University of Liverpool – Question 1: (0 points)
As before, this is a brief summary of the relevant points of material you should already be familiar with. If you need more details, you should revisit your previous modules.
A real function 𝑓 with the properties
Recall that
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E(𝑋)= 𝑥⋅𝑓(𝑥)d𝑥, −
Var(𝑋) = E(𝑋) − (E(𝑋)) . E(𝑎𝑋 + 𝑏) = 𝑎 E(𝑋) + 𝑏,
𝑓(𝑥) ≥ 0, 𝑓(𝑥)d𝑥=1
is called a probability density function (PDF). We use it to calculate
continuous probabilities.
If 𝑋 is a continuous random variable with probability density function 𝑓 then
P(𝑎≤𝑋≤𝑏)= 𝑓(𝑥)d𝑥.
Since the probability space is continuous, the probability of 𝑋 taking any single fixed value is 0, so we may replace either inequality (or both inequalities) with a strict inequality and the same result holds.
The expected value and variance of a continuous random variable 𝑋 with PDF 𝑓 are calculated as
Var(𝑎𝑋 + 𝑏) = 𝑎Var(𝑋) .
If 𝑋 , …, 𝑋 are independent, identically distributed (i.i.d.) variables then
E(𝑋 +⋯+𝑋)=𝑛E(𝑋), Var(𝑋 +⋯+𝑋)=𝑛Var(𝑋).
The normal distribution is a symmetric, bell-shaped distribution with PDF
1 −(−) 𝜙(𝑥) = √2𝜎𝜋 𝑒
The parameters in this definition are chosen in such a way that the required properties of a PDF are satisfied, and such that 𝜇 is the expected value (or mean) of the distribution and 𝜎 is the standard deviation (the square root of the variance).
We will use the notation
𝑋 ∼ N(𝜇, 𝜎)
to indicate that 𝑋 follows a normal distribution with mean 𝜇 and variance 𝜎. We say that 𝑍 follows the standard normal distribution if
𝑍 ∼ N(0,1). Using the results above, we can deduce that setting
provides a transformation between a general normal distribution and the standard normal distribution, so that we only have to make calculations using the standard normal distribution.
It is still a chore to perform the integration, so tables exist to help us quickly find the values we need.
A critical value 𝑧 is a value that satisfies
P(𝑍 ≥ 𝑧) = 𝛼.
Here is a short table of values where 𝛼 ∈ [0, 1] is a probability:
𝛼 𝑧 0.1 1.2816 0.05 1.6499 0.01 2.3263
Question 2: (1 point)
A continuous random variable 𝑍 has probability density function 𝑓, where, for a constant 𝑐,
𝑐= __________
E 𝑍 = __________
𝑓(𝑧)=𝑐𝑧 2≤𝑧≤4, 0 else.
V a r 𝑍 = __________
Question 3: (1 point)
Assume 𝑋 ∼ N(𝜇, 𝜎) with 𝜇 = 62 and 𝜎 = 7. Round your answer to 4 decimal places.
P(𝑋 ≥ 69) = __________
Question 4: (0 points)
We are now venturing into ‘random’ territory, but at first only for a familiar distribution: the normal distribution. And even then, we are initially only assuming randomness during the lead time (hence the ‘quasi’ in ‘quasi-static’).
Let 𝐿 denote the lead time between placing and receiving an order. In the deterministic model, the demand was known and so we knew exactly when to order, and our inventory policy could be expressed as a ‘reorder point’, i.e. a stock level that triggers a reorder, calculated by working backwards from the lead time.
For the non-deterministic problem, if we order at this point then there is a (non- zero) probability that we will run out of stock before the new order arrives. This probability will be denoted by 𝛼.
We denote by 𝐵 the buffer stock size: this is the amount of stock we always aim to keep on hand in order to influence the probability 𝛼.
The idea is to reorder stock at the average rate it is being sold at, and use the buffer stock to cope with the unevenness in the demand.
As 𝐵 gets larger, 𝛼 gets smaller. In practice, we will fix 𝛼 at some maximum acceptable level and determine 𝐵 that satisfies this requirement.
Before completing this calculation, we first choose the reorder point according to the Single-Item Static Continuous Review model, assuming the demand takes its average value.
Question 5: (0 points)
We let 𝑥 denote the number of units that are purchased by customers during the lead time 𝐿. The amount 𝑥 is a random variable. Note that it doesn’t matter
what the probability distribution of the demand is before the stock drops to the reorder point; we just reorder when it gets to this point.
We will assume that 𝜇, the mean of 𝑥, is the reorder point, and hence we will fix 𝛼 and determine 𝐵 using
P(𝑥 ≥𝐵+𝜇)≤𝛼.
That is to say, the demand within the lead time exceeds the stock level at the
reorder point combined with the buffer stock with probability at most 𝛼. We assume for this model that
𝑥 ∼ N 𝜇 , 𝜎 .
Hence this probability can be calculated in terms of the standard normal
distribution 𝑍 ∼ 𝑁(0, 1) after making the substitution 𝑍 = (𝑥 − 𝜇)/𝜎: P⎛𝑍≥ 𝐵⎞≤𝛼.
By defining 𝑧 such that P(𝑍 ≥ 𝑧) = 𝛼, we find that we need
𝐵 ≥ 𝜎 ⋅ 𝑧 .
Notice that this depends only on the standard deviation 𝜎 and the required probability 𝛼 (this mirrors the fact that the reorder point depends only the the demand during the effective lead time).
If we are instead given that the demand per unit time is normally distributed withmean𝐷andstandarddeviation𝜎,then𝜇 =𝐷⋅𝐿and𝜎 =√𝜎⋅𝐿.
Return again to example of the spice reseller, but under the assumption that demand per unit time is normally distributed.
Weknow𝐿=1,andhence𝜎 =𝜎.
Suppose 𝜎 = 3, and we require the maximum probability of running out of stock
tobe5%.Thenwehaveabufferstockofsize𝐵=𝜎 ⋅𝑧 =3×1.64=4.92units. Since 𝛼 is the maximum allowed probability, and 𝛼 decreases as 𝐵 increases, if we have to order whole units then we should take a buffer stock level of 5 units.
Question 6: (0 points)
The Single-Item Quasi-Static Continuous Review model may not be optimal. It is naive to assume that the reorder level should be the same as in the Single-Item Static Continuous Review model. In the model we are now presenting, shortages are not forbidden and we assume backlogging.
Let 𝑓(𝑥) be the probability density function of the total demand 𝑥 during the lead time, and suppose 𝐷 is the expected demand per unit time (earlier it was
constant).
[In principle it is possible to derive a relationship between 𝑓(𝑥) and 𝐷, but this is a complication which we prefer to overlook.]
Let 𝑅 denote the reorder point (the level the stock reaches that triggers a reorder), and let 𝑦 (as usual) be the amount of stock that is reordered.
The setup cost per cycle is 𝐾, and hence the setup cost per unit time is
approximately 𝐷𝐾/𝑦. That is the same formula as earlier, but it is now only an estimate.
The expected level of inventory just before an order arrives is 𝐸(𝑅 − 𝑥) and the expected level after is 𝑦 + 𝐸(𝑅 − 𝑥 ). In the absence of other information, we
assume the stock level between these two times is linear (i.e. we calculate as if the demand is constant).
Hence the average inventory level within the lead time is
(𝑦+𝐸(𝑅−𝑥))+𝐸(𝑅−𝑥) = 𝑦 +𝑅−𝐸(𝑥). 22
The final summand to compute is the expected shortage cost. Shortages occur when the demand 𝑥 during the lead time is greater than the reorder level 𝑅. So
per cycle the expected shortage is
𝑆(𝑅)= (𝑥−𝑅)⋅𝑓(𝑥)d𝑥.
Since this is computed per cycle, we scale by the cycle length 𝑦/𝐷 to get the cost per unit time.
Question 7: (0 points)
The real-life variables over which a retailer has control are the amount of stock to reorder and at what stock level to reorder it. We therefore want to find optimal values for 𝑦 and 𝑅.
Recalling that h is the holding cost and 𝑝 is the shortage cost (both per unit per unit time), we have
𝑇𝐶𝑈(𝑦, 𝑅) = 𝐷𝐾 + h𝑦 + 𝑅 − 𝐸(𝑥) 𝑦2
+ 𝑝𝐷 (𝑥−𝑅)⋅𝑓(𝑥)d𝑥. 𝑦
This function has a single minimum, and the optimal solution may be found by setting its partial derivatives equal to zero:
∂𝑇𝐶𝑈= −𝐷𝐾+h−𝑝𝐷𝑆(𝑅)=0, ∂𝑦 𝑦 2𝑦
∂𝑇𝐶𝑈=h−𝑝𝐷 𝑓(𝑥)d𝑥=0. ∂𝑅𝑦
Hence (after rearranging) the optimal solution satisfies:
𝑦⋆ =2𝐷⋅(𝐾+𝑝⋅𝑆(𝑅⋆)), h
𝑓(𝑥)d𝑥=𝑦⋆ h . ⋆ 𝑝𝐷
Question 8: (0 points)
𝑦⋆ =2𝐷⋅(𝐾+𝑝⋅𝑆(𝑅⋆)), h
𝑓(𝑥)d𝑥=𝑦⋆ h . ⋆ 𝑝𝐷
It is impossible to solve these simultaneous equations directly (at least in general), so we use the Hadley-Whitin algorithm. This is an iterative procedure that generates a sequence of pairs of values for 𝑦 and 𝑅 that (hopefully) converges to the optimal values.
First evaluate the two equations above when 𝑅⋆ = 0. This gives us two ways of calculating 𝑦. We set
𝑦^= 2𝐷⋅(𝐾+𝑝⋅𝐸(𝑥)), h
~𝑦 = 𝑝 𝐷 . h
A sufficient condition for the Hadley-Whitin algorithm to converge is that ~𝑦 ≥ ^𝑦. If it is possible, it is helpful to also compute 𝑆(𝑅) and ∫ 𝑓(𝑥) d 𝑥 as functions of
𝑅 for later use.
The algorithm goes as follows. We take the initial estimate for 𝑦⋆ to be
𝑦 = 2𝐷𝐾. h
Set𝑅 =0.Thenset𝑖=1,and:
use 𝑦 to determine 𝑅 from the first equation;
if𝑅 ≈𝑅− stop,andset𝑦⋆ ≈𝑦,𝑅⋆ ≈𝑅;
otherwise, use 𝑅 to determine 𝑦 + from the second equation, increase 𝑖 by 1, and repeat.
Question 9: (0 points)
Here we give worked examples of using the Hadley-Whitin algorithm, but note that this algorithm is rarely computed by hand; unless all functions are in some sense ‘nice’, approximations to roots must be taken at many points.
In addition, for piecewise-defined probability density functions 𝑓(𝑥) we must
always check that the current value of 𝑅 is compatible with the particular evaluation of the integral we are using: perhaps we may have to compute something again in the case that 𝑅 lies in a different range.
and suppose the probability density function of demand during the lead time is
and hence the Hadley-Whitin method will converge. We compute, for any 𝑅 ∈ [0, 30],
𝐷 = 17, 𝐾 = 3, h = 1, 𝑝 = 2,
, 𝑥∈[0,30] 𝑓(𝑥) =
0, otherwise .
2𝐷(𝐾 + 𝑝 ⋅ 𝐸(𝑥)) ≈ 33.4963, h
~𝑦 = 𝑝 𝐷 = 3 4 ,
𝑆(𝑅) = 30
1d𝑥=1−𝑅,
30 (𝑥 − 𝑅)
𝑦= 17𝑅−68𝑅+1122, 15
𝑅 = 30 − 15 𝑦 . 17
Now we start the algorithm. For the initial value of 𝑅 we take 𝑅 = 0, and for the initial value of 𝑦 we take
𝑦 = 2×17×9 ≈10.0995. 1
d 𝑥 = 60 − 𝑅 + 15. Therefore the two equations for the algorithm simplify to become
Substitute 𝑦 into the formula for 𝑅 to get 𝑅 ≈ 21.0887.
Since 𝑅 ≉ 𝑅, substitute 𝑅 into the formula for 𝑦 to get 𝑦 ≈ 13.8564. Substitute 𝑦 into the formula for 𝑅 to get 𝑅 ≈ 17.7738.
Since 𝑅 ≉ 𝑅, substitute 𝑅 into the formula for 𝑦 to get 𝑦 ≈ 16.4745. Substitute 𝑦 into the formula for 𝑅 to get 𝑅 ≈ 15.4637.
Since 𝑅 ≉ 𝑅, substitute 𝑅 into the formula for 𝑦 to get 𝑦 ≈ 18.4791.
Although we satisfied the requirement for convergence, the convergence is a too slow to perform by hand. However, using Maple (the program takes less than 2 seconds), we find that 𝑅 ≈ 𝑅 (they agree to four decimal places), and so we
take 𝑦⋆ = 29.444, 𝑅⋆ = 4.020. Question 10: (0 points)
This chapter was just an introduction to Inventory Control, and the models we introduced may be overly simplistic for real life scenarios, in the sense that there may be other factors or considerations that need to be taken in account.
Here are some models that add in more details:
Continuous Replenishment: Instead of resupply being immediate, there is
some replenishment rate 𝑟. In this model, we have 𝑇𝐶𝑈(𝑦) = 𝐷𝐾 + h𝑦 ⋅ (1 − 𝐷/𝑟),
h ⋅ (1 − 𝐷/𝑟) As 𝑟 → ∞, this tends to the EOQ for the basic model.
𝑦2 𝑦⋆= 2𝐷𝐾 .
Volume Discounts: Here we must consider the cost per unit of the reorder we purchase, 𝑐 = 𝑐(𝑦). This function may be discontinuous. We have
𝑇𝐶𝑈(𝑦) = 𝐷𝑐(𝑦) + 𝐷𝐾 + h𝑦 . 𝑦2
We can’t explicitly write down the optimal solution in this case due to the possible discontinuity of 𝑐(𝑦). We have to find the economic order quantity for
each connected component of 𝑐(𝑦), and then compare them against each other. Limited Storage: This situation can be addressed using some of the methods
we will cover in later weeks.
Multiple Items: For two items (no shortages, constant demand), the model is simply
𝑇𝐶𝑈(𝑦, 𝑦) = 𝐷𝐾 + h𝑦 𝑦 2
+𝐷𝐾 +h𝑦 𝑦 2
𝑦⋆= 2𝐷𝐾,𝑦⋆= 2𝐷𝐾. h h
However this is more interesting (and more useful) if we include extra constraints such as limited combined storage or restrictions on order quantities.
Again, some of the methods we will discuss later can be needed in this context.
Question 11: (1 point)
For the single-item quasi-static continuous review model, assume that, with our standard notation, 𝐿 = 7.
Assume that, after an order has been placed, demand per unit time, 𝑥, is normally distributed with 𝑥 ∼ N(6, 6), and that demand in each time interval is
independent of that in other time intervals.
a. If 𝑥 is the total demand during the lead time then 𝑥 ∼ N𝜇 , 𝜎 where
𝜇 = __________
𝜎 = __________
b. For the following probabilities 𝛼, find the minimum (integer) buffer stock size to ensure that the probability of running out of stock during the lead time is at most 𝛼.
As the probability 𝛼 decreases, the minimum required buffer stock increases.
minimum integer buffer stock
__________
__________
__________
decreases.
Is this what you would expect?
Question 12: (1 point)
The Hadley-Whitin gives the exact optimal solution
In a particular case it is calculated that ^𝑦 = 43.4 and ~𝑦 = 44.6. Will the Hadley-Whitin algorithm coverge?
sometimes (but not always).
We cannot tell from this information.
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