Exercise Sheet 2
Math269 2021–22
Q1. An interior decorator buys bright pink paint at wholesale prices because some (but not all) of his customers would like it in their homes. The demand (in litres) for bright pink paint each month is normally distributed with mean D = 50 and variance σ = 5. The cost to store the paint is 50p per month per litre, the reorder cost is £20, and it takes 1 month for an order to be delivered. Determine the economic order quantity and the buffer size to ensure there is only a 1% chance of a shortage.
Q2. Consider the single-item continuous review inventory model, with TCU(y) = DK + hy.
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(a) Assume that it costs £100 to place and receive an order, demand is 230 items per week, and the holding cost is h = £1.10 per week. Calculate the economic order quantity y⋆, the daily total cost TCU(y⋆), the average stock level, and the time interval between orders.
What order size would you recommend in practice in this case, considering that the working weeks is 5 days long?
(b) Now assume that some variability is observed in the demand over time. It is established that the demand can be viewed as normally distributed with a mean of 230 items per working week, and a standard deviation of 5 items per working day. The delivery lead time is two working days.
Find the mean and standard deviation of the total demand during the lead time.
Using the single-item quasi-static continuous review model, calculate the size of the buffer stock required to ensure that the probability of running out of stock during the lead time is at most 5%.
Q3. A fast food restaurant uses an expected 7kg of salt per month, according to customers’ demands. It costs only 30p per month to store each kilo of salt, but for each kilo of salt the restaurant is short it costs £1.50 per month. The cost to order is £10, this figure being independent of how much is ordered. During the lead time, the demand has the probability density function
0, x ̸∈ [5, 10].
(i) Verify that the Hadley-Whitin method converges for this example.
(ii) Perform iterations of the Hadley-Whitin method to approximate the reorder point and economic order quantity. Stop when Ri and Ri−1 agree to one decimal place.
1, x∈[5,10], f(x) = 5
Q1. We have
2DK 2×50×20
y⋆ = h = 0.5 = 63.25,
⋆ y⋆ 63.25
t =D= 50 =1.26.
Using the single-item static continuous review model, we know that 2DK
Since the cycle length is greater than the lead time, the effective lead is just the lead time, L = 1. Hence the mean and variance during the lead time are
Hence, since z0.01 = 2.33, we have
μL =DL=50, √
σL= σ2L=σ=5.
B=σL·z0.01 =5×2.33=11.65.
To summarise in words, the economic order quantity is 63.25 litres of paint, and this should be ordered every 1.26 months. Since the lead time is 1 month, the reorder point neglecting the buffer stock is 50 litres of paint. However, taking the buffer stock into account, the reorder point is 61.65 litres of paint.
the variance will be 50 items, giving a standard deviation of
The buffer stock size therefore needs to be at least
of items can be stored, the buffer stock should be at least 12 items.
Q3. WehaveD=7,K=10,h=0.3,p=1.5,μ=7.5. (i) We compute
2D · (K + pμ)
yˆ = h = 31.49,
y ̃ = pD = 35. h
Since y ̃ > yˆ the Hadley-Within method will converge. 2
Paying attention to the units, we see that, in £ per working day, D=46, K=100, h=0.22,
so y⋆ ≈ 204.49 and TCU(y⋆) ≈ 44.99. The average stock level is half the order size, i.e. 102.25, and the optimal time interval between orders, in working days, is
t = D ≈ 4.45.
In practice it would make sense to place one order per week, of 230 items.
(b) In the two days between placing an order and receiving it, the mean demand will be 92 items, and
50 ≈ 7.07.
50·z0.05 ≈ 11.67. Since only an integer number
(ii) We compute
The Hadley-Within equations are
We proceed:
(x−R)·f(x)dx=
10R −2R+10,
0 ≤ R ≤ 5, 5 ≤ R ≤ 10, R ≥ 10.
∞ 1,0≤R≤5,
2 − R5 , 0,
5 ≤ R ≤ 10, R ≥ 10.
2D · (K + pS(R))
f(x)dx = pD. (2) R
Set R0 = 0, y1 = 2KD/h = 21.60.
hence R1 = 6.90. Since R0 ̸≈ R1, we evaluate
2D · (K + pS(R1))
hence R2 = 6.70. Since R1 ̸≈ R2, we evaluate
2D · (K + pS(R2))
f(x)dx= pD =0.62=2− 5, R
Step 2: We solve
y2 = h = 23.12. ∞ hy2 R
f(x)dx= pD =0.66=2− 5, R
y3 = h = 23.30. ∞ hy3 R
hence R3 = 6.67. Since R2 ≈ R3 (to one decimal place, as was requested) the algorithm terminates.
Hence we take 23.3kg as the economic order quantity, and 6.7kg as the reorder point.
Step 3: We solve
f(x)dx= pD =0.67=2− 5, R
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