22-Mec-B4 Integrated Manufacturing Systems · Undated paper
Question 1 of 7: Order Interval and Total Inventory Cost
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examination
16-Mec-B4, Integrated Manufacturing Systems (the archive copy is filed
undated; the printed page header reads May 2019). Three hours, OPEN BOOK,
any non-communicating calculator permitted. Seven questions are
printed; any five constitute a complete paper and only the first five
appearing in the answer book are marked, so each question is worth 20 of the
100 marks and carries about 36 minutes. Some questions require an essay answer,
where clarity and organisation are themselves marked. All seven are worked
below, because the set as a whole is the study resource.
Reference texts for this subject.
Chase & Jacobs, Operations and Supply Chain Management, 16th ed.
— inventory models, break-even and capacity analysis.
Nahmias & Olsen, Production and Operations Analysis, 7th ed.
— the EOQ family and its variants.
Montgomery, Introduction to Statistical Quality Control, 8th ed.
— Shewhart charts, operating-characteristic curves and average run length.
Duncan, Quality Control and Industrial Statistics, 5th ed.
— the classical control-chart and inspection material these questions come from.
Groover, Automation, Production Systems, and Computer-Integrated
Manufacturing, 5th ed. — materials handling, and the data-acquisition
and conversion chapter behind Question 6.
Ebeling, An Introduction to Reliability and Maintainability
Engineering, 3rd ed. — series and active-redundant configurations.
Buffa & Sarin, Modern Production / Operations Management, 8th ed.
— plant layout and materials-handling systems.
Question 1: Order Interval and Total Inventory Cost (20 marks)
Part (a) — ordering charge quoted as a percentage of value
Given.
Quantity
Symbol
Value
Annual value ordered from this supplier
$D_V$
$260,000 per year
Ordering charge, as a fraction of value
$a$
0.01
Carrying charge, as a fraction of average inventory value
$i$
0.18 per year
Weeks in the planning year
—
52
Find. The economic order interval expressed in weeks of
supply — that is, how much material, measured as a length of time's
usage, should be bought on one purchase order.
Approach. Work the whole question in dollars rather than in
units (no unit price is given), form the annual ordering plus carrying cost as a
function of the dollar value ordered at one time, differentiate, and convert the
resulting economic order value into an interval by dividing by the annual
dollar volume.
Check: reading of the ordering charge.
Taken word for word, an order cost that is 1 % of the value of each
order makes the annual ordering cost
$\left(\dfrac{D_V}{Q_V}\right)\!\left(a\,Q_V\right)=a\,D_V$, a
constant with no derivative in $Q_V$; the total-cost curve then decreases
monotonically and the optimum collapses to $Q_V\rightarrow 0$, so the question
as literally written has no answer. The reading intended by the textbook problem
— and the one adopted here under Note 1 of the examination paper
— is that placing an order costs 1 % of the annual value,
$S=a\,D_V$. The confirmation that this is right is that the answer then comes
out exact and independent of the dollar volume, which is why the question can be
answered in weeks with no unit price given.
State the cost per order and the carrying rate.
With the reading declared above, one purchase order costs
$$S = a\,D_V = 0.01 \times 260{,}000 = \textrm{CAD } 2{,}600 \text{ per order,}$$
and holding one dollar of inventory for a year costs $i = 0.18$.
Build the annual total cost as a function of the order value.
Ordering $Q_V$ dollars of material at a time means $D_V/Q_V$ orders a
year, and instantaneous replenishment with steady use makes the average
investment $Q_V/2$, so
$$TC(Q_V) = \frac{D_V}{Q_V}\,S \;+\; i\,\frac{Q_V}{2}.$$
Differentiate and solve for the economic order value.
Setting $\mathrm{d}TC/\mathrm{d}Q_V=0$ gives the classical square-root form,
$$Q_V^{*} = \sqrt{\frac{2\,D_V\,S}{i}}
= \sqrt{\frac{2(260{,}000)(2{,}600)}{0.18}}
= \boxed{\textrm{CAD } 86{,}666.67 \text{ per order.}}$$
Convert the order value into an interval.
The interval is the order value divided by the annual rate at which value is
consumed. Substituting $S=a\,D_V$ before dividing shows that the dollar
volume cancels completely:
$$T^{*} = \frac{Q_V^{*}}{D_V}
= \frac{1}{D_V}\sqrt{\frac{2\,D_V\,(a D_V)}{i}}
= \sqrt{\frac{2a}{i}} = \sqrt{\frac{0.02}{0.18}} = \frac{1}{3}\ \text{year}.$$
In weeks of supply,
$$T^{*} = \tfrac{1}{3}\times 52 = \boxed{17.3\ \text{weeks of supply per order,}}$$
equivalently $\boxed{n = 3\ \text{orders per year.}}$
Close with the cost, and check the EOQ balance.
At the optimum the two cost components must be equal — that identity is the
cheapest possible arithmetic check on any EOQ answer:
$$\frac{D_V}{Q_V^{*}}S = 3 \times 2{,}600 = 7{,}800,\qquad
i\,\frac{Q_V^{*}}{2} = 0.18 \times 43{,}333.33 = 7{,}800,$$
so the relevant annual inventory cost is
$\boxed{TC = \textrm{CAD } 15{,}600 \text{ per year.}}$
The interval depends only on the ratio $a/i$, so the same supplier would be
ordered from every 17.3 weeks whether the annual buy were $26,000 or
$26 million. That insensitivity is the practical message of the
question: policy for a class of purchased material can be set from the two
percentage rates alone.
Part (b) — economic order quantity above a planned minimum
Given.
Quantity
Symbol
Value
Annual demand, used at a steady rate
$D$
40,000 parts per year
Procurement cost per order
$S$
$60 per order
Carrying cost on average inventory
$H$
$0.20 per unit per year
Planned minimum inventory (policy floor)
$B$
500 parts
Replenishment
—
instantaneous, prompt and reliable
Find. The total annual inventory cost under the policy
— which requires the economic order quantity first, because the policy
floor does not change it.
Inventory profile under instantaneous replenishment with a planned minimum of 500 parts. The sawtooth is the ordinary EOQ cycle lifted bodily by B, so the average stock is Q*/2 + B and the buffer adds a constant to the cost without changing the slope of the cost curve.
Approach. Recognise that a constant buffer enters the total
cost additively, so it changes the cost but not the optimising quantity; compute
the EOQ from the usual three data, then add the carrying cost of the buffer.
Write the total cost with the buffer included.
Because replenishment is instantaneous and the policy never plans below $B$, the
stock cycles between $B$ and $B+Q$, so the average is $Q/2 + B$ and
$$TC(Q) = \frac{D}{Q}S + H\left(\frac{Q}{2} + B\right)
= \underbrace{\frac{D}{Q}S + H\frac{Q}{2}}_{\text{depends on } Q} + \underbrace{H B}_{\text{constant}}.$$
Note that the buffer cannot move the optimum.
$H B$ has zero derivative in $Q$, so $\mathrm{d}TC/\mathrm{d}Q = 0$ returns the
ordinary economic order quantity. This is the point of the question: a safety
policy is priced, not optimised away.
Compute the economic order quantity.
$$Q^{*} = \sqrt{\frac{2DS}{H}} = \sqrt{\frac{2(40{,}000)(60)}{0.20}}
= \sqrt{24{,}000{,}000} = \boxed{4{,}899\ \text{parts per order.}}$$
That is $D/Q^{*} = 8.16$ orders a year, one about every 6.4 weeks.
Establish the average inventory the policy actually carries.
$$\bar{I} = \frac{Q^{*}}{2} + B = 2{,}449.5 + 500 = \boxed{2{,}949\ \text{parts.}}$$
Using $Q^{*}$ rather than $Q^{*}/2$ here, or forgetting the buffer, are the two
ways this line is usually lost.
Price the three components and add them.
Ordering, cycle carrying and buffer carrying come to
$$\frac{D}{Q^{*}}S = 489.90,\qquad H\frac{Q^{*}}{2} = 489.90,\qquad H B = 100.00,$$
the first two equal as the EOQ balance requires, so
$$TC = \sqrt{2DSH} + HB = 979.80 + 100.00
= \boxed{TC = \textrm{CAD } 1{,}079.80 \text{ per year.}}$$
The buffer is therefore costing about $100 a year, roughly
9 % of the total, to insure against a supplier whose record is already
described as remarkable. Quantifying that premium — rather than debating
the policy — is what the question is asking a manufacturing engineer to
do.
