Operations Test 2 Math Problems

The following samples capture lengths of machine parts produced by a manufacturing process in a factory. We need variable charts here (both X-bar and R charts). The data and calculations are provided below.The grand mean or X double bar is 15.9970 cm and R-bar is 0.1210 cm.What will be the upper control limit (UCL) of the X-bar or mean chart?

Explanation Sample size (n) = 3 From the given parameter table, A2 = 1.023, D4 = 2.574, D3 = 0 X-bar Chart: X-bar Chart: UCL = 15.997 + 1.023 x 0.121 = 16.1208

The process of making machines parts (for which we developed control charts earlier) has a process mean is 15.99 cm and standard deviation is 0.01 cm. We need to make parts of target length 15.95 cm with +/- 0.05 cm tolerance.What in the chance that machine parts would be longer than 16.00 cm? Use Z-chart, posted on Top Hat / Canvas.

If X = 16.00, then Z = (X - µ) / σ = (16.00 - 15.99) / 0.01 = 1 The area on the left hand side of Z value of 1 is 0.8413 (from Z-chart). Hence, the chance of machine parts being SHORTER than 16.00 cm is 0.8413. The chance of machine parts being LONGER than 16.00 cm is 1 - 0.8413 = 0.1587 (as the whole area under curve is 1).

The process of making machines parts (for which we developed control charts earlier) has a process mean is 15.99 cm and standard deviation is 0.01 cm. We need to make parts of target length 15.95 cm with +/- 0.05 cm tolerance.What in the chance that machine parts would be between 15.97 cm and 16.00 cm? Use Z-chart, posted on Top Hat / Canvas.

If X = 16.00, then Z = (X - µ) / σ = (16.00 - 15.99) / 0.01 = 1 The area on the left hand side of Z value of 1 is 0.8413 (from Z-chart). Hence, the chance of machine parts being less than 16.00 cm is 0.8413. If X = 15.97, then Z = (X - µ) / σ = (15.97 - 15.99) / 0.01 = - 2 (negative two) The area on the LEFT hand side of Z value of 2 is 0.9772 (from Z-chart). The area on the RIGHT hand side of Z value of 2 (positive two) is 1 - 0.9772 = 0.0228 (as the whole area under curve is 1). The area on the LEFT hand side of Z value of - 2 (negative two) is ALSO 0.0228 (by SYMMETRY). Hence, the chance of machine parts being less than 15.97 cm is 0.0228. We already know (from above) that the chance of machine parts being less than 16.00 cm is 0.8413. Thus, the chance of machine parts being in between 15.97 cm and 16.00 cm = 0.8413 - 0.0228 = 0.8185.

Please get a HARD COPY of the blank table from the PRINT TEMPLATE folder BEFORE starting this exercise. It will SAVE time. Please review the Overbooking example provided in the recorded lecture.All pieces of information in that example remain unchanged EXCEPT:Now the revenue from one room is $80 per night (instead of $89).What the expected annual cost (or loss) if the hotel overbooks 3 rooms over its capacity?

$78,450

Based on the information provided below, perform available-to-promise (ATP) calculation using look-ahead procedure. What is the ATP in Week 6? LOOK AT PHONE

ATP - Calculation Review Based on the information provided below, perform available-to-promise (ATP) calculation using look-ahead procedure. What is the ATP in Week 6? Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. Explanation ATP Calculations Week 1: (150+0) - (65+50) = 35 (In the 1st week, beginning inv. = 150, MPS = 0) Week 3: 100 - 60 = 40 Week 4: 100 - (50+50) = 0 Week 6: 100 - (50+0+20) = 30

Daily usage is EXACTLY 60 gallons per day. Lead time is normally distributed with a mean of 10 days and a standard deviation of 2 days. What is the standard deviation of demand during lead time?

60 X 2 Explanation Since only lead time is variable (daily usage does NOT vary), the standard deviation of demand during lead time is the demand times the standard deviation of lead time. See the formula of standard deviation of demand during lead time (Sigma DDLT).

Please get a HARD COPY of the blank table from the PRINT TEMPLATE folder BEFORE starting this exercise. It will SAVE time.The forecast of future demands are given below.Month Jan Feb Mar Apr MayForecast 8500 11500 9500 11000 12000Develop an aggregate plan if the company prefers to match the demand by adjusting workforce size. Each employee has a production capacity of 100 units per month. The payroll cost for each employee is $1,900 per month. The cost of hiring and firing (layoff) per employee are $300 and $800 respectively. The number of employees at the beginning of January is 85.What is the total cost of the plan?

Chase Strategy: The workforce is being changed frequently to match the demand pattern Each month production level will be adjusted to match the demand and employees will be hired or fired depending on the production requirements. Inventory will NOT be carried. Month Jan Feb Mar Apr May Total Forecast 8500 11500 9500 11000 12000 52500 Planned Production 8500 11500 9500 11000 12000 No. of Employees Needed 85 115 95 110 120 No. of Employees at the Beginning of Month 85 85 115 95 110 Hire 0 30 0 15 10 55 Layoffs 0 0 20 0 0 20 No. of Employees after Adjustments 85 115 95 110 120 Payroll Cost 161.5K 218.5K 180.5K 209K 228K $997,500 Hiring Cost $16,500 Layoff Cost $16,000 Total Cost $1,030,000 Number of employees needed in January = 8500 / 100 = 85 Payroll cost = (85 + 115 + 95 + 110 + 120) x 1900 = $997,500 Hiring cost = 55 x 300 = $16,500 Layoff cost = 20 x 800 = $16,000 The total cost of the plan is $1,030,000.

Annual demand for a particular item is 1800 units. Each time items are ordered it costs $25. Annual unit inventory carrying cost is $3. The unit purchase price varies depending on quantity purchased as presented below. Range of Quantity Unit Price 1 to 100 $2.40 101 to 200 $2.35 201 to 300 $2.30 301 to 400 $2.25 401 to 500 $2.20 More than 500 $2.15 Find the optimal order quantity. Use the formula for total annual cost of inventory: TC = DC + (D/Q) x S + (Q/2) x H

D. 301 Explanation This is a case of quantity discount. Annual demand (D) = 1800 Order cost (S) = $25 Unit annual inventory carrying or holding cost (H) = $3 Basic EOQ = SQRT (2 x 1800 x 25 / 3) = 173.21 (approximately 173) Using the following formula total cost (TC) is calculated for different order quantities. TC = DC + (D/Q) x S + (Q/2) x H Q total Cost (TC) Optimal Q 173 4,749.62 201 $4,665.38 301 $4,651.00 401 $4,673.72 501 $4,711.32 NOTE: Since basic EOQ of 173 is outside the first range (0-100) we do NOT needto have any cost calculation for this range. The optimal order quantity is 301 (as it will keep the total annual cost of inventory at the minimum level).

Please review the Grocery Store Checkout Capacity example (posted on Top Hat). All the figures remain the same EXCEPT average service time for regular checkout line is now 12 minutes (instead of 10 minutes). What will be the NEW total capacity of the store on Sundays in PART A (i.e. WITHOUT reduction of hours)?

OLD capacity for regular lines (with average service time of 10 minutes) = 252 customers per day NEW capacity for regular lines (with average service time of 12 minutes) = 252 x (10 / 12) = 210 customers per day The capacity for express line remains unchanged (i.e. 360 customers per day) The NEW total capacity of the store on Sundays = 210 + 360 = 570 customers per day

A chair production facility has a design capacity of making 1500 chairs per week. It is supposed to operate 10 hours a day, 5 days a week. Due to old tools and machinery the operations manager has estimated that effective production capacity is 28 chairs per hour. It is reported to the manager that at present approximately 260 chairs are being produced per day on an average.The utilization of the production facility is:

Output = 260 per day x 5 days per week = 1300 chairs per week Design Capacity = 1500 chairs per week (given) Utilization = 1300/1500 = 0.8667 = 86.67%

A chair production facility has a design capacity of making 1500 chairs per week. It is supposed to operate 10 hours a day, 5 days a week. Due to old tools and machinery the operations manager has estimated that effective production capacity is 28 chairs per hour. It is reported to the manager that at present approximately 260 chairs are being produced per day on an average.The efficiency of the production facility is:

Output = 260 per day x 5 days per week = 1300 chairs per week Effective Capacity = 28 per hour x 10 hours per day x 5 days per week = 1400 chairs per week Efficiency = 1300 chairs per week / 1400 chairs per week = 0.9286 = 92.86%

The following samples capture lengths of machine parts produced by a manufacturing process in a factory. We need variable charts here (both X-bar and R charts). The data and calculations are provided below.The grand mean or X double bar is 15.9970 cm and R-bar is 0.1210 cm.What will be the upper control limit (UCL) of the R or range chart?

Sample size (n) = 3 From the given parameter table, A2 = 1.023, D4 = 2.574, D3 = 0 R Chart: R Chart: UCL = 2.574 x 0.121 = 0.3115 LCL = 0 x 0.121 = 0

The following samples capture lengths of machine parts produced by a manufacturing process in a factory. We need variable charts here (both X-bar and R charts). The data and calculations are provided below.The grand mean or X double bar is 15.9970 cm and R-bar is 0.1210 cm.What will be the lower control limit (LCL) of the X-bar or mean chart?

Sample size (n) = 3 From the given parameter table, A2 = 1.023, D4 = 2.574, D3 = 0 X-bar Chart: UCL = 15.997 + 1.023 x 0.121 = 16.1208 LCL = 15.997 - 1.023 x 0.121 = 15.8732

The process of making machines parts (for which we developed control charts earlier) has a process mean is 15.99 cm and standard deviation is 0.01 cm. We like to determine if the process can be used to make parts of target length 15.95 cm with +/- 0.05 cm tolerance.Can we use process capability ratio (Cp)?

Target = 15.95 cm Process Mean (µ) = 15.99 cm Since the target and process mean are NOT the same, the process is NOT centered. Hence, we can NOT use process capability ratio (Cp).

The available-to-promise (ATP) calculation below is performed using look-ahead procedure. The current time is at the beginning of Week 1. If a new customer order requires 50 units in Week 3, what should be the ATP in Week 1 after accepting the order?

There are 35 units available in Week 1 and 40 units available in Week 3. Order is 50 - 40 = 10 more than the ATP in Week 3. But the ATP of 10 from Week 1 can be used to cover this. Thus, after accepting the order ATP in Week 1 will be 35 - 10 = 25. NOTE: One may think that alternatively we can use all of 35 from Week 1 and 15 from Week 3 to cover 35 + 15 = 50 units in Week 3 (in that case ATP in Week 1 would be zero). But we should NOT do that to keep the flexibility to fulfilling other orders that may appear in Week 1 or 2.

Studies on a machine that molds plastic water pipe indicate that when it is injecting 1-inch-diameter pipe, the process standard deviation is 0.05 inch. The 1-inch pipe has a specification of 1 inch plus or minus 0.10 inch. What is the process capability index (Cpk) if the long-run process mean is 1 inch?

Use the Cpk formula to assess this process' capability. Target = 1 inch Upper Specification (US) = 1 + 0.1 = 1.1 inch Lower Specification (LS) = 1 - 0.1 = 0.9 inch Process Mean (µ) = 1 inch Standard Deviation (σ) = 0.05 inch Cpk = Min { (1.1 - 1)/(3 x 0.05), (1 - 0.9)/(3 x 0.05) } = Min { 0.67, 0.67 } = 0.67

Studies on a bottle-filling machine indicate that it fills bottles to a mean of 16 ounces with a standard deviation of 0.10 ounces. What is the process specification, assuming the Cpk index of 1?

Using the Cpk formula, the data given only allow one to determine that either: (A) lower spec = 15.7 and upper spec >= 16.3; or (B) lower spec <= 15.7 and upper spec = 16.3

Please get a HARD COPY of the blank table from the PRINT TEMPLATE folder BEFORE starting this exercise. It will SAVE time. The forecast of future demands (in 4 months) are given below.Month Jan Feb Mar AprForecast 21000 18000 19500 21500Develop an aggregate plan provided a constant workforce is to be maintained throughout the planning horizon. Each employee has a production capacity of 200 units per month. The payroll cost for each employee is $4,000 per month. The beginning inventory level is 1,000 units and inventory carrying cost is $0.25 per month.What is the total cost of the plan?

see word file Explanation Level Strategy: Constant workforce is maintained Average forecasted demand = (21000 + 18000 + 19500 + 21500) / 4 = 20,000 Each month 20,000 units will be planned to be produced. Month Jan Feb Mar Apr Total Forecast 21000 18000 19500 21500 80000 Planned Production 20000 20000 20000 20000 Beginning Inventory 1000 0 2000 2500 Ending Inventory 0 2000 2500 1000 Average Inventory 500 1000 2250 1750 Inv. Carrying Cost $125 $250 $562.5 $437.5 $1,375 Payroll Cost $1,600,000 TOTAL COST $1,601,375 Inventory carrying cost for January = 500 x 0.25 = $125 Number of employees needed per month = 20,000 / 200 = 100 Payroll cost = 100 employees x 4 months x $4000 / employee / month = $1,600,000 The total cost of the plan is $1,601,375.

Explanation Level Strategy: Constant workforce is maintained Average forecasted demand = (21000 + 18000 + 19500 + 21500) / 4 = 20,000 Each month 20,000 units will be planned to be produced. Month Jan Feb Mar Apr Total Forecast 21000 18000 19500 21500 80000 Planned Production 20000 20000 20000 20000 Beginning Inventory 1000 0 2000 2500 Ending Inventory 0 2000 2500 1000 Average Inventory 500 1000 2250 1750 Inv. Carrying Cost $125 $250 $562.5 $437.5 $1,375 Payroll Cost $1,600,000 TOTAL COST $1,601,375 Inventory carrying cost for January = 500 x 0.25 = $125 Number of employees needed per month = 20,000 / 200 = 100 Payroll cost = 100 employees x 4 months x $4000 / employee / month = $1,600,000 The total cost of the plan is $1,601,375.

Explanation Chase Strategy: The workforce is being changed frequently to match the demand pattern Each month production level will be adjusted to match the demand and employees will be hired or fired depending on the production requirements. Inventory will NOT be carried. Month Jan Feb Mar Apr Total Forecast 21000 18000 19500 21500 80000 Planned Production 21000 18000 19500 21500 No. of Employees Needed 210 180 195 215 No. of Employees at the Beginning of Month 200 210 180 195 Hire 10 0 15 20 45 Layoffs 0 30 0 0 30 No. of Employees after Adjustments 210 180 195 215 Payroll Cost $462K $396K $429K $473K $1,760,000 Hiring Cost $18,000 Layoff Cost $27,000 TOTAL COST $1,805,000 Number of employees needed in January = 21000 / 100 = 210 Payroll cost = (210 + 180 + 195 + 215) x 2200 = $1,760,000 Hiring cost = 45 x 400 = $18,000 Layoff cost = 30 x 900 = $27,000 The total cost of the plan is $1,805,000.

A retailer sells 15 units of a particular product per day on an average with standard deviation of 2 units. Each time order is placed, it costs $104. The average replenishment lead time is 7 days and standard deviation of lead time is 1 day. Annual unit inventory carrying cost is $3. Assume there are 260 working days in a year. Unit purchase price is $45. Inventory is reviewed continuously and the retailer likes to keep the total annual cost of inventory as low as possible. The probability of stock-out should be no more than 5%. The order should be placed when inventory level reaches the reorder point (ROP).How many units should be ordered each time (EOQ)?Formula for EOQ = √(2DS/H)

Explanation EOQ Calculation Average daily demand = 15 Annual demand (D) = 260 x 15 = 3900 Order cost (S) = $104 Unit annual inventory carrying or holding cost (H) = $3 EOQ = √(2DS/H) = √((2 x 3900 x 104)/3) = 520 NOTE: Unit purchase price of $45 is NOT used in any of the above calculations.

An engineer wants to construct a sample mean chart for controlling the service life of a halogen lamp his company produces. He knows from numerous previous samples that this service life is normally distributed with a mean of 500 hours and a standard deviation of 20 hours. On three recent production batches, he tested service life on random samples of four lamps, with the following results:Batch Service Life (hours) 1 495 500 475 500 2 520 515 505 515 3 480 490 470 480He uses upper and lower control limits of 520 and 480 hours. Which one(s) of the following may be TRUE?

Explanation For all three batches, the sample means are within upper and lower control limits of 520 and 480 hours.

Please get a HARD COPY of the blank table from the PRINT TEMPLATE folder BEFORE starting this exercise. It will SAVE time.The forecast of future demands are given below.Month Jan Feb Mar Apr MayForecast 8500 11500 9500 11000 12000Develop an aggregate plan provided a constant workforce is to be maintained during the 5 month period. Each employee has a production capacity of 100 units per month. The payroll cost for each employee is $1,900 per month. The beginning inventory is 0 and inventory carrying cost is $0.20 per month.What is the total cost of the plan?

Explanation Level Strategy: Constant workforce is maintained Average forecasted demand = (8500 + 11500 + 9500 + 11000 + 12000) / 5 = 10,500 Each month 10,500 units will be planned to be produced. Month Jan Feb Mar Apr May Total Forecast 8500 11500 9500 11000 12000 52500 Planned Production 10500 10500 10500 10500 10500 Beginning Inventory 0 2000 1000 2000 1500 Ending Inventory 2000 1000 2000 1500 0 Average Inventory 1000 1500 1500 1750 750 Inv. Carrying Cost $200 $300 $300 $350 $150 $1,300 Payroll Cost $997,500 TOTAL COST $998,800 Inventory carrying cost for January = 1000 x 0.2 = $200 Number of employees needed per month = 10,500 / 100 = 105 Payroll cost = 105 employees x 5 months x $1900 / employee / month = $997,500 The total cost of the plan is $998,800.

Please review the Grocery Store Checkout Capacity example (please see BELOW). All the figures remain the same EXCEPT average service time for express checkout line is now 4 minutes (instead of 3 minutes). What will be the NEW total capacity of the store on Sundays in PART B (i.e. WITH reduction of hours)?

Explanation NEW capacity for express line (with average service time of 4 minutes) = 60 / 4 = 15 customers per hour In PART B, total number of hours the express line is open on Sundays = 18 - 4 = 14 hours In PART B, NEW capacity for express line (with average service time of 4 minutes) on Sundays = 15 x 14 = 210 customers per day In part A, capacity for regular lines (with average service time of 10 minutes) = 252 customers per day In part B, capacity for regular lines (with average service time of 10 minutes) is getting reduced BY = 1 counter less x 2 hours x (60/10) customers per hour = 12 In part B, capacity for regular lines (with average service time of 10 minutes) = 252 - 12 = 240 customers per day In part B (i.e. WITH reduction of hours), the NEW total capacity of the store on Sundays= 210 + 240 = 450 customers per day

The process of making machines parts (for which we developed control charts earlier) has a process mean is 15.99 cm and standard deviation is 0.01 cm. We like to determine if the process can be used to make parts of target length 15.95 cm with +/- 0.05 cm tolerance.What is process capability index (Cpk)?

Explanation Target = 15.95 cm, Tolerance = +/- 0.05 cm Upper Specification (US) = 15.95 + 0.05 = 16.00 cm Lower Specification (LS) = 15.95 - 0.05 = 15.90 cm Process Mean (µ) = 15.99 cm Standard Deviation (σ) = 0.01 cm Process Capability Index (Cpk) = Min { (16.00 - 15.99) / (3 x 0.01) , (15.99 - 15.90) / (3 x 0.01) } = Min {0.3333, 3.00} = 0.3333

Studies on a machine that molds plastic water pipe indicate that when it is injecting 1-inch-diameter pipe, the process standard deviation is 0.05 inch. The 1-inch pipe has a specification of 1 inch plus or minus 0.10 inch. What is the process capability ratio (Cp) if the long-run process mean is 1 inch?

Explanation This is a centered process as the process mean (I inch) and target (1 inch) are identical. Use the Cp formula to assess this process' capability. Target = 1 inch Upper Specification (US) = 1 + 0.1 = 1.1 inch Lower Specification (LS) = 1 - 0.1 = 0.9 inch Process Mean (µ) = 1 inch Standard Deviation (σ) = 0.05 inch Cp = (1.1 - 0.9)/(6 x 0.05) = 0.67

The process of making machines parts (for which we developed control charts earlier) has a process mean is 15.99 cm and standard deviation is 0.01 cm. We like to determine if the process can be used to make parts of target length 15.95 cm with +/- 0.05 cm tolerance. Does the process have the required capability if it is three sigma compliant?

Target = 15.95 cm, Tolerance = +/- 0.05 cm Upper Specification (US) = 15.95 + 0.05 = 16.00 cm Lower Specification (LS) = 15.95 - 0.05 = 15.90 cm Process Mean (µ) = 15.99 cm Standard Deviation (σ) = 0.01 cm Since the target and process mean are NOT the same, the process is NOT centered. Hence, we can NOT use process capability ratio (Cp) - we have to use process capability index (Cpk). Process Capability Index (Cpk) = Min { (16.00 - 15.99) / (30.01) , (15.99 - 15.90) / (30.01) } = Min {0.3333, 3.00} = 0.3333 < 1 For three sigma compliant process the minimum value of Cpk should be one (1). Since the Cpk value is less than 1, the process does NOT have the required capability.

An engineer wants to construct a sample mean chart for controlling the service life of a halogen lamp his company produces. He knows from numerous previous samples that this service life is normally distributed with a mean of 500 hours and a standard deviation of 20 hours. On three recent production batches, he tested service life on random samples of four lamps, with the following results:Batch Service Life (hours)1 495 500 475 5002 525 515 505 5153 480 480 470 480He uses upper and lower control limits of 520 and 480 hours. In which batch(es) (if any) does service life appear to be out of control?

The sample mean of batch 3 is below the lower control limit. (see notes)

Please review the Video Equipment Manufacturer Capacity example (posted on Top Hat). All the figures remain the same EXCEPT the facility is now open for 5 days a week (instead of 6 days a week). How many machines will be needed to meet the demand?NOTE: People work in ONE shift but machines can be used in both (TWO) shifts.

Time available for each machine (per week) = 2 shifts x 8 hours x 60 minutes per hour x 5 days (CHANGED from 6 days) = 4800 minutes in a week Demand or amount of work needed per week (in minutes) = 5000 x 6 + 2500 x 8 = 50000 minutes in a week Machines have 95% efficiency Number of machines needed = 50000 / (4800 x 0.95) = 10.9649 Need to go to the next whole number. So the number of machines needed will be 11.

Please review the Video Equipment Manufacturer Capacity example (please see BELOW). All the figures remain the same EXCEPT the facility is now open for 5 days a week (instead of 6 days a week). How many people (workers) will be needed to meet the demand? NOTE: People work in ONE shift but machines can be used in both (TWO) shifts.

Time available for each worker (per week) = 1 shift x 8 hours x 60 minutes per hour x 5 days (CHANGED from 6 days) = 2400 minutes in a week Demand or amount of work needed per week (in minutes) = 5000 x 6 + 2500 x 8 = 50000 minutes in a week Workers have 80% efficiency Number of workers needed = 50000 / (2400 x 0.8) = 26.0417 Need to go to the next whole number. So the number of workers needed will be 27.

A retailer sells 15 units of a particular product per day on an average with standard deviation of 2 units. Each time order is placed, it costs $104. The average replenishment lead time is 7 days and standard deviation of lead time is 1 day. Annual unit inventory carrying cost is $3. Assume there are 260 working days in a year. Unit purchase price is $45. Inventory is reviewed continuously and the retailer likes to keep the total annual cost of inventory as low as possible. The probability of stock-out should be no more than 5%. The order should be placed when inventory level reaches the reorder point (ROP).Note: Z0.95 = 1.645. The reorder point (ROP) will be:

answer: 132 Mean usage or demand rate (¯u) = 15 units per day Std. dev. of daily demand (σu) = 2 units Mean lead time (¯LT) = 7 days Std. dev. of lead time (σLT) = 1 day The probability of stock-out should be no more than 5%. So the service level (SL) should be at least (1 - 0.05) = 0.95 or 95%. SS and ROP Calculation Standard deviation of demand during lead time (Sigma DDLT or σ_DDLT) = √(¯LT σ_u^2 + ¯u^2 σ_LT^2 ) = √(7 2^2 +〖15〗^2 1^2 ) = 15.906 Safety Stock (SS) = ZSL * σDDLT = Z0.95 * σDDLT = 1.645 x 15.906 = 26.1653 We go to the next whole number. Thus, safety stock (SS) will be 27. Expected demand during lead time = ¯u x ¯LT = 15 x 7 = 105 Reorder point (ROP) = Expected demand during lead time + Safety stock = 105 + 27 = 132