Examples.
- A pharmaceutical distribution service entity. Web server interfaces the mainframe system.
Order statistics:
10,000 orders/day
50% orders are 10-line orders,
30% orders 15-line orders,
20% orders have 20-line orders.
Each line takes 500 msec to be processed
Average time an order takes to be processed
sec
sec
Assuming that each order visits the system once and then exits,
Therefore, total service demand at the mainframe per order sec.
Total busy time for mainframe is at least 10000 6.75 secs = 18.75 hours
System utilization of the mainframe, 125
How much extra it can do? Being fully utilized, it can process at most = 12,800 orders. This means it has capacity to process about extra 2,800 orders a day.
- Customers entering a system for DB search.
Stats:
Total number of arrivals: 39,000
Total number of completions:
Web server busy period, secs
DB server busy period, secs
Observation period, secs
Number of search per completion: 2.3
Visit-count to WS per search = 1.95
Number of visits to DB server per search = 0.85
, Total number of searches = 35000 2.3 = 80500
, service demand at the WS = ==14.9 ms
Similarly,
Service times received per visit are and where
secs. and secs
Arrival rate at the system, = 13 tps
System throughput, tps
Web server utilization, 0.4, 0.67
Device throughputs: tps
=9.92 tps
tps
- Consider an interactive system. Interactions join at the memory queue to receive memory, and if memory is received, it proceeds to CPU queue for CPU burst.
Stats: Avg Think Time at terminal secs
Total number of Clients,
System Throughput, tps
Thus, Response time in the system, sec
Avg. number of xacts in the memory, = 3.2
Avg CPU utilization .
- What is the total service time demanded at CPU per transaction?
sec
- How many clients are not thinking at the terminals?
This includes all those still in the memory queue, outside memory queue but in CPU system or in I/O system. Therefore,
Since 3.2 customers have received memory, (8.5-3.2)= 5.3 customers are still in memory queue waiting for memory.
- How much time is spent in CPU and I/O?
Since , we have
secs
d. How much time is spent in the memory queue waiting for memory? It must be 17-6.4 = 10.7 secs.
- Consider a system with two disks: Fast, and slow. The stats are:
Observation length: 30 mins.
Number of Terminals: 30
Think time: 12 secs
Completed transactions: 1600
Fast disk visits: 28,000 Slow disk visits: 12,000
- Determine the visit count , service time per visit and service demand at each center.
- Give optimistic and pessimistic bounds on throughputs and response times for terminal counts of 4, 5, 10, and 20 , respectively.
Total and
Therefore, for any N, throughput would be bounded as
Thus, the pessimistic and optimistic limits are:
Notice the shift of the bottleneck from a load of 4 customers to 5 customers.
- Consider the following modifications one at a time.
- Move all the files to fast disk.
- Replace the slow disk by another fast disk
- Increase the CPU speed by 50% with original disks
- Increase the CPU speed by 50% and balance the load between two fast disks.
- All the files are moved to fast disk. Therefore, total visit to fast disk is going to be 40,000. This means the visit count per transaction at fast disk is now going to be . Since ,
The total
Notice, now the bottleneck is the fast disk since
Therefore, our throughput would be bounded as
0.16
2. Replace the slow disk with another fast disk.
and , we get 1.88
The first “fast” disk is still the bottleneck. Total service demand 6.93, and the situation is the same as (1).
- Get a CPU twice as fast. 0.338. This means our.