Question

An analyst has estimated that there will be an 84% learning curve for an assembly operation. The first assembly takes 48 minutes and the standard time is set at 6 minutes. a. How long will it take the operator to reach the standard time? b. Unfortunately, the operator falls ill after the first week on the new assembly operation and returns after one week. What will be the estimated time for the operator to reach standard time now?

Answer 1

Answer a=

Learning curve =0.84

Standard time =6 min

N=log0.84/log2=-0.757/0.301 =-0.2515

Standard time=48x-0.2515

6=48x-0.2515

x-0.2515 =6/48=0.125

-0.2515 log x=log(0.125)=-0.9031

log x=3.591

x=antilog (3.591) =3898

Total time = 48*[(3898+0.5)-0.2515+1 - (1-0.5)-0.2515+1)/( -0.2515+1 ) ]=31202.4 minutes

Total time =31202.4/60=520.04 hour

Answer B= Total time =40 hours =40*60=2400 min

Total time = 48*[(x+0.5)-0.2515+1 - (1-0.5)-0.2515+1)/( -0.2515+1 )]

2400=48*[(x+0.5)-0.2515+1 - (1-0.5)-0.2515+1)/( -0.2515+1 )]]

37.425=[(x+0.5)-0.2515+1 - (1-0.5)-0.2515+1)/( -0.2515+1 )

37.425=[(x+0.5)0.7485 -0.595]

36.83=(x+0.5)0.7485

0.7485log (x+0.5)=log 36.83

0.7485log (x+0.5)=1.5662

log (x+0.5)=1.5662/0.7485

log (x+0.5)=2.092

taking antilog

x+0.5 =123.72

x=123.2 units

Cycle time =48*123-0.2515

Cycle time =14.3min

New time for the first unit after the break =[{48+(48-6)*(124-1)}/(1-1398)] =46.66 min

Standard time =46.66*3775n

6=46.66*3775n

3775n =6/46.66=0.12859

n*log3775=log(0.12859)

n*log3775=-0.8908

n*(3.577)=-0.8908

n=-0.2491

Total time=46.66*((3775+0.5)-0.2591+1- (1-0.5-0.2591+1) /(-0.2591+1)

Total time=30112.1

Total time =30112.1/60=501.5 hr