Question

Time Study: Consider the recorded observations of 10 cycles of the cup packaging operation in seconds, shown in the figure below. Notice that the data reflects the continuous method in timing the elements. Determine the select times for each work element, the normal time for the cycle, and the standard time per package. Suppose we want a sample size that gives an average within + 2 percent of the true average 95 percent of the time. Did we make enough observations? If not, how many more should we make? Suppose all we wanted was a precision of £5 percent. How many additional observations would we need? Operation: Coffee Cup Packaging Observer: T. Mann Date: 4/22 Observations 5 6 tbar F RF 1 2 3 4 7 8 9 10 1. Get two cartons 130 231 332 433 0.5 0.95 2. Put liner in carton 34 72 137 178 237 276 340 381 439 480 1.00 1.00 3. Place cups in carton 45 85 149 189 249 288 353 392 452 1.00 0.95 4. Seal carton, set aside r 63 104 169 207 269 309 372 411 472 511 1.00 1.05 Normal Time for cycle: Allowances (% of total time): 12% Standard Time: minutes per piece

Answer 1

SOLUTION:-

The data reflects the continuous method in timing the elements:-

a) Normal Time =

t *F* RF

OP1: 0.476 * 0.5 * 1 = 0.238

OP2: 0.217 * 1* 0.9 = 0.195

OP3 : 0.751 * 1* 1.05 = 0.789

OP4: 1.103 * 1* 0.95 = 1.048

Total normal time = 0.238+0.195+ 0.789+ 1.048

= 2.27 mins

b)

st.dev/t

OP1: 0.109 / 0.476 = 0.229

OP2: 0.163 / 0.317 = 0.749 ____n =

  

(1.96/0.05) + 0.749]>

= 863...need 863 - 10

= 853 more

OP3: 0.202 / 0.751 = 0.268

OP4: 0.455 / 1.103 = 0.412

c)

n= [(1.96/0.1) * 0.749]

= 216 - 10

= 206 more.....

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