Topic 7: Inventory Model: Discount and Multi-Commodity

05115130 Supply Chain Modelling and Optimisation

Lecturer: Sam Wiwatanapataphee

275287a@curtin.edu.au

School of EECMS, Curtin University

11 Jun 2026

Outline

• Inventory Model With Discount

  • Quantity Discounts on All Units
  • Incremental Quantity Discounts

• Multicommodity Inventory Models

  • Aggregating Multiple products in a single order
  • Lot sizing with Multiple products or customers
    • Delivered independently for each product
    • Delivered jointly for all products
    • Delivered jointly for a selected subset of the products
  • Models with capacity constraints
  • Models with joint costs

1. Inventory Model with Discount

• Early 1960s:
  • Research on EOQ with quantity discounts began.
  • Crowther (1964) analysed buyer’s and seller’s perspective.
  • EOQ used as breakpoint quantity for discounts.
• A quantity discount is simply a decreased unit cost for an item when it is purchased in larger quantities.

Important

• To increase sales, many companies offer quantity discounts to their customers.

• Two types of Quantity discounts: quantity discounts on all units & incremental quantity discounts

“Now we move to the inventory model with discount.

In the early 1960s, researchers began studying EOQ models that included quantity discounts. One important contribution was by Crowther in 1964, who analyzed the issue from both the buyer’s and seller’s perspectives. In this context, EOQ was used as a breakpoint quantity for offering discounts.

A quantity discount simply means that the unit cost of an item decreases when the buyer purchases in larger quantities. In other words, the more a customer buys, the lower the price per unit may become.

This is important because many companies use quantity discounts as a strategy to increase sales and encourage customers to order more.

There are two main types of quantity discounts: discounts on all units, where the lower price applies to every unit purchased, and incremental quantity discounts, where the lower price applies only to the units above a certain quantity level.”

1.1 Quantity Discounts on All Units

• \(𝑞_1=0,\, 𝑞_2,\ldots \quad (𝑞_𝑖 < 𝑞_{𝑖+1},\, \, 𝑖=1,2,\ldots)\quad\) are known discount breaks
• \(TC|_{Q=0} = 0 \quad (𝐶_𝑖 > 𝐶_{𝑖+1},\,\, 𝑖=1,2,\ldots)\)

\[ \boxed{TC_i (Q) = C_i D + \frac{D}{Q}A + \frac{Q}{2}h_i,\;\;(i=1,\,2,\,\ldots)} \]

where \(h_𝑖=IC_i, \;(i=1,2,\ldots)\;\) and \(I\) is a % of the purchase cost \(C_i\) or an internal interest rate.

• As q increase c decreases for all units therefore slope change

• Often reordering higher quantities will be more economical than frequently ordering smaller quantities. Thus, the wholesale price of an item C depends on the ordered quantity Q.

“Now we look at quantity discounts on all units.

In an all-units discount, the unit price applies to every unit in the order once the order quantity reaches a discount threshold.

This graph shows total purchasing cost versus order quantity, Q. The blue line, C1, represents the highest unit price for smaller orders. When the order quantity reaches q2, the price drops, so the purchasing cost follows the green line, C2. At q3, another discount is reached, and the cost follows the red line, C3.

The dashed vertical lines show the breakpoints where the price changes. Notice that each lower unit cost has a flatter slope because the cost per unit is lower.

So, when evaluating order quantities, we must calculate the total cost at each feasible price level and compare them to find the minimum total cost.”

Best order quantity

• For each \(C_i\), the best order quantity can be established in the following way:

  • If the wholesale price were \(C_i\), what would be the \(Q_𝑖^*\) ?

  • \(𝐐_𝒊^* < 𝒒_𝒊\): infeasible, as we have to order at least \(𝑏_𝑖\) units to qualify for the price \(C_i\) given this price, it is best to order \(𝑏_𝑖\) units

  • \(𝐪_𝒊 < 𝐐_𝒊^* < 𝒒_{𝒊+𝟏}\) : feasible, the optimal decision is \(𝑄_𝑖^*\)

  • \(𝐐_𝒊* \geq \mathbf{q_{i+1}}\) : inefficient, as we would qualify for an even lower price with this order quantity given this price, we want to order at most \(𝐪_{𝒊+𝟏}−𝟏\) units

Quantity Discount Schedule (QDS)

Discount Number	Discount Quantity	Discount	Discount Cost
1	0 to 999	0%	$5.00
2	1000 to 1,999	4%	$4.80
3	2,000 and over	5%	$4.75

q_i are known discount break (quantity that the discount take into effect)

The major trade-off when considering quantity discounts is between the reduced item cost and the increased carrying cost.

• Buy more – get a discount – more items in the inventory

• For each discount price,
  • Calculate a \(Q^*\) value: \(\displaystyle 𝑄^∗=\sqrt{\frac{2DA}{h_i}}\).
    
  • The unit carrying cost, \(h_𝑖= I \times C_i\).

As a result, the value of \(Q^*\) will be different for each discounted price.

• For any discount level,
  • If the \(Q^*\) value is too low to qualify for the discount, adjust \(Q^*\) upward to the lowest quantity that qualifies for the discount.

Quantity Discount Schedule (QDS)

Discount Number	Discount Quantity	Discount	Discount Cost
1	0 to 999	0%	$5.00
2	1000 to 1,999	4%	$4.80
3	2,000 and over	5%	$4.75

If \(Q^*\) for discount 2 in the QDS Table turns out to be 500 units (green TC), adjust this value up to 1,000 units to get 4% discount (red TC).

\[ \boxed{TC_i(Q) = C_𝑖 D +\frac{DA}{Q} + \frac{Q}{2} h_i}, \]

\[ \quad 𝑞_𝑖 \leq 𝑄 \leq 𝑞_{𝑖+1} \]

“Now let’s look at how total cost behaves under different discount levels.

Each curve represents the total cost for a specific discount tier. The green curve is Discount 1, red is Discount 2, and blue is Discount 3. Notice that each curve has a U-shape, reflecting the tradeoff between ordering cost and holding cost.

However, each curve is only valid within its corresponding quantity range, which is why the graph is divided by vertical lines at the breakpoints—here around 1,000 and 2,000 units.

When we calculate the optimal order quantity, Q, for each discount level, it might not always fall within the valid range for that discount. For example, if Q for Discount 2 is 500 units, that’s below the minimum required quantity of 1,000 units for that discount. So, we adjust it up to the nearest breakpoint—in this case, 1,000 units.

Finally, we evaluate the total cost at each feasible quantity using this formula:

Total cost equals ordering cost plus holding cost plus purchasing cost.

We compute this for each discount level and then select the quantity that gives the lowest overall total cost.”

1.2 Key Concept

\(Q^*\) can be found by the following procedure:

1. Compute, for each \(i\), the EOQ: \(Q_𝑖^*\) using \(C_i\) and \(h_i\)
2. Determine, for each \(i\), the best admissible order quantity \(\hat{Q}_i^*\) as

\[ \hat{Q}_i =\left\{ \begin{array} bb_i & \text{if } Q_i^* < q_i\\ Q_i^* & \text{if } q_i \leq Q_i^* \leq q_{i+1} \\ q_{i+1}-1 & \text{if } Q_i^* > q_{i+1} \end{array} \right. \]

3. Obtain the optimal solution \(Q^*\) as

\[ Q^* = \hat{Q}_i^*| _{TC(\hat{Q}_i^*)}\quad \text{where} \,\, TC(\hat{Q}_i^*) = \min_{i=1,2,\ldots} \{TC(\hat{Q}_i)\} \]

Example 1. Given the following quantity discount schedule table, determine the number of discount level \(𝑚\), discount break \(𝑞_𝑖\) and the corresponding cost \(𝐶_𝑖\).

Discount Number	Discount Quantity	Per-Unit Price
1	0 to 100	$5.00
2	101 to 250	$4.50
3	251 and over	$4.00

• Let \(m\) be the number of discount possibilities.

• In the example in the Table, there are three discount levels, so \(m = 3\).

• From Table,

  • \(q_1=0,\quad q_2=101,\quad q_3=251\)

  • \(C_1=\$5,\quad C_2=\$4.50,\quad C_3=\$4.\)

Algorithm

• Step 1: Set \(j = m\). Compute \(Q_m^* = \sqrt{\frac{2DA}{I \times C_m}}\)

• Step 2: Is \(Q_m^* \geq q_m\)?

  • If yes, \(Q_m^*\) is the optimal order quantity and we are done.
  • If not, the min cost occurs at \(Q = q_m\) owing to the convexity of the cost function.
  • Compute the cost \(TC_\text{min}\) corresponding to \(Q = q_m\).
    • Let \(Q_\text{min} = q_m\) and go to Step 3.

• Step 3: Set \(j = j−1\). Compute the optimal EOQ for the \(j\)th cost curve: \(Q_j^* = \sqrt{\frac{2DA}{I \times C_j}}\)

• Step 4: Is \(Q_j^*\) in \([q_j,\,q_{j+1})\)?

  • If yes, compute \(TC(Q_j^*)\) and compare with \(TC_\text{min}\)
  • If \(TC(Q_j^*) < TC_\text{min}\), \(Q_j^*\) is the optimal order quantity.
  • Otherwise, \(Q_\text{min}\) is the optimal order quantity.

Example 2

An office supplies store sees a uniform demand rate of 10 boxes of pencils per week. Each box costs $5. If the fixed cost of placing an order is $10 and the holding cost rate is 20% per year.

• Suppose that the retailer gets an all-units discount of 5% per box of pencils if he purchases at least 110 pencil boxes in a single order.

• The deal is further sweetened if the retailer purchases at least 150 boxes in which case he gets a 10% discount. Determine the optimal order quantity using the EOQ model.

• Should the retailer change the order quantity?

\[ \begin{aligned} & A=10,\quad I=0.20,\quad C=5,\quad D =(10)(52) = 520. \\ \\ & C_1=5,\quad C_2=5(1−0.05)=4.75,\quad C_3=5(1−0.1)=4.50,\\ \\ & q_1=0,\quad q_2=110,\quad \text{and} \quad q_3=150. \end{aligned} \]

Iteration 1: Initialise \(j=3\)

• Step 1: We compute \(\displaystyle Q_3^* = \sqrt{\frac{2DA}{IC_i}} = \sqrt{\frac{2(520)(10)}{(0.2)(4.50)}}\approx 107.5\)

• Step 2: Since \(Q_3^*\) is not greater than \(q_3=150\), the min cost occuring at \(Q=150\) is

\[ TC_\text{min}\!=\!C_3D\!+\!\frac{D}{Q}A\!+\!\frac{Q}{2}(IC_3)\!=\!(4.50)(520)\!+\!\frac{520(10)}{150}\!+\!\frac{(150)(0.2)(4.50)}{2}\!\approx\!2442.17 \]

• Step 3: Now, we set \(j=2\). We compute \(\displaystyle Q_2^*=\sqrt{\frac{2(10)(520)}{(0.2)(4.75)}}\approx 104.63\)

• Step 4: Once again, \(Q_2^*\approx 104.63 \notin [110, 150]\quad\) is not feasible. The minimum feasible cost thus occurs at \(Q=q_2=110\) and is equal to

\[ \begin{aligned} TC(q_2)&=C_2d + \frac{D}{Q}A+\frac{Q}{2}(I \times C_2)\\ &=(4.75)(520)+\frac{520}{110}10+\frac{110}{2}(0.2)(4.75) \approx 2569.52 > TC_{min} \end{aligned} \]

Iteration 2:

• Step 3: we set \(j=1\) and compute \(\displaystyle Q_1^*=\sqrt{\frac{2(10)(520)}{(0.2)(5.00)}}\approx 101.98\)

• Step 4: \(Q_1^* \in [0, 110)\quad\) is feasible. The corresponding cost is equal to $2701.98, which is higher than \(TC_{min}\)

Thus, the optimal solution is to order 150 units and the corresponding average annual cost of purchasing and managing inventory is equal to $2442.17.

Observe that the algorithm stops as soon as we find a discount for which \(Q_j^*\) is feasible.

1.2 Incremental Quantity Discounts

In the incremental quantity discount, as the quantity per order increases,

• the unit purchasing cost declines incrementally on additional units purchased as opposed to on all the units purchased.

• Assumed that the units are discrete, and the units are infinitely divisible and the purchasing quantity is any real value.

• Let \(q_1 = 0,\, q_2,\, q_3,\,\ldots,\, q_j,\, q_{j+1},\ldots, q_m \quad\) be the order quantities at which the unit purchasing cost changes. The number of discount levels is \(m\).

• The unit purchasing cost is the same for all values of \(Q \in [q_j, q_{j+1})\), and we denote this cost by \(C_j\).

  • By definition, \(C_1>C_2>\ldots>C_j>C_{j+1}> \ldots >C_m\),

Example 3. Let’s say a supplier offers the following incremental discount structure:

Quantity range (units)	Price per unit
0 - 99	$10
100 - 499	$9
500+	$8

• Purchasing 50 units \(\implies 50(10) = \$500\)
• Purchasing 150 units \(\implies 100(10) + 50(9) = \$1,450\)
• Purchasing 600 units \(\implies 100(10) + 400(9) + 100(8) = \$5,400\) \[ \text{Purchasing Cost} = \begin{cases} 10Q, & 0 \le Q < 100, \\[6pt] 100(10) + (Q-100)(9), & 100 \le Q < 500, \\[6pt] 100(10) + 400(9) + (Q-500)(8), & Q \ge 500. \end{cases} \]
• \(m=3, q_1 = 0, q_2 = 100, q_3 = 500\)

import matplotlib.pyplot as plt

# Define key quantity breakpoints
q1, q2, q3, q4 = 0, 3, 6, 9

# Define piecewise cost curve (example values to match shape)
x = [0, q2, q3, q4]
y = [0, 8, 11, 12.5]  # increasing with decreasing slope

# Plot the main curve
plt.plot(x, y)

# Vertical dashed lines
plt.axvline(x=q2, linestyle='--', color='red')
plt.axvline(x=q3, linestyle='--', color='black')
plt.axvline(x=q4, linestyle='--')

# Labels for cost per unit regions
plt.text(q2/2, 4, "C₁/unit", ha='center', fontsize=14)
plt.text((q2+q3)/2, 9.5, "C₂/unit", ha='center', fontsize=14)
plt.text((q3+q4)/2, 12, "C₃/unit", ha='center', fontsize=14)

# Axis labels
plt.xlabel("Order Quantity (Q)", fontsize=14)
plt.ylabel("Total Purchasing Cost", fontsize=14)

# Ticks at q1, q2, q3, q4
plt.xticks([q1, q2, q3, q4], ['q1', 'q2', 'q3', 'q4'])

# Clean up appearance
plt.xlim(0, q4 + 1)
plt.ylim(0, max(y) + 2)

plt.show()

“Now we’re looking at total purchasing cost under an incremental quantity discount structure.

On the x-axis is the order quantity, Q, and on the y-axis is the total purchasing cost.

From q1 up to q2, every unit is purchased at the initial price, C1 per unit, so the cost increases at the steepest rate.

Once we pass q2, the pricing changes—but only for the additional units. The first q2 units are still purchased at C1, and only the extra units beyond q2 are priced at the lower rate, C2. This causes the slope to decrease, but the total cost curve remains continuous—there’s no sudden drop.

The same pattern continues beyond q3. Units up to q2 are priced at C1, units between q2 and q3 are priced at C2, and only the units beyond q3 receive the lowest price, C3. Again, the slope becomes flatter, reflecting the lower marginal cost.

The dashed vertical lines mark the breakpoints where the unit price for additional units changes.

The key takeaway is that, unlike all-units discounts, the price reduction applies only incrementally. This results in a smooth, piecewise-linear cost curve without jumps, and we must account for each pricing tier separately when calculating total purchasing cost.”

Example

The unit purchasing price is $1.00 for every unit up to 200 units. If the order quantity is between 201 and 500, the unit purchasing price drops to $0.98 but only for units numbered 201 through 500.

• The total purchasing cost of 300 units will be (200)(1)+(300−200)(0.98) = $298.

• Similarly, if at least 501 units are purchased, the unit purchasing price decreases to $0.95 for units numbered 501, 502, etc.

  • In this case,

\[ m=3,\, q_1=0,\, q_2 =201,\, q_3 =501,\, C_1 = $1,\, C_2= $0.98,\,C_3 = $0.95 \]

• Let \(Q\) be order quantity in the \(j\) th discount interval, that is, \(Q\) lies between \(q_j\), and \(q_{j+1}\) . The purchasing cost for \(Q\) in this interval is equal to

\[ \begin{aligned} C(Q)&=C_1(q_2-q_1)+C_2(q_3-q_2)+\ldots+C_{j-1}(q_j-q_{j-1})+C_j(Q-q_j)\\ &=R_j +C_j(Q-q_j). \end{aligned} \]

\(\qquad\text{where} \quad R_j= C_1(q_2-q_1)+C_2(q_3-q_2)+\ldots+C_{j-1}(q_j-q_{j-1})\)

• Now the average unit purchasing cost for \(Q\) units is equal to \(\displaystyle \frac{C(Q)}{Q}\), which is equal to

\[ \frac{C(Q)}{Q} = \frac{R_j}{Q}+C_j-C_j\frac{q_j}{Q}. \]

• The average annual purchasing cost when each order consists of \(Q\) units is equal to \(\displaystyle \frac{C(Q)}{Q}d\).

• The fixed order cost remains similar to those in the basic EOQ model, except that the average annual holding cost per unit is now equal to \(\displaystyle I \times \frac{C(Q)}{Q}\)

• The average annual cost of managing inventory is thus equal to

\[ TC(Q) = \frac{C(Q)}{Q}d+\frac{D}{Q}A+\frac{Q}{2}(I\times \frac{C(Q)}{Q}) \]

• For \(q_j \leq Q\leq q_{j+1}\) and \(\frac{C(Q)}{Q}=\frac{R_j}{Q}+C_j-C_j\frac{q_j}{Q},\quad\) we have

\[ TC(Q) = C_jD + (R_j-C_jq_j+A)\frac{D}{Q}+\frac{Q}{2}IC_j+\frac{I(R_j-C_jq_j)}{2}. \]

import numpy as np
import matplotlib.pyplot as plt

# ----- Data -----
Q = np.linspace(0.12, 10, 800)

# Curves chosen to match the shape in the figure
y_red   = 2.8 / Q + 0.42 * Q + 0.1 #1.2
y_blue  = 1.3 / Q + 0.34 * Q + 0.9
y_green = 0.9 / Q + 0.55 * Q + 0.35
y_black = 0.45 / Q + 0.30 * Q + 1.25 #0.95

q1, q2, q3 = 0.8, 2.0, 5.6

# ----- Plot -----
fig, ax = plt.subplots(figsize=(8, 5.6))

ax.plot(Q, y_red,   'r--',  lw=2.8)
ax.plot(Q, y_blue,  'b-.',  lw=2.8)
ax.plot(Q, y_green, color='#00aa55', linestyle='-.', lw=2.8)
# ax.plot(Q, y_black, 'k-',   lw=3.0)

# Vertical dashed markers
ax.vlines(q1, ymin=0, ymax=np.interp(q1, Q, y_blue),  colors='k', linestyles='--', lw=1)
ax.vlines(q2, ymin=0, ymax=np.interp(q2, Q, y_black), colors='k', linestyles='--', lw=1)
ax.vlines(q3, ymin=0, ymax=np.interp(q3, Q, y_black), colors='k', linestyles='--', lw=1)

# Axes limits and labels
ax.set_xlim(0, 11)
ax.set_ylim(0, 12)
ax.set_xlabel('Order Quantity (Q)', fontsize=14, labelpad=12)
ax.set_ylabel('Total\nAnnual\nCost', fontsize=14, rotation=0, labelpad=42, va='center')

# Custom x-ticks
ax.set_xticks([q1, q2, q3])
ax.set_xticklabels(['q₁', 'q₂', 'q₃'], fontsize=14)

# Remove y-ticks to match the figure style
ax.set_yticks([])

# Style the frame
for spine in ax.spines.values():
    spine.set_linewidth(1.2)

ax.tick_params(axis='x', length=0, pad=8)
ax.tick_params(axis='y', length=0)

plt.tight_layout()
plt.show()

• A family of curves (dashed lines), each of which is valid for a given interval.

“Now we move to total annual cost under an incremental quantity discount system.

On the x-axis is the order quantity, Q, and on the y-axis is the total annual cost, which includes ordering, holding, and purchasing costs.

Each curve represents a different price level. The red curve corresponds to the highest unit price, the blue curve to a lower price, and the green curve to the lowest price.

Notice that each curve has a U-shape. This reflects the classic EOQ tradeoff: ordering costs decrease as order quantity increases, while holding costs increase.

The vertical dashed lines—q1, q2, and q3—represent the breakpoints where the unit price changes for additional units.

Because this is an incremental discount system, the cost curves are continuous. There are no sudden drops in total cost when crossing a breakpoint, since only the additional units receive the lower price.

To find the optimal order quantity, we first identify the EOQ for each price level. Then, we check whether that EOQ falls within the valid range for that price tier.

If it does, it’s a feasible solution. If not, we adjust to the nearest breakpoint—either q1, q2, or q3.

Finally, we compare the total annual cost at all feasible points and choose the quantity that gives the lowest overall cost.”

import numpy as np
import matplotlib.pyplot as plt

# ----- Data -----
Q = np.linspace(0.12, 10, 800)

# Curves chosen to match the shape in the figure
y_red   = 2.8 / Q + 0.42 * Q + 0.1 #1.2
y_blue  = 1.3 / Q + 0.34 * Q + 0.9
y_green = 0.9 / Q + 0.55 * Q + 0.35
y_black = 0.45 / Q + 0.30 * Q + 1.25 #0.95

q1, q2, q3 = 0.8, 2.0, 5.6

# ----- Plot -----
fig, ax = plt.subplots(figsize=(8, 5.6))

ax.plot(Q, y_red,   'r--',  lw=2.8)
ax.plot(Q, y_blue,  'b-.',  lw=2.8)
ax.plot(Q, y_green, color='#00aa55', linestyle='-.', lw=2.8)
ax.plot(Q, y_black, 'k-',   lw=3.0)

# Vertical dashed markers
ax.vlines(q1, ymin=0, ymax=np.interp(q1, Q, y_blue),  colors='k', linestyles='--', lw=1)
ax.vlines(q2, ymin=0, ymax=np.interp(q2, Q, y_black), colors='k', linestyles='--', lw=1)
ax.vlines(q3, ymin=0, ymax=np.interp(q3, Q, y_black), colors='k', linestyles='--', lw=1)

# Axes limits and labels
ax.set_xlim(0, 11)
ax.set_ylim(0, 12)
ax.set_xlabel('Order Quantity (Q)', fontsize=14, labelpad=12)
ax.set_ylabel('Total\nAnnual\nCost', fontsize=14, rotation=0, labelpad=42, va='center')

# Custom x-ticks
ax.set_xticks([q1, q2, q3])
ax.set_xticklabels(['q₁', 'q₂', 'q₃'], fontsize=12)

# Remove y-ticks to match the figure style
ax.set_yticks([])

# Style the frame
for spine in ax.spines.values():
    spine.set_linewidth(1.2)

ax.tick_params(axis='x', length=0, pad=8)
ax.tick_params(axis='y', length=0)

plt.tight_layout()
plt.show()

• A solid line (valid portion) constitutes the \(TC(Q)\) function.

Here’s a focused script describing only the black line:

“Let’s focus on the black line, which represents the overall total annual cost under the incremental quantity discount structure.

This curve combines all cost components—ordering, holding, and purchasing—across the different price tiers.

At low order quantities, the total cost is high due to frequent ordering, which drives up ordering costs. As the order quantity increases, the total cost drops rapidly because ordering becomes less frequent.

The curve reaches its minimum point around the feasible optimal order quantity—this is where the balance between ordering and holding costs is most efficient.

Beyond this point, the total cost starts to rise again. This is because holding costs increase as larger quantities are kept in inventory, outweighing the savings from fewer orders and lower incremental prices.

The key takeaway is that the black line represents the actual cost we care about when making a decision, and its lowest point indicates the optimal order quantity under incremental discounts.”

Algorithm

• Step 1. Compute the order quantity that minimises \(TC_j(Q)\) for each \(j\), which is denoted by \(Q_j^*\) and is obtained by setting

\[ \begin{aligned} & \frac{d}{dQ}TC_j(Q)=-(R_j-C_jq_j+A)\frac{D}{Q^2}+\frac{IC_j}{2}=0\\ & Q_j^* = \sqrt{\frac{2(R_j-C_jq_j+A)d}{IC_j}}. \end{aligned} \]

This step gives us a total of m possible order quantities.

• Step 2. Check which one of these potential values for \(Q^∗\) is feasible, that is, \(q_j \leq Q_j^* \leq q_{j+1}.\) Disregard the ones that do not satisfy this inequality.

• Step 3. Calculate the cost \(TC_j(Q_j^*\) corresponding to each remaining \(Q_j^*\). The order quantity \(Q_j^*\) that produces the least cost is the optimal order quantity.

From the preceding example, the retailer is offered an incremental quantity discount. Let us determine the optimal order quantity.

• Step 1. \(R_1=0, R_2=C_1(q_2-q_1)=5(110-0)=550\quad\) and \(\qquad \qquad R_3=R_2+C_2(q_3-q_2) = 550+4.75(150-110)=740.\)

  • We compute

\[ Q_1^*=\sqrt{\frac{2(R_1-C_1q_1+K)d}{IC_1}}=\sqrt{\frac{2(0-0+10)(520)}{(0.2)(5)}} \approx 101.98 \]

\[ Q_2^*=\sqrt{\frac{2(R_2-C_2q_2+K)d}{IC_2}}=\sqrt{\frac{2(550-(4.75)(110))+10)(520)}{(0.2)(4.75)}} \approx 202.61\\ \]

\[ Q_3^*=\sqrt{\frac{2(R_3-C_3q_3+K)d}{IC_3}}=\sqrt{\frac{2(740-(4.5)(150))+10)(520)}{(0.2)(4.5)}} \approx 294.39\\ \]

• Step 2. We disregard \(Q_2^*\) since it does not line in [110, 150). \(Q_1*\) and \(Q_3^*\) are feasible.

• Step 3. We compute the cost corresponding to \(Q_1^*\) and \(Q_3^*\):

\[ \begin{aligned} TC(Q_1^*) & =(5)(520) + (0-0+10)\frac{520}{101.98}+\frac{(0.2)(5)(101.98)}{2}\\ & \quad +\frac{(0.2)(0-0)}{2} \approx 2701.98 \end{aligned} \]

\[ \begin{aligned} TC(Q_3^*) & =(4.5)(520) + (740-(4.5)(150))+10)\frac{520}{294.39}\\ & \quad +\frac{(0.2)(4.5)(294.39)}{2}+\frac{(0.2)(740-(4.5)(150))}{2} \\ & \approx 2611.45 \end{aligned} \]

• Thus, the optimal solution is to order 294.39 units and the corresponding average annual cost of purchasing and managing inventory is equal to $2611.45.

2. Multicommodity Inventory Models

2.1 Aggregating Multiple products in a single order

2.2 Lot sizing with Multiple products or customers

• Delivered independently for each product

• Delivered jointly for all products

• Delivered jointly for a selected subset of the products

2.1 Aggregating Multiple products in a single order

• Goal: To combining the demand for several different items into one joint order rather than ordering each product separately.

• By grouping products together:

  • Fixed ordering costs (e.g., administrative, shipping setup) are shared across items;

  • Transportation and handling become more efficient;

  • Coordination across products improves, especially when items come from the same supplier.

This approach is commonly used in inventory and SC management to reduce ordering costs, take advantage of economies of scale, and simplify logistics.

Example

Demand for the Deskpro computer at Best Buy is 1,000 units per month. Best Buy incurs a fixed order placement, transportation, and receiving cost of $4,000 each time an order is placed. Each computer costs Best Buy $500 and the retailer has a holding cost of 20%. Evaluate the number of computers that the store manager should order in each replenishment lot.

\[ \text{Order cost per lot}\,\, A=\$4000;\quad \text{Unit cost}\,\, C=\$500;\quad I = 0.20\]

• Annual demand \(D = 1000 \times 12 =12000\)

• Annual holding cost per unit \(h = I \times C = 0.2 \times 500 = 100\)

\[ Q^*=\sqrt{\frac{2DA}{IC}}=\sqrt{\frac{(2)(12000)(4000)}{(0.2)(500)}}\approx 979.8\]

• So the store manager should order about 980 computers

• Assume that the demand for each of the four models is 1,000 units per month.
  • If each product manager orders separately, he would order a lot size of 980 units. Across the four models, the total cycle inventory would thus be

\[ (980 \times 4) / 2 = 1960 \,\,\text{units} \]

• If product managers coordinate their purchasing to ensure that all four products arrive on the same truck (aggregating orders),
  • Optimal combined lot size across all four models turns out to be 1,960 units. This is equivalent to 490 units for each model.

This action significantly reduces the cycle inventory as well as cost to Best Buy.

2.2 Lot sizing with Multiple products or customers

• Goal: To determining optimal order quantities when several different items are replenished together, often from the same supplier or through a shared ordering process.

• Instead of calculating an independent order size for each product (like basic EOQ), this approach:

  • Coordinates replenishment cycles across products
  • Shares fixed ordering costs (e.g., transportation, setup, administration)
  • Reduces total system cost by exploiting economies of scale

Inputs

• \(D_i\) : Annual demand for product \(i\)
• \(A\) : Order cost incurred each time an order is placed, independent of the variety of products included in the order
• \(A_i\) : Additional order cost incurred if product \(i\) is included in the order

2.2.1 Lots are Ordered and Delivered Independently for Each Product

Example. Best Buy sells three models of computers, the Litepro, the Medpro, and the Heavypro. Annual demands for the three products are \(D_L = 12,000\) for the Litepro, \(D_M = 1,200\) units for the Medpro, and \(D_H = 120\) units for the Heavypro. Each model costs Best Buy $500. A fixed transportation cost of $4,000 is incurred each time an order is delivered. For each model ordered and delivered on the same truck, an additional fixed cost of $1,000 is incurred for receiving and storage. Best Buy incurs a holding cost of 20 percent. Evaluate the lot sizes that the Best Buy manager should order if lots for each product are ordered and delivered independently. Also evaluate the annual cost of such a policy.

• Demand, \(D_L = 12,000\)/year; \(D_M = 1 ,200\)/year; \(D_H = 120\)/year
• Common order cost, \(A= \$4,000\), Product-specific order cost, \(A_L = A_M = A_H = \$1000\)
• Holding cost, \(I= 0.20\), Unit cost, \(C_L=C_M=C_H=\$500\)
• A fixed ordering cost of \(A+A_i = \$4000 + \$1000 = \$5000\quad\) is incurred for each product delivery.

Metric	Formula	Litepro	Medpro	Heavy pro
Demand per year	\(D\)	12,000	1,200	120
Fixed cost/order	\(A\)	$5,000	$5,000	$5,000
Optimal order size	\(Q^* = \sqrt{\frac{2DA}{h}}\)	1,095	346	110
Cycle inventory	\(\frac{Q^*}{2}\)	548	173	55
Order frequency	\(\frac{D}{Q^*}\)	11.0/year	3.5/year	1.1/year
Average flow time	\(\frac{Q^*}{2D}\)	2.4 weeks	7.5 weeks	23.7 weeks
Annual ordering cost	\(\frac{D}{Q^*} \times A\)	$54,772	$17,321	$5,477
Annual holding cost	\(\frac{Q^*}{2} \times h\)	$54,772	$17,321	$5,477
Annual cost	Holding + Ordering Cost	$109,544	$34,642	$10,954

Metric	Litepro	Medpro	Heavy pro
Demand per year	12,000	1,200	120
Fixed cost/order	$5,000	$5,000	$5,000
Optimal order size	1,095	346	110
Cycle inventory	548	173	55
Order frequency	11.0/year	3.5/year	1.1/year
Average flow time	2.4 weeks	7.5 weeks	23.7 weeks
Annual holding cost	$54,772	$17,321	$5,477
Annual ordering cost	$54,772	$17,321	$5,477
Annual cost	$109,544	$34,642	$10,954

• The annual ordering and holding cost Best Buy incurs if the three models are ordered independently turns out to be $155,140.

2.2.2 Lots are Ordered and Delivered Jointly for All Three Models

• Here all three models are included each time an order is placed. In this case the combined fixed order cost per order is given by

\[ A^* = A + A_L + A_M + A_H \]

• The next step is to identify the optimal ordering frequency. Let \(n\) be the number of orders placed per year. We then have

\[ \text{Annual order cost }= nA^* \]

\[ \text{Lot size of each model }i =\frac{D_i}{n} \]

\[ \text{Annual holding cost }=\frac{D_L I C_L}{2n}+\frac{D_M I C_M}{2n}+\frac{D_H I C_H}{2n} \]

\[ \text{Total Annual cost }TC(n)=nA^*+\frac{D_L I C_L}{2n}+\frac{D_M I C_M}{2n}+\frac{D_H I C_H}{2n} \]

• Taking \(\displaystyle \frac{d}{dn}TC(n)\) and setting it equal to 0. This results in the optimal order frequency \(n^*\), where

\[ n^* = \sqrt{\frac{D_LIC_L + D_MIC_M+D_HIC_H}{2A^*}}.\]

• It can be generalised to the case where there are \(k\) items consolidated on a single order as follows:

\[ n^* = \sqrt{\frac{\sum_{i=1}^k D_i I C_i}{2A^*}}\qquad (*)\]

Example

Products Ordered and Delivered Jointly Consider the Best Buy data in previous Example. The three product managers have decided to aggregate and order all three models each time they place an order. Evaluate the optimal lot size for each model.

• Analysis: Because all three models are included in each order, the combined order cost is

\[ A^* = A + A_L + A_M + A_H = 4000 + 1000 + 1000 + 1000 = \$7000\]

• The optimal order frequency is obtained using Equation (*) and is given by

\[ n^*=\sqrt{\frac{(12000)(100)+(1200)(100)+(120)(100)}{(2)(7000)}}=9.75\]

• In this case the ordering policies and costs are as shown in following Table.

Metric	Litepro	Medpro	Heavypro
Demand per year, \(D\)	12,000	1,200	120
Order frequency per year, \(n^*\)	9.75	9.75	9.75
Optimal order size, \(D/n^*\)	1,230	123	12.3
Cycle inventory, \(\bar{q}=D/(2n^*)\)	615	61.5	6.15
Annual holding cost, \(\bar{q}\times h\)	$61,512	$6,151	$615
Average flow time (weeks), \(\bar{q}\times 52/D\)	2.67	2.67	2.67

• Annual order cost = \(9.75 \times 7,000=\$68,250\)

• Total cost = Annual order cost + Annual holding cost across three models

\[ TC = \$68,250 + (\$61,512 + \$6,151 + \$615) =\$136,528 \]

Example

W.W. Grainger sources from hundreds of suppliers and is considering the aggregation of inbound shipments to lower costs. Truckload shipping costs $500 per truck along with $100 per pickup. Average annual demand from each supplier is 10,000 units. Each unit costs $50 and Grainger incurs a holding cost of 20 percent.

• What is the optimal order frequency and order size if Grainger decides to aggregate four suppliers per truck?

• What is the optimal order size and frequency if each truck has a capacity of 2,500 units?

Inputs

• Demand per product, \(D_i= 10,000\); Holding cost (%), \(I = 0.2\)
• Unit cost per product, \(C_i = \$50\);
  
• Common order cost, \(A= \$500\) and Supplier-specific order cost, \(A_i =\$100\). Then, \(A^* = \$900\) per order

• The optimal order frequency per year is

\[ n^*=\sqrt{\frac{\sum_{i=1}^4 D_iIC_i}{2A^*}} = \sqrt{\frac{(4 \times 10000)(0.2)(50)}{(2)(900)}} \approx 14.91 \]

• The annual ordering cost per supplier is

\[ \text{Annual order cost} = 14.91 \times \frac{900}{4}=\$3,354 \]

• The quantity ordered from each supplier is

\[ = \frac{10,000}{14.91}\approx 671\,\,\text{units per order} \]

• The annual holding cost per supplier is

\[ = IC_i\frac{Q}{2}=0.2 \times 50 \times \frac{671}{2} \approx \$3,355. \]

• This policy, however, requires a total capacity per truck of \(4 \times 671 = 2,684\quad\) units.

• Given a truck capacity of 2,500 units, the order frequency must be increased to ensure that the order quantity from each supplier is \(\displaystyle \frac{2500}{4} = 625\).

• Thus, W.W. Grainger should increase the order frequency to \(\displaystyle \frac{10,000}{625} = 16\).

  • This action will increase the annual order cost per supplier to $3,600 and decrease the annual holding cost per supplier to $3,125.

2.2.3 Lots are Ordered and Delivered Jointly for a Selected Subset of the Products

• Goal: To a coordinated replenishment strategy where only certain related items (not all products) are grouped together in a single order.

Instead of ordering every product together or independently, a subset of products, typically those with similar demand patterns, suppliers, or logistics requirements, is replenished jointly.

How it works:

• A major ordering cost is incurred when placing a joint order for the selected group
• Each product in the subset may also have a minor (item-specific) cost
• Products outside the subset are ordered separately

Step 1. Identify the most frequently ordered product (\(\overline{n}\)) and order frequency of each product (\(\overline{\overline{n}}\))

• For each product \(i\) using Equation (*), evaluate the ordering frequency:

\[ \overline{n}_i = \sqrt{\frac{IC_iD_i}{2(A+A_i)}} \]

Hence, \[ \overline{n}=\max \{\overline{n}_i\}\quad \text{and}\quad \overline{\overline{n}}_i = \sqrt{\frac{IC_iD_i}{2 A_i}} \]

Step 2. Evaluate \(m_i\), the frequency of product \(i\) relative to the most frequently ordered product

• Evaluate the frequency of product \(i\) relative to the most frequently ordered product to be

\[ m_i = \lceil \overline{m}_i \rceil \qquad \text{where} \quad \overline{m}_i=\overline{n}/\overline{\overline{n}}_i \]

Step 3. Recalculate \(n\), the ordering frequency of the most frequently ordered product

\[ n = \sqrt{\frac{\sum IC_i m_i D_i}{2[A+\sum (A_i/m_i)]}}\qquad (**) \]

• Equation \((**)\) reflects the fact that product \(i\) is ordered with frequency \(m_i\) .

Step 4. Evaluate \(m_i\), an order frequency of each product

• For each product, evaluate an order frequency of

\[ n_i = \frac{n}{m_i} \]

where

\[ n = \sqrt{\frac{\sum IC_i m_i D_i}{2[A+\sum (A_i/m_i)]}}\qquad \leftarrow\text{Step 3}\]

\[ m_i = \lceil \overline{m}_i \rceil \qquad \qquad \leftarrow\text{Step 2}\]

Evaluate total cost of such an ordering policy

\[ TC = An + \sum_i A_in_i + \sum_i \frac{IC_iD_i}{2n_i} \]

• Using \(\displaystyle n_i=\frac{n}{m_i}\), this becomes

\[ TC = An + \sum_i A_i\frac{n}{m_i} + \sum_i \frac{IC_iD_im_i}{2n} \]

or equivalently,

\[ TC = n \left[A + \sum_i \frac{A_i}{m_i}\right] + \frac{1}{2n}\sum_i IC_iD_im_i \]

Example

Consider the BestBuy data. Product managers have decided to order jointly, but to be selective about which models they include in each order. Evaluate the ordering policy and costs using the procedure discussed previously.

Analysis

• Recall data
  • Annual Demand: \(D_L = 12,000,\,D_M=1,200,\;D_H=120\)
  • Ordering cost: \(A=\$4000,\, A_L=A_M=A_H=\$1000\)
  • Holding cost: \(I=0.20\)
  • Unit cost: \(C_L=C_M=C_H=\$500\)

Step 1. Obtain

\[ \begin{aligned} \overline{n}_L &= \sqrt{\frac{IC_LD_L}{2(A+A_L)}} = \sqrt{\frac{(0.2)(500)(12000)}{(2)(4000 + 1000)}} \approx 11\\ \overline{n}_M &= \sqrt{\frac{IC_MD_M}{2(A+A_M)}} = \sqrt{\frac{(0.2)(500)(1200)}{(2)(4000 + 1000)}} \approx 3.5\\ \overline{n}_H &= \sqrt{\frac{IC_HD_H}{2(A+A_H)}} = \sqrt{\frac{(0.2)(500)(120)}{(2)(4000 + 1000)}} \approx 1.1\\ \end{aligned}\]

• Clearly, Litepro is the most frequently ordered model. Thus, we set

\[ \overline{n}=\max\{\overline{n}_L,\overline{n}_M,\overline{n}_H\}=11.\]

Step 2. Determine how often Medpro and Heavypro are jointly ordered with Litepro.

• Using \(\displaystyle \overline{\overline{n}}_i = \sqrt{\frac{IC_iD_i}{2 A_i}}\), we first obtain

• \(\overline{\overline{n}}_M = \sqrt{\dfrac{IC_MD_M}{2A_M}}=\sqrt{\dfrac{(0.2)(500)(1200)}{(2)(1000)}}\approx 7.7\)

• \(\overline{\overline{n}}_H = \sqrt{\dfrac{IC_HD_H}{2A_H}}=\sqrt{\dfrac{(0.2)(500)(120)}{(2)(1000)}}\approx 2.4\)

• Next using \(\overline{m}_i = \overline{n}/\overline{\overline{n}}_i\) and \(\overline{n}=11\), we evaluate

  • \(\overline{m}_M = \overline{n}/\overline{\overline{n}}_M = 11/7.7 =1.4\)
  • \(\overline{m}_H = \overline{n}/\overline{\overline{n}}_H = 11/2.4 =4.5\)
  • Then, we have
    • \(m_M = \lceil \overline{m}_M \rceil = \lceil 1.4 \rceil = 2\)
    • \(m_H = \lceil \overline{m}_H \rceil = \lceil 4.5 \rceil = 5\)

Thus,

• \(m_M =2\) means that Medpro is included in every other order

• \(m_H=5\) means that Heavypro is included in every fifth order

• Litepro, the most frequently ordered model, is included in every order.

Ordering frequency of each model

• Apply Step 3 to recalculate the ordering frequency of the most frequently ordered model as \(n = 11.47\)

• Thus, the Litepro is ordered 11.47 times per year.

• Next we apply Step 4 to obtain an ordering frequency per year for each product:

  • \(n_L = n =11.47\)
    
  • \(n_M = n/m_M = 11.47/2 \approx 5.74\)
    
  • \(n_H = n/m_H = 11.47/5 \approx 2.29\)

Metric	Litepro	Medpro	Heavypro
Demand per year (\(D_i\))	12,000	1,200	120
Order frequency per year (\(n_i\))	11.47	5.74	2.29
Order size (\(D/n_i\))	1,046	209	52
Cycle inventory (\(\bar{q}=D_i/(2n_i)\))	523	104.5	26
Annual holding cost (\(IC_i\bar{q}\))	$52,307	$10,461	$2,615
Average flow time (weeks) (\(\bar{q}\times 52/D\))	2.27	4.53	11.35

• The annual holding cost of this policy is $65,383.00.

• As \(A_L=A_M=A_H=1000,\quad\) the annual order cost is given by

  • \(nA + n_LA_L +n_MA_M + n_HA_H\)

\[ =11.47(4000) + 1000 (11.47 +5.74 + 2.29) =\$65,380.00 \]

• The total annual cost is thus equal to $130,763.00.

• Tailored aggregation results in a cost reduction of $5,761 (about 4 %) compared to the joint ordering of all models.

2.3 Models with capacity constraints and joint costs

• When several commodities are kept in stock, their inventory policies are intertwined because of common constraints and joint costs, as we now discuss in two separate cases.

  • In the first case, a limit is placed on the total investment in inventories, or on the warehouse space.

  • In the second case, commodities share joint ordering costs.

• For the sake of simplicity, both analyses will be performed under the EOQ model hypotheses

2.3.1 Models with Capacity Constraints

• Let \(n\) be the number of commodities in stock
  • \(Q_j,\;j=1,2,\dots, n\quad\) be the amount of commodity j ordered at each replenishment.
• The inventory management problem can be formulated as follows:

\[ \begin{aligned} \text{minimise}\quad TC(Q_1,Q_2,\ldots,Q_n) & \hspace{6em}(1) \\ \text{subject to} \quad g(Q_1,\ldots,Q_n) \leq b, & \hspace{6em}(2)\\ Q_1,\ldots,Q_n \geq 0. & \hspace{6em}(3) \end{aligned} \]

• The objective function (1) is the total average cost per time unit.

  • \(TC(Q_1,Q_2,\ldots,Q_n) = \sum_{j=1}^n TC_j(Q_j),\)

  • \(TC_j(Q_j) = \frac{D_j}{Q_j}A_j+ C_jD_j +\frac{Q_j}{2}h_j,\quad j=1,\ldots,n\)

• Equation (2): \(g(Q_1,\ldots,Q_n) \leq b\) is referred to as a capacity constraint representing both a budget constraint or a warehouse constraint.

• It can usually be considered as linear,

• \(\displaystyle \sum_{j=1}^n a_j Q_j \leq b,\hspace{6em}(4)\)

• As a result, problem (1)-(3) has to be solved through iterative methods for NLP problems, such as the conjugate gradient method.

• Alternatively, the following simple heuristic can be used if the capacity constraint is linear and the interest rates are identical for all the commodities:

  • \(I_j = I, \quad j=1,\ldots,n\)

Step 1.

• Using the instantaneous resupply model, compute the EOQ order sizes

\[ Q_j = \sqrt{\frac{2 D_j A_j}{I \times C_j}} \quad j=1,\ldots,n\hspace{3em}(5) \]

• If the capacity constraint (4), \(\sum_{j=1}^n a_jQ_j \leq b\quad\), is satisfied, STOP, the optimal order size for each product \(j=1,\ldots,n\quad\) has been determined.

Step 2.

• Increase the interest rate \(I\) of a \(\delta\) quantity to be determined. Then, the order sizes become

\[ Q_j(\delta) = \sqrt{\frac{2D_jA_j}{(I + \delta) C_j}}\quad j=1,\ldots,n \hspace{3em}(6) \]

• Determine the optimal value \(\delta^*\) satisfying the relation,

\[ \sum_{j=1}^n a_jQ_j(\delta^*) = b \]

\[ \therefore \quad \delta^* = \left(\frac{1}{b} \sum_{j=1}^n a_j \sqrt{\frac{2D_jA_j}{C_j}} \right)^2 - I\hspace{3em}(7) \]

• Substituting \(\delta^*\) in Equation (6) in order to determine the order sizes

\[ \hat{Q}_j,\;\,j=1,\dots,n \]

Example

New Frontier distributes knapsacks and suitcases in most US states. Its most successful models are the Preppie knapsack and the Yuppie suitcase. The Preppie knapsack has a yearly demand of 150,000 units, a value of $30 and a yearly holding cost equal to 20% of its value. The Yuppie suitcase has a yearly demand of 100,000 units, a value of $45 and a yearly holding cost equal to 20% of its value. In both cases, placing an order costs $250. The company’s management requires that the average capital invested in inventories does not exceed $75,000. This condition can be expressed by the following constraint,

\[ 30\frac{Q_1}{2}+45\frac{Q_2}{2}\leq 75000,\]

where it is assumed, as a precaution, that the average inventory level is the sum of the average inventory levels of the two items.

• The EOQ order sizes, given by Equation (5),

\[ \begin{aligned} Q_1' &= \sqrt{\frac{(2)(250)(150000)}{(0.2)(30)}}=3535.53\quad\text{units} \\ Q_2' &= \sqrt{\frac{(2)(250)(100000)}{(0.2)(45)}}=2357.02\quad\text{units}\\ & \text{do not satisfy the budget constraint} \end{aligned} \]

• Applying the conjugated gradient method starting from the initial values \((Q_1,\, Q_2)=(1,1),\)

• The following solution is obtained after 300 iterations,

\[ \hat{Q}_1 = 2500\quad\text{units}\quad \hat{Q}_2 = 1666.66\quad \text{units} \]

whose total cost is $9,045,000.

from scipy.optimize import minimize

# Data
D1, D2 = 150000, 100000      # annual demand
v1, v2 = 30, 45              # unit value
i = 0.20                     # holding cost rate
K1 = K2 = 250                # ordering cost
budget = 75000

h1 = i * v1                  # annual holding cost per unit
h2 = i * v2

# Total annual cost
def total_cost(x):
    Q1, Q2 = x
    if Q1 <= 0 or Q2 <= 0:
        return float("inf")

    purchase_cost = D1 * v1 + D2 * v2
    ordering_cost = K1 * D1 / Q1 + K2 * D2 / Q2
    holding_cost = h1 * Q1 / 2 + h2 * Q2 / 2

    return purchase_cost + ordering_cost + holding_cost

# Constraint:
# 30*Q1/2 + 45*Q2/2 <= 75000
# scipy wants g(x) >= 0
def constraint_budget(x):
    Q1, Q2 = x
    return budget - (v1 * Q1 / 2 + v2 * Q2 / 2)

x0 = [2000, 1500]
bounds = [(1e-8, None), (1e-8, None)]
constraints = [{"type": "ineq", "fun": constraint_budget}]

result = minimize(
    total_cost,
    x0=x0,
    method="SLSQP",
    bounds=bounds,
    constraints=constraints
)

Q1_opt, Q2_opt = result.x

print(f"Optimal Q1 = {Q1_opt:.2f}")
print(f"Optimal Q2 = {Q2_opt:.2f}")
print(f"Minimum total cost = ${result.fun:,.2f}")
print(f"Budget used = ${v1 * Q1_opt / 2 + v2 * Q2_opt / 2:,.2f}")
print(f"Success = {result.success}")
print(result.message)

Optimal Q1 = 2500.61
Optimal Q2 = 1666.26
Minimum total cost = $9,045,000.00
Budget used = $75,000.00
Success = True
Optimization terminated successfully

Applying the heuristic procedure,

• We use Equation (7), \(\displaystyle \delta^* = \left(\frac{1}{b} \sum_{j=1}^n a_j \sqrt{\frac{2D_jA_j}{C_j}} \right)^2 - I\)

\[ \delta^* =\left[\frac{1}{75000}\left( \frac{30}{2}\sqrt{\frac{(2)(250)(150000)}{30}}+\frac{45}{2}\sqrt{\frac{(2)(250)(100000)}{45}} \right)\right]^2 - 0.2 \approx 0.2 \]

• Hence, \(\displaystyle Q_j(\delta) = \sqrt{\frac{2D_jA_j}{(I + \delta) C_j}}\quad j=1,\ldots,n\),

\[ \hat{Q}_1 = \sqrt{\frac{(2)(250)(150000)}{(0.2+0.2)\times 30}} \approx 2500\quad\text{units} \]

\[ \hat{Q}_2 = \sqrt{\frac{(2)(250)(100000)}{(0.2+0.2)\times 45}} \approx 1666\quad\text{units} \]

2.4 Models with Joint Costs

• For the sake of simplicity, we assume that only two commodities are kept in the inventory.

• Let \(A_1\) and \(A_2\) be the fixed costs for reordering the two commodities at different moments in time,

• \(A_{1-2}\) be the fixed cost for ordering both commodities at the same time and

\[ A_{1-2} < A_1 + A_2 \]

import matplotlib.pyplot as plt

# Figure
fig, ax = plt.subplots(figsize=(8, 5.5))

# Parameters
Q1 = 1.0
Q2 = 2.0
T1 = 1.2
T2 = 3 * T1   # big base = sum of 3 small bases

x0 = 0.7
x1 = x0 + T2

blue = "#5b93cf"
lw = 2.8

# -------------------------
# Big synchronized cycle
# -------------------------
ax.plot([x0, x0], [0, Q2], color=blue, lw=lw)          # jump up
ax.plot([x0, x1], [Q2, Q2], color=blue, lw=lw)         # top horizontal
ax.plot([x0, x1], [Q2, 0], color=blue, lw=lw)          # large decline

# second big cycle
ax.plot([x1, x1], [0, Q2], color=blue, lw=lw)
ax.plot([x1, x1 + T2], [Q2, 0], color=blue, lw=lw)

# -------------------------
# Small equal triangles
# -------------------------
small_starts = [x0 + i * T1 for i in range(6)]  # 3 in first big cycle, 3 in second

for s in small_starts:
    e = s + T1
    ax.plot([s, s], [0, Q1], color=blue, lw=lw)              # vertical jump
    ax.plot([s, e], [Q1, 0], color=blue, lw=lw)              # solid slope
    ax.plot([s + 0.10, e + 0.10], [Q1, 0],                   # dashed slope
            color=blue, lw=lw, ls=(0, (5, 4)))

# -------------------------
# Axis arrows
# -------------------------
ax.arrow(0, 0, 8.0, 0, length_includes_head=True,
         head_width=0.06, head_length=0.15, fc="black", ec="black", lw=2)
ax.arrow(0, 0, 0, 2.35, length_includes_head=True,
         head_width=0.12, head_length=0.08, fc="black", ec="black", lw=2)

# Tick marks and labels
ax.plot([-0.12, 0.25], [Q1, Q1], color="black", lw=1.3)
ax.plot([-0.12, 0.25], [Q2, Q2], color="black", lw=1.3)
ax.text(-0.22, Q1, r"$Q_1$", ha="right", va="center", fontsize=17)
ax.text(-0.22, Q2, r"$Q_2$", ha="right", va="center", fontsize=17)

# -------------------------
# T1 and T2 annotations
# -------------------------
y_t1 = 0.82
ax.annotate("", xy=(x0 + T1, y_t1), xytext=(x0, y_t1),
            arrowprops=dict(arrowstyle="->", color="gray", lw=1.8))
ax.text(x0 + T1/2, y_t1 + 0.08, r"$T_1$", ha="center", va="bottom", fontsize=18)

y_t2 = 2.08
ax.annotate("", xy=(x1, y_t2), xytext=(x0, y_t2),
            arrowprops=dict(arrowstyle="->", color="gray", lw=1.8))
ax.text((x0 + x1)/2, y_t2 + 0.02, r"$T_2 = 3T_1$", ha="center", va="bottom", fontsize=18)

# Axis label
ax.text(-0.05, 2.42, r"$q(t)$", ha="center", va="bottom", fontsize=18)

# Caption
fig.text(0.14, 0.03,
         "Fig 1. Inventory level as a function of time in the case of\nsynchronised order",
         fontsize=15, family="serif")

# Layout
ax.set_xlim(-0.35, 8.1)
ax.set_ylim(-0.15, 2.45)
ax.axis("off")

plt.show()

• Let \(T_1\) and \(T_2\) be the time lapses between consecutive replenishments of commodities 1 and 2, respectively.

“Here, the vertical axis shows the inventory level, (q(t)), and the horizontal axis represents time.

We consider two commodities. Commodity 1 is replenished every (T_1), while commodity 2 is replenished every (T_2). In this example, (T_2 = 3T_1), meaning commodity 2 is ordered once for every three replenishment cycles of commodity 1.

The sawtooth pattern shows how inventory decreases gradually as demand occurs, then jumps back up when a new order arrives. The smaller cycles correspond to commodity 1, with inventory rising to (Q_1). The larger cycle corresponds to commodity 2, with inventory rising to (Q_2).

Because the orders are synchronized, the replenishment of commodity 2 occurs at the same time as one of the replenishments of commodity 1. This coordination can help reduce ordering costs and improve inventory management.”

• Let \(T_1\) and \(T_2\) be the time lapses between consecutive replenishments of commodities 1 and 2, respectively.

\[ Q_1 = D_1T_1 \hspace{6em}(8) \]

\[ Q_2 = D_2T_2 \hspace{6em}(9) \]

• The periodicity of a joint replenishment policy is

\[ T=\max(T_1,T_2). \]

• In each period \(T\), the orders issued for the two items are

  • \(N_1 = T/T_1,\quad N_2=T/T_2\)
  • \(N_1=3,\qquad N_2=1\)

• During each period \(T\), two items are ordered simultaneously exactly once.

• Moreover, \(N_j-1\) single orders are placed for each item \(j, \quad (j=1,2)\)

• Hence, the total average cost per time unit is

\[ \begin{aligned} TC(T,N_1, N_2) &= \frac{A_{1-2}+(N_1-1)A_1 +(N_2-1)A_2}{T}\\ & \quad + C_1D_1+C_2D_2+\frac{h_1D_1T}{2N_1}+\frac{h_2D_2T}{2N_2}\qquad \hspace{6em}(10)\end{aligned} \]

• By solving the equation,

\[ \frac{\partial}{\partial T}TC(T,N_1,N_2)|_{T=T^*}=0. \]

• The value \(T^*(N_1,N_2)\) that minimises \(TC(T,N_1,N_2)\) is obtained,

\[ T^*(N_1,N_2)=\sqrt{\frac{2N_1N_2\,[\,A_{1-2}+(N_1-1)A_1+(N_2-1)A_2 \,]}{h_1D_1N_2+h_2D_2N_1}}\hspace{4em}\qquad (11) \]

as a function of \(N_1\,\) and \(N_2\)

Example

Shamrock Microelectronics Ltd is an Irish company which assembles printed circuit boards (PCBs) for a number of major companies in the appliance sector. The Y23 PCB has an annual demand of 3000 units, a value of €30 and a holding cost equal to 20% of its value. The Y24 PCB has an annual request of 5000 units, a value of €40 and a holding cost equal to 25% of its value. The cost of issuing a joint order is €300 while ordering a single item costs €250. If no joint orders are placed, the order sizes are,

\[ Q_1^* = \sqrt{\frac{(2)(250)(3000)}{0.2\times 30}} = 500\quad\text{units}\quad Q_2^* = \sqrt{\frac{(2)(250)(5000)}{0.25\times 40}} = 500\quad\text{units} \]

• From Equations (8) and (9), \(Q_1 = D_1T_1.\quad Q_2 = D_2T_2\quad\):

\[ T_1^* = \frac{500}{3000}=\frac{1}{6},\quad T_2^* = \frac{500}{5000}=\frac{1}{10}\]

• This means that Shamrock would issue

\[ \begin{aligned} \frac{1}{T_1^*}&=6\,\,\text{orders per year of Y23 PCB}\\ \frac{1}{T_2^*}&=10\,\,\text{orders per year of Y24 PCB} \end{aligned}\]

• If a separate order is placed,

\[ TC_1(Q_1^*) = \frac{250 \times 3000}{500}+30\times 3000 + \frac{0.2 \times 30 \times 500}{2}= 93000\,\,\text{euros}\]

\[ TC_2(Q_2^*) = \frac{250 \times 5000}{500}+40\times 5000 + \frac{0.25 \times 40 \times 500}{2}= 205000\,\,\text{euros}\]

\[ \text{Average annual cost } = \frac{TC_1+TC_2}{2} = 298000 \,\text{euros}\]

• We use equation (11):

\[ T^*(N_1,N_2)=\sqrt{\frac{2N_1N_2\,[\,A_{1-2}+(N_1-1)A_1+(N_2-1)A_2 \,]}{h_1D_1N_2+h_2D_2N_1}}\]

• If a joint order is placed and \(N_1=1, N_2=2,\,\,\) the periodicity of joint orders is, according to Equation (11),

\[ T^*= \sqrt{\frac{(2)(1)(2)(300+250)}{(0.2)(30)(3000)(2)+(0.25)(40)(5000)(1)}} = 0.16 \]

• Shamrock would issue \(\displaystyle \frac{1}{T^*}=6.25\,\,\)joint orders per year.

We use equation (10)

\[ \begin{aligned} TC(T,N_1, N_2) &= \frac{A_{1-2}+(N_1-1)A_1 +(N_2-1)A_2}{T}\\ & \quad + C_1D_1+C_2D_2+\frac{h_1D_1T}{2N_1}+\frac{h_2D_2T}{2N_2}\end{aligned} \]

• The annual average cost, computed through Equation (10), is equal to

\[ \begin{aligned} TC(T^*,1,2) &= \frac{300+250}{0.16}+30\times 3000+40 \times 5000 \\ \\ & \qquad + \frac{(0.2)(30)(3000)(0.16)}{2}+\frac{(0.25)(40)(5000)(0.16)}{2\times 2}\\ \\ & =296877.5 \,\text{euros}\end{aligned}\]