴54 4 600y r r

#10B:12B:12

15

Starting in Week 6

Order Horizon

Total Holding cost

12

0$136 (0.8*170)

K=360

76 6 380y r r

#10

Week 1 2 3 4 5 6 7

Time-phased net reqts 290 180 150 320 280 210 170

S-M 620 0 0 600 0 380 0

Ending Inventory 330 150 0 280 0 170 0 930

Total Costs = 3*($320) + ($0.8)*(930) = $1,704 ( P.P.B.)

B:12B:12

16

#11 7 7

1 1j j

j j

C r

Yes

Week

Req.Capacities

(C-R)lot-for-lotInitial Solution

(C-R’)

1 2 3 4 5 6 7

290290 180180 150 320 280280 210 170

450 500 100100 100100 450450 100100 100100

160 320320 (50)(50) (220)(220) 170170 (110)(110) (70)(70)

290290 460460 100100 100100 450450 100100 100100

160 40 0 0 0 0 0

R’

Alternative #1- ( Not optimal ! ) K=$225 ; h=$0.8

(A) Week 4th lot : by Week 2 2*($0.8)*4040=$64by Week 1 3*($0.8)*6060=$144

Saves $17

Saves $65

(B) Week 3rd lot : by Week 1 2*($0.8)*100100=$160

Total savings $82

&

B:9B:9

B:10:B:10:

17

Alternative #2

(A) Week 3rd lot :

(B) Week 4th lot :

by Week 2 $(0.8)(4040)= $32by Week 1 $(0.8)(2)(6060)=$96

by Week 1 3($0.8)(100)=$120

the best Total Savings $97

X

Total costs:

Initial SolutionInitial Solution : ($225225)*77+($0.80.8)[280+230+10+180+70]= $2,191$2,191

Optimal SolutionOptimal Solution : ($225)*66+($0.8)[60+380+230+10+180+70]=$2,094]=$2,094

Week

Req.Capacities

1 2 3 4 5 6 7

(C-R’)

R’

Optimal Solution

60 380 230 10 180 70 0Inventories

#12

290290 180180 150 320 280280 210 170

450 500 100100 100100 450450 100100 100100

290290 460460 100100 100100 450450 100100 100100

160 40 0 0 0 0 0

K=$225 ; h=$0.8

350350 500500 00 100100 450450 100100 100100

B:10B:10

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The End