IE 376, Spring 2018 Homework #3 Due on April 06, 2018 by 5 pm Question 1: Consid
ID: 324900 • Letter: I
Question
IE 376, Spring 2018 Homework #3 Due on April 06, 2018 by 5 pm Question 1: Consider the following jobs to be scheduled on a single machine. Assume that all jobs are availablc and the machinc is idle at tie 0 Processing lime Duc Date 10 a) What are the total tardiness, total flow time and maximum tardiness of the schedule that is b) Start from schedule that you find in part a) and obtain a schedule such that total tardiness is c What are the total tardiness, total flow time and maximum tardiness of the schedule that is di Use the Apparent Urgency rule or R&M; rule discussed in class to obtain a schedule. Use the rule obtaincd using EDD rule? minimized using the greedy approach explained in class obtained using critical ratio rule? dynamically to dispatch one job at a time. What is the resulting total tardiness? Assume that the weights arl cqual o and use k-2.0. Assume e average value of processing times is dynamically calculated. Question 2: A chair manufacturer produces two types of chairs, chair X and chair Y. There are three main components to manufacture chairs. Chair X consists of components A and B, one unit each. Chair Y uses 1 unit of A, and 2 units of C. C consists of 1 unit of B MPS PERIODS 60 10 30 30 60 30 10 20 20 30 OPERATION LOT SIZE WORK CENTER SETUP HOURS RUN HOUR/UNIT LEAD TIME 100 100 200 200 200 300 1-1 30 10 20 20 10 10 0.5 0.7 0.2 0.4 0.4 0.5 1.5 1-1 1-1 a) Generate the bill of capacity for the two items.Explanation / Answer
Answer 1 (a)
In the earliest due date method, step 1 is to arrange all the jobs as per ascending order of due dates. The sequencing would be as follows
B – E – C – D – A (we have basically arranged the jobs as per due dates, starting with earliest due date to max due date)
Step 2 is to find out Flow In and Flow Out times. We write the jobs in the same sequence as above
Jobs
Flow In
Flow Out
Due Date
Lateness
Earliness
Tardiness
B
0
3
E
3
7
C
7
8
D
8
14
A
14
16
Flow in for the first job is 0 and flow out will be basically the sum of flow in date and processing time. Hence Flow In for B is 0 and since processing time for B is 3, hence Flow Out is 3. The Flow In for E (next job) will be 3 and Flow Out for E would 7 (3+4) and so on and so forth.
Next step is to write the due dates as given in the problem in the above table.
Jobs
Flow In
Flow Out
Due Date
Lateness
Earliness
Tardiness
B
0
3
4
-1
E
3
7
6
1
C
7
8
7
1
D
8
14
10
4
A
14
16
12
4
To find out Lateness, we simply subtract Due Date from Flow Out. If the result is negative, we write it under Earliness and if the answer is positive, then we write the number under Tardiness.
Total Flow Time = Sum of Flow Out = 3+7+8+14+16 = 48
Total Tardiness = Simply the sum of tardiness = 1+1+4+4 = 10
Maximum Tardiness = 4
1 (c)
Critical Ratio Method
Critical Ratio (CR) = (Due Date – Date Now)/processing time
Step 1 is to find the critical ratios for each job
Jobs
Processing Time
Due Date
Critical Ratio
A
2
12
6
B
3
4
1.3
C
1
7
7
D
6
10
1.6
E
4
6
1.5
CR for A = (12 – 0)/2 | CR for B = (4 – 0)/3 | CR for C = (7-0)/1 and so on and so forth. We are taking Due date as 0 now since the sequencing has not been done yet.
Now, the job with lowest CR is the first job that will be taken up. In this case, it is B.
Since processing time for B is 3, on day 3, B will be completed.
Jobs
Processing Time
Due Date
Critical Ratio
A
2
12
4.5
C
1
7
4
D
6
10
1.16
E
4
6
0.75
Now in the above table critical ratio will be again calculated. But at this stage, “Date Now” in the CR formula will be 3 (processing time of job with lowest CR). So CR for A = (12-3)/2, CR for C = (7-3)/1 so on and so forth.
Here, lowest CR is 0.75, i.e. Job E. Now repeat the same process again. “Due date” in CR formula will now be Processing time for B + Processing time for E = 7.
Once the entire sequence is done, the answer can be obtained in the same fashion as has been shown in 1 (a)
Jobs
Flow In
Flow Out
Due Date
Lateness
Earliness
Tardiness
B
0
3
E
3
7
C
7
8
D
8
14
A
14
16