Ch 8 Select HW Solutions

Ch 8 Select HW Solutions - 8.2 All oustanding orders at a...

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 8.2 All oustanding orders at a time t must arrive by time HI. and no order placed after I can arrive by t+L. Therefore,_ the net stock at time t+L must be equal to the inventory position at time rminus any demand x0 in t to r+L. Now the inventory position at time ris s+j,j=1,2,..,,Q with probability IIQ (i) If pm(nolj) is the probability that the net stock at r+L takes on the value no given an inventory position of s+j at rI then (n I j: ppo(s+j—no|fL),s+j—n020 pm 0 J Ootherwise The expected on-hand inventory, given the inventory position s+j is 181 fj = inoPMG‘o I "0‘0 = g(5+j'xo)Ppo(xo '37:.) In=0 Thus, Q11"? i=2— (s+j—no)ppo(xo IJ‘cL) using (1). 1:1 qu-D Furthermore, the probability that a particular demand required backordering is equal to the probability that the net stock is zero or less. That is, a demand is not satisfied prob given an inventory position = z pmmo l j) s + j at time t ""‘m = :Ppobe I fit.) xo=s+j Over all initial inVentory positions, prob{a demand is not satisfied] = 2 lamb:a IEL) i=1 xu=s+j The expected shortage costs per unit time are CI = (cost per shortage) * (expected demand per unit time) * prob {a demand is not satisfied} = ENDié :Ppob‘o 13;) i=1 xn=s+1 Expected total relevant costs per unit time are EIFC(S) = fvr-i- C, 132 1 a; i=1 + Bny‘ i=1 ETRC(S +1) — ETRCU) = O a . = wfi i=1 => rippos(s+jta)=321) 1—] ) i Q xo=s+j 1 :+' I a;(+l)ppo(xo 13L)+ 0} E(S+j—xo)Ppo(xo Iii.) 19:0 :Ppooco IfI.) 1 + semi—{— ppo(s + j I m}: 0 1=1 :Ppucs'i'le—EL) F1 3.8 ESPRC = 0.2(0) + (O.8)(8)Gu[ ESPR C Q 0.02 = 6.4G“ [5 "840 ) /100 Gu["'340]=.3125 0—102): 5—40 8 194 J From Table GR(0.19) = .3125 "40 =0.19 s = 41.52 say 42 units b. SS = s — it, = 42 — 0.8(40) = 10 units 8.15 8.16 a. The total cost per year of the current item with standard deviation of lead time demand = 100 is $871.78,_ from Problem 8.3. By reducing O'L to 70, there is no change to the EOQ or to k. However, the reorder point 5 = 500 + 2.1300) = 649, and the total cost using the formula given in the solution to Problem 8.3 reduces to $685 .24, a saving of $186.54. This saving would continue until the underlying parameters change again. Other factors include ease of forecasting, higher service level, the possibility of using the new system for other items. So the minimum the company would be willing to pay is $186, but they would probably pay significantly more because of the other factors. ZAD vr EOQ = = 120 units. Air Alberta is using a B, shortage cost, so the safety factor is determined as in the appendix to Chapter '7. Check that A = 5.32 >1 w/ZJIQVO'LI' Thus set k = «Din 5.32 = 1.83 Hence 3 = J’EL +k0'L = 150+ 1.83(60) = 260 units. 13. The expected total relevant costs per year are Emc=%+{9+k0L }r+%mz(kl 2 Q (1) 193 c. He is right—there is a small drop in ETRC. The EOQ minimizes the sum of replenishing and carrying costs. Increasing Q above the EOQ raises the total of 8.17 8.18 = 240+(60+110)4+ = $1027.527yr. these costs but also reduces Kl pflUc), and this reduction outweighs the increase (at first). Check: Q = 1.05 EOQ = 126 units DB 1920 * 200 k = 1.80 (proceeding-as in part a.) (0.0336) and ETRC = $1021.98/yr., a small reduction. a. Equation 8.3 shows that vn’D = 20 and D = 8, so use an order quantity of 1. Eq. 8.4’s right hand side is 0.0521. The left hand side is 0.2353 when s = 0 and is 0.0349 when r = 1. Therefore the optimal s = 1. b. Chapter '1' methods yield an EOQ of 1 (rounded up). Qi-i’DB2 = 0.0521, as in part a. So I: = 1.62 and s = 8(2 weeks)7'52 weeks per year + 1.62(0.55) = 1. The result is the same in this case. Beginning with the EOQ = 165, we iterate using the two equations in the problem. Note that there is a slight error in the equation that should be pointed out to students: The numerator of the equation for Q should read 2(AD + DBzvolijD. This solution uses the correct formula. Iteration 1 2 3 G (k) Q 165.380 0.0971 172.901 0.1027 173.327 0.103 173.352 0.103 173.353 0.103 173.353 199 k 0.9186 0.88119 0.8861 0.8860 0.8860 0.8860 0.1792 0.1873 0.1878 0.1873 0.1878 0.1878 Therefore, Q = 173 and s = 32 + 0.886042) = 45. 8.19 a. With order quantity Q = 1, the inventory position is always equal to the value 5 + 1. If p;i (ho) is the probability that a backorder of size be is outstanding at a random point in time, then px(s+1+bo),bo >0 = 3+1 p3(b0) sz(xo)rbo=0 15:0 where p, (x0) is the probability that the iead time demand takes on the value x0. Hence the expected backorder of a random point in time is E(bo)= Zbap3(ba) bad) i (x -— (s +1))px(xo) xo=s+l 'b. Recall that the average stock on hand equals Inventory position - Emu-order) + E(bo) =(s+1)+JT:L+E(bD) since Q = 1. Thus the expected total relevant costs as a function of s are 5mm) = [(s + 1) — 52L + 5(3)o )]vr + B3vE(bD) =(s+1)vr— am 3941+ r) 2(x—(s+1))px(xo) x°=s+l One is indifferent between re-order levels s and s+1 when 200 01' ETRC(s) _= ETRC(S +1) (5 +1)vr— iLvr + Bzv(1+ r) :(x -— (s +1))px(xo) x°=s+l = (s + 2)vr -- iLvr + Bzv(1+ r) i (x — (S + 2)),0; (x0) xn=c+2 which implies that ‘Bzv(1+r) i px(x0)=vr—232v(l+r) 2px(x0) xn=5+1 Xo=$+2 " vr => x x = —— 10:2,}: ( a) Bzv(1+ r) => P» (5 +1) r = Bz(l+r) ...
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 .

Page1 / 9

Ch 8 Select HW Solutions - 8.2 All oustanding orders at a...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online