Online Homework System
Assignment Worksheet
2/4/22 – 11:23:32 AM GMT
Section #: ____________________________ Assignment:Week 1
Copyright By PowCoder代写 加微信 powcoder
Name: ____________________________ Class #: ____________________________ Instructor:
Question 1: (0 points)
Consider a car retailer who sells just two models: a very popular model, and a slightly less popular model. There is a cost associated with restocking the cars, and there may be a volume discount for buying more at once, which means the retailer is encouraged to maximise the number of cars they buy in one order.
However, there is a cost associated with storing the cars, which encourages the retailer to minimise the number of cars they buy in one order. There may also be constraints they need to work within, such as how much space they have available for storing the cars.
How does the retailer decide when and how many cars to order to minimise expenditure, maximise profit, and guarantee (or at least minimise the probability) that a customer is never disappointed by a car being out of stock?
The retailer needs an Inventory Control Policy that takes into account the demand (deterministic or random) for the products.
The car manufacturer, knowing how many cars the retailer will (or is likely to) buy, will have their own inventory control policy for ordering the components for the cars. Similarly, the component manufacturers will have an inventory control policy for purchasing the raw materials and the raw materials supplier will have an inventory control policy.
In this module we will only take stocks of one product into consideration. We will start by considering deterministic demand scenarios before looking at probabilistic (random) demand.
Question 2: (0 points)
The total cost of inventory can be calculated as
Purchasing
Inventory = Cost + Cost ()()()
Purchasing Cost: The price per unit of a good, as a function of the size of an order. This need not be linear or even continuous as it is common to receive discounts for larger orders. In all the cases we will consider, we assume the purchasing cost is constant. With this assumption, we can ignore the purchasing cost, as it will not alter the optimal solution.
Setup Cost: The fixed cost of placing an order. This may include postage or delivery charges and staffing costs to place and receive the order.
Holding Cost: The cost of holding stock as a function of the amount of stock being held. This should include staffing and handling costs, and may include fixed costs such as warehouse rental. It need not be linear, and may not be continuous, e.g. if holding stock over a certain quantity requires opening a new warehouse. The holding cost also includes:
Salvage Cost: The amount any unsold stock is worth to the company, e.g. if it is sold at a discounted rate due to low demand.
Shortage Cost: This reflects the loss to the company as a result of not being able to satisfy demand. If revenue is not considered by the model, loss of revenue is included in this cost. It also includes estimated future losses, e.g. due to decreased customer loyalty. There are two scenarios:
Backlogging: Customer demand is not lost, but is satisfied on the next shipment. Losses here are mainly due to decreased goodwill.
No Backlogging: Customer demand is either satisfied by a priority shipment (at increased cost), or is lost.
Notice that we do not typically include the sale price of the good; regardless of the sale price, to maximise profit we should minimise the inventory cost.
(Holding ) (Shortage ) + Cost + Cost .
Question 3: (0 points)
The goal of the inventory control policy is to determine
how much to order; when to order
in order to minimise the total inventory cost.
Various features of the retail context can affect what the optimal policy is.
Types of demand:
Deterministic and Constant: The demand does not change over time.
Deterministic and Dynamic: The demand changes, but changes predictably (e.g. with seasons). Probabilistic: The demand is random but follows some known or estimated probability distribution.
Types of review:
Continuous review: The reorder is triggered whenever a stock falls to a specified level.
Periodic review: The stock level is checked at discrete intervals, and reordered when the stock falls to or below a prescribed level, even if this may introduce shortages.
Lead Time: The delay between ordering and receiving a shipment. It is often assumed to be fixed, so a company can calculate when to order so that stock arrives just in time.
Question 4: (0 points)
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.
Typically in OR the aim is to optimise a particular function. That is, we are often looking for a global maximum or minimum. Generally the search for a global maximum or minimum has two steps:
1. create a list of possible locations for the extremum;
2. compare the value of the function at each of the candidate points.
Essentially the first step is a process of identifying as many points as possible that we know are not an extremum, using general theory.
Consider a function f : [a, b] → R. We know that, for a point c [a, b], if c is an extremum and the two-sided derivative of f exists at c then f′(c) = 0.
We can use this to rule out points with a non-zero derivative. Note that we can never rule out the endpoints a and b in this way. In addition to a list of points where no two-sided derivative exists, we now have a (hopefully short) list of points c where f ′ (c) = 0.
We can look at the second derivative at these points to help us identify the behaviour of the function there. If f ′ (c) = 0 and f ′′ (c) > 0 then c is a minimum.
If f ′ (c) = 0 and f ′′ (c) < 0 then c is a maximum.
If f ′ (c) = f ′′ (c) = 0 then we cannot immediately draw a conclusion; the point c might be a maximum, a minimum or a point of inflection. To establish the behaviour of the function at this point we need to investigate further, for example by checking the value of f at nearby points either side of c.
Find examples of functions f : [a, b] → R where the global maximum occurs at
the lower endpoint a;
at the upper endpoint b;
at an interior point of the interval.
Find examples of functions f : [a, b] → R with a point c a maximum;
a minimum;
(a, b) where f ′ (c) = f ′′ (c) = 0 and c is
an inflection point.
How many of these features can you make happen in the same function?
Question 5: (1 point)
Find the coordinates of the stationary point of the function y = x3 + 18 x2 + 108 x + 8, giving your answers to 3 decimal places if needed.
Use brackets and a comma to enter the point, not a vector.
The stationary point is at (x, y) = __________
The point is a:
(a) Maximum
(b) Minimum
(c) Point of Inflexion
Question 6: (1 point)
ex/4 Considerthefunctiony= −x2−9.
This function has two stationary points.
i. Give the coordinates (in exact form) of a stationary point with negative x coordinate. (x, y) =
__________
Use brackets and a comma to enter your answer, not a vector.
ii. Classify this stationary point. (a) Maximum
(b) Minimum
(c) Point of Inflexion
i. Give the coordinates (in exact form) of a stationary point with positive x coordinate. (x, y) =
__________
Use brackets and a comma to enter your answer, not a vector.
ii. Classify this stationary point. (a) Maximum
(b) Minimum
(c) Point of Inflexion
Question 7: (0 points)
We first introduce the most basic model, called the Single-Item Static Continuous Review Model, and will refine or extend it
later. This model is also called the EOQ Model, or Economic Order Quantity Model. This model is for a single item. Assumptions of this model:
1. Items are sold from inventory at a known constant rate D, i.e. the demand is static.
2. Inventory is immediately replenished when it reaches zero by a quantity y (i.e. we assume a constant or zero lead time with
continuous review). The optimal value y of y is known as the Economic Order Quantity. 3. Shortages are not permitted.
The rest of the parameters are as follows.
We assume no volume discounts and so do not include the purchase cost in our model. The profit/loss from each unit is independent from the inventory cost.
The setup cost of each batch of y units will be denoted by K.
The holding cost is h per unit stock per unit time.
Hence the inventory level as a function of time follows this graph.
Question 8: (0 points)
Let's calculate all associated costs for this model per order cycle. The setup cost is K, and there are no shortages permitted, so we need only calculate the holding cost.
The average inventory level per cycle is y/2 since demand is linear, hence the cost per unit time is hy/2. Multiplied by the length of hy2
the cycle, the holding cost per cycle is 2D . Hence the total cost per cycle is
and the total cost per unit time is
To find the minimum value of this function, we need to look for static points (specifically a minimum point). We compute the
derivatives
Solving dTCU = 0 gives a single candidate point (with y > 0) for where the minimum total cost occurs: dy
Since d2TCU > 0, this point is indeed a minimum. dy2
The corresponding optimal cycle length is
y =√2DK. h
t =y =√2K. D Dh
hy2 K + 2D ,
K + hy2 TCU(y) = 2D
= DK + hy.
dTCU= −DK+h dy y2 2
d2TCU = 2DK. dy2 y3
These formulas reflect the real life scenario well:
if the setup cost K increases, we order more items less frequently;
if the storage cost h increases, we order fewer items more frequently; if the demand D increases, we order more items more frequently.
Question 9: (0 points)
A spice reseller sells dried basil in 1kg jars, and has a regular customer base to whom the reseller predictably sells 30 jars per day. Postage is fixed at £5, regardless of the quantity ordered. Taking into account spoilage, past experience indicates that it costs 6p per jar per day to store the jars.
In terms of the variables previously defined,
D = 30, K = 5,
y =√2×30×5≈70.71, 0.06
t =yD≈2.36.
So, to minimise inventory cost, the reseller should resupply their stock by 70.71kg every 2.36 days. Suppose the lead time L = 1, so that orders must be placed 24 hours before delivery. Then they must make the reorder once the stock drops to L D = 30 units, as this is the amount they will sell during the lead time.
Notice in the above example that the lead time L is less than the cycle length t . If this were not the case, then they would have to order several cycles in advance.
The effective lead time is the time between reaching the reorder point and replenishment; that is, we take the lead time and subtract sufficiently many multiples of the cycle length to get into the correct range (i.e. below the length of one cycle).
Question 10: (0 points)
In practice, ordering 70.71kg of dried basil every 2.36 days is not going to be a realistic solution. How should this policy be modified to fit with the constraints of real life? What would happen if we simply rounded either of these figures up or down? How much impact will those modifications have on the total inventory cost?
Sensitivity analysis is a process of examining how the output of a mathematical model (in this case the total inventory cost) is affected by changes in the inputs. That is, it is an analysis of how sensitive the model is to the variation in the parameters.
Let’s look at how sensitive TCU is to changes in y. We evaluate the ratio TCU(y) = (DK/y) + (hy/2)
DK/√h + h√h /2
√DK y√h = y√2h + 2√2DK
= 2y + 2√2DK/h
1(y y) =2+y, y
which measures the relative difference in cost when using a value y instead of the optimal value y . Suppose, in some context, we have K = 100, D = 20, h = 4. This gives
y ≈ 316.2.
What if we may only order multiples of 25? Do we take y = 300 or y = 325? Well,
TCU(300) 1 316.2 300 TCU(y )=2(300 +316.2)
( 2DK)(2DK )
= 1.0014, TCU(325) 1 316.2 325
TCU(y )=2(325 +316.2) = 1.0003,
which shows that y = 325 is slightly better than y = 300, but both are very, very small relative differences. What if the company is satisfied with a relative difference of up to 5%? We solve
TCU(y) TCU(y )
12 ( y + y ) yy
2 ( y + 316.2 )
0.0016y2 − 1.05y + 158.1 230.78, 433.24.
So for any y
Why would taking y = 230 or y = 434 not be suitable?
[231, 433], the relative increased cost will be within 5% of the optimal cost.
Question 11: (0 points)
Now suppose that the demand has been incorrectly calculated as D ′ instead of D. We use y incorrect value D ′ and make a substitution:
for the true value D, and y for the
1(√ ′ √′ )
For example, if we suppose that we have overestimated demand by 100%, i.e. D ′ = 2D, we find
TCU(y) = 12 (√D/2D + √2D/D) TCU(y )
= 12 ( √ 1 / 2 + √ 2 )
so this overestimation has resulted in a 6% relative increased cost.
See the exercise sheet for an example of working the other direction with this equation. Exercise:
What policy would you recommend to the spice reseller? Why?
2DK 2D′K y =√h ,y=√h
TCU(y) 1 (√2DK/h ′
TCU(y ) = 2 √2D K/h + √2DK/h =2 D/D+ D/D.
Question 12: (0 points)
If the storage cost is high relative to the shortage cost, then it may be sensible to allow planned shortages in our model. These will be ‘backlogged’ shortages.
Assumptions of this model:
1. Items are sold from inventory at a known constant rate D.
2. Inventory is immediately replenished when it reaches a fixed constant value by a quantity y.
3. Shortages are permitted, where p is the cost of the shortage per unit stock and per unit time, and w is the size of the shortage that is permitted. Since this model includes backlogging, we may therefore think of having −w stock just before replenishment, so that after replenishment the stock level will be y − w.
Here is the graph of inventory level over time.
The associated costs per cycle are as follows.
The purchase cost will again be denoted by K.
The holding cost calculation is exactly the same as for the basic model, but only the positive portion of the cycle
h(y−w)2 contributes: 2D .
Similarly, the shortage cost is 2D , this being the same calculation but for the negative part of the cycle.
Question 13: (0 points)
The total cost per cycle is
and hence the total cost per unit time is
h(y − w)2 pw2 2D + 2D ,
TCU(y,w) =
h(y−w)2 pw2 K+ 2D +2D
DK h(y − w)2 pw2
=y+ 2y +2y. For each individual value for y, we can find the optimal choice of w in terms of y:
∂TCU = −h(y − w) + pw = 0 ∂w y y
w=h. y h+p
Notice that there is only one candidate for w for a given value of y.
At this point we have a choice of how to find the optimal value for y.
We can now think of w as fixed, and search for candidate values of y in terms of w:
∂TCU = − DK + h(y − w) ∂y y2 y
h(y − w)2 pw2 2y2 −2y2 DK − (h + p)w2
2hy2 −2hwy−hy2 +2hwy
2y2 DK − (h + p)w2
(h + p)w2 h 2y2 =2.
We can then substitute into this the value we have already identified for w.
Alternatively, we can substitute the formula for w in before differentiating:
2 TCU(y,w(y))= y +2(h+p)
py h 2 + 2 (h+p),
dTCU(y, w(y)) = − DK + h p2
dy y2 y2 = 2DK(h + p).
hp Either method gives the optimal values of y, w, y − w and t as
2DK h+p y=√h√p,
2 (h + p)2
+ 2 (h + p)2 = 0
2DK h w=√√,
2DK p y−w=√√,
2K h+p t =y /D=√√ .
Let’s discuss how changing the parameters affects the optimal solution.
If h is constant, then as p increases w tends to 0, meaning that as the cost of shortages rises we prefer no shortages. As
p → ∞, the parameters tend to their values in the basic EOQ model.
However, if p is fixed and h increases then y − w decreases so that as the cost of holding rises we prefer paying the
shortage costs rather than the holding costs.
Look again at the example of the spice reseller, but now allow shortages with a shortage cost of p = 0.09. Plugging everything into the formulas, we get
y = 115.5, w = 46.2, y −w =69.3,
Hence the optimal order size is 115.5kg, with planned shortages of 46.2kg, with reorders every 3.9 days. Notice how the parameters are such that it is optimal to include shortages.
Question 14: (1 point)
Consider a Single-Item Static Continuous Review Model, with parameters:
D: a known constant rate of demand,
y: the size of the re-ordered, which is replenished the instant inventory reaches zero, K: the setup cost of each order,
h: the holding cost per unit, per unit time.
a. Enter the formula for the Total Cost per Unit ordered per unit time. Remember to include * for multiplication.
TCU(y) = __________
b. Enter the optimal value of y for this model.
y= __________
Question 15: (1 point)
A company assumes a basic EOQ model for a certain item, and estimates the parameters:
D = 30 K = 10 h = 0.01
a. Compute the Economic Order Quantity.
Enter a correct number to two decimal places, and use this number in subsequent calculations.
y= __________
b. Assume now that the company may only order in multiples of 10 units and that the implemented TCU needs only to be with 5% of the optimal TCU. Enter the smallest and largest order quantities that satisfy this requirement.
Smallest: __________
Largest: __________
c. The company commissioned a market intelligence report that said the estimate for D may be incorrect by up to 10 units either way (i.e. D [20, 40]). To the nearest percent, by how much might the implemented TCU be greater than the optimal TCU?
__________ %
Question 16: (1 point)
Assume a Single-Item Static Continuous Review Model with Planned Shortages, using the standard notation from the course materials.
Give a condition on h and p for the amount of backlogged shortages to be at least twice as much as the stock level after replenishment.
You can enter ≤ by typing <= .
__________
程序代写 CS代考 加微信: powcoder QQ: 1823890830 Email: powcoder@163.com