INSTRUCTIONS:
1. If you submit on Moodle, name your file: LASTNAME_FIRSTNAME_HW1
2. Please show all your work! You will NOT be given credit if you just write the answer.
3. This homework assignment is worth 25 points in total. The breakdown of points for each question is given adjacent to each question.
1) Short answer essay questions from The Goal (no more than 3 sentences per answer please!)
a. What is the goal and why can’t Alex achieve it by focusing on efficiency? (2 POINTS)
The goal is to make money and Alex can achieve it by focusing on efficiency because he’s not focusing on the importance of throughput, instead of getting distracted with the efficiency of each individual process. For example, if there was a bottleneck, that would be more easily identifiable if Alex was looking at the throughput and what it would take to get the orders fulfilled.
b. Why don’t Alex’s robots actually increase productivity? (2 POINTS)
Although these robots increase the productivity of specific process steps, they almost highlight the inefficiencies of the slower processes. For example, this process will only be as strong as its weakest component. Even if every step took little time at all, if there was a bottleneck that slowed the whole process, throughput would diminish and not actually increase productivity.
c. List the two important phenomena that Alex illustrates via the ‘matchstick’ game that he invents on the scouting trip? (2 POINTS)
The two important phenomena are dependent events and statistical fluctuations. Dependent events are events that must happen before another can happen. In the game, each of the boys output was dependent on the output of the others. Statistical fluctuations are information that caries from one instance to the next. In the game, the the amount that could be passed on was determined by a die. After these phenomena, it was clear that throughput was lower than a balanced system.
3) Sellinger Business School information technology service (ITS) is considering a new process to refurbish older computers in order to save on costs of buying new computers. The five steps to the process are:
Step
Description
Time per computer
1
Move the machine to their work area
8.0 minutes
2
Scrub and clean hardware
12.0 minutes
3
Strip off old operating system
6.0 minutes
4
Install new operating system & software
9.0 minutes
5
Move machine back to its previous location
10.0 minutes
One member of ITS will be assigned to each step. Each member will work a 40-hour week (excluding breaks) and rotate jobs each week. The goal is to refurbish 250 computers per week but ITS is unsure that they can deliver on the promise.
a. What is the bottleneck? (1 POINT)
A bottleneck is the production step that limits the process capacity. In this process the bottleneck is to scrub and clean the hardware. Since this production step takes the most amount of time as compared to the other steps, it is limiting the capacity of the process.
b. Assuming the process runs as designed, what’s the maximum weekly output? (2 POINTS)
40 hours * 60 minutes = 2400 Minutes
Step 1: 2400 / 8 = 300
Step 2: 2400 / 12 = 200
Step 3: 2400 / 6 = 400
Step 4: 2400 / 9 = 266.7
Step 5: 2400 / 10 = 240
The maximum weekly output is 200 units.
c. After 8 hours, how many units of work-in-process inventory will be at station 2 if all stations are working at their maximum capacity? (2 POINTS)
8 hours * 60 minutes = 480 minutes
Step 1: 480 / 8 = 60 units
Step 1: 480 / 12 = 40 units
Since 60 units leave step 1, and 40 units are completed in step 2, there will be 20 units in work-in-progress inventory at station 2 if all the others are working at capacity.
2) Various financial data for InMotion Manufacturing for 2012 & 2013 follow (in 1000s).
2012
2013
Output:
Sales
$1,500
$1,750
Inputs:
Labor
$40
$41
Raw Materials
$45
$61
Energy
$9
$11
Capital Expenses
$150
$188
Other
$3
$6
What is the percentage change in InMotion’s total productivity measure between 2012 & 2013? (3 POINTS)
2012:
Productivity = 1500 / (40 + 45 + 9 + 150 + 3) = 6.07
2013:
Productivity = 1750 / (41 + 61 + 11 + 188 + 6) = 5.70
Percent Change:
(5.70 – 6.07) / 6.07 = – 6.1%
4. An operations manager is deciding on the level of automation for a new process. The fixed cost for automation includes the equipment purchase price, installation, and initial spare parts. The variable costs per unit for each level of automation are primarily labor related. Each unit can be sold for $80. As in many cases, you have the default alternative of doing nothing ($0 fixed cost, $0 variable costs). Hint: Please consider the “Do Nothing” option as a viable option when making your decision.
Alternative
Fixed Costs
Variable Costs per Unit
A
$100,000
$54
B
$280,000
$38
C
$560,000
$20
a. Calculate the break-even quantities for each alternative. If the projected demand is 3,200 units, what should you do? (2 POINT)
Break-Even for A: 100,000 / (80 – 54) = 3,846.15 (3,846)
Break-Even for B: 280,000 / (80 – 38) = 6,666.67 (6,667)
Break-Even for C: 560,000 / (80 – 20) = 9,333.33 (9,333)
Since the alternatives will make more than 3,200 units, there won’t be any profit for the alternative so we should take the “Do Nothing” option.
b. For each alternative, at what specific volume range is it the most attractive? Please specify the volume ranges for each alternative, including the “Do Nothing” alternative. (3 POINTS)
Volume Range “Do Nothing”: 0 – 3,846 units
Volume Range “A”: 3,847 – 11,250 units
Upper Range: (280,000 – 100,000) / (54 – 38) = 11,250
Volume Range “B”: 11,251 – 15,556
Upper Range: (560,000 – 280,000) / (38 – 20) = 15,556
Volume Range “C”: 15,557 +
5. A pizza place is considering opening a drive-through window for customer pickups. It is estimated that at peak times (6-9pm and 1-3am), customers will arrive at a rate of 11 per hour. The clerk who will staff the window can serve customers at the rate of one every three minutes. For the peak times, what is the:
a. Average number of customers in the waiting line (in the queue)? (2 POINT)
λ = 11.0 / hour
μ = 1 / 3 minutes = 20 / hour
Avg. number of customers waiting = 11^2 / 20* (20 – 11) = 0.67
b. Average customer waiting time in line (in minutes)? (2 POINT)
λ = 11.0 / hour
μ = 1 / 3 minutes = 20 / hour
Avg. waiting time = 11 / 20* (20 – 11) = 0.061 hours
c. Probability that more than two customers will be in the system at one time? (2 POINTS)
= (1) (1 – .55) + (.55) (1 – .55) + (.30) (1 – .55) + (.17) (1 – .55) = 0.9085
1 – 0.9085 = 0.0915 = 9.15%