Practice Exam Questions - Georgia Institute of Technology



Final Exam

ISyE 6203

Spring 2007

Vande Vate

You have 2 hours and 50 minutes to complete the exam. Watch your time. If you are having difficulty with a question, you might wish to pass over it and return to it later if you have enough time at the end of the exam.

There are 5 questions and an Extra Credit question. To get full credit you only need to answer 4 questions. If you answer all 5 you can earn 125 points. If you answer all 5 and the extra credit question you can earn 135 points.

Be sure to put your name on your exam. Show your work – that will allow me to award partial credit. Make sure your answers are legible and clearly marked.

The exam is open book: you may use any notes or text material.

Note: For your convenience, models given in this exam are shown in two ways: AMPL and standard mathematical representation, e.g.,

/* The objective: Minimize Transportation Cost in $/year */

Minimize TransportCost:

Sum{(orig, dest) in LANES}TruckCosts[orig, dest]*Trucks[orig, dest];

Minimize [pic]

You may provide answers using either notation. Be sure your answers are clear and unambiguous.

The exam is to be done individually, without collaboration with anyone else.

NAME___________Key___________________________

1. (25 points) Another approach BMW might take to deciding how much of each engine type to ship to the plant via ocean each week is to view the decision as a News Vendor problem. This perspective on the problem is a blend of the two versions we discussed in class since it is intended to trade off the costs of carrying inventory if we send too much against the costs of expediting if we send too little. For simplicity assume we make weekly ocean shipments and that we operate 50 weeks each year.

A. From past experience we pay $200 in airfreight for each engine we expedite. The value of an engine is about $2,000 and we estimate our inventory carrying charge to be about 20%. Given these values, suggest the appropriate target probability P, i.e., we should choose the shipment quantity Q so that P is the probability that the demand in the weeks after the shipment arrives is less than or equal to Q:

Prob(demand ≤ Q) = P.

If we send too much this time, the excess will sit in inventory. We don’t know how long they will sit exactly, but a reasonable approximation is to assume they will sit there for one week. And so the cost of sending one too many engines is 20%*$2000/50 = $8 per engine

If we send too few engines this time, we will have to expedite the balance at a cost of $200 per engine.

Doing our marginal analysis, if we ship Q engines, the last engine we ship saves us $200 with probability equal to the probability demand exceeds Q. On the other hand, we will carry that last engine in inventory for a week with probability equal to P, the probability demand is smaller than Q.

So we want these two to balance:

$200*(1-P) = $8*P

And so P = 200/208 = 96.15%

B. Since these costs remain unchanged from week to week, the probability P will remain unchanged from week to week. Will this mean that the shipment quantity recommended by this strategy will remain the same from week to week?

Either answer is acceptable.

C. If your answer to part B was “yes”, what advantages does the ship-to-average strategy we discussed enjoy over this approach? If your answer to part B was “no”, explain why the recommended shipment quantity should change each week.

The value of P doesn’t depend on the distribution. We calculated it from the various costs. Translating this probability into a shipment quantity does require a distribution and the appropriate distribution to use is the distribution of demand from the time this shipment arrives to the time the next shipment arrives minus the inventory on hand when this shipment arrives.

If your answer to part B was “yes”, the reason should be that this distribution changes depending on the current inventory at the plant, the forecast of demand and how much you sent in previous shipments. These changes to the distribution will change the shipment quantity you obtain from the fixed level of P.

If you answer was “no”, then you must have assumed we would calculate a sort of steady state distribution for the difference between demand and inventory and so arrived at one unchanging value for the shipment quantity. In this case, the policy has the clear disadvantage that the actual usage at the plant is changing and unpredictable and so if we ship the same quantity every time, the only tool we have left to control the inventory is expediting. Inventory will wander without any upper limit. Our ship-to-average policy exerts a modest control to keep inventory within reasonable ranges.

2. (25 Points) GE Appliances faced the decision of whether to use rail or truck to for transport from its plant in Tennessee to its distribution center on the east coast. The company is trying to maintain a 98% level of service at the east cost DC.

Relevant Information

| |Rail |Truck |Units |

|Average Transit Time |14 |2.5 |days |

|Std Deviation in Transit |3 |0.5 |days |

|Operates |7 |5 |days per week |

|Freight Cost |$8 |$15 |per unit |

|Frequency |7 |5 |Days per week |

| | | | |

| | | | |

|Average Production/Demand |100 |units per week day |

|Std Deviation in Demand |5 |units per day |

|Product value |500 |$ per unit | |

|Holding cost |20% |per year | |

|Weeks per year |50 | |

|Service Level Target |98% | | |

Since this is but one of many products we ship between these two locations, simply assume we will ship an average quantity at the indicated frequency.

A. Estimate the total cost involved in using each mode exclusively.

| |Rail |Truck |Units | | | |

|TransAverage Transit Time |14 |2.5 |days | | | |

|Std Deviation in Transit |3 |0.5 |days | | | |

|Operates |7 |5 |days per week | | |

|Freight Cost |$8 |$15 |per unit | | | |

|Frequency |7 |5 |Days per week |

| | | | | | | |

| | | | |

|Cycle | $ 7,143 | $ 10,000 |First calculate the average shipment size using weekly demand and frequency of shipment | | |

| | | |then calculate the inventory holding cost for this | | |

| | $ 368,841 | $ 420,396 |The total | | | |

| | | | | | | |

|2.05 | | | | | | |

| | | | | | | |

| | | | | | | |

| |Rail |Truck |Units | | | |

|TransAverage Transit Time |14 |2.5 |days | | | |

|Std Deviation in Transit |3 |0.5 |days | | | |

|Operates |7 |5 |days per week | | |

|Freight Cost |$8 |$15 |per unit | | | |

|Frequency |1 |5 |Days per week |

| | | | | | | |

| | | | | | | |

|Average Production/Demand |100 |100 |units per week day | | | |

|Std Deviation in Demand |5 |5 |units per day | | | |

|Product value |500 |500 |$ per unit | | | |

|Holding cost |20% |20% |per year | | | |

|Weeks per year |50 |50 | | | | |

| | | | | | | |

|Service Level Target |98% |98% | | | | |

|Freight | $ 200,000 | $ 375,000 |Cost per unit * units per week day * weeks per year * days per week = $ / year |

|Pipeline | $ 100,000 | $ 25,000 |Annual Demand/Days per year gives volume per day. Multiply this by the transit |

| | | |time to get the number of units in the pipeline. Multiply this by the annual holding |

| | | |cost to get the pipeline inventory cost |

|Cycle | $ 50,000 | $ 10,000 |First calculate the average shipment size using weekly demand and |

| | | |frequency of shipment then calculate the inventory holding cost for this |

|SafetyStock | $ 61,741 | $ 10,360 |Calculate the standard deviation in lead time demand remembering that this is now a |

| | | |Periodic Review model and so we should consider T + E[L] when calculating the |

| | | |Std Deviation. Then calculate the safety stock and its inventory holding cost |

| | $ 411,741 | $ 420,360 |The total | | | |

So the main changes are in the Cycle Stock and a small increase in Safety Stock.

3.(25 Points) We move products from suppliers to customers and developed the following standard capacity and inventory planning model to help.

In AMPL

param Demand{p in Products, c in Customers, t in TimePeriods};

param LeadTime{p in Products, s in Suppliers, c in Customers};

param Capacity{p in Products, s in Suppliers, t in TimePeriods};

param TransitCost{p in Products, s in Suppliers, c in Customers};

param HoldingCost{p in Products};

var Inventory{p in Products, c in Customers, t in TimePeriods}>= 0;

var Ship{p in Products, s in Supplier, c in Customer, t in TimePeriods} >= 0;

minimize TotalCost:

sum{p in Products, s in Suppliers, c in Customers, t in TimePeriods}

TransitCost[p, s, c]*Ship[p, s, c, t] +

sum{p in Products, c in Customers, t in TimePeriods}

HoldingCost[p]*Inventory[p, c, t];

subject to LimitedCapacityAtEachSupplier{p in Products, s in Suppliers, t in TimePeriods}:

sum{c in Customers} Ship[p, s, c, t] 1}:

Inventory[p, c, t] = Inventory[p, c, t-1] +

sum{s in Suppliers: t-LeadTime[p, s, c]> 0}

Ship[p, s, c, t-LeadTime[p, s, c]] – Demand[p, c, t];

In Mathematical Notation:

Minimize TotalCost:

[pic] +

[pic]

s.t. LimitedCapacityAtEachSupplier

[pic]

for each p in Products, s in Suppliers and t in TimePeriods

s.t. Inventory BalanceAtEachCustomer

Inventory[p, c, t] = Inventory[p, c, t-1]+

[pic]

We want to address the issue that each customer allocates a limited space for inventories of our products but, at some additional cost, we can procure additional space by using transportation assets eg (trailers or containers) to hold inventory temporarily. At times, we may not be able to meet all the needs of our customers and incur penalties for this. Some customers impose a penalty for each item that is late each period. Other customers impose a fixed penalty if anything is late in a period.

Suppose Customer c allocates MaxSpace[c] cubic ft of storage for our products and that one unit of product p requires Cube[p] cubic ft. We can rent additional space at Rent[c] $/cubic ft per time period.

We distinguish between customers that charge per unit of missed demand and those that charge a fixed cost when anything is late via the parameter ChargesByUnit?[c] = 1 if the customer charges for each unit that is late, in which case we pay UnitLateFee[c] for each item that is late each period and be the cost and ChargesByUnit?[c] = 0 if the customer charges a fixed fee if anything is late, in which case we pay FixedLateFee[c] in each period in which we fail to meet the customer’s demands.

Adjust the standard capacity and inventory planning model to address these issues.

Add

var BackOrder{p in Products, c in Customers, t in TimePeriods} >= 0;

var TempSpace{c in Customers, t in TimePeriods} >= 0;

var AnythingLate{c in Customers, t in TimePeriods: ChargesByUnit?[c] = 0} binary;

to represent the units of product p that are late to customer c in period t.

Add the constraints:

s.t. CustomerSpaceLimit{c in Customers, t in TimePeriods}:

sum{p in Products} Cube[p]*Inventory[p, c, t] 0}

Ship[p, s, c, t-LeadTime[p, s, c]] – Demand[p, c, t];

To allow both early and late shipments to the customer.

Change the objective to

minimize TotalCost:

sum{p in Products, s in Suppliers, c in Customers, t in TimePeriods}

TransitCost[p, s, c]*Ship[p, s, c, t] +

sum{p in Products, c in Customers, t in TimePeriods}

HoldingCost[p]*Inventory[p, c, t] +

sum{ c in Customers, t in TimePeriods}

Rent[c]*TempSpace[c, t] +

sum{ c in Customers, t in TimePeriods: ChargesByUnit?[c] = 0}

FixedLateFee[c]*AnythingLate[c, t] +

sum{ p in Products, c in Customers, t in TimePeriods: ChargesByUnit?[c] = 1}

UnitLateFee[c]*BackOrder[p, c, t];

4. (25 pts) In the Children’s Hospital project we saw that service level was the key driver and that the OmniCell provided what amounts to POS (Point-of-Sale) data telling us exactly what items are used when and by whom.

This information simplifies the task of determining how much of each item we should stock in the warehouse by giving accurate and timely information about average demand and the variability in demand.

One way to ensure high service levels from the OmniCells is to keep them full and, if we head down that path, a cost effective approach to keeping an OmniCell filled (assuming we have already decided which items it will hold) is to focus on two issues:

i. Setting reorder points for each item that ensure the desired level of service given the rate and variability of its usage and the mean and standard deviation in the replenishment lead time (the time from when we request a replenishment to the time it is completed)

ii. Ensuring that each item reaches its reorder point at approximately the same time so that we restock as much as possible each time we do replenish the OmniCell and so minimize the replenishment costs

Assume the average replenishment lead time is 2 hours and the std deviation is 30 minutes.

Assume every SKU (stock keeping unit) is the same size (for simplicity) and so the capacity of the OmniCell can be described by the number of units it can hold. Assume the OmniCell can hold 1,000 units and that this is the only constraint on its capacity, i.e., we can allocate those 1,000 units in any way we want across the different SKUs.

Illustrate how to implement this “Keep the OmniCell Full” strategy assuming only the following three SKUs are stored in the OmniCell:

|SKU Number |Average Usage (units/hr) |Std Deviation in Usage |

| | |(units/hr) |

|1 |10 |10 |

|2 |100 |50 |

|3 |6 |2 |

This requires specifying

• The portion of the 1,000 units of the OmniCell capacity that we will dedicate to each SKU – whenever we replenish the SKU, we will bring it back to this level, and

• The reorder point for each SKU – whenever any SKU reaches its reorder point, we will replenish all the SKUs

|SKU Number |Allocation of OmniCell |Reorder Point |

| |Capacity | |

|1 |105 |67 |

|2 |847 |468 |

|3 |48 |25 |

See spreadsheet for details.

5. (25 points) In the Coca Cola project, there were two dimensions to the capacity of a vehicle: the number of cases it could hold and the number of stops it could make. Let’s ignore the issue of stops and focus on the issue of cases. One team recognized that the distribution of daily demand (in cases) during a given month fits well with a normal distribution.

Suppose that:

• The amortized capital cost of owning vehicles is C $/case/day, i.e., if there are 30 days in a month and we own enough vehicles to deliver 100 cases in a day, then we can think of the capital charge as a monthly lease payment that will be:

30 days * C $/case/day * 100 cases = $3,000 C

• The operational cost of using that capacity to deliver a case is v $/case, i.e., if we only deliver 50 cases one day then we pay 50 v for the driver, fuel, wear-and-tear, etc. that day.

• The cost of having a third party deliver a case is r $/case, if the next day the demand is 150 cases, we will pay the lease cost, deliver 100 cases with our own trucks at an operating cost of 100 v and pay the third party to deliver the last 50 cases at a cost of 50 r

To simplify the process (this is an exam after all) let’s imagine we are making a decision just for this month and that this month is 30 days long. We want to decide how much capacity in cases/day we should own in order to minimize the expected total cost for the month in terms of lease costs, operating costs and third party costs.

A. What type of problem is this?

i. EOQ

ii. Safety Stock

iii. Ship-to-Average

iv. News Vendor

v. Linear Programming

B. Describe how to solve it. Be specific.

.

Let V be the capacity in cases/day we purchase. Then the expected total cost each day is

C * V + [pic]

You can use calculus or common sense to solve this. The common sense approach looks at the last unit of capacity we purchase. It incurs the lease cost C every day. We will use it and pay the operating cost v with probability that the demand on that day exceeds V, let’s call that probability (1-P) so that P is the probability our last unit is idle. Finally, it will save us from hiring a unit of 3PL capacity whenever we do have to hire 3PL capacity, which again has the probability 1-P

So, we want these costs to balance:

C + v (1-P) = r (1-P)

Which means we want P to satisfy

P = 1 – C/(r-v)

This has the right features: P is 1 if the 3PL cost r is less than our own operating cost v – in that case, it is better to let them do everything. P decreases as the lease cost increases – the more expensive the asset is to lease, the less willing we are to let it be idle. P increases as the 3PL’s premium over our operating cost increases – but we are willing to let those assets idle more if having them can save us more.

Now we simply look up the value V in the normal distribution that has the cumulative probability P

V = NormInv(P, mean, stddev)

Where mean is the average daily volume for the month and stddev is the standard deviation in daily volume for the month.

Extra Credit (10 pts). In our discussion of the Sport Obermeyer case, we used return on investment as a surrogate measure of risk with the idea that riskier investments demand a higher return.

A. Propose a better measure of risk that can be computed from data available in the case.

B. Explain the advantages of your measure over the measure used in class

Give partial credit liberally. The measure should be specific and the data to compute it should be available.

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