Chapter4 Ans Nahmias

Solutions To Sample Problems From Chapter 4

4.1

When to order and how much to order.

4.4

a) Sum total of ending inventories is 26 + 38 + 31 + 22 + 13 + 9 + 16 + 5 = 160. Monthly holding cost h = Ic where I = monthly interest rate and c = value of item. In this case: I = (.20 + .03 + .02)/12 = .02083 . . . per month c = $8,000 Hence, h = Ic = $166.67/truck/month. ⇒ total holding cost = (160)(166.67) = $26,666.67.

b)

Since we observed a cost of $26,666.67 for 8 months, the average annual cost would be approximately (26,666.67) 2/3

4.5

= $40,000 per year

Total cost in 5 years = 12,500 + 5(2,000) = $22,500.

Let s = yearly savings as fraction of sales. Then s solves (80,000)(s)(5) = $22,500 s = 05625 (answer 5.625% per year)

4.7

a) 400 per year b)

200 6 wks Jan

6 wks Jun

Note: This diagram is not to scale. Avg

=

(200)(6) + (200)(6) (200)(12) = = 46.1538 52 52

c) γ = 6 weeks = 6/52 years. From part (a), λ = 400/year. Hence λγ = (6/52)(400) = 46.1538. (Same as the answer to part (b).) 4.9

The demand rate is 720 per year. a)

1 One truck per week implies shipments of 14 housings each week. Therefore the cost is (300 + 160)(52) = $23,920. 2. Since each truck holds 40 housings, it follows that this policy results in 720/40 = 18 shipments of one full truckload. Cost = (300 + 160)(18) = $8,280. 3. 720/120 = 6 shipments per year of 3 truckloads each shipment. Cost = 6(300) + 18(160) = $4,680.

b) The highest annual cost corresponds to policy 1. If demand is highly variable and/or unpredictable, shipping one truck per week will minimize the likelihood that stockouts are incurred. If stockouts of the casings are very costly, this policy could be optimal in a systems sense.

4.12

λ = (1250)(.18) = 225 c = $18.50 I = .25 h = Ic = (.25)(18.50) = 4.625 K = 28 a) Q* =

2 Kλ (2)(28)(225) = = 52 (.25)(18.50) h T = Q*/λ = 52/225 = .2311 yrs.

b) T = 12.02 weeks Hence, r = λγ = (225/52)(6) = 25.96 ≈ 26 units

c) If Q = 225, the average inventory level is Q/2 = 225/2 = 112.5. The annual holding cost is (112.5)(4.625) = $520.31. At the optimal solution, the annual holding cost is (52/2)(4.625) = $120.25. The excess holding cost is $400.06 annually. The annual holding and set-up cost incurred by this policy is $520.31 + 28 = $548.31 since there is only one set-up annually. The average annual holding and set-up cost at the optimal policy is

2Kλ h

=

(2)(28)(225)(4.65)

= $241.40.

Therefore the annual difference

= $306.91

4.14

a)

K = 100 I = .25

Hex Nuts c λ

For hex nuts:

= .15 = 20,000

*

Q

*

*

Q

T2 b) 1.

2

c λ

(2)(100)(20,000) (.25)(.15)

=

1

T1 = . Q For molly screws:

Molly Screws

1 /λ

=

= 10,328

= 5164 years (2)(100)(14, 000) (.25)(.38)

*

Q 2/ λ

=

= .38 = 14,000

=

=

5,4229

.3879 years

Average annual cost when ordered separately: (2)(100)20,000)(.25)(.15)

+

(2)(100)(14,000 )(.25).38)

= 387.30 + $515.75 = $903.05 2

If both products are ordered when the hex nuts are ordered (every .5164 yrs.), then hex nut cost is the same. Molly screw cost is only the holding cost. Qmolly = (λ) (γ1) = (14,000)(.5164) = 7230. Holding cost = (7230/2)(.25)(.38) = $343.43 Total cost of this policy = $387.30 + 343.41 = $730.73 (a savings of $172.34 annually from ordering separately).

3. If both products are ordered when the molly screws are normally ordered (every .3878 yrs.), then the lot size for the hex nuts is: Qhexnuts = λT2 = (20,000)(.3878) = 7756 Holding cost = (7756/2)(.25)(.15) = $145.43 The total cost of this policy is $515.75 + $145.43 = $661.18 which represents a savings of $241.87 over ordering separately.

4.19

P λ K c I ′ h

= = = = = =

150/month = 1800/year 720/year 700 85.00 .28 Ic(1 -λ /P) = (.28)(85)(1-720/1800) = 14.28

a)

Q* =

b)

T1 T

= =

T2 = c)

2Kλ (2)(700)(720) = = 266 h' 14.28 Q*/P = 266/1800 = .1478 years (up time) Q*/λ = 266/720 = .3694 years (cycle time) T - T1 = .3694 - .1478 = .2216 years (down time)

Maximum inventory level = H = Q*[1 - λ/P] = 266[1-720/1800] = 159.60 Maximum dollar investment = (159.60)(85.00) = $13,566.00.

4.21

a)

Optimal number of single rolls to purchase

Q(0) =

(2)(1)(4) =8 (.25)(.45)

where λ = 4 I = .25 K = 1 c0 = .45 b)

If you buy in single packs, the average annual cost is λ c0 +

2Kλ Ic0

= (4)(.45) + (2)(1)(4)(.25)(.45) = $2.75 yearly

If you buy in 12 packs, the average annual cost is λ c1

Ic1Q Kλ (.25)(.42)(12) + = (4)(.42) + 2 Q 2 which is slightly less expensive.

+

+

(1)(4) = $2.64 12

c) It may be inconvenient to carry and store the tissue. A 12 pack requires three years before it is completely consumed. The tissue may become brittle during that time.

4.26 First we find the space consumed by lettuce and zucchini. .45/.29 = 1.55 ⇒ lettuce consumes (.5)(1.55) = .775 .25/.29 = .862 ⇒ zucchini consumes (.5)(.862) = .431 Computing the respective EOQs, we have

EOQtom

=

(2)(100)(850) = 1531 (,.25)(.29)

EOQlettuce

=

(2)(100)(1280) = 1509 (.25)(.45)

EOQzucchini

=

(2)(100)(630 = 1420 (.25)(.25)

Next we find the value of the multiplier m.

m =

W 1000 1000 = = ∑ wi EOQi (.5)(1531) + (.775)(1509) + (.431)(1420) 2547

= .3926

Hence, the optimal order quantities are:

Qtomatoes

= (1531)(.3926) = 601

Qlettuce

= (1509)(.3926) = 592

Qzucchini

= (1420)(.3926) = 558

Checking that the constraint is satisfied:

(.5)(601) + (.775)(592) + (.431)(558) = 999.8.

4.29

The input data for this problem are: λ 2500 5500 1450

P 45000 40000 26000

h 2.88 3.24 3.96

h′ 2.7200 2.7945 3.7392

K 80 120 60

*

a) First we compute T . T

*

(2)(80 + 120 + 60) (2.72)(2500) + (2.7945)(5500) + (3.7392)(1450)

=

= .1373

years b) and c) The order quantities for each item are given by the formula

Qj

= λ T j.

Q1 Q2 Q3

= 343.21 = 755.05 = 199.06

Obtain:

The respective production times are given by Tj = Qj/Pj. Substituting, one obtains:

T1 T2 T3

= .007626 = .018876 = .007656

It follows that the total up time each cycle is the sum of these three quantities which gives: total up time = .03416. The total idle time each cycle is .1373 .0342 = .1031. The percentage of each cycle which is idle time is thus .1031/.1373 = 75%.

d) Using the formula n

G(T) = ∑ (K

j

/ T + hj ' λ j T / 2)

j =1

one obtains, G(T) = $3787.82 annually.

4.35

First find optimal policy for supplier A. λ = (40)(12) = 480

G0 G1

500

G2

1000

All Units

Q(2)

=

(2)(150)(480 (..23)(1.10)

Q(1)

=

(2)(150)(480) (.23)(1.20)

= 754

= 722

not realizable.

OK

Compare costs at 1000 and 722.

At 1000: G2(1000)=(480)(1.10)+

(150)(480) (.223)(1.10)(1000) + 1000 2

= $726.50

At 722:

G1(722) = (480)(1.20) + (2)(150)(480)(.23)(1.20) = $775.36

If they use supplier A, then standing order should be 1000 units at cost $726.50 However, if they use supplier B: Slope = 1.05

875 Slope = 1.25

700

First find EOQs:

at $1.25

Q

at $1.05: Q

*

* 0

*

Cost at Q 1:

1

= =

(2)(150)(480) (.23)(1.25)

= 708

(2)(480)[150 + 875 − 735] (.23)(1.05)

not realizable

= 1074 OK.

λc1 + λ [K + R1 − c1 q1 ]+ IR1 + c1 (Q * −q1 )

= (480)(1.05) +

Q

2

2

480 .23 [150 + 875 − 735]+ [875 + 1.05(374)] 1074 2

= $779.39 Use supplier A at cost of $726.50 annually. Standing order should be 1000 units.

4.45

Type Q A B C D E F G H

a), b) and c).

Lambda

25900 64833 42000 105135 14400 36046 46000 115148 12500 31290 75000 187741 30000 75097 18900 47311

Prod Rate

Cost

h

h'

K

h'lam

1125000

0.003

0.00063

0.000615

120

15.94

1375000

0.002

0.00042

0.000407

80

17.10

825000

0.008

0.00168

0.001650

160

23.77

800000

0.002

0.00042

0.000395

80

18.21

450000

0.01

0.0021

0.002041

60

25.52

975000

0.005

0.00105

0.000969

120

72.69

725000

0.004

0.00084

0.000805

20

24.16

300000

0.007

0.00147

0.001377

60

26.03

700 T* =

223.42

2.503218

Average Annual cost Computation: Set-up 47.9382 31.9588 63.9177 31.9588 23.9691 47.9382 7.98971 23.9691 279.640 d)

Holding 19.95233 21.40398 29.75042 22.79067 31.94210 90.98235 30.23542 32.58272 279.6400

Total Average Annual Cost: $559.28

A rotation cycle time of 2.5 years will clearly be too long to satisfy their obligation of making three shipments per year for each type of button. They would need to reduce the rotation cycle time to 4 months and adjust the lot sizes down accordingly. The actual lot sizes would depend on the contractual obligations the firm has with their customers.