Question

Problem 3. The figure below shows a typical process chart. a. Complete the summary (top right) portion of the chart. b. Step 1 takes 2.5 minutes, so the maximum number of units that can be processed is 60/2.5 24 per hour. Step 2 takes 2.0minutes, so the maximum output is 30 per hour. By going similarly through each step, determine the maximum output of the whole process (this is the bottleneck). Assume the process variable cost is $25 each time the process is performed. The process is c performed 15 times per hour (on average) and this factory is open 350 days a year, 18 hours a day. What is the total labor cost associated with the process? d. The process chart also suggests areas for improvements. Discuss Inputs Solver-Process Charts Enter data in yelow shaded areas Process Typical process chat Summary Subject Assembly Number Time Distance Beginning: Fetch components Advty (mn of Steps Ending Cary unt to store Operation Transport Inset Step Inspect Append Step Delay Remoe Step Store Step Time Dstance No. (min) 2 50 50 0 Step Description Fetch camponents Put cemponents on machine Start machine Fecthe subassembly Wat for machine to stop Unload machine Inspect resut Joint subassembly Move unit to machine Load machine and stan Wat for machine to stop Unload machine Cary unt to inspaction anea Inspect and test Cany unt to Ensh area Finish unt Finish inspection Cary unt to stere 200 120 3 n 5.20 40.0 200 150 500 250 200 5.00 25.0 10 11 12 140 13 200 25.0 14 5 20 1 140 5 50 20.0 16 17 350 18 530 450

Answer 1

activity	Number of steps	Time	Dist ft
operations	7	19.1
Transport	6	16.7	205
Inspect	3	10.2
Dely	2	10.2
Store	0	0

Step	Time	Output
1	2.5	24
2	2	30
3	1.2	50
4	3	20
5	5.2	11.53846
6	2	30
7	1.5	40
8	5	12
9	2.5	24
10	2	30
11	5	12
12	1.4	42.85714
13	2	30
14	5.2	11.53846
15	1.4	42.85714
16	5.5	10.90909
17	3.5	17.14286
80	5.3	11.32075

Bottleneck is step 16 with output 10.9.

c. Yearly process cost = 15x25x350x18 = 2362500

Number of outputs = 10.9 x18x350 = 68670

labour cost per output = 34.4

d. Delay time of 10.2 hours which is due to waiting can be reduced by effective scheduling.

Transportation can be minimised by using proper layout strategy. It will also reduce time and thus outputs.

By reducing finish time, carrying time of unit and other time values which are more than 5 min per step can increase the output significantly