Ch 9 Select HW Solutions

Ch 9 Select HW Solutions - 9.1 9.2 9.3 a. D = 50 units/yr....

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Unformatted text preview: 9.1 9.2 9.3 a. D = 50 units/yr. Chapter 9 Answers to End-of-Chapter Problems v = $0.40/unit Dv = $20/yr A From Table 9.1 Dv = 20 => 12 months time supply Q = D = 50 units b. IQ = 3-D = 12.5 units >10 12 normal TBS = 20 + 12 month time supply From Table 9.2 k = 1.64 Demand is Poisson 0' L = 2L = 3.54 s = 52L + kO'L = 12.5+1.64(3.54) =‘_- 18 units R is the time supply of an order = 12 months (from Prob. 9.1) " - 12D = 62.5 units xR+L — 12 normal TBS = 20 + 12 month time supply k = 1.64 (Table 9.2) S = 2M + key“ = 62.5 +1.64 - 7.91 = 75.472 units a. 1 Simply place an extra order for 5 units. - 28, that is 7 units less than the sales b. At the next ordering point, order 35 less than 7 units, reduce the following since the last review time. If the sales were order quantity by the remainder. 205 9.4 9.5 a. From Eq. 9.9 we have: (1) ’ ATR = a2/2b = 1102/2(45) = 134 litres ' 2AD VI“ ___ ’2 ° 25 ° 1 10 (4.5)(0.3) = 64L. Current EOQ = 1.3 ~EOQ=112L. * ATR > 1.3EOQ Therefore use Q = EOQ = 64L. b. When the Q of 64 is used up, ATR = 134 #64 = 70L. To compute the new EOQ, we need the demand level at that time. we can use Eq. 1 as follows: ‘ 87 = a2/2-45=> a : 88.5 Then EOQ = = 57L 1.3EOQ = 74.4 ATR < 1.3EOQ Thus use one last replenishment of 70L. W: I— EOQ — D(v — g)/vr I = 20 year supply = 20(D) = 10,000 beakers W: 8110 beakers should be sold. 206 9.6 5.4.202 E(i) E(i) (0" ' 6’) Equation 9.19 is c, > , . 2 Equation 9.20 is E(t)0',r > -E—(f:)-(o'm _. as) .65] Dim E(i) 2A,E(t) 0'er(1') TRC,(E0Q)=1’2-4%%-G—’C+ci+Das ' l . TRCM = Am/E(i) + Dam ‘ EOQ if the item is stocked = If cs > Am/Eo') + D(0'm —_ 0") then TRCm < TRCS. - - - £91 _ Le. 1fc, >Am/E(l)+ Ea) (Gui 6:) This is equation 9.19 For TRCM to be less than TRCS in general, 327+ Dam < + cs +06, E(l) e0) 2 , or m > _. as) _. cs) 5(1) ‘ 5(1) . 2 or E(t)o"r > 53.2w“ — o") - q] 2A, E0) E0) which is equation 9.20. r 207 9.7 V 9.8 9.9 , j . ' 2 We do not stock if either (1) c, > A/E(i) or (2) E(t)o“r > £92 -—A—— cs , 2A E(i) (1) => c, > 2'50 =13 10/52 10/52 250 _ 2. (2) => (1.4)(4.7)(.24) > 2.2410/52 a] 1.5792 > -Z%[13 — c,]2 =3. c, 2 6.60 (1) or (2) will hold if c, 2 6.60 and at that point it will he unattractiVe to stOck the item. D = 18 units/yr ' A = $3/unit A = $1.20 mD ' Q = quantity ordered = ? = where m is an integer r = .24/year m = 2 =>‘ order 4 month supply. 93: = 3m(3)(.24) 2 , # of replenishments of C item/year = 6/m Inventory costs/year = = 1:08," ' p TRC(m)=1.08m + (6/m)(1.20) r = 1.08m + m Pick the integer value of m that minimizes this. m = 3 Order a 6 month time supply. Let v = cost per unit, A = replenishment cost, D = demand per unit time, r = holding cost/unit/unit time, B = shortage cost per unit, L = lead time, R = review period, S = order up to level, Q = reorder quantity; 3 = reorder level 1 Assume demand is normally distributed with mean D + variance 02. pdf = f (x, I) At any one time there is only 1 order outstanding with (s, Q). C v 208 In (R, S) let A include the cost of reviewing the stock status (physical stock count for C items). ‘ x r - I ‘ r The cost/unit time for(R, \S) is' E(R,S) = Dv+%+-:—rDR+r(S- D(R+L))+%]:(x—S)f(x,R+ L)dx) S ’ The near optimal solution for this is I 32AD ‘ A Q Q Q: , pu2((s—DL)/0JZ)=E_5= DUES)- ,7 For C items it is not necessary to improve on these solutions. If pu2(k) = % = then the optimal (near) values for the cost/unit time become _ E(R,S) r"- Dvi+%+‘Efl)—__ r7 l—<I>(k) - 'where 4) and d3 are the p.d.f. and c.d.f. of a normal variate. _ 2AD r¢(k)oJf ‘ E(s,Q)—va _ Q +——.1_q)(k) +A where A is the difference the cost of operating the (s, Q) and the (R,S) systems for an item. (This cost could include the intangible cost of the two—bin (s, Q) system—there is achance that the signal for reorder would not be given when the reserve bin is opened). the (s, Q) system will yield lower costs if r0'[¢(k)/1—L<I>(k)][«/R + L — JZ] > A where .l — (I>(k)r= pu2(k) = R/ TBS and assuming that the cost of counting-the stock for (R,S) is very small. If it isn’t a slightly different inequality from the above will result. In general, itemsusing the (s, Q) system of control will have high unit cost, high , variance of demand, high shortage cost, low lead time in "cemparisonwith R. 209 9.10 9.11 The disposal decision is a mirror image of the specific opportunity to buy. We have a special opportunity to dispose at a unit value of g instead of acquiring at v,-. , Equation (9.13) can be rewritten as: I—W=EOQ+v;g(—lI-:-) (r1) I-Wis the desired ending position after disposal. Comparison of (1) and (5.23) shows a close resemblance. Qop, in (5.23) is the desired position after taking advantage of the special opportunity to buy. Note also that the derivations of (5.23) and (9.13) are very similar. ’ a. log((Vc +a/i) / (Vu +a/i))/log(D/(D+i)) = log((0.5 + 0.09/0.12) / (0.7 + 0.09/0.12)) / log(20 / (20 + O. 12)) = 24.81 So the firm should save 24 units. If current inventory is 50 units, they should dispose of 26 units. b. log((Vc +a/i)/(Vu +a/1))/log(D/(D+i)) = log((0.8 + 0.09/O.12) / (0.7 + 0.09/0.12)) / log(20 / (20 + 0.12)) = -11.15 So the firth should save -12 units. Because the current salvage value is higher than the ultimate salvage value, the firm should net save’any units. It is desirable to dispose of all inventory. _ ' ’ C. log((Vc +a / i)/(Vu +a/i)) / log(D/ (0+0) = log((0.7 + 0.09/0.12) / (0.7 + 0.09/o.12)) / log(20 / (29 + 0.12)) = o ' So the firm should save 0 units. Because the current salvage value is the same as * the ultimate salvage value, the firm should not save any units. It is again desirable to dispose of all inventory. d. log((Vc+a/i)/(Vu +a/i)) _ log((O.5+0.9/0_.12)/(0.7+0.9/0.12)) _ 62 14 log(qD/ (D + i) + (l - (1)) log(0.4(20) / (20 + 0.12) + (1 - 0.4)) 210 9.12 So the firm should save 62 units. Because the batch size is larger now (averaging 2.5), the firm should save more units than in case (a). Note: The instructor may wish to provide students with the following hint: Hint: Suppose thatthe current level is a and that n replenishments (including the‘ one needed now) are used at times t; (i=1,2,3. . .n) where t1=0 (current time). Let H =a/b define the moment (horizon) where the demand rate becomes zero. Then show that the normalized total relevant costs as a function of n and the ti’s depend 2 only upon M = Abs . vra Solution: , n ’i4-l TRC(n,t,.'s)=nA+erj(t—t,)x,dt r 7 t ’ (1) ' i=1 n Note: tn+1 = g = H Consider the quantity: n ‘M J=2I(t-ti)x,dt / i=1 i H t :13 1m = Inga—24. Ix, 0 H 1,- ‘14-1 = }t(a —bt)dt—:t,._[ 0 i=1 4 (a - bt)dt Integration and simplification leads to: 3 :1 J = H b—k t..{(H-t.-)2-(H-t.-+.)2} 6 2i=1 Substitution in (1): 3 ‘ 3 n TRC(n,t,.'s) =sz+vr|:H6b - H2], Emil-“.02 -(1-ui+1)2}] i=1 t- . . . . where ui = , Le. we redefine the tlmes as fractions of the horizon. TRC(n, ti' s) Define ' NTRC(n.t,-'S)= 3 H bvr 211 Then NTRC(n,t.-I' S): '3‘";th "' WY ’(1’ “1+1)2}]‘ = "M +[%’%§“1kl_ut)2 " (1'— “i+1)2}] Thus the only parameter, Other‘tha‘n therdecisionqvariables is M. I 9. 13 a. Equation 9.5 is L in months D in units/year. 32L = DL = 60(1.5)'/ 12 = 7.5 0L=JDL=J75 =2.74 I . We know that the reorder point is half of 60': 30. So, using Equation 9.3: 30 = 7.5 + k(2.74), or k = 0.91. 9 D(TBS) 2:5 D pu2(0.91) = 0.181, N From Equation 9.2 pu2(k) = 0.5 TBS = ————- = 2.77 years 0.181 b. ' 2L = DL =15 CL = 3.87 k = w = -1.29 3.87 ' . . Q 1 D F mE uato 9.2 a k =' or -1.29 =—-——, =— m q l n p“2( ) D(TBS) p"z( ) 2TBS Q 2 0.902 = —1—; :5 TBS = 0.55 years 2TBS 212 c. D=30 L=1.5 :2, =3.75 0L,=1.94 k =M=0545 _ 1.94 As above, using Equation 9.2 TBS =% pu2(0.645)=-21-/0.259 = 1.93 years, " ’ ' 9.14 . . . 2AD , . 7 - ' Wlth derivative demand EOQ = , I, the inventory level, goes from EOQ down vr ‘ to 0 in a linear manner. (The time from I: EOQ to I = 0 isthe length of a, cycle). ‘ Coverage = C0 = % (eqn. 9.13) C0 varies from 12‘ to 0, linearly. - Dvr r 9.17 a. 7 pr = prob{no Poisson demands in T years}: e'” 0.9 = e‘” — DT = In 0.9 DTV=V 1n 1/.9 = .10536 the graph is ahyperbola DT= .10536 ,T D b. v , pr = prob{m or more demands in 1 year} °° D’e‘” "‘“ D’e'” 1:»: j! i=0 1' c. As c1 increases, T and m should both increase, since it is less desirable to deactivate or activate an item. d. Using a Bayesian argument, start with a prior distribution on D values.- Specify a probability distribution of times of changes (eg. Poisson) and sizesof changes. This becomes intractable Without some'assumptions, such! as a gamma prior distribution for D. Or else assume that the only possible changes are to O or away from 0 to a D from a gamma distribution. Another approach would be to simulate the behavior and costs using actual demand data for different pairs of T and m. - 215 g 9.18 - Let P(E) = prob{st0ckout in a lead time}. As in the development of Equation 9.2 we want P(E) to satisfy TBS = 94.1.... A D P(E) OI' P(E) D (ms) .(1) Ifs = DL+ kJD—L . ’ 1 ' (2)1 Then in theory we need “to find the value of k such that a Poisson variable with mean DL, exceeds DL + kJDL with probability p(S). For a C item this is too sophisticated. Instead, we use a normal approximatiOn to the Poisson which leads to Equation 9.2, namely select k so as to satisfy . Q D(TBS) P.2(k) = (3) For a given pair of Q/D and TBS k is uniquely determined by Eq. (3). Then Eq. (2) gives a plot of 3 versus 'DL (fOr the fixed k value). Graphically we have ‘— k2 for 2nd pair of Q/D-TBS <— k1 for one pair on/D-TBS DL For D = 0.5 yrs. and TBS .= 10 yrs. Equation (3) gives: _ 0.5 _: k =———=0.05 pu2( ) . from Table B.1 k = 1.645 216 I A, Then Equation (1) gives s = 3.7 +1.645«/3.7 = 6.86 say 7 units. Note: Using the exact Poisson formula, with a mean of 3.7 ' prob(x > 6) '= 0.82 r (x is demand in the lead time) prob(x > 7 = 0.035 From Equation (1) we want p(E) = 0.05. The smallest s that does at least-this well is seen to be s = 7 which is the exagt same result as that obtained above using the normal approximation. ...
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Ch 9 Select HW Solutions - 9.1 9.2 9.3 a. D = 50 units/yr....

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