Exam 2 Solution - Production Planning Invent Control | ISE 4204, Exams of Systems Engineering

Download Exam 2 Solution - Production Planning Invent Control | ISE 4204 and more Exams Systems Engineering in PDF only on Docsity! _ ae ISE 4204 Exam IT Name: | (5 pts.) ‘L What is the meaning of independent demand and dependent demand in inventory (7) systems? Explain each term and give an example. Inde vides de avd cnvertan: + slew < ‘ demand cs exe neil to ‘aod crdependent: oe eycleu. Cemand Loe end thems Ce oC ndependert — bres delermened using forecaeting method s. Dependent deurand cavenke rs eysleass | demand ¢s dependent. upen other deus for which Lume-phased requctenedts are kuoun excl. Demand foe lower- level ¢ s ¢S - dependent — found throu h MRP _ exploston cabulus 2. Every fall, Myer’s Department Store purchases artificial Christmas trees to sell to its customers over the Christmas season. The trees are purchased from a national distributor for $25 and are sold for $49.99 each. Units not sold are marked-down in price 80% after Christmas: this usually results in all leftover units being sold. Demand for the trees over the season is as follows: No. of trees: 3000__ 3500, 4000 4500__—- 5000 Probability of occurrence: 0.12 0.18 0.35 0.21 = 0.14 How many trees should Myer’s order for the upcoming Christmas season? Cus underage cost = 49,99-29~- 2499 Co.= overage cof-= 15- C-,8)49.99 = 15.0 Cu a 2499 _ 25 Cattle mang = O6 Gnd @ zt FO cs mon. vale 2 O.6Z5 Q F@ BOCO Onn 2300 0.30 £000 065 &~* , 4500 0.86 = Ge 4000 Erees ee S080 1.00 2 ———— ISE 4204 Exam II Name: ’ / An assembly ceil is faced with six jobs ready for immediate processing. Processing times and due dates for the jobs are as follaws: Job Processing time Due date A 9 ul B 10 24 ¢ iL 14 D 3 5 EB 8 31 E 7 18 (5) (a) Determine the order sequence which minimizes maximum job lateness. What is the maximum lateness? EDD Segueuce minlniczes hwnax Sequence here. 7 CAGABE 4 ewan” gt dD A Cc a B Ee di 5S i 4 18 24 «3B Pi SB 3 Th z° 0 8 GZ in 23 30 40 48 Les BL 9 rR 1B t TT: 3 mcathS gy Ke 8 Tt? at hs lyr pat” gs ce Faso nents 2 >. Le: b279252* 0.678 X= Gvye3sy 21929 p> #5 /peck My + T Mourn” 258 OD * UF Meus * 43.3 a a PE [eee . afi7 VeI2 0.678 p. 4- F(R) Qh. GIO » pr sd axed " O, O27 Frou Table AOS? S* 2.02 Ro > Wy + S* 25+(2.02X434) * 342,47 = 35 Q.: L(z* 202) * 0, 008 nle)*% L¢2) ~ 43. 3>¢0.008) 2 O, 3464 0 . 2h LK en t® : a) Lier) L304 Aso 4 \ hk O67 ey = sa7. a= 507 R. 1- FR) Qh. @DG - ) e2 05) (1020) ~ 0.022 Tible At = 2 29| Feera R, = Ud + Oe > 25st 63,292.21) * 342037 3+3 Q. L (e+ 2ol) : 2.0083 n(R>* % aL - . ve 336 227 = 0.3594 peewee 27 = SCR 00G = 508 0.475 Oy b , Gowles® * soty(0.0T) 0.0224 ye FORD RI ) ort Tp dom) Fron Table At = z:20) 2(253)+ G5 (2.0) 342,03 = 34> RL My + OF S Ig-@l =| J cae step p,-R! =! | (OR) * (508, 342) Q. 31. a492 I P 4. in ts + > a AY me, c - ( ) be’ weed orde pleceres $ 2u 103 | oder Q eh tor ctely every é men Sefety Steck level 7 $= R - uo Lar nl)): ascagtt dyed eo Sy ths ISE 4204 Exam 0 Name: 6. A-small, family-owned manufacturing outfit produces small pull carts, Each cart is made up of a box assembly, a handle assembly, and two wheel assemblies, Each wheel assembly in turn consists of a shaft and two wheels. One week is required for assembling a box assembly, handle assembly, and two wheel assemblies into a finished cart. The lead time is one week each for box assemblies and handle assemblies, and two weeks for wheel assemblies. It takes one week to machine the shafts required for the wheel assemblies. Wheels are ordered from a local vendor: the order lead time is three weeks. Predicted demand for the carts over a six-week period is as follows: F Week| 5 6 7 | 8 9_| 10 | Demand] 45 | 60 so_| 75 | 60 | 40 and the on-hand inventory and scheduled receipts for each item are as follows: [ tem On-Hand Inventory Scheduled Receipts Cart 18 0 Wheel Assembly 40 150 in period 6 ~ Box Assembly 45 20 in period 6 Handle Assembly 0 Q Wheel 35 50 in period 7 Shaft 65 50 in period 3 (5) (a) Draw the product structure diagram for the cart. Include lead times in your diagram. Cart LT= 4 week an I | Wheel Assy @ Box Asch 4 Havdle Ass'y @) LT= 2 weeks . LT=4 week uT= 1 week Whee (2 Shatk UT= 3 weeks T= 4 weed ISE 4204 Exam II Name: (10) (b) Determine the planned order releases for carts, whee! assemblies, and shafts. Assume carts and wheel assemblies are produced using lot-for-lot, and shafts are produced in lots of size 50. Cart +,2[a3ltats{]el7]alol to Gross Req'ts L 45144 | 2017s eo! 40 Scheduled Receipts | On-Hand Inventory hs [i Slis 1S |1S}0 Net Rea'ts L 46 160 |Go0 175 | £0146 Time-Phased Net fleq'ts a (29 léo lea] 75 lé0.]40 Planned Order Releases 30160 |$0 1715 145 140 Planned Order Deliveries 30. |40 1 £0 Tle 40 Proj. Ending Inventov [1S [is [is [is fis fo fa lo fo la lo Wheel Assembly _ 1 2 3 4 5 | 6 {--7 |. 8 | .9 J] 19 Gross Req'ts 60 [120] 1e6| isa 20] 30 Scheduled Receipts 1d { Qn-Hand Inventory _ 140 40} 40 [40 | O| a o-. | 50 lo oO Net Req'ts 26/1201 © [toolizo [80 Time-Phased Net Req'ts 20 [20] 0 | Ico |1% | oO Planned Order Releases 201120) 0 | 100/122 | 30 Planned Order Deliveries 20 120 | 10 | 122 | 8O Proj. Ending Inventory! 40] 40 [a0 [aol oo | olo lo Shaft rfel[silatstef[r7{[aiol io Gross Req'ts 201120 [9 |:100 | 125 | 8 Scheduled Receipts $0 | On-Hand Inventory I 65/65 145 10 ° gloids Net Req’ts © 125/19 {leo | ro | Time-Phased Net Req'ts © 125/90 jfeo lize | 8O Planned Order Releases Oo (80/9 Noa |ico 1H Planned Order Deliveries © 150} 0 [100] 100} 100 Proj, Ending Inventory | 6S [65 (48/26 |2¢[25| 8 lage  ISE 4204 Exam 0 Name: (10) (c) Consider a different production scenario, where the time-phased net requirements for the box assembly are calculated to be as follows: © Week 5 6 7 8 9 10 Time-Phased Net Reg'ts 50 70 55 10 15 45 The holding cost per box assembly per period is $0.60, and it costs $75 to setup for producing the assemblies. Determine the planned order release schedule for the box assemblies using the Least Unit Cost method. uc . . —_— z % tq S10 SO 7d So Oo tS 435 b=S: C= Kirg =75]40 =1.S CA = CerheaylGrg +re) = 154.60) (04d 0.975 C12) = C4S+.6048) +.400688)](S0470 459) =105 C@)> cz) = {ortss% = 120 b=}. c= TW38/95 = 1.36 CLD = (1S+ 2G6)/ (954 1) = 1.246 CB) = 1S 40041208) estou) = 1.237 CA) = C15 £6} +1.209)+1,849) f(sstons+48)— [44 CA) >0(3) == 42 = a 4lg+Q@— BO Ea. yor 4S <0 ue hwe PoRs y = Q20,0, 80,0,0,49) eal 10