Nahmias+Solutions+Chapter+4

SOLUTIONS TO SELECTED PROBLEMS FROM NAHMIAS’ BOOK

CHAPTER 4

EOQ

4.14 a) K = 100 Hex Nuts Molly Screws I = .25

c = .15 c = .38 λ = 20,000 λ = 14,000

For hex nuts: Q*

1 =

(2)(100)(20,000)

(.25)(.15) = 10,328 T1 = . Q*1 /λ = 5164 years

For molly screws: Q*2 =

(2)(100)(14, 000)

(.25)(.38) = 5,4229 T2 = Q*2/λ = .3879 years

b) 1. 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 = 150/month = 1800/year λ = 720/year K = 700 c = 85.00 I = .28 h′ = Ic(1 -λ /P) = (.28)(85)(1-720/1800) = 14.28 a) Q* = 2Kλ h' = (2)(700)(720) 14.28 = 266

b) T1 = Q*/P = 266/1800 = .1478 years (up time) T = Q*/λ = 266/720 = .3694 years (cycle time)

T2 = T - T1 = .3694 - .1478 = .2216 years (down time) c) 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) _{= 8} ) 45 )(. 25 (. ) 4 )( 1 )( 2 ( ₌ where λ = 4 I = .25 K = 1 c0 = .45

b) If you buy in single packs, the average annual cost is

λc0 + 2KλIc0 = (4)(.45) + (2)(1)(4)(.25)(.45)

= $2.75 yearly

If you buy in 12 packs, the average annual cost is λ_c 1 + Ic1Q 2 + Kλ Q =( 4)(.42) + (.25)(.42 )(12) 2 + (1)(4) 12 = $2.64

which is slightly less expensive.

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.

4.23 Incremental schedule for source B. C(Q) = 2.55Q for Q ≤ 3,000 C(Q) = (2.55)(3,000) + 2.25(Q - 3,000) for Q ≥ 3,000 = 7650 + 2.25Q - 6750 = 900 + 2.25Q for Q ≥ 3,000 C(Q)/Q = 2.55 for Q 3,000≤ = 900/Q + 2.25 for Q 3,000≥ G(Q) = [20,000] 900 Q +2.25+ (100 )( 20, 000 ) Q +(.20) 900 Q +(2.25) Q 2 = 18,000,000 Q + 2,000,000 Q + (.20)(2.25)Q 2 +$45,090 = 20,000,000 Q +0.225Q+$45,090

The minimizing Q occurs where Q = 20,000,000

.225 = 9428 cost = 2 0 , 0 0 0 , 0 0 0

9 4 2 8 + (.225)(9428) + $45,090 = $49,332.64

Since the annual cost for source A exceeds $50,000, now source B should be used.

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)

EOQlettuce = (2)(100)(1280) (.25)(.45) =1509 EOQzucchini = (2)(100)(630 (.25)(.25) =1420 Next we find the value of the multiplier m.

m = W

wiEOQi

∑

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.

4.27 If the vegetables are purchased at different times, then larger lot sizes could be used without violating the space constraint, since the maximum inventory levels of Q1

would not be reached simultaneously.

4.29 The input data for this problem are:

λ P h h′ K 2500 45000 2.88 2.7200 80 5500 40000 3.24 2.7945 120 1450 26000 3.96 3.7392 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 = λTj. Obtain:

Q1 = 343.21 Q2 = 755.05 Q3 = 199.06

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

T1 = .007626 T2 = .018876 T3 = .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

G(T) =

j=1 n

∑

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

4.31 Given information: λ = 60 per month = 720 per year K = $20. c = $2.80 I = .22 (giving h = (.22)(2.8) = .616. a) EOQ = 2Kλ h = (2)(20)(720) .616 =216 T = Q/λ = 216/720 = 0.3 years (3.6 months). b) R = λτ = (60 per month)(3/4 months) = 45.

c) G(Q*) = 2Kλh= (2)(20)(720)(.616) = $133.19 annually.

d) It is possible that the store does not have enough room to store 216 boxes of the

4.34 K = 200 (both cases) λ = (200)(52) = 10,400 c = 75 locally

= 65 overseas

a) Average annual cost at the optimal solution is G(Q) = λc +

2KλIc in both cases. Substituting the appropriate costs we obtain: locally G(Q) = (10,400)(75) + (2)(200)(10,400)(.2)(75) = $787,899.37 overseas G(Q) = (10,400)(65) + (2)(200)(10,400)(.2)(65) = $683,353.91 Difference = $104,545 annually.

b) The average pipeline inventory is λγ . Hence:

locally: (200)(52)(1/12) = 866.67 units overseas: (200)(52)(6/12) = 5,200 units The value of the local inventory is (866.67)(75) = $65,000 Value of overseas is (5,200)(65) = $338,000

Difference = $273,000

Assuming 20% rate of interest, this difference would cost the firm (273,000) (.2) = $54,600 annually, which is less than the savings realized in holding and set-up cost realized by producing overseas. Hence on the basis of cost alone, overseas production is still preferred.

c) The differences in cost of $50,000 yearly could easily be outweighed by the firm's competitive advantage at being able to be more responsive to market demands due to the shortened production lead time locally.