David Chiu
IEOR 130
HW#1 Assignment
5 September 2014
1.)
a.) Wafer Fab: A fabrication plant that carries out wafer fabrication process and wafer probe.
b.) Die Sort: First special electrical structures are tested to verify the proper electrical characteristics on a wafer. Afterwards each of the individual chips on each wafer is tested with a fully functional test to sort the good and failed chips on a wafer.
c.) Die Bank: An inventory point of unassembled sorted die. The place where they hold product pending specific customer demand for finished goods.
d.) Assembly: Dice up the completed wafer and package the chips in plastic or ceramic housings with electrical leads
e.) Test: Full functional test of the packaged device at various temperatures (“device test”), branding device ID, high-temperature operation (“burn-in”) followed by re-test, packing for shipment
f.) Line Yield: The fraction of wafers surviving the fab process flow
g.) Die Yield: The fraction of chips on a completed wafer that function at wafer probe.
h.) Overall Equipment Efficiency (OEE): The “should-take” time for the work actually completed divided by the total time.
i.) Die or chip: a small block of semiconducting material, on which a given functional circuit is fabricated. Integrated circuit on a wafer.
j.) Cycle Time: The time it takes manufacturing lots to pass through the entire production process.
h.) Bottleneck: a phenomenon where the performance or capacity of an entire system is limited by a single or limited number of components or resources.
2.) Fabs A and B are identical fabs (i.e., they have exactly the same set of equipment) that operate only one process flow that has identical specifications in each fab. Fab A has 85% line yield (LY), 90% die yield (DY), and 85% overall equipment efficiency (OEE) of the bottleneck equipment type. Fab B has 82% line yield, 98% die yield, and 60% overall equipment efficiency of the bottleneck equipment type. Which fab produces more output per unit time? Explain briefly. What if the fabs were not identical, i.e., either the process specifications were different and/or the equipment sets were different?
To find out the factory’s output per unit time (OUT): OUT=DYOEE, with the LY have an effect depending on where it occurred.
Situation 1 (if ALL of the LY occurs after the last bottleneck step of the process: OUT=(DY)(OEE)(LY)
Factory A, OUT=DYOEELY=.90.85.85=0.65025
Factory B, OUT=.98.60.82=0.48216
Factory A is producing more output per unit time for situation 1.
Situation 2 (if ALL of the LY occurs before the 1st bottleneck step of the process: If that is the case, then the appropriate number of wafers can be created to replace the ones that were lost and be able to proceed to the next step after the bottleneck.
OUT=(DY)(OEE)
Factory A, OUT=DYOEE=.90.85=0.765
Factory B, OUT=.98.60=0.588
Factory A is producing more output per unit time for situation 2.
If the Fabs are different, the process specifications were different and/or the equipment sets were different, then the OEE values of the two FABs cannot necessarily be comparable to one another. As such in this case, no distinct conclusion can be said with the current information.
3. Explain briefly why the managements of fabs producing DRAMs (dynamic random access memory) or Pentium chips traditionally did not emphasize reduction of flow times ("cycle times"), while the managements of ASIC fabs (i.e., fabs producing products designed by customers) traditionally have strongly emphasized reduction of flow times. More recently, fabs producing DRAMs and Pentiums have starting placing more emphasis on fast cycle time. Why do you think that is happening?
DRAM or Pentium producers produce chips to forecast, while ASIC producers produce chips to order. For the first kind of producer, the customers do not see the factory flow time, since they buy from finished goods inventory. For the second kind of producers, the customers must wait for the factory flow time. So lower flow time for the second kind of producer is a competitive advantage. Even when there is no competitive issue, reducing cycle time affords higher average selling prices in the case that prices are declining with time. Perhaps DRAM and Pentium prices go down faster than they used to, encouraging managements to emphasize cycle time reduction.
Another benefit of cycle time reduction concerns die yield. Losses due to equipment or process “excursions” are exacerbated by long cycle time. Moreover, time to complete process experiments leading to yield improvement depends on cycle time. So if yield losses to excursions are increasing, or if the number of process changes required to reach high yields is increasing, then cycle time takes on increased importance.
4. In a fab producing microprocessor chips, assume line yield is 90%, and die yield is 85%.
Assume the fab starts 1,000 wafers per day. Assume there are 100 die (chips) per wafer, and that the average selling price per chip is $500 per chip.
(a) Calculate how much revenue per day (RPD) the fab generates.
RPD=LYDY#wafersday#chipwafer(price per chip)
RPD=.90.851000wafersday100chipswafer$500 per chip=$3.825*107/day
(b) How much would revenue increase if die yield rose to 90%?
RPD=.90.901000wafersday100chipswafer$500 per chip=$4.05*107/day
Revenue would increase by $2,250,000/day
(c) How much would revenue increase if the fab starts rose to 1,100 per day?
RPD=.90.851100wafersday100chipswafer$500 per chip=
$4.2075*107/day
(d) With no other changes, what would happen to the fab cycle time in (c)?
Since the machines are taking more time to producing more wafers (10001100 wafers), the utilization of the machines would increase resulting in a longer waiting time. As a result the fab cycle time would subsequently increase
If the prices are going down with time this could potentially counteract the gain calculated above assuming the prices remain the same.
5.) The scheduled output of a fab in a particular week was as follows:
Product 1 1,000 units
Product 2 2,000 units
Product 3 20,000 units
Product 4 10,000 units
Product 5 5,000 units
The actual output in the week was as follows:
Product 1 900 units
Product 2 1,900 units
Product 3 21,000 units
Product 4 1,000 units
Product 5 5,500 units
The LIPAS (line-item performance against schedule) is an on-time delivery metric defined as the fraction of items with scheduled output in a time period whose output in the period meets or exceeds the scheduled output. What is the LIPAS score for that week?
Answer:
Product 1 = NO, not on-time
Product 2 = NO
Product 3 = YES, on-time
Product 4 = NO
Product 5 = YES
LIPAS score = products that met or exceeded scheduled outputtotal products during the expected scheduled output=25=0.4
6.) The LIPAS metric for on-time delivery presented in class (and also widely used in industry) is the fraction of all product types that were delivered on time. (A product type is considered on-time if actual output equals or exceeds scheduled output.) This metric is insensitive to the amount of shortage of each product
Develop a new metric for on-time delivery that reflects the amount of shortage for each product. The metric should have the properties that (1) a score of 1.0 means all product types were on time, (2) no credit is given for excess production of any product type, but there is no penalty for excess production, either, and (3) if the shortage of any product increases while shortages of all other products are held constant, then the metric score decreases.
1st Case: On-time delivery in a single time period fi=fraction of product i delivered on time di=scheduled output quantity of product i pi=actual output quantity of product i
