運(yùn)營管理課后習(xí)題答案-文庫吧資料
2025-06-25 21:06本頁面
【正文】 2495TC1,000 =1,000($) +4,900($50) + $(4,900) = $25,49021,000TC4,000 =4,000($) +4,900($50) + $(4,900) = $27,99124,000TC6,000 =6,000($) +4,900($50) + $(4,900) = $29,62626,000 A = 4StaplerTop AssemblyBase AssemblyCoverSpringSlide AssemblyBaseStrike PadRubber Pad 2SlideSpringb. 4.Master ScheduleDayBeg. Inv.1234567Quantity100150200Table Beg. Inv.1234567Gross requirements100150200Scheduled receiptsProjected on handNet requirements100150200Plannedorder receipts100150200Plannedorder releases100150200Wood SectionsBeg. Inv.1234567Gross requirements200300400Scheduled receipts100Projected on hand 100100Net requirements100300400Plannedorder receipts100300400Plannedorder releases400400Braces Beg. Inv.1234567Gross requirements300450600Scheduled receiptsProjected on hand 60606060Net requirements240450600Plannedorder receipts240450600Plannedorder releases240450600Legs Beg. Inv.1234567Gross requirements400600800Scheduled receiptsProjected on hand 120120120120888871Net requirements280600800Plannedorder receipts308660880Plannedorder releases96888010.Week1234Material40806070Week1234Labor hr.160320240280Mach. hr.120240180210a. Capacity utilizationWeek1234Labor%%80%%Machine60%120%90%105%b. Capacity utilization exceeds 100% for both labor and machine in week 2, and for machine alone in week 4.Production could be shifted to earlier or later weeks in which capacity is underutilized. Shifting to an earlier week would result in added carrying costs。 D = 10。 H = 1。 3UCL = , LCL = 4Range Chart: UCL = D4= () = 5LCL = D3= 0() = 06[Both charts suggest the process is in control: Neither has any points outside the limits.]6. n = 200 Control Limits = Thus, UCL is .0234 and LCL bees 0.Since n = 200, the fraction represented by each data point is half the amount shown. ., 1 defective = .005, 2 defectives = .01, etc.Sample 10 is too large. 7. Control limits: UCL is , LCL bees 0. All values are within the limits.14. Let USL = Upper Specification Limit, LSL = Lower Specification Limit, = Process mean, s = Process standard deviationFor process H: For process K:Assuming the minimum acceptable is , since , the process is not capable.For process T:Since = , the process is capable.Chapter 10 Aggregate Planning and Master Scheduling7. a. No backlogs are allowedPeriodMar.Apr.MayJun.JulyAug.Sep.TotalForecast50445560504051350OutputRegular40404040404040280Overtime888883851Subcontract2031220019Output Forecast04–4003–3InventoryBeginning0040003Ending0400030Average022007Backlog00000000Costs:Regular 3,2003,2003,2003,2003,2003,2003,20022,400Overtime 9609609609609603609606,120Subcontract 28004201,680280002,660Inventory 02020
環(huán)評(píng)公示相關(guān)推薦
鄂ICP備17016276號(hào)-1