This preview shows pages 1–15. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 5.1 5 .‘2 Chapter 5
Answers to EndofChapter Problems a. Lat A1 and EOQl be the old values of A and EOQ. Then, {2141) r A ’2A'D (12.80 1 1
BO = = _ = 0 = Q vr Al vr 3.20 Q Q Thus, i) EOQ = 800 units,
ii) vEOQ = $0.4oxunit X 800 units = $320.00 121)
iii) TEOQ = ﬁ = 4 month supply Increasing the ﬁxed replenishment cost A should result in less frequent (hence
longer) orders, as we have just found. E EOQ‘ = 0.394E0Q1. b. As before, EOQ =
0.30 Thus, i) EOQ = 358 units,
ii) vEOQ = $143.20
iii) TEOQ = 1.79 month supply One expects that an increase in the inventory carrying charge : would tend to
decrease the average inventory carried, and result in more frequent orders. D = 250 unitsfwk
v = $ IOIunit
m = shipment size Total shipment cost = TSC = $50 + 52m. Because TSC has a ﬁxed component of
$50, and varied directly by $2lunit, a (ﬁxed cost per order) is $50, and v becomes 10 + 2 = $12funit. 91 5.3 MOP, =EOQ =112AD
vr __ l2x50><250
12x0.004 M0,, = 721.?
= 722 unitsfshipment If the firm loads and packages its own ﬁlm, it would be in the situation of having
a ﬁnite replenishment rate, and the total annual cost would be: TRC = Jl—DImdZADvr +Dv =$12,634 where A = $20, v = $1.23froll and. 2 = ﬂ = 0.055.
m 365(500) If the ﬁrm purchases the film prepackaged, the annual cost would be: TRC(EOQ) = VZAIDv'r + Di)1 = $12,735 where A1 = $3.00 and v1 = $1.251roll. Based solely on inventory costs, the ﬁrm should load and package its own ﬁlm.
However, from the production standpoint, other considerations would include the
possible need to invest in equipment, the need to keep stocks of bulk ﬁlm, and the
need to train manpower for the job. From the standpoint of marketing, it may be
that the “Discount” brand ﬁlm is better known and would sell more. On the other
hand, producing one’s Own ﬁlm might guarantee a more dependable source of
supply, and if the ﬁrm can sell processing with the ﬁlm, it could garner extra
business. Also, we should consider whether or not it is likelythat the supplier will
raise prices in the near future. A side issue is the meaning of the phrase “valued
at $1.23 per roll”—one really doesn’t know if the ﬁrm values at full cost,
marginal cost, sales price, etc. Any of these factors could affect the ﬁnal decision. 5.5 Inventory
hwd Q—s Q—s s
+ _
2 ° D 0'1.)=(QSJ2
9 29
D
Average backorder is
oQ“‘+ii 2
D 2 D=S_
2 29
D The number of backorders per replenishment cycle is of course 3 units. 93 (Q_S) vr+—;A2+B(Bzv)s b. TRC(Q,S)= 2Q Q Q Setting the derivatives of TRC (Q,s) equal to zero gives the conditions 2
2 AD+DBzvs+S W 3TRC 2 .....1 8Q vr
and
0=5TRC=>Q=5+DBZV ...... ..2 as vr Unfortunately, substituting 2 into 1 give the result that B: = lat:
Dv
which is completely independent of s and Q, and certainly need not hold. In fact, the minimum occurs at one of the boundary values of 5, either 0 or no.
Comparing the two cases, we have s = O Q = EOQ
and TRC (s = 0) = JZADvr and s = no Q = s (all demand are backordered) and TRC (s = co) = arm The solution is therefore to use:
s = 0, Q = EOQ if JZADvr <DBgv
s = no if «IZADvr > D321; anyOSssco 2A0 s(s+ 21332)‘ . if JzAnw = my
vr , 94 c. Now the cost equation becomes AD (Q—var D .s'2
TRC , =_ _ __ _ I (Q5) Q + 29 +9392!) _ 5rch * 2 _ 2AD+(s*)2v(r+33)
0— am =(Q_) — vr "'1
andO=aTRC=ﬂ= Q“ ...2 as (Bz+r) Substituting 2 into 1 gives
9* = B3 + r . PAD
B3 vr
and 5* = ZADr
v33 (r+ .33) Thus, TRC(Q*,5*) = 213L1983 vr
33 + r 5.9 a. Step 1: Find EOQ based on the discount price ZAD E0 (1‘ t = ——
Q( 15001111) “JO—d” D = 40x5é = 2080 units/yr EOQ(discount) = W = 203 units
_ V (9.7)(026) Step 2:
Is the EOQ (discount) from Step 1 feasible? No. (203 < Qb) Go to Step 3 97 Step 3: Calculate TRC at feasible EOQ’s. and breakpoints. TRC(EOQ) = ,/2(25)(2080)(10)(0.25) + 2080><(10) = $21320! yr
300x9.7><0.26 + 25x2080 2 300
= 320727.63 x yr < TRC(EOQ) TRC(Q,,) = + 2080X9.7 Use Q = Q}: 300 units b. Let v1 = unknown (maximum) unit cost 500 x v1 x026 + 25 x 2080 TRC(500) = +2080><vl = 2145*»?1 +104 The mining company will be indifferent when
TRC(300) = TRC(SOO)
This is ctrrently the best quantity
20727.63 = 2145191 + 104
v' = $9.61funit = maximum acceptable cost 5.10 a.
TRC1(Q)=Q:°r+%(A+QvD) O<Q<Qh
TRC2(Q)=g%—a2:‘—i)—r+£QJ[A+Qvo(l+d)] QbsQ b. A term by term comparison of the above equations, shows that for the same
value of Q TRC; is below TRCz. 98 Case 1: Case 2: Case 3: TRC2  The best order quantities for the 3 cases are Q5, EOQ; (with no d), and EOQ;
(with (I). Step 1: Find EOQ with no discount EOQ(no discount) = 1’ 2A0 = 2236 units
vr Step 2: Is the EOQ from Step 1 valid?
No 2236 > (2;, Go to Step 3. Step 3: The best solution is either Q5 — 8 = 1499
or EOQ (with discount) TRC1(1499)V= 1499(1'.0)(0.2) /2 + 50,000f_1499 X (10 +1499 x 1.0) There are 3 possible cases for the best order quantity.
c
 = $50,483/yr. TRC1(EOQ with discount): 1/2ADvD (1+ d)r + Dvo(1+ d) = 1}2><10><50,000><1.Ox 1.005X0.2 + 50,000x1.0x1.005 = $50,698.33/yr > TRC. We use 1499 units as Q. 99 5.15 3.. The EOQ is EOQ= ’ZXSOXIOO = 31
4OX0.25 104 .6' 5.16 ,5on lﬂ= (wig;
_ rv (0.25)(40) b. To ﬁnd the value of A, set the total cost at Q = 1 equal to the total cost at
Q = 2. AD vr AD 2w
+ — + — Q(=1) 3'Qc=2> 2 Substituting and solvinggives A = $0.10! The issue here is that there is a shared resource in each case. The container must
be shared among several items in the latter example, and the production machine
must be shared in the former. Taking machine time, or container space, with one
item necessarily means that less will be available for other items. This is an
example of the joint replenishment problem that will be covered in Chapter 1 1.
Often, there is a large ﬁxed cost for the container, or to setup the machine for a
family of items; then there is a smaller cost to add an item to the container or to
the production schedule. 5.19 Strictly Speaking, neither person is correct. The most appropriate strategy is to
reduce the set up cost to $25 and use the EOQ based on the new value. _ 92: ﬂ
TRC(Q) — 2 + Q
Under the currenrA value
_TRC(Q) = w... Lm_)§902 = 0_28Q + 502000 EOQ (current A) = 1336 units
TRC(EOQ) = $748.33/yr TRC(500) = $1140.00!yr‘ 107 Under the reduced A value. 125,000
Q TRC’(Q) = 0.28Q + EOQ(new A) = 668 units
TRC'(new_ EOQ) = $374.17fyr TRC’(500) = $390.00!yr To summarize: A value Q used Costsr’yr
$100 EOQ for A=100 $748.33fyr
$100 500 units $1140.001yr
$25 EOQ for A=25 $374.17lyr
$25 500 units $390.00/yr We can see that the foreman’s suggestion to reduce A and stay with Q = 500 is
better than using the EOQ with the higher A value. 5.22 For D=10000
Q: = 2(25)(10000) ___ 577
U (3)(5)
2(25)(10000) 4.9 = —_ = 583 Q U (3)(49)
_ (20300000) = Q‘” _ (.3)(4.75) 592 Because the EOQ with discount is less than both breakpoints, we compare the
total cost at each breakpoint with the total cost at the EOQ. TRC(EOQ) = AD IQ + Qvon’ 2 + Dvﬁ = 25(10,000) 1577 + 577(5)(0.3) I 2 +10,000(5)
= $50,866 TRCUOOO) = AD {Q + Qvaa —d)r/'2 + Dvuﬂ — d) = 25(10,000)I1000+1000(5)(1_— 0.02)(0.3);2 +
10,000(5)(1— 0.02) = $49,935 TRC(2000) = AD IQ + Qva (l  d)ra’2 + Dvoﬂ  d) = 25(10,000)!2000 + 2000(5)(1— 0.05)(0.3) f 2 +
10,000(5)(1  0.05) = $49,050 So, the best order .quantity is 2000. 2(25)(1000)
= ———=183
9’ V (36)
_ [2(25)(1000) =
Q“°_ (.3)(4.9) 185 For D=1000 110 _ 2(25)(1000) =
Q4'”_V (3004.75) 187 The total cost at the EOQ = $5,274, at the breakpoth of 1000 = $5,660, and the
' breakpoint of 2000 = $6187. Hence, the EOQ with no discount is best.
5 23
l
 For D=4000, the same logic is applied. The EOQ is 365, and the EOQs at the two
breakpoints, respectively, are 369 and 375. The total relevant cost at the three
points (EOQ, 1000, and 2000) are $20,548, $20,435, and $20,475. Thus, the best
order quantity is at the breakpoint of 1000. For D=130,000, the same logic is again applied. The EOQ is 2082, and the EOQs
at the two breakpoints, respectively, are 2103 and 2136. The total relevant cost at
the three points (EOQ, 1000, and 2000) are $653,123, $640,985, and $620,550.
Thus, the best order quantity is at the breakpoint of 2000. i ,ny,
1 70.71
2 56.57
3 28.28 Total 155.56 TCS_§
N r (TCS)(N) =%(2\/17,v,)2 = 12099 Current TCS =2%= 125 111 .....__W._.—...._.._~_...m__..__,__._ _._._... .....__..... __._ .. __
. u 2500.0 2000.0 1500.0
Currant Operating Point 1000.0 500.0 0.0 For TCS=112S, (TCS)(N) =12099 => N =10.76 Then TCS =3!i = E = 104.6 N r r  Similarly N = 12 => i = 84.0
r Thus for any A in 84.0 < i < 104.6 improvements are possible on both
1' r dimensions. ...
View
Full
Document
This note was uploaded on 11/03/2008 for the course IE 251 taught by Professor Wilson during the Spring '08 term at Lehigh University .
 Spring '08
 Wilson

Click to edit the document details