SOLUTIONS TO SELECTED PROBLEMS FROM NAHMIAS’ BOOK CHAPTER 4 EOQ 4.14
a)
K = 100 I = .25
Hex Nuts c λ
For hex nuts:
Molly Screws
= .15 = 20,000
Q*1 =
(2)(100)(20,000) (.25)(.15)
T1 = . Q*1 /λ = For molly screws:
Q *2 = T2
b) 1.
=
c λ
= .38 = 14,000 = 10,328
5164 years
(2)(100)(14, 000) (.25)(.38)
Q*2/λ
=
=
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* =
2K λ (2)(700 )(720) = = 266 h ' 14.28
b)
T1
Q*/P = 266/1800 = .1478 years (up time)
T
= =
T2 =
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
c)
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 λc0 +
2KλIc0
= (4)(.45) + = $2.75 yearly
(2)(1)(4)(.25)(.45)
If you buy in 12 packs, the average annual cost is λc
1
+
Ic1Q K λ + = ( 4)(.42) 2 Q
+
(.25)(.42 )(12) 2
+
(1)(4) = $2.64 12
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 (100 )( 20, 000 ) 900 Q + 2.25 + + (.20) + (2.25) Q Q Q 2
=
18,000,000 2,000,000 (.20)(2.25)Q + + + $45,090 Q Q 2
=
20,000,000 + 0.225Q+ $45,090 Q
The minimizing Q occurs where Q =
cost =
2 0,0 0 0,0 0 0 9428
20,000,000 = 9428 .225
+ (.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) = 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 = = = .3926 ∑ wi EOQi (.5)(1531) + (.775)(1509) + (.431)(1420) 2547
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 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* = years b) and c)
(2)(80 + 120 + 60) (2.72)(2500) + (2.7945)(5500) + (3.7392)(1450)
= .1373
The order quantities for each item are given by the formula Qj
= λTj.
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 =1
j
/ T + hj ' λ j T / 2)
one obtains, G(T) = $3787.82 annually.
4.31
Given information: λ K c I
a)
EOQ =
= 60 per month = 720 per year = $20. = $2.80 = .22 (giving h = (.22)(2.8) = .616.
2Kλ = h
(2)(20)(720) = 216 .616
T = Q/λ = 216/720 = 0.3 years (3.6 months). b)
R = λτ
c)
G(Q*) =
= (60 per month)(3/4 months) = 45. 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 filters, thus requiring a smaller standing order.
4.34
K λ c
a)
= = = =
200 (both cases) (200)(52) = 10,400 75 locally 65 overseas
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) + = $787,899.37
(2)(200)(10,400)(.2)(75)
overseas G(Q) = (10,400)(65) + $683,353.91
(2)(200)(10,400)(.2)(65) =
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.
