ECONOMICS 3600
ANSWERS TO Exercise Set 8-B:
Because it is more general and much easier to use to find the Long-Run equilibrium, I will use only
equation (8-8) to get the answers.
(8-8)
[Dn-1•h•(wC– wB)] = (OCC + DCC), where OCC = ohC•wB.
1. For this Problem, h=2000, n=20,=0, so Dn-1=19, ohC=1000 and, initially, DCC=0.
a) 19•2000•(wC– wB) = 1000wB + 0  (38,000wC – 38,000wB) = 1000wB 
38,000wC = 39,000wB  wC = (39,000/38,000)wB  wC = 1.0263wB .
X’ = 1.0263; Z’ = 0
C
B
B
C
B
b) 19•2000•(w – w ) = 1000w + $10,000  (38,000w – 38,000w ) = (1000wB + $10,000) 
38,000wC = (39,000wB + $10,000]  wC = [(39,000/38,000)wB + $10,000/38,000] 
wC = 1.0263wB + $.263.
X’ = 1.0263; Z’ = $.263
c) 14•2000•(wC– wB) = 1000wB + $10,000  (28,000wC – 28,000wB) = (1000wB + $10,000) 
28,000wC = (29,000wB + $10,000)  wC = [(29,000/28,000)wA + $10,000/28,000] 
wB = 1.0357wA + $.357.
X’ = 1.0357; Z’ = $.357
d) Each X’ and each Z’ is larger than the equivalents in Exercise 8-A because the C-costs are the
same as the B-costs, but the wage differential, (wC– wB) is received for one fewer period than
is (wB– wA). The two LHSs must be equal because the two RHSs are equal. The LHS for C
vs B has one fewer time period in than the LHS for B vs A, so the wage differential must be
larger to keep the LHS equal.
2. One again, X’and Z’ are long-run equilibrium values. The long-run wage differential won’t
change, because none of the variables that determine X’ or Z’ – h, ohC , Dn-1 , or DCC – has changed.
So wC = 1.0263wA + $.263.
X’ = 1.0263; Z’ = $.263
For all below, wB = $11.00/hour
3. From the introductory information, you know that h=2000 hours. Training for C takes a year of
working 1000 hours, so ohC=1000 and OCC=1000wB. n=5, so when =0, Dn-1=4. DCC=0.
a) 4•2000•(wC – $11) = (1000•$11 + 0)  (8,000wC – $88,000) = $11,000 
8,000wC = $99,000  wC = ($99,000/8,000)  wC = $12.375 Differential = $1.375
b) 3•2000•(wC – $11) = (1000•$11 + 0)  (6,000wC – $66,000) = $11,000 
6,000wC = $77,000  wC = ($77,000/6,000)  wC = $12.833 Differential = $1.833
Note that in both of these, compared to number 3. on Exercise 8-A, (wC– wB) is always greater than
(wB– wA) there. Why? This is the result that the age/wage profile is steeper for higher skill levels.
4. Now DCC=$10,000.
a) 4•2000•(wC – $11) = (1000•$11 + $10,000)  (8,000wC – $88,000) = $21,000 
8,000wC = $109,000  wC = ($109,000/8,000)  wC = $13.625 Differential = $2.625
b) 3•2000•(wC – $11) = (1000•$11 + $10,000)  (6,000wC – $66,000) = $21,000 
6,000wC = $87,000  wC = ($87,000/6,000)  wC = $14.50 Differential = $3.50
Note that in both of these, compared to number 5. on Exercise 8-A, (wC– wB) is always greater than
(wB– wA) there. Why? This is the result, again, that the age/wage profile is steeper for higher skill
levels.
5. You are now given an equilibrium wage at C. You have to work backwards to find the
others. The easiest way is to begin with the numbers from 3-a) but rewritten slightly to find
wB:
4•2000•(wC– wB) = (1000•wB + 0)  (8000•$18.00 – 8,000wB) = 1000wB 
9,000wB = $144,000  wB = ($144,000/9,000)  wC = $16.00 [Note that the differential is
higher. Why?]
Now repeat, using the numbers from 3-a) from Exercise set 8-A to find wA :
5•2000•(wB– wA) = (1000• wA + 0)  (10,000•$16.00 –10,000wA) = 1000wA 
11,000wA = $160,000  wA = ($160,000/11,000)  wA = $14.55 [Note that the differential is
higher. Why?]