Ch 5 Select HW Solutions

Ch 5 Select HW Solutions - 5.1 5 .‘2 Chapter 5 Answers to...

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Unformatted text preview: 5.1 5 .‘2 Chapter 5 Answers to End-of-Chapter 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—DIm-dZADvr +Dv =$12,634 where A = $20, v = $1.23froll and. 2 = fl = 0.055. m 365(500) If the firm purchases the film pre-packaged, 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.)=(Q-SJ2 9 29 D Average back-order is o-Q“‘+i-i 2 D 2 D=S_ 2 29 D The number of back-orders 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 back-ordered) and TRC (s = co) = arm The solution is therefore to use: s = 0, Q = EOQ if JZADvr <DBgv s = no if «IZADvr > D321; anyOSssc-o 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)(4-9) _ (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. ...
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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 .

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Ch 5 Select HW Solutions - 5.1 5 .‘2 Chapter 5 Answers to...

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