Name:______ID:______
Operations Management II 73-431 Winter 2008
Odette School of Business
University of Windsor
Midterm Exam II Solution
Thursday, January 31, 2:30 pm – 3:50 pm
Odette Building OB B04
Instructor: Mohammed Fazle Baki
Aids Permitted: Calculator, straightedge, and a one-sided formula sheet.
Time available: 1 hour 20 min
Instructions:
- This exam has 20 pages including this cover page, four bank pages and 8 pages of tables.
- Please be sure to put your name and student ID number on each odd numbered page.
- Show your work.
- State results up to four decimal places.
- Do not return tables and formula sheet.
Grading:
QuestionMarks:
1/10
2/13
3/16
4/13
5/13
Total:/65
Question 1: (10 points) Circle the most appropriate answer
1.1(i) Functional products include the staples that people buy in a wide range of retail outlets such as grocery stores and gas stations. (ii) We should match functional products with a responsive supply chain.
- (i) true (ii) false
- (i) false (ii) true
- Both true
- Both false
1.2(i) Supply chain includes suppliers and their supplier, manufacturers and their manufacturers. (ii) The Supply-Chain Management includes managing information, materials and services from raw-materials suppliers through factories and warehouses to the end customer.
- (i) true (ii) false
- (i) false (ii) true
- Both true
- Both false
1.3Suppose that the locations of warehouse = (0,0), customer 1 = (3,0) and customer 2 = (0,4). What is savings (1,2)?
- 2 Correct choice
- 3
- 4
- 5
1.4(i) Saving matrix method guarantees an optimal solution. (ii) Nearest neighbor method guarantees an optimal solution.
- (i) true (ii) false
- (i) false (ii) true
- Both true
- Both false
1.5For which of the following a single period model is the most appropriate?
- pen
- pencil
- paper
- computer
1.6Suppose that and If demand is Normally distributed with mean , the order size
- Correct choice
- Non of the above
1.7Notation used to denote the expected number of units stock-out per
- cycle or order
- order or period
- period or cycle
- year
1.8Consider the iterative procedure used to find an optimal policy without service. The iterative procedure stops when
- Both
- None
1.9Type 2 service is
- the proportion of demand not met from stock
- the proportion of demand met from stock
- the probability of no stockout
- the probability of stockout?
1.10Consider the procedure used to find an optimal policy with Type I service. The optimal order size
- EOQ
- EOQ
- EOQ
- None of the above
Question 2: (13 points)
A large national producer of canned foods plans to purchase a combine that will be customized for its needs. One of the parts used in the combine is a replaceable blade for harvesting corn. Spare blades can be purchased at the time the order is placed for $130 each, but will cost $530 each if purchased at a later time because a special production run will be required. It is estimated that the number of replacement blades required by the combine over its useful lifetime can be closely approximated by a normal distribution with mean 20 and standard deviation 5. The combine maker agrees to buy back unused blades for $30 each. How many spare blades should the company purchase with the combine?
- (2 points) Compute the underage cost 530-130=$400
- (2 points) Compute the overage cost 130-30=$100
- (6 points) How many spare blades should the company purchase with the combine?
, 0.845,20+0.84(5)=24.2
- (3 points) How would you change your answer to part c if the company purchases 4 combines? How many spare blades should the company purchase with 4 combines?
20(4)=80,=10,80+0.84(10)=88.4
Question 3: (16 points)
An automotive warehouse stocks a variety of parts that are sold at neighborhood stores. One particular part, a popular brand of oil filter, is purchased by the warehouse for $3 each. It is estimated that the cost of order processing and receipt is $25 per order. The company uses an inventory carrying charge based on a 20 percent annual interest rate. The monthly demand for the filter follows a normal distribution with mean 200 and standard deviation 25. Order lead time is assumed to be four months. Assume that if a filter is demanded when the warehouse is out of stock, then the demand is back-ordered, and the cost assessed for each back-ordered demand is $1.00. Determine the average annual cost of holding, setup, and stock-out associated with this item assuming
200(12)=2400,200(4)=800,50,0.20(3)=$0.60/unit/year
- (4 points) Compute the average annual holding cost.
(0.60)+(900-800)(0.60)=180+60=$240
- (2 points) Compute the average annual setup cost.
$100
- (4 points) Compute the average annual stock-out cost.
2, 50(0.0085)=0.425
1.70
- (4 points) Compute the probability(no stock-out)
0.9772
- (2 points) Compute the fill rate (up to four decimal places)
0.9993
Question 4: (13 points)
Consider Question 3 again. Find an Optimal policy assuming no service constraint.
Iteration 1
Step 1: 447.21
Step 2: 0.1118
1.22 (Table A-4)
800+1.22(50)=861
Step 3: 0.0538 (Table A-4)
50(0.0538) = 2.69
Step 4: 470.66 (not near 447.21, more iterations needed)
Step 5: 0.1177
1.185 (Table A-4)
800+1.185(50)=859.25
Iteration 2
Step 3: 0.05785 (Table A-4)
50(0.05785) = 2.8925
Step 4: 472.38 (near 470.6, may stop after finding )
Step 5: 0.1181
1.185 (Table A-4)
800+1.185(50)=859.25
andconverge. An optimal policy is Q=472 units, R=859 units
Question 5: (13 points)
The manager at Albertson’s, a grocery store selling on-line, has 5 orders that are to be delivered to customers. The location and order size for each customer are shown below:
X-coordinate / Y-coordinate / Order sizeWarehouse / 0 / 0
Customer 1 / 0 / 5 / 50
Customer 2 / 7 / 1 / 60
Customer 3 / 2 / 10 / 70
Customer 4 / 4 / 5 / 80
Customer 5 / 9 / 2 / 40
The Albertson’s fulfillment store has two trucks, each capable of carrying upto 220 units. Using the savings matrix and nearest neighbor methods devise a suitable delivery schedule. If there is any route that intersects itself, improve the route. State the route of each vehicle.
The distance matrix is shown below:
W / C1 / C2 / C3 / C4 / C5W / 0
C1 / 0
C2 / 0
C3 / 0
C4 / 0
C5 / 0
The savings matrix is shown below:
C1 / C2 / C3 / C4 / C5C1 / 0
C2 / 0
C3 / 0
C4 / 0
C5 / 0
Rank savings: (2,5), (3,4), (1,3), …
Grouping: Vehicle 1: Vehicle 1: (2,5), Vehicle 2: (1,3,4)
Nearest neighbour tours: W-2-5-W, W-1-4-3-W (contains intersection, see plot below)
Eliminate intersection:
1. Remove (1,4), (3,W)
2. Reverse (4,3)
3. Add (1,3), (4,W)
Intersection free tours: W-2-5-W, W-1-3-4-W
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