exam 2 practice problems

Café Industrial Relightment (CIR), a coffee company, uses 90 bags of whole bean coffee every month.  CIR’s supplier charges $105 for every shipment, regardless of order size.  The monthly inventory holding cost is $0.75.  Assume a month has 30 days. 

What is the economic order quantity (EOQ) in bags per order?

158.7451 bags/order

R = 90 [bags/month]

K = 105 [dollars/order]

h = 0.75 [dollars/month]

 

EOQ = √((2*K*R)/h) = √((2 90 [bags/month] 105 [dollars/order]) / 0.75 [dollars/month]) = 158.7451 [bags/order]

Café Industrial Relightment (CIR), a coffee company, uses 90 bags of whole bean coffee every month.  CIR’s supplier charges $105 for every shipment, regardless of order size.  The monthly inventory holding cost is $0.75.  Assume a month has 30 days. 

If CIR orders EOQ every time, then how many times per year does CIR order?

Here is the step-by-step breakdown of how to solve this problem:

Step 1: Identify the Given Variables

First, pull out all the numbers provided in the prompt and ensure they are on a consistent time frame (monthly):

• Monthly Demand ($R$): $90$ bags per month

  
  

• Annual Demand ($D$ or $R_{\text{annual}}$): $90 \times 12 = 1,080$ bags per year

  
  

• Setup/Ordering Cost ($K$): $\$105$</span> per shipment (flat fee regardless of size)</p><p></p></li><li><p><strong>Monthly Holding Cost (</strong><span style="background-color: rgba(0, 0, 0, 0);"><strong>$h$</strong></span><strong>):</strong> <span style="background-color: rgba(0, 0, 0, 0);">$\$0.75$ per bag per month

  
  

Step 2: Calculate the Economic Order Quantity ($\text{EOQ}$ or $Q^*$)

Using the standard EOQ formula from your coursework:

$$\text{EOQ} = \sqrt{\frac{2KR}{h}}$$

Plug in the monthly numbers:

$$\text{EOQ} = \sqrt{\frac{2 \times 90 \times 105}{0.75}}$$

$$\text{EOQ} = \sqrt{\frac{18,900}{0.75}}$$

$$\text{EOQ} = \sqrt{25,200} \approx \mathbf{158.7451 \text{ bags per order}}$$

Step 3: Calculate the Number of Orders Per Year

To find out how many times CIR places an order annually, divide the total annual demand by the batch size ($\text{EOQ}$) you just calculated:

$$\text{Number of Orders per Year} = \frac{\text{Annual Demand}}{\text{EOQ}}$$

$$\text{Number of Orders per Year} = \frac{90 \text{ bags/month} \times 12 \text{ months/year}}{158.7451 \text{ bags/order}}$$

$$\text{Number of Orders per Year} = \frac{1,080}{158.7451} \approx \mathbf{6.8034 \text{ orders/year}}$$

Answer:

The correct option is 6.8034 orders/year.

Café Industrial Relightment (CIR), a coffee company, uses 90 bags of whole bean coffee every month.  CIR’s supplier charges $105 for every shipment, regardless of order size.  The monthly inventory holding cost is $0.75.  Assume a month has 30 days. 

What is the monthly ordering cost if CIR orders EOQ?

$$\text{Total Order Cost} = \left(\frac{R}{Q}\right) \times K$$

• r/q gives you the exact number of orders placed per time period (Demand divided by your batch size/EOQ).

  
  

• K multiplies that frequency by the fixed cost per order.

• Q is 158.7451 bags per order

  
  

Plugging in your numbers yields $59.53 per month, which matches the EOQ cost-balancing equilibrium where monthly ordering costs equal monthly holding costs.

Café Industrial Relightment (CIR), a coffee company, uses 90 bags of whole bean coffee every month.  CIR’s supplier charges $105 for every shipment, regardless of order size.  The monthly inventory holding cost is $0.75.  Assume a month has 30 days. 

What is the monthly holding cost when CIR orders EOQ?

Here is the step-by-step breakdown explaining the exact solution provided in your prompt:

$$\text{Holding Cost} = \left(\frac{Q}{2}\right) \times h$$

Step 1: Identify the Given Variables

• $R$ (Monthly Demand): $90 \text{ bags/month}$

  
  

• $K$ (Ordering/Setup Cost): $\$105 \text{ per order}$</span></p><p></p></li><li><p><span style="background-color: rgba(0, 0, 0, 0);"><strong>$h$</strong></span><strong> (Monthly Holding Cost):</strong> <span style="background-color: rgba(0, 0, 0, 0);">$\$0.75 \text{ per bag/month}$

  
  

Step 2: Calculate the EOQ ($Q$)

First, plug your variables into the Economic Order Quantity formula to find the optimal batch size:

$$\text{EOQ} = \sqrt{\frac{2KR}{h}}$$

$$\text{EOQ} = \sqrt{\frac{2 \times 90 \times 105}{0.75}} = \sqrt{25,200} \approx 158.7451 \text{ bags/order}$$

Step 3: Apply the Holding Cost Formula

The formula for holding cost at EOQ multiplies your average inventory ($\frac{Q}{2}$) by your holding cost rate ($h$):

$$\text{Holding Cost} = \left(\frac{\text{EOQ}}{2}\right) \times h$$

Substituting the numbers directly from the solution:

$$\text{Holding Cost} = \left(\frac{158.7451}{2}\right) \times 0.75 = 79.37255 \times 0.75 = \mathbf{\$59.53 \text{ per month}}$$

(This final number proves the EOQ cost balance: your monthly holding cost perfectly matches your monthly ordering cost of $59.53).

Café Industrial Relightment (CIR), a coffee company, uses 90 bags of whole bean coffee every month.  CIR’s supplier charges $105 for every shipment, regardless of order size.  The monthly inventory holding cost is $0.75.  Assume a month has 30 days. 

If CIR orders EOQ, then what is the monthly cost of ordering and holding?

Here is the step-by-step explanation of the solution provided in your prompt for finding the total monthly cost of ordering and holding at the Economic Order Quantity ($\text{EOQ}$):

The Formula

$$\text{Total Monthly Cost} = \left(\frac{R}{Q} \times K\right) + \left(\frac{Q}{2} \times h\right)$$

(Total Monthly Cost = Monthly Ordering Cost + Monthly Holding Cost)

Step 1: Identify the Given Variables

• Monthly Demand ($R$) = $90$ bags/month

  
  

• Setup/Ordering Cost ($K$) = $\$105$</span> per order</p><p></p></li><li><p><strong>Monthly Holding Cost (</strong><span style="background-color: rgba(0, 0, 0, 0);"><strong>$h$</strong></span><strong>)</strong> = <span style="background-color: rgba(0, 0, 0, 0);">$\$0.75$ per bag/month

  
  

• Economic Order Quantity ($\text{EOQ}$ or $Q$) = $158.7451$ bags/order (calculated via $\sqrt{\frac{2KR}{h}}$)

  
  

Step 2: Calculate the Monthly Ordering Cost (r/q x k)

This part calculates how many orders you place per month and multiplies it by the fixed cost per order:

$$\text{Ordering Cost} = \left(\frac{90}{158.7451}\right) \times 105 \approx \mathbf{\$59.5294 \text{ per month}}$$

Step 3: Calculate the Monthly Holding Cost (q/2 x h)

This part calculates your average inventory ($\text{EOQ} / 2$) and multiplies it by the monthly holding rate:

$$\text{Holding Cost} = \left(\frac{158.7451}{2}\right) \times 0.75 = 79.37255 \times 0.75 \approx \mathbf{\$59.5294 \text{ per month}}$$

Step 4: Add Them Together

Combine both costs to get the total monthly inventory cost:

$$\text{Total Cost} = \$59.5294 + \$59.5294 = \mathbf{\$119.0588 \text{ per month}}$$

(Notice that because you are operating at the exact $\text{EOQ}$, your ordering cost and holding cost are perfectly equal at $59.53 each, which is the mathematical hallmark of an optimized EOQ cost curve!)

 Café Industrial Relightment (CIR), a coffee company, uses 90 bags of whole bean coffee every month.  CIR’s supplier charges $105 for every shipment, regardless of order size.  The monthly inventory holding cost is $0.75.  Assume a month has 30 days. 

If CIR orders 100 bags of coffee every time it makes an order, then what is the order cycle of CIR?.

Here is the step-by-step breakdown of how to calculate the order cycle (how often CIR places an order) when the order quantity is fixed at 100 bags:

The Formula

$$\text{Order Cycle} = \frac{\text{Batch Size } (Q)}{\text{Demand Rate } (R)}$$

Step 1: Identify the Given Variables

• Order Quantity ($Q$) = $100$ bags per order

  
  

• Monthly Demand ($R$) = $90$ bags per month

  
  

Step 2: Calculate the Order Cycle in Months

Divide the order size by the monthly demand rate:

$$\text{Order Cycle} = \frac{100 \text{ bags/order}}{90 \text{ bags/month}} \approx \mathbf{1.11111 \text{ months per order}}$$

Optional: Convert to Days

Since the prompt notes that a month has 30 days, you can also express the order cycle in days by multiplying the result by 30:

$$1.11111 \text{ months} \times 30 \text{ days/month} = \mathbf{33.33 \text{ days between orders}}$$

Café Industrial Relightment (CIR), a coffee company, uses 90 bags of whole bean coffee every month.  CIR’s supplier charges $105 for every shipment, regardless of order size.  The monthly inventory holding cost is $0.75.  Assume a month has 30 days. 

 

If CIR orders 100 bags of coffee every time it makes an order, then how many times does it order per month?

The Formula

$$\text{Orders per Month} = \frac{\text{Monthly Demand } (R)}{\text{Order Quantity } (Q)}$$

Step 1: Identify the Given Variables

• Monthly Demand ($R$) = $90$ bags per month

  
  

• Order Quantity ($Q$) = $100$ bags per order

  
  

Step 2: Calculate Orders per Month

Divide the total monthly demand by the batch size:

$$\text{Orders per Month} = \frac{90 \text{ bags/month}}{100 \text{ bags/order}} = \mathbf{0.9 \text{ orders/month}}$$

(This means CIR places slightly less than one order per month—specifically, one order every $1.11$ months or about every $33.3$ days).

Café Industrial Relightment (CIR), a coffee company, uses 90 bags of whole bean coffee every month.  CIR’s supplier charges $105 for every shipment, regardless of order size.  The monthly inventory holding cost is $0.75.  Assume a month has 30 days. 

If CIR buys 100 bags of coffee per order, what is the average flow time of bags at CIR? (Hint: First find the average inventory (I). Then use Little’s Law to find the flow time (T) in months. Note that the flow rate (R) is 90 [bags/month].)

Average Inventory = I = Q/2

Here is the step-by-step breakdown of how the solution provided in your prompt calculates the average flow time when CIR orders in batches of 100 bags:

Step 1: Calculate Average Inventory ($I$)

Using the standard average inventory formula ($Q/2$), where $Q$ is the order batch size:

$$I = \frac{Q}{2} = \frac{100 \text{ bags}}{2} = \mathbf{50 \text{ bags}}$$

Step 2: Identify the Flow Rate ($R$)

From the problem prompt, the monthly demand (throughput or flow rate) is:

$$R = \mathbf{90 \text{ bags/month}}$$

Step 3: Apply Little’s Law to Find Flow Time ($T$)

Little's Law states that Inventory equals Flow Rate multiplied by Flow Time:

$$I = R \times T$$

Rearranging the formula to solve for Flow Time ($T$):

$$T = \frac{I}{R}$$

Substitute your numbers into the equation:

$$T = \frac{50 \text{ bags}}{90 \text{ bags/month}} = \frac{5}{9} \approx \mathbf{0.5556 \text{ months}}$$

(To convert this to days, you can multiply $0.5556$ months by 30 days/month, which equals roughly $16.67$ days that an average bag of coffee spends in inventory before being used).

Café Industrial Relightment (CIR), a coffee company, uses 90 bags of whole bean coffee every month.  CIR’s supplier charges $105 for every shipment, regardless of order size.  The monthly inventory holding cost is $0.75.  Assume a month has 30 days. 

If CIR orders 200 bags of coffee per order, then what is the monthly cost of ordering and holding?

Here is the step-by-step breakdown explaining how to calculate the total monthly cost of ordering and holding when CIR places orders in batches of 200 bags:

The Formula

$$C(Q) = \left(\frac{R}{Q} \times K\right) + \left(\frac{Q}{2} \times h\right)$$

(Total Monthly Cost = Monthly Ordering Cost + Monthly Holding Cost)

Step 1: Identify the Given Variables

• Monthly Demand ($R$) = $90$ bags/month

  
  

• Order Quantity ($Q$) = $200$ bags/order

  
  

• Setup/Ordering Cost ($K$) = $\$105$</span> per order</p><p></p></li><li><p><strong>Monthly Holding Cost (</strong><span style="background-color: rgba(0, 0, 0, 0);"><strong>$h$</strong></span><strong>)</strong> = <span style="background-color: rgba(0, 0, 0, 0);">$\$0.75$ per bag/month

  
  

Step 2: Calculate the Monthly Ordering Cost (r/q x k) This part calculates how many orders are placed per month (Demand divided by batch size) and multiplies it by the fixed cost per order:

$$\text{Ordering Cost} = \left(\frac{90}{200}\right) \times 105$$

\text{Ordering Cost} = 0.45 \times \105 = \mathbf{\47.25 \text{ per month}}

Step 3: Calculate the Monthly Holding Cost (q/2 x h)

This part calculates the average inventory held in the warehouse ($\frac{Q}{2}$) and multiplies it by the monthly holding rate:

$$\text{Holding Cost} = \left(\frac{200}{2}\right) \times 0.75$$

\text{Holding Cost} = 100 \times \0.75 = \mathbf{\75.00 \text{ per month}}

Step 4: Add Them Together

Combine both costs to get the total monthly inventory cost:

$$\text{Total Cost} = \$47.25 + \$75.00 = \mathbf{\$122.25 \text{ per month}}$$

RACHEL’S ORGANIC SOAP

Rachel’s Organic Soap Company makes organic liquid soaps. She purchases organic Palm oil to make her soaps. She needs 750 kgs of Palm oil per day on average. The supplier charges a $75 delivery fee per order (which is independent of the order size). Rachel’s annual holding cost is $1.125 per kg of palm oil. Assume there are 365 days a year. If Rachel wants to minimize inventory holding and ordering costs, how much Palm oil should she purchase with each order (in kgs)? (Answer to 4 decimal places).

To use the EOQ formula, all time-based rates (demand and holding costs) must match. Since demand is given in days, the holding cost must also be converted to a per day basis:

• Setup/Ordering Cost ($K$) = $\$75$</span> per order</p></li><li><p><strong>Daily Demand Rate (</strong><span><strong>$R$) = $750 kgs.day

• Annual Holding Cost (h annual) = $1.125 per kg/year</span></p></li><li><p><strong>Daily Holding Cost (</strong><span><strong>$h$) = 1.125/365 days/year= 9/2920= 0.003082 dollars/kg/day

Step 2: Plug the Variables into the EOQ Formula

$$\text{EOQ} = \sqrt{\frac{2KR}{h}}$$

Substitute the values into the equation:

$$\text{EOQ} = \sqrt{\frac{2 \times 75 \times 750}{0.00308219}}$$

$$\text{EOQ} = \sqrt{\frac{112,500}{0.00308219}}$$

$$\text{EOQ} = \sqrt{36,500,000} \approx \mathbf{6,041.5229 \text{ kgs/order}}$$

(Note: Depending on whether you use the exact fraction $\frac{9}{2920}$ or the rounded decimal $0.0031$ for the daily holding cost, the precise decimal will calculate out to approximately $6,024.145$ or $6,041.5229$ kgs per order).

Rachel’s Organic Soap Company makes organic liquid soaps. She purchases organic Palm oil to make her soaps. She needs 750 kgs of Palm oil per day on average. The supplier charges a $75 delivery fee per order (which is independent of the order size). Rachel’s annual holding cost is $1.125 per kg of palm oil. Assume there are 365 days a year. If Rachel wants to minimize inventory holding and ordering costs, how much Palm oil should she purchase with each order (in kgs)?

What is the daily cost of order and holding if Rachel orders EOQ? (Answer to 4 decimal places.)

The Formula

$$C(Q) = \left(\frac{R}{Q} \times K\right) + \left(\frac{Q}{2} \times h\right)$$

(Total Daily Cost = Daily Ordering Cost + Daily Holding Cost)

Step 1: Identify the Given Variables

• Daily Demand ($R$) = $750 \text{ kgs/day}$

  
  

• Ordering/Delivery Fee ($K$) = $\$75 \text{ per order}$</span></p><p></p></li><li><p><strong>Daily Holding Cost (</strong><span style="background-color: rgba(0, 0, 0, 0);"><strong>$h$</strong></span><strong>)</strong> = <span style="background-color: rgba(0, 0, 0, 0);">9/2920 = 0.0031 dollars</span></p><p></p></li><li><p><strong>Economic Order Quantity (</strong><span style="background-color: rgba(0, 0, 0, 0);"><strong>$Q$</strong></span><strong> or </strong><span style="background-color: rgba(0, 0, 0, 0);"><strong>$\text{EOQ}$) = $6,024.145 kgs/order

  
  

Step 2: Calculate the Daily Ordering Cost (r/q x k)

This calculates how many orders are placed per day (Daily Demand divided by batch size) and multiplies it by the fixed delivery fee per order:

$$\text{Ordering Cost} = \left(\frac{750}{6024.145}\right) \times 75$$

$$\text{Ordering Cost} = 0.1244988 \times 75 \approx \mathbf{\$9.3374 \text{ per day}}$$

Step 3: Calculate the Daily Holding Cost (q/2 x h)

This calculates the average inventory held in the warehouse (q/2) and multiplies it by the daily holding rate:

$$\text{Holding Cost} = \left(\frac{6024.145}{2}\right) \times 0.0031$$

$$\text{Holding Cost} = 3012.0725 \times 0.0031 \approx \mathbf{\$9.3374 \text{ per day}}$$

Step 4: Add Them Together

Combine both costs to get the total daily inventory cost:

$$\text{Total Cost} = \$9.3374 + \$9.3374 = \mathbf{\$18.6749 \text{ per day}}$$

(Just like the coffee problem, because Rachel is operating at the exact $\text{EOQ}$, her daily ordering cost and daily holding cost are perfectly balanced and equal to each other!)

Again, assume that Racel needs 750 kgs of Palm oil per day. Sarah’s supplier is willing to sell Palm oil to her at a 10% discount on its selling price if she purchases 6,000 kgs at a time. If Sarah takes the offer and orders 6,000 kg at a time, how much will be the resulting daily cost of ordering and holding? (Assume that the fixed cost of ordering is $75 per order and the per unit inventory holding cost is 0.0031 per day per kg of Palm oil.)

The Formula

$$C(Q) = \left(\frac{R}{Q} \times K\right) + \left(\frac{Q}{2} \times h\right)$$

(Total Daily Cost = Daily Ordering Cost + Daily Holding Cost)

Step 1: Identify the Given Variables

• Daily Demand ($R$) = $750 \text{ kg/day}$

  
  

• Ordering Fee ($K$) = $\$75 \text{ per order}$</span></p><p></p></li><li><p><strong>Daily Holding Cost (</strong><span style="background-color: rgba(0, 0, 0, 0);"><strong>$h$</strong></span><strong>)</strong> = <span style="background-color: rgba(0, 0, 0, 0);">$\$0.0031 \text{ per kg/day}$

  
  

• New Order Quantity ($Q$) = $6,000 \text{ kg/order}$

  
  

Step 2: Calculate the Daily Ordering Cost ($\frac{R}{Q} \times K$)

This calculates how many orders are placed per day (Daily Demand divided by the new batch size) and multiplies it by the fixed order fee:

$$\text{Ordering Cost} = \left(\frac{750}{6,000}\right) \times 75$$

$$\text{Ordering Cost} = 0.125 \times 75 = \mathbf{\$9.375 \text{ per day}}$$

Step 3: Calculate the Daily Holding Cost ($\frac{Q}{2} \times h$)

This calculates the new average inventory held in the warehouse ($\frac{Q}{2}$) and multiplies it by the daily holding rate:

$$\text{Holding Cost} = \left(\frac{6,000}{2}\right) \times 0.0031$$

$$\text{Holding Cost} = 3,000 \times 0.0031 = \mathbf{\$9.30 \text{ per day}}$$

Step 4: Add Them Together

Combine both components to get the total daily inventory cost:

$$\text{Total Cost} = \$9.375 + \$9.30 = \mathbf{\$18.675 \text{ per day}}$$

Again, assume that Rachel needs 750 kgs of Palm oil per day. Sarah’s supplier is willing to sell Palm oil to her at a 10% discount on its selling price if she purchases 6,000 kgs at a time. The current selling price for Palm oil before the quantity discount is $8 per kg. If Sarah takes the offer and orders 6,000 kg at a time, how much does she save on the cost of Palm oil each day? (Note that the daily cost of Palm oil is determined by the daily demand multiplied by the selling price.)

The Logic

To find out how much money is saved each day, you need to calculate the value of the 10% discount applied to the total daily volume of Palm oil purchased.

Step 1: Identify the Given Variables

• Daily Demand ($R$) = $750 \text{ kg/day}$

  
  

• Original Unit Price = $\$8.00 \text{ per kg}$

  
  

• Discount Percentage = $10\%$ (or $0.10$)

  
  

Step 2: Calculate the Savings Per Kilogram

Find how much money is saved on each individual kilogram of Palm oil:

$$\text{Savings per kg} = \text{Original Price} \times \text{Discount Percentage}$$

\text{Savings per kg} = \8.00 \times 0.10 = \mathbf{\0.80 \text{ saved per kg}}

Step 3: Calculate the Total Daily Savings

Multiply the savings per kilogram by the total number of kilograms purchased and consumed each day ($750 \text{ kg/day}$):

$$\text{Daily Savings} = \text{Daily Demand} \times \text{Savings per kg}$$

\text{Daily Savings} = 750 \text{ kg/day} \times \0.80/\text{kg} = \mathbf{\600.00 \text{ per day}}

(Alternatively, following the exact equation from your prompt: 750 \text{ kg/day} \times \8.00/\text{kg} \times 10\% = \mathbf{\600 \text{ per day}}).

Spring Water

The demand for coconut water at SLC WalMart is 600 gallons per week. The setup cost for placing an order to replenish inventory is $25. Coconut water loses its freshness when stored, and SLC Walmart incurs an annual holding cost of $2.6/gallon. Assume that there are 52 weeks a year.
Determine SLC WalMart’s Economic Order Quantity in [gallons].

Step 1: Identify and Align the Variables

To use the EOQ formula, both your demand rate ($R$) and your holding cost ($h$) must be on the same time frame (either both weekly or both annual). Using a weekly basis:

• Setup/Ordering Cost ($K$) = $\$25 \text{ per order}$</span></p><p></p></li><li><p><strong>Weekly Demand (</strong><span style="background-color: rgba(0, 0, 0, 0);"><strong>$R$) = $600 \text{ gallons/week}$</span></p><p></p></li><li><p><strong>Annual Holding Cost</strong> = <span style="background-color: rgba(0, 0, 0, 0);">$\$2.60 \text{ per gallon/year}$

  
  

• Weekly Holding Cost ($h$) = 2.60 per year/52 weeks/year= 0.05 per gallon/week

  
  

Step 2: Plug the Variables into the EOQ Formula

$$\text{EOQ} = \sqrt{\frac{2KR}{h}}$$

Substitute the weekly values into the equation:

$$\text{EOQ} = \sqrt{\frac{2 \times 25 \times 600}{0.05}}$$

$$\text{EOQ} = \sqrt{\frac{30,000}{0.05}}$$

$$\text{EOQ} = \sqrt{600,000} \approx \mathbf{774.5967 \text{ gallons/order}}$$

Answer:

The Economic Order Quantity is approximately $774.60$ gallons (which falls right inside your answer range of $765 – 785$).

SARAH’S ORGANIC SOAP

Sarah’s Organic Soap Company makes organic liquid soaps. She purchases organic Palm oil to
make her soaps. She needs 1,000 kgs of Palm oil per day on average. The supplier charges a $60
delivery fee per order (which is independent of the order size). Sarah’s annual holding cost is $1.825 per kg of palm oil. Assume there are 365 days a year. If Sarah wants to minimize inventory holding and ordering costs, how much Palm oil should she purchase with each order (in kgs)?

What is the daily cost of ordering and holding when EOQ is ordered?

Step 1: Identify and Align the Variables

To use the EOQ formula, both your demand rate ($R$) and holding cost ($h$) must be on the exact same time frame (daily):

• Daily Demand ($R$) = $1,000 \text{ kgs/day}$

  
  

• Setup/Ordering Cost ($K$) = $\$60 \text{ per order}$</span></p><p></p></li><li><p><strong>Annual Holding Cost</strong> = <span style="background-color: rgba(0, 0, 0, 0);">$\$1.825 \text{ per kg/year}$

  
  

• Days per Year = $365 \text{ days}$

  
  

Step 2: Convert Annual Holding Cost to a Daily Rate ($h$)

Divide the annual holding cost by the number of days in the year:

h = \frac{\1.825 \text{ per year}}{365 \text{ days}} = \mathbf{\0.005 \text{ per kg/day}}

Step 3: Calculate the Economic Order Quantity ($\text{EOQ}$ or $Q$)

Using the standard EOQ formula:

$$\text{EOQ} = \sqrt{\frac{2KR}{h}}$$

Substitute the daily values into the equation:

$$\text{EOQ} = \sqrt{\frac{2 \times 60 \times 1,000}{0.05}}$$

$$\text{EOQ} = \sqrt{\frac{120,000}{0.005}}$$

$$\text{EOQ} = \sqrt{24,000,000} \approx \mathbf{4,898.979 \text{ kgs/order}}$$

Step 4: Calculate the Daily Cost of Ordering and Holding ($C(Q)$)

Using the total cost formula:

$$C(Q) = \left(\frac{R}{Q} \times K\right) + \left(\frac{Q}{2} \times h\right)$$

1. Daily Ordering Cost:

   
   
   

   $$\left(\frac{1,000}{4,898.979}\right) \times 60 \approx \mathbf{\$12.2475 \text{ per day}}$$

2. Daily Holding Cost:

   
   
   

   $$\left(\frac{4,898.979}{2}\right) \times 0.005 \approx \mathbf{\$12.2475 \text{ per day}}$$

3. Total Daily Cost:

   $$\$12.2475 + \$12.2475 = \mathbf{\$24.4949 \text{ per day}}$$

Spring Water

The demand for coconut water at SLC WalMart is 600 gallons per week. The setup cost for placing an order to replenish inventory is $25. Coconut water loses its freshness when stored, and SLC Walmart incurs an annual holding cost of $2.6/gallon. Assume that there are 52 weeks a year.

If SLC WalMart buys 750 gallons of coconut water per order, what is the number of orders placed
per week?

Here is the step-by-step breakdown of how to calculate the number of orders placed per week when SLC WalMart orders 750 gallons at a time:

The Formula

$$\text{Orders per Week} = \frac{\text{Weekly Demand } (R)}{\text{Order Quantity } (Q)}$$

Step 1: Identify the Given Variables

• Weekly Demand ($R$) = $600 \text{ gallons/week}$

  
  

• Order Quantity ($Q$) = $750 \text{ gallons/order}$

  
  

Step 2: Calculate Orders per Week

Divide the total weekly demand by the batch size:

$$\text{Orders per Week} = \frac{600 \text{ gallons/week}}{750 \text{ gallons/order}} = \mathbf{0.8 \text{ orders/week}}$$

(This means SLC WalMart places slightly less than one order per week—specifically, one order every $1.25$ weeks or every $8.75$ days).

Again, assume that Sarah needs 1,000 kgs of Palm oil per day. Sarah’s supplier is willing to sell
Palm oil to her at a 10% discount on its selling price if she purchases 10,000 kgs at a time. If Sarah takes the offer and orders 10,000 kg at a time, how much will be the resulting daily cost of ordering and holding? (Assume that the fixed cost of ordering is $60 per order and the per unit inventory holding cost is 0.005 per day per kg of Palm oil.)

The Formula

$$C(Q) = \left(\frac{R}{Q} \times K\right) + \left(\frac{Q}{2} \times h\right)$$

(Total Daily Cost = Daily Ordering Cost + Daily Holding Cost)

Step 1: Identify the Given Variables

• Daily Demand ($R$) = $1,000 \text{ kg/day}$

  
  

• Ordering Fee ($K$) = $\$60 \text{ per order}$</span></p><p></p></li><li><p><strong>Daily Holding Cost (</strong><span style="background-color: rgba(0, 0, 0, 0);"><strong>$h$</strong></span><strong>)</strong> = <span style="background-color: rgba(0, 0, 0, 0);">$\$0.005 \text{ per kg/day}$

  
  

• New Order Quantity ($Q$) = $10,000 \text{ kg/order}$

  
  

Step 2: Calculate the Daily Ordering Cost ($\frac{R}{Q} \times K$)

This calculates how many orders are placed per day (Daily Demand divided by the new batch size) and multiplies it by the fixed order fee:

$$\text{Ordering Cost} = \left(\frac{1,000}{10,000}\right) \times 60$$

$$\text{Ordering Cost} = 0.1 \times 60 = \mathbf{\$6.00 \text{ per day}}$$

Step 3: Calculate the Daily Holding Cost ($\frac{Q}{2} \times h$)

This calculates the new average inventory held in the warehouse ($\frac{Q}{2}$) and multiplies it by the daily holding rate:

$$\text{Holding Cost} = \left(\frac{10,000}{2}\right) \times 0.005$$

$$\text{Holding Cost} = 5,000 \times 0.005 = \mathbf{\$25.00 \text{ per day}}$$

Step 4: Add Them Together

Combine both components to get the total daily inventory cost:

$$\text{Total Cost} = \$6.00 + \$25.00 = \mathbf{\$31.00 \text{ per day}}$$

Imagine a production process.  During the milling stage two separate parts are made, part A and part B.  It takes 1 minute to produce 1 unit of part A, and it takes 1 minute to produce 1 unit of part B during the milling stage.  Before switching to produce parts A and B, the milling machine undergoes a 60 minute setup time, and this happens before beginning to mill both parts A and B.  The batch size is 50 units.  After producing the batch of 50 for both parts A and B, the assembly stage can begin.  The assembly stage take 4 minutes per each complete unit made of parts A and B.

What is the utilization of the milling stage? (It might help to draw out the process in a chart).

(Answer in percentage notation to 4 decimal places).

 

To find the utilization of the milling stage, we break down the total time required to produce a batch of 50 complete units (where each unit requires 1 unit of Part A and 1 unit of Part B) and compare the active processing time against the total time spent by the machine.

1. Breaking Down the Milling Stage Time (Total Cycle Time)

The milling stage involves setup and production for both Part A and Part B. According to the problem description:

• Part A Setup: 60 minutes

• Part A Production: 50 units \times 1 minute/unit = 50 minutes

• Part B Setup: 60 minutes

• Part B Production: 50 units \times 1 minute/unit = 50 minutes

Summing these up gives the total time required for the milling stage per batch:

\text{Total Milling Time} = 60 + 50 + 60 + 50 = 220 \text{ minutes}

2. Determining Active Processing Time

The active time spent actually producing the parts (excluding setup time) is the sum of the production times for both parts:

\text{Active Processing Time} = (50 \times 1) + (50 \times 1) = 100 \text{ minutes}

3. Calculating Milling Stage Utilization

Utilization is defined as the fraction of total time that the resource is actively processing units (or the ratio of active time to total cycle time):

\text{Utilization} = active processing time/total milling time =100/220

Expressed in percentage notation to 4 decimal places:

0.454545

Summary of Stages & Capacities

• Milling Stage Capacity / Utilization: Active time (100 mins) divided by total cycle time (220 mins) yields 45.4545%.

• Assembly Stage: Takes 4 minutes per complete unit for a batch of 50, totaling 50 \times 4 = 200 minutes. Its capacity relative to the batch is 50/200 = 0.25 units/min (or 25%), making assembly the bottleneck stage with the lowest capacity.

Imagine a production process. During the milling stage two separate parts are made, part A and part B. It takes 1 minute to produce 1 unit of part A, and it takes 1 minute to produce 1 unit of part B during the milling stage. Before switching to produce parts A and B, the milling machine undergoes a 60 minute setup time, and this happens before beginning to mill both parts A and B. The batch size is 50 units. After producing the batch of 50 for both parts A and B, the assembly stage can begin. The assembly stage take 4 minutes per each complete unit made of parts A and B.

What is the utilization of the assembly stage? (It might help to draw out the process in a chart).

To find the utilization of the assembly stage, we compare its capacity or active working time against the system's bottleneck or total cycle time.

1. Breaking Down the Assembly Stage Time

• Batch Size: 50 complete units

• Assembly Time: 4 minutes per complete unit

• Total assembly time per batch = 50 units x 4 minutes per unit = 200 minutes

2. Determining Stage Capacities (Units per Minute)

In operations management, the capacity of each stage is determined by how many complete units it can produce per unit of time (over the total cycle time required for a batch):

• Milling Stage Total Time: 60 (set up A) + 50 (run A) + 60 (setup B) + 50 (run B) = 220 minutes

• Milling capacity 50 units/220 minutes = 0.2273 units/min

• Assembly Stage Total Time: 200 minutes

• Assembly capacity = 50 units/200 minutes = 0.2500 units/min

3. Calculating Assembly Stage Utilization

The bottleneck of the process is the milling stage because it has the lowest capacity (0.2273 < 0.2500).

In a paced production system, non-bottleneck resources (like assembly) operate at a utilization determined by the ratio of the bottleneck capacity to their own capacity (or equivalently, the ratio of their processing time to the bottleneck/total cycle time):

Assembly utilization = milling capacity/assembly capacity = 200/220 =0.909

Expressed in percentage notation to 4 decimal places:

\mathbf{90.9091\%}

(Alternatively, expressed as a fraction of the total milling cycle time: \frac{200}{220} = 90.91\%.)

Imagine a production process. During the milling stage two separate parts are made, part A and part B. It takes 1 minute to produce 1 unit of part A, and it takes 1 minute to produce 1 unit of part B during the milling stage. Before switching to produce parts A and B, the milling machine undergoes a 60 minute setup time, and this happens before beginning to mill both parts A and B. The batch size is 50 units. After producing the batch of 50 for both parts A and B, the assembly stage can begin. The assembly stage take 4 minutes per each complete unit made of parts A and B.

How much idle time is in the process? (It might help to draw out the process in a chart). Explain this solution step by step

Milling Stage = 60 + 50 + 60 + 50 = 220 mins

Assembly Stage = 4 * 50 = 200 mins

Assembly stage must wait idle for 20 minutes

Imagine a production process.  During the milling stage two separate parts are made, part A and part B.  It takes 1 minute to produce 1 unit of part A, and it takes 1 minute to produce 1 unit of part B during the milling stage.  Before switching to produce parts A and B, the milling machine undergoes a 60 minute setup time, and this happens before beginning to mill both parts A and B.  The batch size is 100 units.  After producing the batch of 100 for both parts A and B, the assembly stage can begin.  The assembly stage take 4 minutes per each complete unit made of parts A and B.

What is the capacity of the bottleneck of the process? (It might help to draw out the process in a chart).

Your calculations are completely correct! You have accurately accounted for the setup times, run times, and batch sizes to find the capacity of each stage.

Here is the step-by-step breakdown confirming your analysis:

1. Milling Stage Capacity

• Batch Size (Q): 100 units

• Total Time per Batch:

  • Setup for Part A: 60 mins

  • Production for Part A (100 \times 1): 100 mins

  • Setup/Switch for Part B: 60 mins

  • Production for Part B (100 \times 1): 100 mins

  • Total Time: 60 + 100 + 60 + 100 = 320 minutes

• Milling Capacity: 100 units/320 minutes= 0.3125 units per minute

2. Assembly Stage Capacity

• Batch Size (Q): 100 units

• Total Time per Batch:

  • Assembly time (100 \times 4): 400 mins

  • Total Time: 400 minutes

• Assembly Capacity: 100 units/400 minutes=0.25 units per minute

3. Bottleneck Identification

Process Stage	Capacity Calculation	Output Rate
Milling Stage	100 / 320	0.3125 units/min
Assembly Stage	100 / 400	0.25 units/min

Since the Assembly Stage has the lowest capacity (0.25 units per minute compared to Milling's 0.3125 units per minute), it is the bottleneck of the production process.

Imagine a production process.  During the milling stage two separate parts are made, part A and part B.  It takes 1 minute to produce 1 unit of part A, and it takes 1 minute to produce 1 unit of part B during the milling stage.  Before switching to produce parts A and B, the milling machine undergoes a 60 minute setup time, and this happens before beginning to mill both parts A and B.  The batch size is 100 units.  After producing the batch of 100 for both parts A and B, the assembly stage can begin.  The assembly stage take 4 minutes per each complete unit made of parts A and B.

What stage has idle wait time in this process? (It might help to draw out the process in a chart).

Step-by-Step Breakdown of Batch Timing

1. Milling Stage Duration: Takes 320 minutes total (60 + 100 + 60 + 100) to produce the batch of 100 units for both parts. Milling finishes at minute 320.

2. Assembly Stage Duration: Takes 400 minutes total (100 \times 4) to assemble the batch.

3. Synchronization: Assembly cannot start until Milling completes at minute 320. Assembly then runs from minute 320 to minute 720.

Idle Time Summary

• Assembly Stage Idle Time: Assembly is starved of parts for the first 320 minutes of the cycle while waiting for Milling to finish.

• Milling Stage Idle Time: Milling finishes its batch at minute 320, but because Assembly takes until minute 400 to process that batch, Milling sits idle for 80 minutes per batch cycle (400 - 320 = 80).

Verification Checklist

• [x] Confirm Milling batch completion time: 320 minutes

• [x] Confirm Assembly batch completion time: 400 minutes

• [x] Calculate Milling idle time per cycle: 80 minutes (400 - 320)

• [x] Calculate Assembly startup idle time: 320 minutes

Imagine a production process.  During the milling stage two separate parts are made, part A and part B.  It takes 1 minute to produce 1 unit of part A, and it takes 1 minute to produce 1 unit of part B during the milling stage.  Before switching to produce parts A and B, the milling machine undergoes a 60 minute setup time, and this happens before beginning to mill both parts A and B.  The batch size is 100 units.  After producing the batch of 100 for both parts A and B, the assembly stage can begin.  The assembly stage take 4 minutes per each complete unit made of parts A and B.

How much idle time is in this process? (It might help to draw out the process in a chart).

Resource Idle Time Summary

• Milling Stage Idle Time (Blocking): 80 minutes per batch. Milling finishes at minute 320, but the slower assembly bottleneck takes until minute 400 to process the batch, forcing the milling machine to wait.

• Assembly Stage Idle Time (Starvation): 320 minutes at the start of each batch cycle. Assembly must wait for the entire milling stage to complete before it can begin processing.

Process Verification Checklist

• [x] Milling Batch Duration: 320 minutes (60 \text{ setup} + 100 \text{ run} + 60 \text{ setup} + 100 \text{ run})

• [x] Assembly Batch Duration: 400 minutes (100 \times 4)

• [x] Milling Idle / Blocking Time: 80 minutes (400 - 320)

• [x] Assembly Starvation Time: 320 minutes (Duration of milling batch)

Imagine a production process.  During the milling stage two separate parts are made, part A and part B.  It takes 1 minute to produce 1 unit of part A, and it takes 1 minute to produce 1 unit of part B during the milling stage.  Before switching to produce parts A and B, the milling machine undergoes a 60 minute setup time, and this happens before beginning to mill both parts A and B.  After producing the batch for both parts A and B, the assembly stage can begin.  The assembly stage take 4 minutes per each complete unit made of parts A and B.

What batch size eliminates any idle time?

Step-by-Step Verification

1. Milling Stage Time Formula:$\text{Setup A} + (\text{Run A} \times x) + \text{Setup B} + (\text{Run B} \times x)$ = 60 + x + 60 + x = 120 + 2x

2. Assembly Stage Time Formula: assembly rate * x = 4x

3. Balancing the Stages (Zero Idle Time): 120 + 2x = 4x 120 = 2x

4. x = 60 units

5. Time Check at Batch Size = 60:

   • Milling Stage: 60 +$(60 \times 1) + 60 + (60 \times 1) = \mathbf{240 \text{ minutes}}$

   • Assembly Stage: $60 \times 4 = \mathbf{240 \text{ minutes}}$

At a batch size of 60 units, both stages take exactly 240 minutes, perfectly synchronizing the process with zero idle time between them.