MGT 3110: Exam 3 Study Guide

DO NOT RELY ONLY ON THIS STUDY GUIDE ALONETO PREAPRE FOR THE EXAM

Discussion questions1. Define independent and dependent demand items. 2. What is Master Production Schedule? 3. What is Bill of Materials? 4. What is Low-Level coding and what how is it used?5. What are the benefits of MRP? 6. What are the inputs required for MRP? 7. What is “Lot Sizing” in MRP? 8. What are the reasons for using a lot sizing method other than Lot-for-lot?9. What does the part-period balancing lot-sizing technique attempt to do in deciding the lot

sizes?10. Explain the terms flowtime and lateness.11. What are the advantages and disadvantages of shortest processing time (SPT) rule?12. What is the Critical Ratio? To what jobs does Critical Ratio give priority to?13. What can be said about the jobs if CR < 1, or =1, or > 1?14. What is input-output control?15. Define maintenance.16. Define reliability.17. What is FR(%)? Define it.18. Define FR(N). How is computed?19. Define MTBF. How is related to FR(N)?20. What is redundancy?21. What is the impact on system reliability of adding parts or components in parallel?22. Increasing the number of parts or components in a product tends to reduce its reliability. Why

is this true only when adding components in series?23. Explain carefully how redundancy improves product reliability.24. What is breakdown maintenance?25. Is there an optimal amount of preventive maintenance? What caution should be exercised

before calculating this optimal amount?

Problems1. Consider the following Solver model for an aggregate planning problem.

a. Determine the excel formula for the following cells:

B18B19B20B23E23F23B24B31C31F31G31H31B32B37B38B39B40B41B42B43

b. What is the Solver Target cell?c. What are the Solver changing cells?d. What are the Solver constraints?e. What options of Solver must be checked?

2. A Bill of Materials is desired for a bracket (A) that is made up of a base (B), two springs (C) and four clamps (D). The base is assembled from one clamp (D) and two housings (E). Each clamp has one handle (F) and one casting (G). Each housing has two bearings (H) and one shaft (I). a. Develop a product structure tree.b. The lead time for the parts are given below. Develop a time-phased product structure.c. The available inventory for each part is given in the table below. Determine the net

requirement quantities of all parts required to assemble 50 units of bracket A.

Item Lead timeAvailabl

eA 1 5B 2 5C 3 10D 2 20E 1 50F 2 150

G 1 50H 1 5I 2 0

3. A product (A) consists of a base (B) and a casting (C). The base consists of a plate (P) and three fasteners (F). The lead time, current on-hand inventory and scheduled receipts are given below. All components are lot for lot. The MPS requires start of production of 100 units of product A in week 4 and 150 in week 6. Produce the MRP for the upcoming six weeks. Produce a list of all planned order releases.

Part Lead time On-hand Scheduled receiptsB 1 100 50 in week 1C 3 30 20 in week 1, 30 in week 2P 2 0 50 in week 1F 4 0 30 in week 1, 40 in week 3

4. For the following item the inventory holding cost is $0.80 per week and the setup cost is $300. Determine the lot sizes and total cost for this item under (i) Lot-for-Lot, (ii) EOQ, and (iii) PPB methods

Item LT = 1Week: 1 2 3 4 5 6 7 8Gross requirement 100 250 200 150 250 200 200 150Scheduled receiptsProjected on-hand 100Net RequirementPlanned receiptsPlanned order releases

5. Consider the following planned and actual hours of input and output.

Week ending 1 2 3 4 5 6 7 8Planned input 500 800 700 600 600 800 700 800Actual input 700 700 700 800 600 500 500 800Planned output 650 650 650 650 650 700 700 700Actual output 600 700 800 700 650 500 600 800

Prepare the Input/Output Control chart for this workstation. Assume an initial actual backlog of 120 hours.

6. A company wishes to assign a set of jobs to a set of machines. The following table provides data on the profit margin of each job when performed on a specific machine. Setup an Excel Solver model to determine the set of assignments that maximizes production value.

a. Fill out the Excel formulas for the following cells:

G13

G14

G15

G16

C17

D17

E17

F17

G19

b. What is the Solver Target cell?c. What are the Solver changing cells?d. What are the Solver constraints?e. What options of Solver must be checked?

R = 98% R = 99% R = 98% R = 95%

R = 98% R = 92% R = 90%

R = 90%

R = 90%

R = 90% R = 94%

R = 90%

R = 90%

R = 94%

R = 94%

R = 95%

R = 95%

R = 95%

7. The following jobs are waiting to be processed on day 250

Job Date Job received Production days needed Date job dueA 215 30 290B 220 20 415C 225 40 375D 240 50 315E 245 20 420F 250 35 380

Sequence the jobs in the order of SPT, EDD, and Critical Ratio, and compute (i) Average flow time, (ii) Average lateness, (iii) Average no. of jobs in the system, and (iv) Utilization, for each of the three schedule of jobs.

8. A control rod mechanism is a nuclear reactor has 15 critical components with an average reliability of 99.5% for each component. Determine the overall reliability of the mechanism.

9. Determine the system reliability for the following systems.

(a)

(b)

10. Fifty components of a safety system were tested for reliability, each for 200 hours of operation. Of the 50, 2 failed after 50 hours of operation, 3 after 75 hours, 1 after 120 hours, and 1 after 150 hours. The rest of the components did not fail. Determine the following failure rate statistics.

a. Percentage of failures.b. Number of failures per unit-hour of operation.c. Mean time between failures.d. What is the expected number of failures over 1000 hours of operation?

11. In a mining operation an electoral exhaust system is used to pump out the fumes. The break down data over the past 200 weeks is shown in the table below. Each time the system breaks down the mining operation must be shut down. The estimated cost to repair the system and the lost production amounted to $12,000. Routine maintenance service for the exhaust system may be purchased at a cost of $6,000 per week. With this contract the number of breakdowns is expected to average only 1.2 per week. Then, the cost to bring the system back on line is $5,000 in lost production. Determine whether it is economical to purchase the preventive maintenance contract.

No. of breakdowns No. of weeks that many breakdowns occurred (Frequency)

0 75

1 35

2 50

3 20

4 10

5 10

Answers to discussion questions

1. Define independent and dependent demand items.Finished products whose demand is independent of production decisions are called “Independent demand” items. Items for which demand can be directly calculated from production decisions are called “Dependent demand” items. These are raw-materials and parts required for the production of the finished goods.

2. What is Master Production Schedule?Master Production Schedule specifies production quantities of each Independent Demand item for a planning horizon of 12 to 15 weeks. Total of MPS quantities must be in accordance with the aggregate production plan.

3. What is Bill of Materials?Bill of materials is structured list of components, ingredients, and materials needed to make an end product. Items needed to produce a given part are called components or “children”. The part into which the components go us called “Parent”. The BOM also gives the number of units of a child item needed to produce one unit of the parent item.

4. What is Low-Level coding and what how is it used?A level code starting from zero at the top of the BOM tree and incremented by 1 going down each level of the BOM tree is assigned. Then, the lowest level at which an item appears is called Low-Level code. The MRP computations are processed one level at a time, starting from level zero.

5. What are the benefits of MRP? Better response to customer orders Faster response to market changes Improved utilization of facilities and labor Reduced inventory levels

6. What are the inputs required for MRP? Master Production Schedule Bill of Materials Inventory status

7. What is “Lot Sizing” in MRP?The process of combining net requirements into production lots is called lot sizing.

8. What are the reasons for using a lot sizing method other than Lot-for-lot? Lot-for-lot often requires too many lots that may not be economically justifiable Sometime lot-for-lot generates absurdly small lots

9. What does the part-period balancing lot-sizing technique attempt to do in deciding the lot sizes?It balances the setup and holding costs. PPB uses additional information by changing the lot size to reflect requirements of the next lot size in the future.

10. Explain the terms flowtime and lateness.

Flow time is the length of time a job is in the system; lateness is completion time minus due date.

11. What are the advantages and disadvantages of shortest processing time (SPT) rule?SPT minimizes the average flow time, average lateness, and average number of jobs in the system. It maximizes the number of jobs completed at any point. The disadvantage is that long jobs are pushed back in the schedule.

12. What is the Critical Ratio? To what jobs does Critical Ratio give priority to?The Critical Ratio (CR) is an index number computed by diving the time until due date by the working time remaining. The CR gives priority to jobs that must be done to keep shipping on schedule.

13. What can be said about the jobs if CR < 1, or =1, or > 1?If CR < 1, then the job has fallen behind, the work remaining exceeds the time until due date. If CR = 1, then the job is on schedule, the work remaining exactly equals the time until due date. If CR > 1, then there is slack, the time until due date exceeds the work remaining.

14. What is input-output control?Input/output control keeps track of planned versus actual inputs and outputs, highlighting deviations and indicating bottlenecks using cumulative backlog.

15. Define maintenance.Maintenance consists of all activities involved in keeping a system's equipment in working order.

16. Define reliability.Reliability is the probability that a machine part or product will function properly for a specified time under stated conditions.

17. What is FR(%)? Define it.FR(%) refers to product failure rate. It measures the percent of failures among the total number of products tested.

18. Define FR(N). How is computed?FR(N) refers to the number of failures during a period of time. It measures the number of failures over the total unit-hours the products operated without failing. it is the ratio of failed units to total operating hours.

19. Define MTBF. How is related to FR(N)?MTBF stands for Mean time between failures. It is the expected time between a repair and the next failure of a component, machine, process, or product. MTBF is the reciprocal of FR(N)

20. What is redundancy?Redundancy is the use of a component in parallel to raise reliabilities.

21. What is the impact on system reliability of adding parts or components in parallel?

This will increase the reliability of the system by introducing redundancy.

22. Increasing the number of parts or components in a product tends to reduce its reliability. Why is this true only when adding components in series?Adding parts in series involves an additional multiplication by a value less than one, so that reliability must fall. Adding parts in parallel (the redundancy concept) increases reliability because only one part of the parallel system must function.

23. Explain carefully how redundancy improves product reliability.A redundant part or component is connected in parallel with the primary part or component. "In parallel" means that either the original part or its backup needs to work, not that both must work at the same time. Redundancy increases reliability by providing an additional path (through the redundant part) to provide system reliability.

24. What is breakdown maintenance?Breakdown maintenance is the remedial maintenance that occurs when equipment fails and must be repaired on an emergency or priority basis.

25. Is there an optimal amount of preventive maintenance? What caution should be exercised before calculating this optimal amount?Too little preventive maintenance causes breakdown costs to rise sharply, adding more to cost than is saved by less preventive maintenance; too much preventive maintenance reduces breakdowns, but by an amount insufficient to offset the added cost of preventive maintenance. Operations managers should assure that all costs of breakdowns have been properly included in the calculations. There is a history of not including indirect and subjective breakdown cost elements, which leads to performing too little preventive maintenance.

A

B C2

D4

F

E2

H2 I

D1

GF G

Answers to problems

1.

B18 =B14/B13b. B43

c. C23:D26, D31:E34

d. D31:D34 <= H31:H34G31:G34 >= E15C23:D26 = Integer (if needed)D31:E34 = Integer (if needed)

e. Assume linear modelAssume non-negative

B19 =B14*B5B20 =B13*B6B23 =E13E23 =B23+C23-D23F23 =E23*F6*$B$18B24 =E23B31 =E14C31 =F23F31 =E6G31 =B31+SUM(C31:E31)-F31H31 =C31*$B$15B32 =G31B37 =SUMPRODUCT(E23:E26,F6:F9)*B19B38 =D35*B20B39 =E35*B7B40 =C27*B9B41 =D27*B10B42 =G35*B8B43 =SUM(B37:B42)

2.

F

D

G

B

H

E

I

C A

F

D

G

1 2 3 4 5 6 7

Lead time = 7 weeks

Part Gross Available NetA 50 5 50 – 5 = 45B 1 x A = 45 5 45 – 5 = 40C 2 x A = 2 x 45 = 90 10 90 – 10 = 80D 4 x A + 1 x B = 4 x 45 + 40 = 220 20 220 – 20 = 200E 2 x B = 80 50 80 – 50 = 30F 1 x D = 200 150 200 – 150 = 50G 1 x D = 200 50 200 – 50 = 150H 2 x E = 2 x 30 = 60 5 60 – 5 = 55I 1 x E = 30 0 30 – 0 = 30

3. A has releases of 100 in 4, 150 in 7; B has a release of 150 in 6, but on hand inventory accounted for all other needs; C has releases of 150 in 4, 70 in 1; P has a release of 150 in 5 and F has a release of 450 in 5 (the beginning inventory of B leads to no other gross requirements of P or F).

1 2 3 4 5 6MPS start for A 100 150

Item B Lead time = 1Week: 1 2 3 4 5 6Gross requirement 0 0 0 100 0 150Scheduled receipts 50Projected on-hand 100 100 150 150 150 50 50Planned receipts 0 0 0 0 0 100Planned order releases 0 0 0 0 100 0

Item C Lead time = 3Week: 1 2 3 4 5 6Gross requirement 0 0 0 100 0 150Scheduled receipts 20 30Projected on-hand 30 30 50 80 80 0 0Planned receipts 0 0 0 20 0 150Planned order releases 20 0 150 0 0 0

Item P Lead time = 2Week: 1 2 3 4 5 6Gross requirement 0 0 0 0 100 0Scheduled receipts 50Projected on-hand 0 0 50 50 50 50 0Planned receipts 0 0 0 0 50 0Planned order releases 0 0 50 0 0 0

Item F Lead time = 4Week 1 2 3 4 5 6Gross requirement 0 0 0 0 300 0Scheduled receipts 30 40Projected on-hand 0 0 30 30 70 70 0Planned receipts 0 0 0 0 230 0Planned order releases 230 0 0 0 0 0

4.

(i) L-4-L

Item LT = 1Week: 1 2 3 4 5 6 7 8Gross requirement 100 250 200 150 250 200 200 150Scheduled receiptsProjected on-hand 100 100 0 0 0 0 0 0 0Net Requirement 0 250 200 150 250 200 200 150Planned receipts 0 250 200 150 250 200 200 150Planned order releases 250 200 150 250 200 200 150 0

No. of setup = 7Carrying cost = 0Setup cost = 7 x $300 = 2100Total cost = 2100

(ii) EOQ:Total demand for 8 weeks = 1500D (1 year) = (1500/8) x 52 weeks/year = 9750H (for 52 weeks) = $0.80/week x 52 weeks = $41.60S = 300

Q = √ 2 (9750 )30041.6

= 375

Item LT = 1Week: 1 2 3 4 5 6 7 8Gross requirement 100 250 200 150 250 200 200 150Scheduled receipts 0 0 0 0 0 0 0 0Projected on-hand 100 100 0 125 300 150 275 75 250Net Requirement 0 250 75 0 100 0 125 0Planned receipts 0 375 375 0 375 0 375 0Planned order releases 375 375 0 375 0 375 0 0

Annual setup cost = (D/Q) S = (9750/375) x 300 = 7800Annual holding cost = (Q/2)H per year = (375/2) x 41.60 = 7800Annual cost = Annual setup cost + Annual holding cost = 7800 + 7800 = 15600Weekly cost = Annual cost/52 = 15600/52 = 300Cost for 8 weeks = Cost per week x 8 weeks = 300 x 8 = 2400

(iii) PPB

EPP = 300/0.80 = 375

Item LT = 1

Week: 1 2 3 4 5 6 7 8Gross requirement 100 250 200 150 250 200 200 150Scheduled receipts 0 0 0 0 0 0 0 0Projected on-hand 100 100 0 350 150 0 200 0Net Requirement 0 250 250 200Planned receipts 0 600 450Planned order releases 600 450

Periods combined

Quantities combined

Periods brought forward – last qty

New Part-periods

Total combined part-periods

Lot #1 Receipt in week #2

2 250 0 0 02, 3 250 + 200 = 450 1 1x 200 = 200 0 + 200 = 200

2, 3, 4 450 + 150 = 550 2 2 x 150 = 300200 + 300 = 500

(Close to 375)Lot #2 Receipt in week #55 250 0 0 0

5, 6 250 + 200 = 450 1 1 x 200 = 2000 + 200 = 200(Close to 375)

5, 6, 7 450 + 200 = 650 2 2 x 200 = 400 200 + 400 = 600Lot #3 Receipt in week #77 200 0 0 07, 8 200 + 150 = 350 1 1x 150 = 150 0 + 150 = 150

Summary:

Lot # in Week Lot size Lot PP Setup cost = 3 setups x $300 = $900Total Lot PP = 850Total holding cost = 850 x $0.80 = $680

Total cost = 900 + 680 = $1,580

Lot #1 Receipt in week #2 550 500

Lot #2 Receipt in week #5 450 200

Lot #3 Receipt in week #7 350 150

Total cost = 700 + 460 + 420 = $1,580

#5.Week ending 1 2 3 4 5 6 7 8Planned input 500 800 700 600 600 800 700 800Actual input 700 700 700 800 600 500 500 800Cumulative deviation 200 100 100 300 300 0 -200 -200Planned output 650 650 650 650 650 700 700 700Actual output 600 700 800 700 650 500 600 800Cumulative deviation -50 0 150 200 200 0 -100 0Backlog 120 220 220 120 220 170 170 70 70

#6.

G13 =SUM(C13:F13)

G14 =SUM(C14:F14)

G15 =SUM(C15:F15)

G16 =SUM(C16:F16)

C17 =SUM(C13:C16)

D17 =SUM(D13:D16)

E17 =SUM(E13:E16)

F17 =SUM(F13:F16)

G19 =SUMPRODUCT(C5:F8,C13:F16)

Solver parametersSet Target Cell G19 MAX

Changing cells C13:F16Constraints G13:G17 <= 1

C17:F17 = 1

Options: □ Assume linear model □ Assume non-negative

#7. SPT

JobProcessing time (Days)

Days till due date

Completion time (Flowtime) Lateness

B 20 165 20 0E 20 170 40 0A 30 40 70 30F 35 130 105 0C 40 125 145 20D 50 65 195 130

195 575 180

Average flow time = 95.833Average lateness = 30Average WIP = 2.949Utilization = 0.339

EDD

JobProcessing time (Days)

Days till due date

Completion time (Flowtime) Lateness

A 30 40 30 0D 50 65 80 15C 40 125 120 0F 35 130 155 25B 20 165 175 10E 20 170 195 25

195 755 75

Average flow time = 125.83Average lateness = 12.50Average no. of jobs in the system = 3.87Utilization = 0.26

CR

JobProcessing time (Days)

Days till due date

Completion time (Flowtime) Lateness CR

D 50 65 50 0 1.30A 30 40 80 40 1.33C 40 125 120 0 3.13F 35 130 155 25 3.71B 20 165 175 10 8.25E 20 170 195 25 8.50

195 775 100

Average flow time = 129.17Average lateness = 16.67Average no. of jobs in the system = 3.97Utilization = 0.25

#8. 0.99515 = 0.9257569 or 92.7569%

#9.(a)RRow 1= 0.98 x 0.92 x {1 – (1-.9) (1-.9) (1-.9)} = 0.900698RRow 2 = 0.98 x 0.92 x 0.98 x 0.95 = 0.903256RS = 1 – (1 - RRow 1)( 1 - RRow 2) = 1 – (1 – 0.900698)(1 – 0.903256) = 0.990393, or 99.0393%

(b) R1 = 1 – (1 – 0.95) (1 – 0.95) (1 – 0.95) = 0.999875R2 = 1 – (1 – 0.90) (1 – 0.90) (1 – 0.90) = 0.999R3 = 1 – (1 – 0.94) (1 – 0.94) (1 – 0.94) = 0.998659

Rs = R1 x R2 x R3 = .998659 = 99.8659%

#10.

(a) Failed = 7FR(%) = 14.0%Total tested unit-hours = 10000

FailedOperating time

Non-operating time/unit

Non-operating time

2 50 150 3003 75 125 3751 120 80 801 150 50 50

Total non-operating time = 805Operating time = 10000 – 805 = 9195

(b) 7/9195 = 0.000761283(c) 1/0.000761283 = 1313.57(d) 1000* 0.000761283 = 0.761283306

#11.

No. of breakdowns

No. of weeks that many breakdowns occurred

(Frequency)Relative

frequencyExpected

breakdowns0 75 0.375 0.0001 35 0.175 0.1752 50 0.250 0.5003 20 0.100 0.3004 10 0.050 0.2005 10 0.050 0.250

200 1.000 1.425

Expected number of failures/week = 1.425Average breakdown with routine maintenance = 1.2

Repair cost/breakdown without routine maintenance = 12000Routine maintenance cost/week = 6000

Repair cost/breakdown with routine maintenance = 5000

Cost without routine maintenance/week = 1.425 x $12,000 =$17,10

0

Cost with routine maintenance/week = 6000 + 5000 x 1.2 =$13,20

0

Total cost/week with routine maintenance is less expensive compared to without it. Therefore, purchase routine maintenance contract.