Exercise Sheet 1
Q1. A business sells storage units at a constant rate of 7 per day. It costs £38 to place an order, and 40p per day to store each unit. It takes two days for a shipment to arrive after ordering.
(a) First assume no shortages are allowed.
(i) What are the parameters D, K and h for the Single-Item Static Continuous Review Model, in pounds per day?
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(ii) Calculate the optimal values y⋆ and t⋆.
(iii) What are the lead time and the reorder point?
(b) Now assume that shortages are allowed.
(i) Calculate the optimal values y⋆, w⋆, y⋆ − w⋆ and t⋆ when p takes the values 0.01, 0.1, 0.4, 1, 10, 100.
(ii) How does the optimal strategy change as the ratio between h and p varies? Does this fit with what you would expect to happen?
(iii) What happens as p → ∞? What happens as p → 0?
Q2. A 24-hour fruit shop buys individual apples for 15p each, and sells them for 35p each. The shop sells 50 apples a day. Their delivery has a fixed cost of £20, independently of how many apples they order, a delivery takes 10 days to arrive, and the storage space costs 1p per day for each apple. Assume that no shortages are permitted.
(i) Determine the optimal number of apples to order, and the reorder point (i.e. the level that will trigger a reorder).
(ii) If the shop may only order apples in multiples of 10, how many apples should be ordered to minimise the increased cost relative to the cost associated with the economic order quantity.
(iii) If the shop may only order apples in multiples of 10, state an interval such that if the number of apples ordered is contained in this interval, the increased cost relative to the cost associated with the economic order quantity is at most 5%.
(iv) The shop implemented the inventory policy derived in part (ii) and, after a week, realised that the policy was not suitable. Why is this model not suitable for this problem?
Q3. A large frozen food shop sells popular frozen ready meals at a rate of 72 per day. Each delivery costs £64. It costs 16p per day to hold each ready meal, but only 9p per day for each ready meal that the shop is short. Assume that shortages are backlogged.
(i) State the economic order quantity, the corresponding optimal maximal shortage, and the cycle length.
(ii) Are any of our assumptions unrealistic?
Q4. Using the EOQ model, assume that the demand in a particular situation has been estimated as D′ = 100. Find the range of integer values D such that if D is the real demand then modelling using the estimate D′ increases costs by at most 10%. Is it safer to under-estimate or over-estimate the demand?
Q1. (a) (i) (ii)
(iii) (b) (i)
WehaveD=7,K=38,h=0.4. y⋆ = 2×7×38 ≈ 36.469
t⋆ = 2×38 ≈ 5.210
The lead time is two days. Since 14 units will be sold within two days, this is the reorder point.
p y⋆ w⋆ y⋆−w⋆ t⋆ 0.01 233.517 228.821 5.700 33.360
0.1 81.548 65.238 0.4 51.575 25.788 1 43.151 12.329 10 37.191 1.430
100 36.542 0.146
16.310 25.788 30.822 35.761 36.396
11.650 7.368 6.164 5.313 5.220
When the shortage cost is very low relative to the holding cost, the preferred strategy is to build up a big backlog of orders, then get a lot of stock delivered that can quickly be passed on to the customers. On the other hand, if shortages are much more expensive than holding stock, the preferred strategy is to order more frequently to avoid running out of stock too quickly. When p = h we find that the two sides are balanced.
Hint: How long will a customer wait from the point they order an item to when they receive it?
that the purchase and sale cost of the apples is irrelevant as without quantity discounts this (i) We have,
Q2. Note first
does not affect the inventory cost.
2DK 2×50×20 h = 0.01
⋆ y⋆ 447.2 t=D= 50 =8.9.
Since the lead time 10 is larger than the cycle, we should reorder during the previous cycle. I.e. the effective lead time is L = 10 − 8.9 = 1.1. Hence the reorder point occurs when the inventory level is LD = 1.1 × 50 = 55.
(ii) We use the formula,
to compute
TCU(y) 1 y y⋆
TCU(y⋆)=2 y⋆ + y (1)
TCU(440) = 1.00013, TCU(y⋆)
TCU(450) = 1.00001. TCU(y⋆)
So it makes very little difference, but an order of 450 apples makes for a marginally cheaper inventory cost.
(iii) We set Equation 1 to 1.05 and solve, to get
y = 326.4, 612.8.
Hence, remembering that we may only order multiples of 10, we should order an amount in the interval [330, 610].
(iv) Hint: What is the longest time between when an apple is delivered and when it is sold? (i) These are,
2DKp+h 2×72×640.09+0.16
y⋆ = h p = 0.16 0.09 = 400,
2DK h 2×72×64 0.16
w⋆ = p p+h = 0.09 0.09+0.16 =256,
t = D = 72 = 5.6.
(ii) Hint: How far in advance do you order a ready meal? We set
1 D 100
2 100 + D = 1.1.
This gives us the quadratic equation
We know that D ≥ 0. Write d =
D, so the equation becomes d + 10 = 2.2.
which has solutions d = 11 ±
21. From this we find
D ≈ 242.8 and D ≈ 41.2.
d2 +100−22d=0,
Therefore the integer values of D that mean costs are increased by at most 10% are D ∈ [42, 242].
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.
(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
Exercise Sheet 3
Q1. Alice is planning an active city break and is looking at discounted passes to attractions. She has decided that she wants to visit 3 ‘gold’ attractions and 4 ‘silver’ attractions, and to take 3 cruises. An ‘ultra’ pass grants access to 2 gold attractions, 2 silver attractions and 1 cruise, and costs e50; a ‘mega’ pass grants access to 1 gold attraction, 4 silver attractions and 2 cruises, and costs e40. This is summarised in the following table.
Gold Silver Cruise Cost Ultra 2 2 1 50 Mega 1 4 2 40
Required 3 4 3 Formulate (but do not solve) a linear program to minimise Alice’s cost.
Q2. Alice’s tourist agency buys its tickets from the management company of the attractions. By taking into account Alice’s demands, formulate (but do not solve) a linear program (in terms of the price per attraction) that maximises the price at which the management company may sell Alice’s tickets to the agency.
Q3. Solve the following linear programs graphically, indicating any binding or redundant constraints. (a) maxz=x1+3×2 subjectto
(b) maxz=x2+2×2 subjectto
(c) min w = y2 subject to
4×1 + 5×2 0, 8×1 3×2 56,
x1, x2 0.
x2 12, x1 + x2 19, 9×1 + x2 99,
x1, x2 0.
y1 + 2y2 15, y1 + 4y2 15,
y1 + y2 6, 2y1 + 11y2 55.
Q4. Solve the following linear programs using the primal simplex algorithm in tableaux form. (a) Maximise z = 3×1 + 4×2 subject to
x1 + 2×2 4, 2×1 x2 5, 3×1 5, x1, x2 0.
x1 + x2 + x3 4, 3×1 + x3 10,
x1, x2, x3 0. Max z = 2×1 + cx2
x1 + x2 d, x1 + kx2 5, x1 5, x1, x2 0.
(b) Maximise z = x1 2×2 + x3 subject to
Q5. Consider the following linear program:
subject to
Sketch the feasible region for c = 3, d = 1, k = 2, and answer the following questions using your sketch.
(a) For the given values of c, d, k, what are the optimal values of x1, x2, z?
(b) Keeping c = 3, k = 2 and increasing d, what is the smallest value of d such that x1 +x2 d becomes redundant?
(c) Keeping c = 3,d = 1 and decreasing k, what is the largest value of k such that x1 +ky 5 becomes redundant?
(d) Keeping d = 1, k = 2, for what range of c is the optimal solution (x1 , x2 ) unchanged?
Q1. Let y1 be the number of ‘ultra’ passes bought, and y2 be the number of ‘mega’ passes. The required program is
subject to
min w = 50y1 + 40y2
2y1 + y2 3, 2y1 + 4y2 4, y1 + 2y2 3, y1, y2 0.
Q2. Let x1, x2, x3 be the respective prices for Gold, Silver and Cruise attractions. The required program is max z = 3×1 + 4×2 + 3×3,
subject to
2×1 +2×2 +x3 50, x1 +4×2 +2×2 40,
x1, x2, x3 0.
The inequalities come from the fact that agency will not buy the product at a higher price than it sells
Q3. (a) The feasible region is as follows, with a z = constant line shown in red.
7 6 5 4 3 2 1
Hence the linear program is optimised at (x1,x2) = (10,8), with z = 34. We see that both of the
constraints
4×1 + 5×2 0, 8×1 3×2 56
are binding, and the redundant constraint is x1 0. 3
1 2 3 4 5 6 7 8 9 10
(b) This time, the feasible region with a z = constant line in red are as shown. x2
9 8 7 6 5 4 3 2 1
x1 1 2 3 4 5 6 7 8 9 10 11
According to the gradient of the z = constant line, the optimal solution is (x1,x2) = (7,12) with z = 31. The constraints
x2 12, x1 + x2 19
are binding. There are no redundant constraints (even though the constraint 9×1 + x2 99 is not binding, it still defines one of the boundary components of the feasible region so it is not redundant).
(c) Finally we have
6 5 4 3 2 1
1 2 3 4 5 6 7 8 9 10 11 12 13
Since we are minimising, the optimal solution is (y1 , y2 ) = (9, 3) with w = 3. The constraints y1 + 2y2 15,
y1 + y2 6, are binding. The constraint 2y1 + 11y2 55 is redundant.
On each iteration we mark the pivot in grey and proceed. On the first iteration, after choosing our pivoting column (with the most negative entry in the z row), the ratio test (dividing the value in the right-column and by the value in the pivoting column and seeing which is minimal) has one one legal possibility since only one entry in this column is positive.
x1 x2 x3 x4 x5
z 3 4 0 0 0 0 x3 1 2 1 0 0 4 x4 2 1 0 1 0 5 x5 3 0 0 0 1 5
The variable x2 becomes basic, the variable x3 becomes non-basic, so the first row operation is R2 := 12 R3 to force the pivot to equal 1. All other row operations are of the form Rj := Rj ↵iR3, where ↵i is chosen in such a way that all other entries in the pivoting row becomes 0 so that the new tableau is in canonical form. This time we have something to check with the ratio test: since 5/3 < 7/(5/2) < 2/(1/2), we get our pivot.
x1 x2 x3 x4 x5
z 1 0 2 0 0 8 x2 1/2 1 1/2 0 0 2 x4 5/2 0 1/2 1 0 7 x5 3 0 0 0 1 5
Then after similar operations to put the next tableau into canonical form we get
x1 x2 x3 x4 x5
z 0 0 2 0 1/3 29/3
x2 0 1 1/2 0 1/6 7/6 x4 0 0 1/2 1 5/6 17/6 x1 1 0 0 0 1/3 5/3
Since there are no more negatives in the z row, the optimal solution is (x1,x2) = (5/3,7/6) with z = 29/3.
Details are omitted from these calculations, but the algorithm is almost identical to the previous solution. However, we have more than one choice for the first step. Did you try to same as below? If so, try the other one. Do you get the same answer?
x1 x2 x3 x4 x5
z 1 2 1 0 0 0 x4 1 1 1 1 0 4 x5 3 0 1 0 1 10
x1 x2 x3 x4 x5 z030104 x3 1 1 1 1 0 4 x5 2 1 0 1 1 6
the given constants, the feasible region follows.
Hence we have compute that z = 4 is optimal, and this occurs when (x1, x2, x3) = (0, 0, 4). Does it occurs for any other value? Notice that x1 is a nonbasic variable, but its entry in the z row is 0. This means that increasing x1 shouldn’t change the answer. Try to make x1 a basic variable (replacing x3 or x5) and see what happens.
The optimal values are (x1, x2) = (3, 4), z = 6.
The line x1 + 2x2 = 5 (bounding a constraint) meets the axis at (0, 5/2). Hence if d increases to the point where x1 + x2 = d meets the axis at the same point, the corresponding constraint becomes redundant. That is, d = 5/2.
The lines x1 + x2 = 1 and x1 = 5 (bounding corresponding constraints) intersect at the point (x1,x2) = (5,6). If the value of k is reduced to the point where x1 +kx2 = 5 meets this intersection, then the corresponding constraint becomes redundant. That is, when k = 5/3.
As c decreases past the point that z =constant is parallel to x1 + x2 = 1, the optimal solution changes. This happens when c = 2. As c increase past the point that z =constant is parallel to x1 + 2x2 = 5, the optimal solution changes. This is when c = 4. Hence, for c 2 [2, 4], (x1 , x2 ) = (3, 4) is optimal. Note that z changes whenever c changes.
Exercise Sheet 4
Q1. Solve the following linear programs using the Big M Method (and the primal simplex method with tableaux). In both cases you will need to multiply something by 1 before proceeding.
(i) Maximise subject to
(ii) Minimise subject to
Q2. The linear program subject to
is not feasible. Show this
(i) graphically,
(ii) using the Big M Method.
z = x1 + x2 + x3,
x1 +x2 +x3 10, x1 + x2 5,
x1, x2, x3 0. z = x1 + x2
x1 + 2x2 3, 3x1 x2 4, x1, x2 0.
Max z = 3x1 2x2,
3x1 + 7x2 49, 6x1 + x2 20, 2x1 + x2 13,
x1, x2 0
Q3. A craft shop sells beaded necklaces. For 1.50, they sell a necklace consisting of 10 blue beads, 10 red beads and 10 yellow beads. For 2.00, they sell a necklace consisting of 25 blue beads and 5 red beads. On a particular day, they take delivery of 500 blue beads, 150 red beads and 50 yellow beads.
(i) How many of each necklace should the craft shop make in order to maximise its income (assuming there is su cient demand for all of its products)?
(ii) The shop intends to increase its order of one colour by a small amount. Which colour should the shop order more of?
(iii) Use an appropriate shadow price to estimate the increase in income for 100 more yellow beads. Solve the linear program assuming 100 more yellow beads. Why do you get a di↵erent answer than your estimate?
(iv) Give an interpretation of the reduced cost of x1 and x2.
Q4. Consider the linear program
subject to
Max z = x1 x2
x1 + x2 15, 2x1 + 3x2 17,
x1, x2 0.
(i) Find the optimal solution using the primal simplex method with tableaux.
(ii) Hence determine the optimal solution when the objective function is replaced by z1 = (1 )x1 x2, for all real values of .
(iii) Dothesameforz2 =x1+( 1 )x2.
This becomes subject to
That is, subject to
The tableaux follow.
Max z = x1 + x2 + x3,
x1 +x2 +x3 10, x1 x2 5, x1, x2, x3 0.
Max z = x1 + x2 + x3 Mx6,
x1 +x2 +x3 +x4 10, x1 x2 x5 +x6 5, x1,...,x6 0.
x1 x2 x3 x4 x5 x6
z 1 1 1 0 0 M 0 x4 1 1 1 1 0 0 10
x6 1 1 0 0 1 1 5
Since this is not in canonical form, we rewrite by setting Rz := Rz M Rx6 .
x1 x2 x3 x4 x5 x6
z M+1 M 1 1 0 M 0 5M
x4 1 1 1 1 0 0 10 x6 1 1 0 0 1 1 5
x1 x2 x3 x4 x5 x6
z 0 0 1 0 1 M 1 5
x4 0 2 1 1 1 1 5 x1 1 1 0 0 1 1 5
x1 x2 x3 x4 x5 x6 z02012M 20 x3 0 2 1 1 1 1 5 x1 1 1 0 0 1 1 5
Hence the optimal solution is z = 0 which occurs when (x1, x2, x3) = (5, 0, 5).
(ii) Writing z0 = z, this becomes: subject to
So we get:
Max z0 = x1 x2
x1 + 2x2 3, 3x1 x2 4, x1, x2 0.
Max z0 = x1 x2 Mx4 Mx6 3
subject to
And in the tableaux:
x1 +2x2 x3 +x4 =3, 3x1 x2 x5 +x6 =4, x1,...,x6 0.
x1 x2 x3 x4 x5 x6 z110M0M0 x4 1 2 1 1 0 0 3 x6 3 1 0 0 1 1 4
Butthisisnotincanonicalform,sowesetRz =Rz MRx4 MRx6.
x1 x2 x3 x4 x5 x6
z 4M+1 M+1 M 0 M 0 7M
x4 1 2 11003 x6 3 1 0 0 1 1 4
x1 x2 x3x4 x5 x6
z 0 7M/3+4/3 M 0 M/3+1/3 4M/3 1/3 5M/3 4/3
x4 0 7/3 1 1 1/3 1/3 x1 1 1/3 0 0 1/3 1/3
ary of the ‘ ’ constraint is drawn in red. x2
x1x2x3 x4 x5 x6
0 0 4/7 M 4/7 1/7 M 1/7
The following graph indicates the feasible region bounded by the two ‘’ constraints. The bound-
x2 0 1 3/7 3/7 1/7 1/7 5/7 x1 1 0 1/7 1/7 2/7 2/7 11/7
Since the point satisfying the ‘ ’ constraint lie above and to the right of the red line, the feasible region is empty and
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