Question

You have six scanner techs, three work first shift (D); three work second shift (A) and each is assigned to a specific scanner noted below. A scanning backlog is growing over the past week, so you have decided to evaluate their productivity. They each recorded their daily productivity over the past two weeks. Since they do not have other department responsibilities, all their time is spent on scanning. One staff member, Teresa, worked a half day on one of the days, and one person, John, had a personal day; otherwise the techs all worked their full 8-hour shift daily for the two-week period under study.

1. Calculate each tech’s hourly productivity as well as an overall group productivity

Scanner 1(D) John 112,892

Scanner 2(D) Teresa 160,588

Scanner 3(D) Larry 193,248   

Scanner 1(A) Mary 138,800

Scanner 2(A) Jenny 182,160  

Scanner 3(A) Tom 184,960  

2. Determine if the individuals and group are meeting the benchmark of 1,200–2,400scanned images per hour

3. What conclusions can you draw from this study?

4. Recommend a new productivity standard based on the information gathered abovein order to reduce the back log

Tip:

Labor Productivity=       Completed work_____

                                   Hours worked for total output

Answer 1

1. ANS:

Scanner 1(D) John       112,892/72= 1,568

Scanner 2(D) Teresa    160,588/76= 2,113

Scanner 3(D) Larry       193,248/80= 2, 416

Scanner 1(A) Mary       138,800/80= 1,735

Scanner 2(A) Jenny     182,160/80= 2,227

Scanner 3(A) Tom        184,960/80= 2,312

Therefore

Total scans: 112,892 + 160,588 + 193,248+ 138,800+ 182,160+ 184,960= 972,648

Total hours: 72+ 76+ 80+80+80+80 = 468

Overall productivity= 972,648/ 468 = 2078.31 Scans/ hour

2. ANS: Yes, all individuals as well the group are meeting the benchmark.

3. ANS: while they are all meeting the benchmark, some are on the low end. It appears that the lowest recording were form the same machine, so it might be a technical issue such as an old machine or one that needs maintenance.

4. ANS: create a stretch product standard, using the differences between the group average and the highest individual average.

2,416- 2,078= 338/2 = 169 + 2078 = 2247