# Basic Inventory Models

A coffee shop needs to decide how much coffee beans to purchase and how frequently should they purchase. Factors to consider include:

• Beans could expire
• Storage cost
• Uncertainty in demand / supply. There could be a surge in customers
• Possible discounts for bulk purchase
• Contracts with bean providers (amortize fixed ordering)

• Everything is deterministic
• Demand is constant (always the same amount of customers)
• Only 1 product (1 type of coffee)
• No backorders (coffee bean provider is always available)
• Zero load time (supply instantly appears when ordered)
• No discounts (coffee is always flat rate)

d

Total annual demand (in this case pounds of coffee needed per year)

f

Fixed cost of an order (shipping & handling, etc.)

c

Unit cost of each item (price per pound of coffee)

h

Holding / Storage cost per item per year

Q Decision Variable

Quantity

Since orders are fulfilled instantly, we only order a batch of coffee beans when stock goes to 0.

We can compute the following:

\begin{aligned}
\text{Order Cost} &= f\cdot\overbrace{\frac dQ}^{\text{\# orders}} + \overbrace{cd}^\text{actual coffee cost}\\[10pt]
\text{Holding Cost} &= h\cdot\underbrace{\frac Q2}_{\small\substack{\text{avg.}\\\text{inventory size}}}
\end{aligned}

Then the total cost \text{TC} is:

\text{TC}(Q) = f\frac dQ + cd + h\frac Q2

Let the purchase quantity that minimizes total cost be Q^*:

\begin{aligned}\frac{d}{dQ}(\text{TC}) &= -\frac{fd}{Q^2} + \frac h2 = 0\\[10pt]
Q^* &= \sqrt{\frac{2fd}{h}}
\end{aligned}

The optimal total cost can be found by plugging in Q^* into \text{TC}:

\text{TC}(Q^*) = \sqrt{2fdh} + cd

\begin{aligned}
\text{TC}(Q^*) &= f\frac{d}{Q^*}+cd + h\frac Q2\\
&= \frac{fd}{\sqrt{\frac{2fd}{h}}} + cd + h\frac{\sqrt {\frac{2fd}{h}}}{2}\\
&= fd\frac{\sqrt{\frac{2fd}{h}}}{\frac{2fd}{h}} + \frac h2 \sqrt{\frac{2fd}{h}} + cd\\
&= h\sqrt{\frac{2fd}{h}} + cd\\
&= \sqrt{2fdh} + cd
\end{aligned}

We can find the optimum Q^* when we graph the total cost at the intersection of holding cost y(Q)=\frac {hQ}{2} and ordering cost y(Q) = \frac{fd}{Q}.

Since Q^* is likely not an integer, we are interested in what will happen if we pick a nearby integer.

In particular, we could find a range of Q such that \text{TC}(Q) is within some range around the optimal total cost.

\text{TC}(Q)\leqslant (1 + 0.01)\cdot \text{TC}(Q^*)

A newspaper vendor needs to decide how many newspapers to buy from the news publisher. Newspapers expire in a day. Expired newspapers can be salvaged (returned to vendor) for a lower price.

d

Discrete R.V., the uncertain demand of newspapers

P (unused)

The probability distribution of newspaper demand. d\sim P

p

Sale price of each newspaper

c

Cost of each newspaper

s

Salvage price, sale price for expired newspapers. (e.g. recycling the paper)

B Decision Variable

The number of newspapers to buy from the supplier

The 3 prices have the following inequality:

p > c > s

The total profit \text{TP} is given by:

\text{TP}(d, B) = \begin{cases}
(p-c)\cdot B & \text{if }d\geqslant B\\
(p-c)\cdot d + (s-c)\cdot (B-d) & \text{if }d < B
\end{cases}

• We either sold all the newspapers or have to salvage some of them.

The objective is to maximize the expected value of \text{TP} by picking the best B.

\max_{B}\,\Bbb E[\text{TP}]

Since d is a discrete random variable, we have a probability mass function:

\Bbb P(d=i) = p_i

For simplicity assume that i is finite, so i=0,1,2\dots, M for some M\in\N.

Then the expectation is:

\Bbb E[d] = \sum_ii\cdot \Bbb P(d=i)

Fix d=i. If i\geqslant B, then:

\text{TP}(B)=(p-c)\cdot B

If i < B, then:

\text{TP}(B)=(p-c)\cdot i + (s-c)(B-i)

Combining the 2 cases we have the expected profit:

\Bbb E[\text{TP}] =\sum^B_{i=0}p_i \big((p-c)i + (s-c)(B-i)\big) + \sum^M_{i=B+1}p_i(p-c)B

For simpler notation let f(B) be the expected profit \Bbb E[\text{TP}] when we purchase B newspapers.

Rearrange:

\begin{aligned}
f(B) ={}& \sum^B_{i=0}p_i(pi + s(B-i) - cB)\\
&+\fcolorbox{green}{none}{$\displaystyle(1-\sum^B_{i=0}p_i)$}\cdot(p-c)\cdot B && \color{green} \small\text{See E.1}\\
={}&p\cdot \left[(\sum^B_{i=0}p_i i) + (1-\sum^B_{i=0}p_i)\cdot B\right] &&\small\blue{\Bbb E[\text{Sales}]}\\
&+s\sum^B_{i=0}p_i(B-i) &&\small\blue{\Bbb E[\text{Salvage}]}\\
&-cB&&\small\blue{\Bbb E[\text{Cost}]}

\end{aligned}

\begin{aligned}
\sum^M_{i=B+1}p_i(p-c)B &= (p-c)B\sum^M_{i=B+1}p_i\\ &= (p-c)B(1-\Bbb P(i \leqslant B))\\
&= (p-c)B(1-\sum^B_{i=0}p_i)
\end{aligned}

Now we consider the finite difference of f(B+1) - f(B).

\begin{aligned}f(B+1)
={}&p\cdot \left[(\sum^{B+1}_{i=0}p_i \cdot i) + (1-\sum^{B+1}_{i=0}p_i)\cdot (B+1)\right]\\
&+s\sum^{B+1}_{i=0}p_i(B+1-i) \\
&-c(B+1)\\
f(B+1)-f(B) ={}&p\cdot(1-\sum^B_{i=0}p_i) + s\sum^B_{i=0}p_i - c\\
={}& (p-c)-(p-s)\cdot\fcolorbox{green}{none}{$\displaystyle\sum^B_{i=0}p_i$}

\end{aligned}

Notice that we got the c.d.f. of d to appear in the difference:

Q(d) = \sum^d_{i=0}p_i

Plug in Q:

f(B+1) - f(B)=(p-c)-(p-s)Q(B)

If the finite difference is positive, that means we are making more money by buying more newspapers. Otherwise we are losing money or not making more.

\begin{aligned}
(p-c)-(p-s)\cdot Q(B) &\geqslant 0\\
\frac{p-c}{p-s}&\geqslant Q(B)\\\\
\hdashline\\
(p-c)-(p-s)\cdot Q(B) &< 0\\
\frac{p-c}{p-s}&< Q(B)
\end{aligned}

Q(B)\geqslant\frac{p-c}{p-s}

Suppose the salvage price s increases and all else equal.

s\uparrow\implies p-s\darr \implies\frac{p-c}{p-s}\uarr\implies B^*\uarr

Key Example

Suppose we want to sign a contract with a lighting company to purchase their maintenance services. The amount of services to put on the contract will be fixed. But we don’t know how many services we will actually need.

• Contract too much: We get refunded for remaining services at a less price
• Contract too little: We need to pay more for the extra services on the fly

D

A discrete random variable, the uncertain number of services we actually need

c

Price of each service on the contract

s

Refund price

p

Price of extra services

q(d)

PMF of D, same as \Bbb P(D=d), in this case d = 0,1,2,\dots

Q(d)

CDF of D, same as \Bbb P(D\leqslant d)

B Decision Variable

The number of services to buy from the contracting company

The objective is to minimize the expected total cost \text {TC} by picking the optimal B.

\min_B\Bbb E[\text {TC}]

The expression for \Bbb E[{\text {TC}}] is:

\begin{aligned}
\Bbb E[{\text {TC}}] = \underbrace{cB}_{\substack{\text{Contract}\\\text{Price}}} + \underbrace{p\sum_{d\geqslant B+1}q(d)\cdot (d-B)}_\text{Expected cost of extra services}-\underbrace{s\sum_{d < B}q(d)(B-d)}_\text{Expected total refund}
\end{aligned}

Here we notice that d-B and B-d can be expressed as positive & negative parts.

\Bbb E[\text{TC}] = cB + p\sum_{d\geqslant 0}q(d)(d-B)^+ + s\sum_{d\geqslant 0}q(d)(d-B)^-

where (d-B)^+ = \max(d-B, 0) and (d-B)^- = \min(d-B, 0).

Strat. Critical Ratio Method still works because \Bbb E[\text{TC}] is dual of \Bbb E[\text{Profit}] from the newsvenfor problem.

• Minimizing \Bbb E[\text{TC}] is the same as maximizing \Bbb E[\text{Profit}].