Problem Sheet (1)

PROBLEM: INVENTORY MANAGEMENT (Chapter 12)Problem-01: Fred Jacobs, assistant manager of Home and Hearth, is reviewing the cost associated with the stores best-selling frying pan. The data available to Mr. Jacobs concerning this frying pan follow. Demand = 20units/weekOrder cost = $75/ orderHolding cost = $15/ unit/yr.Home and Hearth operates 50 weeks per year.a) The current order quantity is 125 units per order. What is the annual ordering Cost, holding cost and total cost?b) Jacobs has recently learned of the economic order quantity in his Operations management class. If Home and Hearth implemented the EOQ, what are the changes Jacob found in ordering cost, holding cost and total cost?

Problem-02: (Solved problem 3 page 436 krajewski)A regional warehouse purchases hand tools from various suppliers and than distribute them on demand to retailers in the region. The warehouse operates 5 days per week, 52 weeks per year. The following data are available for a particular item.Average daily demand = 100 unitsStandard deviation of daily demand = 30 unitsLead time = 3 daysHolding cost = $ 9. 4/unit/yearOrdering cost = $ 35/orderCycle-service level = 92 % (Z = 1.4 )Current on-hand inventory is = 380 units with no open orders or backordersa) What is the economic order quantity (EOQ)? What would be the average time between orders?b) What is the desired safety stock level if the company has a policy of maintaining a 92 percent cycle-service level?c) What is the reorder point if the company decides to have a 92 percent Cycle-service level?d) Current on-hand inventory is = 380 units with no open orders or backorders and an inventory withdrawal of 10 units was just made. Is it time to reorder?e) If on-hand inventory is 40 units, there is an open order for 440 units and there are no back orders, should a new order be placed?

Problem-03: (Solved problem 4 page 437 krajewski)

Suppose that a periodic review system is used at the warehouse, but otherwise the data are the same as in problem 3a) What is the value of period P on the basis of EOQ.b) What is the desired safety stock in P system?c) What is the value of the target inventory level?d) Its time to review the item. On-hand inventory is 40 units. There is a schedule receipt of 440 units, and there is no backorders. How much should be reordered?

Problem-04: (Exercise problem 7 page 441 krajewski)

Sams Cat Hotel operates 52 weeks per year, 6 days per week and uses a continuous review system. It purchases kitty litter for $ 11.7 per bag. The following information is available about these bags.Demand = 90 bags/weekOrder cost = $54/ orderAnnual Holding cost = 27% of the cost.Desired Cycle-service level = 80 %Lead time = 3 weeksStandard deviation of weekly demand= 15 bagsCurrent on hand inventory is 320 bags, with no open orders or back orders.a) What is the economic order quantity (EOQ)? What would be the average time between orders?b) What is the desired safety stock level?c) What is the reorder point?d) An inventory withdrawal of 10 bags was just made. Is it time to reorder?e) The store currently uses a lot size of 500 bags. What is the annual holding cost, ordering cost and total cost of this policy?f) What would be the annual cost saved by shifting from 500 bags lot size to EOQ?Problem-05: (Exercise problem 16 page 442 krajewski)

Suppose that Sams Cat Hotel in problem 4uses a P system instead of Q system, but otherwise the data are the same as in problem 4.a) What P and T should be used? Round to the nearest week.b) How much more safety stock is needed than with a Q system?c) It is time for periodic review. How much kitty litter should be re-ordered?Problem-06: (Exercise problem 17 page 442 krajewski)

Your firm uses a continuous review system and operates 52 weeks per year. One of the SKUs has the following characteristics.

Demand= 20,000 units/yearOrdering and set-up cost = $ 40 per orderHolding cost = $ 2 per unit per yearLead time =2 weeksCycle service level = 95%Demand is normally distributed with a Standard deviation of weekly demand of 100 units.Current on-hand inventory is 1040 units, with no scheduled receipt and no backorders.

a) Calculate the items EOQ. What is the average time between orders in weeks?b) Find the safety stock and re-order point that provides a 95% cycle service level.c) For these policies, what are the annual cost of (i) holding the cycle inventory and (ii) placing orders.d) A withdrawal of 15 units just occurred, is it time to re-order?

Problem-07: (Exercise problem 18page 442 krajewski)

Suppose that your firm uses a periodic review system but otherwise the data are the same as in problem 8.a) Calculate the P that gives approximately the same number of orders per year as the EOQ. Round your answer to the nearest weekb) Find the safety stock and the target inventory level that provide a 95% cycle service level.c) How much larger is the safety stock than with a Q system?


5. A Golf specialty wholesaler operates 50 weeks per year. Management is trying to determine an inventory policy for its 1-irons which have the following characteristics:

Demand = 2000 units/year Ordering and set-up cost = $ 40 per orderHolding cost = $ 5 per unit per yearLead time =4 weeksStandard deviation of weekly demand = 3 unitsCycle service level = 90% (Z = 1.4)a) If the company uses a periodic review system, what should P and T be? Round P to the nearest weeks.b) If the company uses a continuous review system, what should R be?


Sams Cat Hotela. Economic order quantity

= 90/weekD = (90 bags/week)(52 weeks/yr) = 4,680S = $54Price = $11.70H = (27%)($11.70) = $3.16

= 399.93, or 400 bags.Time between orders, in weeks

b. Reorder point, RR = demand during protection interval + safety stock

Demand during protection interval = L = 90 * 3 = 270 bagsSafety stock = zdLT When the desired cycle-service level is 80%, .

= 15 = 25.98 or 26 Safety stock = 0.84 * 26 = 21.82, or 22 bags

c. Initial inventory position = OH + SR BO = 320 + 0 0320 10 = 310.Because inventory position remains above 292, it is not yet time to place an order.

d. Annual holding cost Annual ordering cost

When the EOQ is used these two costs are equal. When , the annual holding cost is larger than the ordering cost, therefore Q is too large. Total costs are $789.75 + $505.44 = $1,295.19.e. Annual holding cost Annual ordering cost

Total cost using EOQ is $1,263.60, which is $31.59 less than when the order quantity is 500 bags.

Problem-05: (Exercise problem 16 page 442 krajewski)Sams Cat Hotel with a P systema. Referring to Problem 7, the EOQ is 400 bags. When the demand rate is 15 per day, the average time between orders is (400/15) = 26.67 or about 27 days. The lead time is 3 weeks 6 days per week = 18 days. If the review period is set equal to the EOQs average time between orders (27 days), then the protection interval (P + L) = (27 + 18) = 45 days.For an 80% cycle-service levelz = 0.84

= 41.08

Safety stock = = 0.84(41.08) = 34.51 or 35 bagsT = Average demand during the protection interval + Safety stockT = (15*45) + 35 = 710b. In Problem 7, the Q system required a safety stock of 22 bags to achieve an 80% cycle-service level. Therefore, the P system requires a safety stock that is larger by (35 22) = 13 bags.c. From Problem 7, inventory position, IP = 320. The amount to reorder is T IP = 710 320 = 390.Problem-06: (Exercise problem 17 page 442 krajewski)Continuous review system.a. Economic order quantity. or 894 unitsTime between orders (TBO) = Q/D = 894/20,000 = 0.0447 years = 2.32 weeksb. Weekly demand = 20,000/52 = 385 unitsFor a 95% cycle-service level, z = 1.65

Safety stock: = (1.65)(100) = 233.34, or 233 unitsNow solve for R, as

R = L + Safety stock = 385(2) + 233 = 1,003 units c. cost i.Annual holding cost of cycle inventory

ii. Annual ordering cost

d. With the 15-unit withdrawal, IP drops from 1,040 to 1,025 units. Because this level is above the reorder point (1,025 > 1,003), a new order is not placed.

Problem-07: (Exercise problem 18page 442 krajewski)

Periodic review systeme. From Problem 17, or 894 unitsNumber of orders per year = = 20,000/894 = 22.4 orders per year. weeksP is rounded to 2 weeks.f. For a 95% cycle-service level, z = 1.65. ThereforeSafety stock

= 200 unitsSafety stock = 1.65(200) = 330 units,T = Average demand during the protection interval + Safety stockT = (385 * 4) + 330 = 1,870 unitsg. In Problem 17, with a Q system the safety stock is 233 units. Therefore, (330 233) = 97 more units of safety stock are needed.

2. Problem-08: (Exercise problem 22 page 443 krajewski)Golf specialty wholesalera. Periodic Review System or 179 1-irons or 4.0 weeksWhen cycle-service level is 90%, z = 1.28.Weekly demand is (2,000 units/yr)/(50 wk/yr) = 40 units/wkL = 4 weeksSafety stock:

z = (1.28) = 10.86, or 11 irons

T = (P+L) + Safety stock = 40(4+4) + 11 = 331 irons.b. Continuous review system

Safety stock = = (1.28)(3) = 1.28(3)(2) = 7.68, or 8 irons

R = L + Safety stock = 40(4) + 8 =168 iron

PROBLEM: LEAN MANUFACTURING SYSTEM (Chapter 8)

The number of Kanban card is

d = Expected daily demandw=Average waiting time during processingP = Average processing time per containerc = Quantity in a std. container = A policy variable that reflects the efficiency of the workstations

Problem-01: (Example 8.1 page 309 krajjewski)The westerville Auto parts company produces rocker-arm assemblies for use in the steering and suspension system of four wheel drive trucks. A typical container of parts spends 0.02 day in processing and 0.08 day in material handling and waiting during its manufacturing cycle. The daily demand for the part is 2000 units. Management believes that demand for the rocker-arm assembly is uncertain enough to warrant a safety stock equivalent of 10 percent of its authorized inventory.(a) If there are 22 parts in each container, how many Kanban cards sets (production order and withdrawal cards) should be authorized?(b) Suppose that a proposal to revise the plant layout would cut materials handling and waiting time per container to 0.06 day. Hoe many card sets would be neededProblem-02: (Solved problem page 316 krajjewski)A company using a kanban system has an inefficient machine group.. For example, the daily demand for part L105A is 3000 units. The average waiting time for a container of parts is 0.8 day. The processing time for a container of L105A is 0.2 day and a container holds 270 units. Currently 20 containers are used for this item.a) What is the value of the policy variable, ?b) What is the total planned inventory?c) Suppose that the policy variable, was 0. How many containers would be needed now? What is the effect of the policy variable in this example?

Problem-03: (Exercise problem 2 page 317 krajjewski)A fabrication cell at Spradleys Sprockets uses the pull method to supply gears to an assembly lines. George Jitson is in charge of the assembly line which requires 500 gears per day. Containers typically waits 0.2o day in the fabrication cell. Each container holds 20 gears and one container requires 1.8 days in machine time. Setup times are negligible. If the policy variable for unforeseen contingencies is set at 5 percent, how many containers should Jitson authorize for the gear replenishment system?Problem-04: (Exercise problem 3 page 317 krajjewski)You are asked to analyse the kanban system of LeWin, a French manufacturer of gaming devices. One of the workstations feeding the assembly line produces part M670N. The daily demand for M670N is 1,800 units. The average processing time per unit is 0.003 day. LeWins record show that the average container spends 1.03 days waiting at the feeder workstation. The container for M670N can hold 300 units. Twelve containers are authorized for the part.a) Find the policy variable, , that expresses the amount of implied safety stock in this systemb) Use the implied value of from part (a) to determine the required reduction in waiting time if one container was removed. Assume that all other parameters remain constant.Problem-05: (Exercise problem 4 page 317 krajjewski)

An assembly line requires two components: gadjits and witjits. Gadjits are produced by center 1 and widjits by center 2. Each unit of end item called a jit-together, requires 3 gadjits and 2 widjits. The daily production quota on the assembly line is 800 jit-togethers.The container of gadjit s holds 80 units. The policy variable for center is set at 0.09. The average waiting time for a container of gadjits is 0.09 day and 0.06 day is needed to produce a container. The container for widjits holds 50 units and the policy variable for center 2 is 0.08. The average waiting time for container of widjits is 0.14 day, and the time required to process a container is 0.20 day.a) How many containers are needed for gadjits?b) How many containers are needed for widjits?Problem-06: A toy company uses a Kanban system to make a metal fastener of several toys. The waiting time for a container of the fasteners during production is 0.25 day; average processing time is 0.15 day per container. Each container holds 200 fasteners. The daily demand of the fastener is 2000 units. Using an efficiency factor of 5%, calculate the number of Kanban card sets needed for the fasteners.

Problem: 03Spradleys Sprockets

k =

k = k = 52.5Problem: 04LeWina. Solving for implied policy variable,

k =

b. Reduction in waiting time

The reduction in waiting time is:

Problem: 05Gadjits and Widjitsa. Containers for gadjits

k =

k = = 4.905k = 5b. Containers for widjits

k =

k = = 11.750k = 12

PROBLEM: SUPPLY CHAIN MANAGEMENT / DESIGN (Chapter 9)Problem- 01: (Excercise problem 5 page 346 krajewski): Jack Jones, the materials manager at Precision Enterprises is beginning to look for ways to reduce inventories. A recent accounting statement shows the following inventory investment by category: raw materials $3,129,500; work-in-process $6,237,000; and finished goods $2,686,500. This years cost of goods sold will be about $32.5 million. Assuming 52 business weeks per year, express total inventory as:a) Weeks of supplyb) Inventory turnsProblem- 02: (Advanced problem 9 page 347 krajewski): 3. Sterling Inc. operates 52 weeks per year and its cost of goods sold last year was $ 6,500,000. The firm carries eight items in inventory: four raw materials, two work-in-process items and two finished goods. Following table shows last years average inventory levels for these items along with their unit values.a. What is the average aggregate inventory value?b. How many weeks of supply does the firm have?c. What was the inventory turnover last year?

CategoryPart NumberAverage Inventory (units)Value per unitRaw materialsRM-120,000$ 1RM-2 5,000$ 5RM-3 3,000$ 6RM-4 1,000$ 8Work-in-processWIP-1 6,000$10WIP-2 8,000$12Finished GoodsFG-1 1,000$65FG-2 500$88

Problem- 03: (Solved problem 2 page 345 krajewski): A firms cost of goods sold last year was $3,410,000 and the firm operates 52 weeks per year. It carries seven items in inventory: three raw materials, two work in process and two finished goods. The following table contains last years average inventory level for each item, along with its value.a) What is the average aggregate inventory value?b) How many weeks of supply does the firm maintain?c) What was the inventory turnover last year?CategoryPart NumberAverage Inventory (units)Value per unitRaw materialsRM-115,000$ 3RM-2 2,500$ 5RM-3 3,000$ 1Work-in-processWIP-1 5,000$14WIP-2 4,000$18

Finished GoodsFG-1 2,000$48FG-2 1,000$62

Problem- 01: (Exercise problem 5 page 346 krajewski):

Precision Enterprises. Average aggregate inventory value= Raw materials + WIP + Finished goods= $3,129,500 + $6,237,000 + $2,686,500= $12,053,000

a. Sales per week= Cost of goods sold/52 weeks per year= $32,500,000/52= $625,000

Weeks of supply= Average aggregate inventory value/Weekly sales= $12,053,000/$625,000= 19.28 wk

b. Inventory turnover= (Annual sales at cost)/(Average aggregate inventory value)= $32,500,000/$12,053,000= 2.6964 turns/year

Problem- 02: (Advanced problem 9 page 347 krajewski): Sterling Inc.a.Average

Part NumberInventory (units)Value ($/unit)Total Value ($)

RM-120,0001.0020,000

RM-25,0005.0025,000

RM-33,0006.0018,000

RM-41,0008.008,000

WIP-16,00010.0060,000

WIP-28,00012.0096,000

FG-11,00065.0065,000

FG-250088.0044,000

Average aggregate inventory value: $336,000

b. Average weekly sales at cost= $6,500,000/52= $125,000Weeks of supply = $336,000/$125,000= 2.688 weeks.

c. Inventory turnover = Annual sales (at cost) /Average aggregate inventory value= $6,500,000/$336,000= 19.34 turns.

PROBLEM: RESOURCE PLANNING (CHAPTER 15)Problem-01: Exercise Problem 1 page 568 KrajewskiProblem-02: Exercise Problem 2 page 568 KrajewskiProblem-03: Exercise Problem 3 page 568 KrajewskiProblem-04: Exercise Problem 16 page 572 KrajewskiProblem-05: Exercise Problem 17 page 572 KrajewskiProblem-06: Exercise Problem 20 page 574 Krajewski

Problem-01: Exercise Problem 1 page 568 KrajewskiBill of materials, Fig. 15.24a.Item I has only one parent (E). However, item E has two parents (B and C). b.Item A has 10 unique components (B, C, D, E, F, G, H, I, J, and K). c.Item A has five purchased items (I, F, G, H, and K). These are the items without components. d.Item A has five intermediate items (B, C, D, E, and J). These items have both parents and components. e.The longest path is IECA at 11 weeks.

Problem-02: Exercise Problem 2 page 568 Krajewski2.Item A. The bill of materials for item A is shown following.

Problem-03: Exercise Problem 3 page 568 Krajewski3.Lead time is determined by the longest path, CBA, at 13 weeks.

Problem-04: Exercise Problem 16 page 572 KrajewskiCompleted inventory recordsa.FOQ of 50 unitsItem:Drive shaftLot Size:50 units

Lead Time: 3 weeks

Week12345678

Gross requirements3525152040405050

Scheduled receipts80

Projected on hand1055301545 5151515

Planned receipts50505050

Planned order releases50505050

b.L4LItem:Drive shaftLot Size:L4L

Lead Time: 3 weeks

Week12345678



Projected on hand10553015 0 0 0 0 0

Planned receipts 540405050

Planned order releases 540405050

c.POQ with P = 4Item:Drive shaftLot Size:POQ, P = 4

Lead Time: 3 weeks

Week12345678

Gross requirements 352515 2040405050

Scheduled receipts 80

Projected on hand10 5530151309050 0 0



Problem-05: Exercise Problem 17page 572 Krajewski17.Rear wheel assembly, Fig. 15.33a.Fixed order quantity = 300Item:MQ09Lot Size:FOQ: 300

Description:Rear wheel assemblyLead Time: 1 week

Week12345678

Gross requirements205130 85 70 60 95


Projected on hand10019519565280280 210150 55

Planned receipts300


b.Lot-for-lotItem:MQ09Lot Size:L4L


Week12345678



Projected on hand100195195 65 0 0 0 0 0


Planned order releases 20 706095

c. Period order quantity, P = 4

Item:MQ09Lot Size:POQ, P = 4


Week12345678

Gross requirements205130 85 706095


Projected on hand100195195 65130130600 0



Problem-06: Exercise Problem 20 page 574 KrajewskiMRP for Fig. 15.36

Data CategoryItem

CDEF

Lot-size ruleFOQ = 220L4LFOQ = 300POQ (P = 2)

Lead time3 weeks2 weeks3 weeks2 weeks

Scheduled receipts280 (week 1)None300 (week 3)None

On-hand inventory250150600

Item:CLot Size:FOQ = 220

Description:Lead Time: 3 weeks

Week12345678



Projected on hand25305305135135135155155155

Planned receipts220


Item:DLot Size:L4L


Week12345678

Gross requirements 85180100

Scheduled receipts

Projected on hand0 00 0 0 0 000

Planned receipts 85180100


Item:ELot Size:FOQ = 300


Week12345678



Projected on hand15065 65185 85 25252525

Planned receipts300


Item:FLot Size:POQ = 2


Week12345678


Scheduled receipts

Projected on hand600430130200 00000

Planned receipts430


Two notices are identified: order 85 units of D and 430 units of F. Please note that action notices were not asked for in the problem.

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Problem Sheet (1)