Business Decision Analytics under Uncertainty 33:136:400
Solutions to Final Exam Practice Questions
Product testing
First, we define the events
A = product acceptable without burn-in
G = test says “good”.
We are given
⇒
P{A} = 0.95
P Ā = 0.05
P G | Ā = 0.15
P{G | A} = 0.96.
So we may compute
A
A
G
G
P{AG} = P{A} P{G | A} = (.95)(.96) = .912
P AG = .95 − .912 = .038
.95
P AG = P A P G | A = (.05)(.15) = .0075 P A G = .05 − .0075 = .0425 .05
P G = .038 + .0425 = .0805
P{G} = .912 + .0075 = .9195
(Note that this uses a slightly different method based on the table above that I did not discuss in
class, but it is still using the same formulas, i.e. P {G} = P {G|A}P {A} + P {G|Ā}P {Ā} and so
on.)
(a) P{G} = .9195
P AG
.0075
P A|G =
=
≈ .00816
P{G}
.9195
P AG
.0425
P A|G = =
≈ .5279.
.0805
P G
(b) The decision tree is shown on the next page. The optimal policy is to perform the test and
then ship the unit if the test result is “good”. If the test result is “suspect”, then one should
1
perform burn-in before shipping the unit. The EMV is $3,877.13.
0.991843
Accepted
Ship
0
0
3935.53
0.91950
Good
0
3935.53
1
3960
$ 3,960
0.008157
Returned
$ (3,000)
960
$
960
Burn-In
$
(750)
$ 3,210
3210
Apply Test
$ 3,960
0.47205
Accepted
3877.125
Ship
0
0 2376.149
0.08050
Suspect
0
3210
2
3960
$ 3,960
0.52795
Returned
$ (3,000)
960
$
960
Burn-In
3877.125
1
$
(750)
3210
$ 3,210
0.95000
Accepted
Ship
0
$ 4,000
3850
4000
4000
0.05000
Returned
$ (3,000) $ 1,000
$ 1,000
Burn-In
$ 3,250
$ 3,250
3250
(c) From the tree, we calculate
EVSI = (EMV with free test) − (Best EMV with no test)
= (3877.13 + 40) − 3850
= $67.13
(d) To compute the EVPI, we first need the EMV with a free and perfect test, which we calculate
to be $3,962.50 from the following tree:
2
Good
Ship
0.95
4000
4000
-750
3250
-3000
1000
Burn In
Ship
0.05
4000
4000
1
3962.5
Bad
0
3250
2
Burn In
-750
3250
4000
3250
1000
3250
We may then calculate
EVPI = (EMV with free perfect test) − (Best EMV with no test)
= 3962.50 − 3850
= $112.50
Stocking Holiday Decorations
This is a “vanilla” critical fractile problem. We calculate
G = Gain from an extra order if it turns out to be needed
= $48 − $36
= $12
L = Loss from an extra order if it turns out not to be needed
= $36 − $18
= $18.
The critical fractile is then
p=
12
12
G
=
=
= 0.4.
G+L
12 + 18
30
We then calculate the cumulative distribution of the demand random variable D until we reach of
exceed the critical fractile p:
d P{D = d}
0
0.05
1
0.10
2
0.25
3
0.20
3
P{D ≤ d}
0.05
0.15
0.40
0.60
The first demand level that meets or exceeds the critical fractice is 2, so it is optimal to just order
2 units. However, since P{D ≤ 2} = 0.40, which is exactly equal to the critical fractile, we are
actually indifferent in and EMV sense between ordering 2 or 3 units, so 3 is also an optimal solution
in this case, and an answer of either “2” or “3” would be correct. This kind of phenomenon only
occurs when the cumulative distribution has a point where it exactly equals the critical fractile.
Bulk Orders
First note that G and L are completely unchanged in this scenario. Nothing in this scenario will
change them (you can try recomputing them if you want). What will change is the distribution of D
and therefore where the cut-off is will change. Let B be the event that you receive a bulk order of 5
decorations. Then P [B] = 0.2. The original table is the probability distribution of customers/orders
under B̄, i.e. the event that no bulk order is made. It is the distribution P [D = d|B̄]. If the bulk
order is made, you would get these 5 orders plus any from the usual customers. Thus the distribution
P [D = d|B] is just P [D = d|B̄] shifted up by 5 units. Therefore for 0 ≤ d ≤ 4, P [D = d|B] = 0
because you must get at least 5 units when B happens.
What you need is the “marginal” distribution which is not conditioned on B. You need
P [D = d] = P [D = d|B̄]P [B̄] + P [D = d|B]P [B]
= 0.8P [D = d|B̄] + 0.2P [D = d|B].
Thus for every d = 0, 1, 2, . . . we can calculate the marginal as follows
d P D = d|B̄
P{D = d|B} P{D = d} P{D ≤ d}
0
0.05
0
0.04
0.04
1
0.10
0
0.08
0.12
2
0.25
0
0.2
0.32
3
0.20
0
0.16
0.48.
Thus the solution is to order three units.
Production Planning
This is a deterministic dynamic programming problem:
ˆ There are 4 stages, corresponding to the 4 plants.
ˆ The state is how many batches of chemical still need to be produced.
The values we need to calculate take the form
ft (i) = Lowest attainable cost by producing i batches among plants t through 4.
4
We start with stage 4, which we can essentially just read off the table (in millions of dollars)
f4 (0) = 0
f4 (1) = 1500
f4 (2) = 2100
f4 (3) = 2800
f4 (4) = 3600
f4 (5) = 5100.
Next, we consider stage 3, calculating
f3 (0) = 0 + f4 (0) = 0 + 0 = 0∗
(
Produce 0 : 0 + f4 (1) = 0 + 1500 = 1500
f3 (1) = min
Produce 1 : 900 + f4 (0) = 900 + 0 = 900∗


Produce 0 : 0 + f4 (2) = 0 + 2100 = 2100
f3 (2) = min Produce 1 : 900 + f4 (1) = 900 + 1500 = 2400


Produce 2 : 1800 + f4 (0) = 1800 + 0 = 1800∗

Produce



Produce
f3 (3) = min

Produce



Produce
0:
1:
2:
3:
0 + f4 (3) = 0 + 2800 = 2800
900 + f4 (2) = 900 + 2100 = 3000
1800 + f4 (1) = 1800 + 1500 = 3300
2700 + f4 (0) = 2700 + 0 = 2700∗


Produce





Produce
f3 (4) = min Produce



Produce



Produce
0:
1:
2:
3:
4:
0 + f4 (4) = 0 + 3600 = 3600∗
900 + f4 (3) = 900 + 2800 = 3700
1800 + f4 (2) = 1800 + 2100 = 3900
2700 + f4 (1) = 2700 + 1500 = 4200
4000 + f4 (0) = 4000 + 0 = 4000

Produce





Produce



Produce
f3 (5) = min

Produce





Produce



Produce
0:
1:
2:
3:
4:
5:
0 + f4 (5) = 0 + 5100 = 5100
900 + f4 (4) = 900 + 3600 = 4500∗
1800 + f4 (3) = 1800 + 2800 = 4600
2700 + f4 (2) = 2700 + 2100 = 4800
4000 + f4 (1) = 4000 + 1500 = 5500
4600 + f4 (0) = 4600 + 0 = 4600
5
We now proceed to stage 2, calculating
f2 (0) = 0 + f3 (0) = 0 + 0 = 0∗
(
Produce 0 : 0 + f3 (1) = 0 + 900 = 900∗
f2 (1) = min
Produce 1 : 1200 + f3 (0) = 1200 + 0 = 1200


Produce 0 : 0 + f3 (2) = 0 + 1800 = 1800∗
f2 (2) = min Produce 1 : 1200 + f3 (1) = 1200 + 1800 = 2100


Produce 2 : 1900 + f3 (0) = 1900 + 0 = 1900∗

Produce



Produce
f2 (3) = min

Produce



Produce
0:
1:
2:
3:
0 + f3 (3) = 0 + 2700 = 2700
1200 + f3 (2) = 1200 + 1800 = 3000
1900 + f3 (1) = 1900 + 900 = 2800
2700 + f3 (0) = 2600 + 0 = 2600∗


Produce





Produce
f2 (4) = min Produce



Produce



Produce
0:
1:
2:
3:
4:
0 + f3 (4) = 0 + 3600 = 3600
1200 + f3 (3) = 1200 + 2700 = 3900
1900 + f3 (2) = 1900 + 1800 = 3700
2600 + f3 (1) = 2600 + 900 = 3500∗
4100 + f3 (0) = 4100 + 0 = 4100

Produce





Produce



Produce
f2 (5) = min

Produce





Produce



Produce
0:
1:
2:
3:
4:
5:
0 + f3 (5) = 0 + 4500 = 4500
1200 + f3 (4) = 1200 + 3600 = 4800
1900 + f3 (3) = 1900 + 2700 = 4600
2600 + f3 (2) = 2600 + 1800 = 4400∗
4100 + f3 (1) = 4100 + 900 = 5000
4700 + f3 (0) = 4700 + 0 = 4700
Finally, in stage 1 we only need to compute f1 (i) for i = 5, since we know we have 5 representatives
to start with. So, we calculate

Produce 0 : 0 + f2 (5) = 0 + 4400 = 4400





Produce 1 : 1000 + f2 (4) = 1000 + 3500 = 4500



Produce 2 : 1800 + f (3) = 1800 + 2600 = 4400
2
f1 (5) = min

Produce 3 : 2500 + f2 (2) = 2500 + 1800 = 4300∗





Produce 4 : 4300 + f2 (1) = 4300 + 900 = 5200



Produce 5 : 5000 + f2 (0) = 5000 + 0 = 5000
6
We immediately conclude that the optimal production cost is $4,300. Reviewing the “*” markers
in the calculation above, this maximum is acheived by producing
ˆ 3 batches in plant 1 (examining the calculation of f1 (5))
ˆ 0 batches in plant 2 (examining the calculation of f2 (2))
ˆ 2 batches in plant 3 (examining the calculation of f3 (2))
ˆ 0 batches in plant 4 (the only possibility in f4 (0)).
More succinctly, we produce 3 batches in plant 1, and 2 batches in plant 3.
Managing a Repair Facility
This is a stochastic dynamic programming problem:
ˆ Stage t will denote the beginning of week t = 1, . . . , 4. We will also include a stage 5 to denote
the end of the time horizon, or equivalently the end of week 4.
ˆ The state i will represent the number of machines awaiting repair at the start of the week
(before any machines break down for that week). This value can range from 0 to 3.
Consequently, the values we intend to calculate are
ft (i) = Minimal expected cost from the start of week t to the end of the time horizon, given that
i units are awaiting repair.
As usual, we begin the calculations at the end of the time horizon. The problem says to assume
that we eventually repair each machine left at the end of the time horizon at a cost of $4000 each,
so we have
f5 (0) = 0
f5 (1) = 4000
f5 (2) = 8000
f5 (3) = 12000.
We next proceed to stage 4, and make the following observations:
If there are 0 machines awaiting repair, the only option is to repair 0 machines.
If there is 1 machine awaiting repair, we may choose not to repair it, incurring the opportunity cost of $1000, or we may repair it at a cost of $4000.
If there are 2 machines awaiting repair, we must repair at least one of them because otherwise we could run out of room if two more machines break down. Repairing one machine
costs $4000, but will also incur a $1000 opportunity cost because one machine will remain
unrepaired at the end of the week. We also have the option of repairing two machines at a
cost of $10000, with no opportunity cost.
If there are 3 machines awaiting repair, our only option is to repair two of them (at a cost of
$10000) because otherwise there is a high probability of running out of room. But that still
leaves one machine unrepaired at the end of the week, so we also incur the $1000 opportunity
cost.
7
We accordingly calculate:
f4 (0) = 0 + 0.2f5 (0) + 0.5f5 (1) + 0.3f5 (2)
= 0 + 0.2 · 0 + 0.5 · 4000 + 0.3 · 8000
= 4400 ∗
(
Repair 0: 1000 + 0.2f5 (1) + 0.5f5 (2) + 0.3f5 (3)
f4 (1) = min
Repair 1: 4000 + 0.2f5 (0) + 0.5f5 (1) + 0.3f5 (2)
(
Repair 0: 1000 + 0.2 · 4000 + 0.5 · 8000 + 0.3 · 12000
= min
Repair 1: 4000 + 0.2 · 0 + 0.5 · 4000 + 0.3 · 8000
(
Repair 0: 1000 + 800 + 4000 + 3600 = 9400
= min
Repair 1: 4000 + 0 + 2000 + 2400 = 8400 ∗
(
Repair
f4 (2) = min
Repair
(
Repair
= min
Repair
(
Repair
= min
Repair
1: 4000 + 1000 + 0.2f5 (1) + 0.5f5 (2) + 0.3f5 (3)
2: 10000 + 0.2f5 (0) + 0.5f5 (1) + 0.3f5 (2)
1: 4000 + 1000 + 0.2 · 4000 + 0.5 · 8000 + 0.3 · 12000
2: 10000 + 0.2 · 0 + 0.5 · 4000 + 0.3 · 8000
1: 4000 + 1000 + 800 + 4000 + 3600 = 13400 ∗
2: 10000 + 0 + 2000 + 2400 = 14400
f4 (3) = 10000 + 1000 + 0.2f5 (1) + 0.5f5 (2) + 0.3f5 (3)
= 10000 + 1000 + 0.2 · 4000 + 0.5 · 8000 + 0.3 · 12000
= 10000 + 1000 + 800 + 4000 + 3600 = 19400 ∗
In each of these calculations, we are taking an expected value over the number of machines that
break down during the period, which is random. Stage 3 involves a similar set of calculations:
f3 (0) = 0 + 0.2f4 (0) + 0.5f4 (1) + 0.3f4 (2)
= 0 + 0.2 · 4400 + 0.5 · 8400 + 0.3 · 13400
= 9100 ∗
(
Repair 0: 1000 + 0.2f4 (1) + 0.5f4 (2) + 0.3f4 (3)
f3 (1) = min
Repair 1: 4000 + 0.2f4 (0) + 0.5f4 (1) + 0.3f4 (2)
(
Repair 0: 1000 + 0.2 · 8400 + 0.5 · 13400 + 0.3 · 19400
= min
Repair 1: 4000 + 0.2 · 4400 + 0.5 · 8400 + 0.3 · 13400
(
Repair 0: 1000 + 1680 + 6700 + 5820 = 15200
= min
Repair 1: 4000 + 2200 + 4200 + 4020 = 13100 ∗
8
(
Repair
f3 (2) = min
Repair
(
Repair
= min
Repair
(
Repair
= min
Repair
1: 4000 + 1000 + 0.2f4 (1) + 0.5f4 (2) + 0.3f4 (3)
2: 10000 + 0.2f4 (0) + 0.5f4 (1) + 0.3f4 (2)
1: 4000 + 1000 + 0.2 · 8400 + 0.5 · 13400 + 0.3 · 19400
2: 10000 + 0.2 · 4400 + 0.5 · 8400 + 0.3 · 13400
1: 4000 + 1000 + 1680 + 6700 + 5820 = 19200
2: 10000 + 880 + 4200 + 4020 = 19100 ∗
f3 (3) = 10000 + 1000 + 0.2f4 (1) + 0.5f4 (2) + 0.3f4 (3)
= 10000 + 1000 + 0.2 · 8400 + 0.5 · 13400 + 0.3 · 19400
= 10000 + 1000 + 4200 + 6700 + 5820 = 25200 ∗
Stage 2 comes next, with another similar string of calculations
f2 (0) = 0 + 0.2f3 (0) + 0.5f3 (1) + 0.3f3 (2)
= 0 + 0.2 · 9100 + 0.5 · 13100 + 0.3 · 19100
= 14100 ∗
(
Repair 0: 1000 + 0.2f3 (1) + 0.5f3 (2) + 0.3f3 (3)
f2 (1) = min
Repair 1: 4000 + 0.2f3 (0) + 0.5f3 (1) + 0.3f3 (2)
(
Repair 0: 1000 + 0.2 · 13100 + 0.5 · 19100 + 0.3 · 25200
= min
Repair 1: 4000 + 0.2 · 9100 + 0.5 · 13100 + 0.3 · 19100
(
Repair 0: 1000 + 2620 + 9550 + 7560 = 20730
= min
Repair 1: 4000 + 1820 + 6550 + 5730 = 18100 ∗
(
Repair
f2 (2) = min
Repair
(
Repair
= min
Repair
(
Repair
= min
Repair
1: 4000 + 1000 + 0.2f3 (1) + 0.5f3 (2) + 0.3f3 (3)
2: 10000 + 0.2f3 (0) + 0.5f3 (1) + 0.3f3 (2)
1: 4000 + 1000 + 0.2 · 13100 + 0.5 · 19100 + 0.3 · 19400
2: 10000 + 0.2 · 9100 + 0.5 · 13100 + 0.3 · 19100
1: 4000 + 1000 + 2620 + 9550 + 5820 = 24730
2: 10000 + 1820 + 6550 + 5730 = 24100 ∗
f2 (3) = 10000 + 1000 + 0.2f3 (1) + 0.5f3 (2) + 0.3f3 (3)
= 10000 + 1000 + 0.2 · 13100 + 0.5 · 19100 + 0.3 · 19400
= 10000 + 1000 + 2620 + 9550 + 5820 = 30730 ∗
Finally, for stage 1 we only need to consider state 1, so we calculate
(
Repair 0: 1000 + 0.2f2 (1) + 0.5f2 (2) + 0.3f2 (3)
f1 (1) = min
Repair 1: 4000 + 0.2f2 (0) + 0.5f2 (1) + 0.3f2 (2)
9
(
Repair
= min
Repair
(
Repair
= min
Repair
0: 1000 + 0.2 · 18100 + 0.5 · 24100 + 0.3 · 30730
1: 4000 + 0.2 · 14100 + 0.5 · 18100 + 0.3 · 24100
0: 1000 + 3620 + 12050 + 9219 = 25889
1: 4000 + 2820 + 9050 + 7230 = 23100 ∗
We conclude that the optimal expected cost is $23,100 and (by examining the “*” markers) that
the optimal policy is as follows:
Week 1: Repair 1 unit
Week 2: Repair as many units as possible in each state
Week 3: Repair as many units as possible in each state
Week 4: Repair 1 unit if 1 or 2 are broken (if 0 or 3 are broken, the only choices are to repair none
or 2, respectively).
Waiting in Line at the Car Wash
We know
Average interarrival time = 1/λ = 15 minutes, so
Arrival rate = λ = 1/15 cars/minute
σ ≈ 0 (because the service time variance is negligible).
This is single server queue with presumably memoryless arrivals, so we can use the PollaczekKhinchin formulas. To meet the 5-minute average wait requirement, we need
5 minutes = Wq =
λ2 σ 2 + ρ2 1
Lq
=
· .
λ
2(1 − ρ) λ
We may drop the λ2 σ 2 term from the numerator, because we know that σ ≈ 0. Since we know
1/λ = 15 minutes, we have
5 minutes =
ρ2
· (15 minutes) ,
2(1 − ρ)
and hence, dividing through by 15 minutes,
1
ρ2
=
3
2(1 − ρ)
⇔
⇔
2
(1 − ρ) = ρ2
3
0 = ρ2 + (2/3)ρ − (2/3)
We can solve this equation with the quadratic formula, which says that
The solutions of ax + bx2 + c = 0 are given by x =
10
−b ±
√
b2 − 4ac
.
2a
Setting a = 1, b = 2/3, and c = −(2/3), this formula yields
p
p
−(2/3) ± (2/3)2 − 4(−2/3)
−(2/3) ± (2/3)2 + 4(2/3)
ρ=
=
.
2
2
The “minus” option of the ± will yield a negative value of ρ, which makes no sense, so we can focus
on the “plus” option, obtaining
p
1 p
−(2/3) + (2/3)2 + 4(2/3)
1 p
=
ρ=
4/9 + 8/3 − 2/3 =
28/9 − 2/3
2
2
2
1 √
√
1
(2/3) 7 − 2/3 =
7 − 1 ≈ 0.549.
=
2
3
So we need ρ to be about 0.549 to get the desired performance. Since ρ = λ/µ, it must be that the
service time, which is 1/µ, obeys the equation 1/µ = ρ/λ (we obtain this by dividing through by
λ). So, we need
ρ
1
= ≈
µ
λ
0.549
1
15 minutes
= (0.549)(15 minutes) ≈ 8.23 minutes .
So we need a service time of about 8.2 minutes.
Response Time of a Query System
We begin by calculating
E [S] = 0.2 · 1 sec + 0.3 · 2 sec + 0.4 · 3 sec + 0.1 · 4 sec = 2.4 sec
E S 2 = 0.2 · 12 sec2 + 0.3 · 22 sec2 + 0.4 · 32 sec2 + 0.1 · 42 sec2
= 0.2 · 1 sec2 + 0.3 · 4 sec2 + 0.4 · 9 sec2 + 0.1 · 16 sec2
= 6.6 sec2 .
We immediately conclude that µ = 1/ E [S] = 1/(2.4 sec). Furthermore, since σ 2 is the variance of
S, we have
σ 2 = E S 2 − (E [S])2 = 6.6 sec2 − (2.4 sec)2 = 0.84 sec2 .
We are now in a position to answer the questions:
(a) In this case, we have 1/λ = 10 sec, and hence λ = 0.1 sec−1 . We then calculate
1/µ
2.4 sec
λ
=
=
= 0.24
µ
1/λ
10 sec
σ 2 λ2 + ρ 2
0.84(0.1)2 + (0.24)2
Lq =
=
≈ 0.0434
2(1 − ρ)
2(1 − 0.24)
Lq
0.0434
Wq =
≈
= 0.434 sec
λ
0.1 sec−1
W = Wq + Ws = Wq + E [S] ≈ 0.434 sec + 2.4 sec = 2.83 sec .
ρ=
11
(b) Now we instead have 1/λ = 5 sec, and hence λ = 0.2 sec−1 . So now we get
λ
1/µ
2.4 sec
=
=
= 0.48
µ
1/λ
5 sec
σ 2 λ2 + ρ2
0.84(0.2)2 + (0.48)2
Lq =
=
≈ 0.254
2(1 − ρ)
2(1 − 0.48)
0.254
Lq
≈
Wq =
= 1.27 sec
λ
0.2 sec−1
W = Wq + Ws = Wq + E [S] ≈ 1.27 sec + 2.4 sec = 3.67 sec .
ρ=
Checkout Times at a Convenience Store
(a) We are given that λ is 20 customers/hour, or 1/3 customers per minute. Thus, the average
interarrival time 1/λ must be 3 minutes. Since the cashier is busy in 65% of the pictures in
our sample, we surmise that ρ ≈ 0.65, there being only one server. Similarly, since there are
an average of 1.4 customers in each picture, we may surmise that L ≈ 1.4.
(b) We know ρ = λ/µ ≈ 0.65. Multiplying both sides by 1/λ, we have ρ · (1/λ) = 1/µ, so
E [S] = 1/µ = ρ · (1/λ) = (0.65)(3 minutes) = 1.95 minutes .
(c) We know that L = Lq + Ls = Lq + ρ, the second equality being because this system has only
one server. Subtracting ρ from both sides, we have
Lq = L − ρ = 1.4 − 0.65 = 0.75 .
(d) We use Little’s law on the queue part of the system, which says that Lq = λWq , hence
Wq = Lq · (1/λ) = (0.75)(3 minutes) = 2.25 minutes .
(e) Since there is only one server and arrivals are presumably memoryless, we may apply the
Pollaczek-Khinchin formula,
λ2 σ 2 + ρ2
.
Lq =
2(1 − ρ)
We already know every variable in this formula except σ, so we can simply solve for it. This
yields
2Lq (1 − ρ) = λ2 σ 2 + ρ2
2Lq (1 − ρ) − ρ2 = λ2 σ 2
1
2
·
2L
(1
−
ρ)
−
ρ
= σ2
q
2
λ q
1
· 2Lq (1 − ρ) − ρ2 = σ.
λ
So, we may calculate
q
p
1
σ = · 2Lq (1 − ρ) − ρ2 = (3 minutes) 2(0.75)(0.35) − (0.65)2 ≈ 0.960 minutes .
λ
12
Part replacement and maintenance
ˆ The stages are the four days
ˆ In each stage, the state is the condition of the part, excellent (E), satisfactory (S), or broken
(B)
ˆ In each stage/state combination, the decision is whether to replace the part, perform maintenance (or repair, in the case of a broken part), or do nothing.

200
 Replace:
f4 (E) = max Maintenance: 800

Do nothing: 1000 ∗

200
 Replace:
f4 (S) = max Maintenance: 800

Do nothing: 1000 ∗

200
 Replace:
400 ∗
f4 (B) = max Repair:

Do nothing: 0

 Replace:
f3 (E) = max Maintenance:

Do nothing:

 Replace:
= max Maintenance:

Do nothing:

 Replace:
= max Maintenance:

Do nothing:
200 + f4 (E)
800 + 0.70f4 (E) + 0.25f4 (S) + 0.05f4 (B)
1000 + 0.40f4 (E) + 0.50f4 (S) + 0.10f4 (B)

 Replace:
f3 (S) = max Maintenance:

Do nothing:

 Replace:
= max Maintenance:

Do nothing:

 Replace:
= max Maintenance:

Do nothing:
200 + f4 (E)
800 + 0.30f4 (E) + 0.50f4 (S) + 0.20f4 (B)
1000 + 0.60f4 (S) + 0.40f4 (B)
200 + 1000
800 + 0.70 · 1000 + 0.25 · 1000 + 0.05 · 400
1000 + 0.40 · 1000 + 0.50 · 1000 + 0.10 · 400
1200
1770
1940 ∗
200 + 1000
800 + 0.30 · 1000 + 0.50 · 1000 + 0.20 · 400
1000 + 0.60 · 1000 + 0.40 · 400
1200
1680
1760 ∗
13

 Replace:
f3 (B) = max Repair:

Do nothing:

 Replace:
= max Repair:

Do nothing:

 Replace:
= max Repair:

Do nothing:
200 + f4 (E)
400 + 0.20f4 (E) + 0.60f4 (S) + 0.20f4 (B)
0 + f4 (B)
200 + 1000
400 + 0.20 · 1000 + 0.60 · 1000 + 0.20 · 400
0 + 400
1200
1280 ∗
400

 Replace:
f2 (E) = max Maintenance:

Do nothing:

 Replace:
= max Maintenance:

Do nothing:

 Replace:
= max Maintenance:

Do nothing:
200 + f3 (E)
800 + 0.70f3 (E) + 0.25f3 (S) + 0.05f3 (B)
1000 + 0.40f3 (E) + 0.50f3 (S) + 0.10f3 (B)

 Replace:
f2 (S) = max Maintenance:

Do nothing:

 Replace:
= max Maintenance:

Do nothing:

 Replace:
= max Maintenance:

Do nothing:
200 + f3 (E)
800 + 0.30f3 (E) + 0.50f3 (S) + 0.20f3 (B)
1000 + 0.60f3 (S) + 0.40f3 (B)

 Replace:
f2 (B) = max Repair:

Do nothing:

 Replace:
= max Repair:

Do nothing:

 Replace:
= max Repair:

Do nothing:
200 + 1940
800 + 0.70 · 1940 + 0.25 · 1760 + 0.05 · 1280
1000 + 0.40 · 1940 + 0.50 · 1760 + 0.10 · 120
2140
2662
2784 ∗
200 + 1940
800 + 0.30 · 1940 + 0.50 · 1760 + 0.20 · 1280
1000 + 0.60 · 1760 + 0.40 · 1280
2140
2518
2568 ∗
200 + f3 (E)
400 + 0.20f3 (E) + 0.60f3 (S) + 0.20f3 (B)
0 + f3 (B)
200 + 1940
400 + 0.20 · 1940 + 0.60 · 1760 + 0.20 · 1280
0 + 1280
2140 ∗
2100
1280
14

 Replace:
f1 (S) = max Maintenance:

Do nothing:

 Replace:
= max Maintenance:

Do nothing:

 Replace:
= max Maintenance:

Do nothing:
200 + f2 (E)
800 + 0.30f2 (E) + 0.50f2 (S) + 0.20f2 (B)
1000 + 0.60f2 (S) + 0.40f2 (B)
200 + 2784
800 + 0.30 · 2784 + 0.50 · 2568 + 0.20 · 2140
1000 + 0.60 · 2568 + 0.40 · 2140
2984
3336.4
3396.8 ∗
From these calculations, we deduce the following optimal policy:
Day 1. Do nothing
Day 2. Replace the part if it is broken, and otherwise do nothing
Day 3. Repair the part if it is broken, and otherwise do nothing
Day 4. Repair the part if it is broken, and otherwise do nothing.
The expected value from this policy is $3,396.80.
Sizing a Temporary Worker Pool
If we hire an extra worker into the pool and turn out to be able to use them, we avoided a cost of
$280 for a temporary worker, but incurred a cost of $170, so G = 280 − 170 = 110.
On the other hand, if we hire an extra worker into the pool whom we are not able to use, we
incurred an expense of $65, so L = 65.
Therefore,
G
110
110
=
=
≈ 0.629.
G+L
110 + 65
175
Since we have observed exactly 100 past days, the “number of days” column in the table may treated
as a percentage probability in an empirical demand distribution. Calculating the corresponding
cumulative distribution,
Workers Needed
0
3
4
6
7
8
9
10
11
13
14
15
17
Probability (%)
2
4
4
3
5
4
6
8
3
4
7
8
6
15
Cumulative Probability
0.02
0.06
0.10
0.13
0.18
0.22
0.28
0.36
0.39
0.43
0.50
0.58
0.64 *
17 workers is the first level whose cumulative probability level exceeds 0.629, so we should hire 17
workers into the pool.
Operating a Credit Union Office
(a) E [S] = (0.5)(2 + 2) + (0.5)(3 + 2) = (0.5 · 2 + 0.5 · 3) + 2 = 4.5 hours.
Therefore, µ = 1/ E [S] = 1/4.5 = 2/9 ≈ 0.222.
(b) The arrival rate is λ = 1/5 = 0.2 customers per hour.
The utilization factor ρ is thus ρ = λ/µ = 0.2/(2/9) = 0.9.
(c) E [S 2 ] = (0.5)(2 + 2)2 + (0.5)(3 + 2)2 = 0.5 · 16 + 0.5 · 25 = 20.5 hours2
Therefore, σ 2 = E [S 2 ] − E [S]2 = 20.5 − 4.52 = 20.5 − 20.25 = 0.25 hours2
(d) Because the bank has a large number of customers, the time between arrival may be assumed
to be exponentially distributed. Since there is only one server for this queue (the single
employee), the exponential interarrival times mean that we may use the Pollaczek-Khinchin
formula, which predicts that the average number of customers waiting in queue is
Lq =
λ2 σ s + ρ2
(0.2)2 0.25 + 0.92
0.01 + 0.81
0.82
=
=
=
= 4.1.
2(1 − ρ)
2(1 − 0.9)
2(0.1)
0.2
(e) Applying Little’s law to the queue portion of the system, we have Lq = λWq , and hence
Wq = Lq /λ. This yields an average queuing time of Wq = Lq /λ = 4.1/0.2 = 20.5 hours.
Finally, the average time until the applicant sees a response is W = Wq + Ws = Wq + E [S] =
20.5 + 4.5 = 25 hours.
(f) We recompute
E [S] = (0.5)(2 + 0.5) + (0.5)(3 + 0.5) = (0.5 · 2 + 0.5 · 3) + 0.5 = 3 hours.
µ = 1/ E [S] = 1/3 ≈ 0.333
ρ = λ/µ = 0.2/(1/3) = 0.6
2
E S = (0.5)(2 + 0.5)2 + (0.5)(3 + 0.5)2 = 0.5 · 2.52 + 0.5 · 3.52 = 9.25 hours2
σ 2 = E S 2 − E [S]2 = 9.25 − 32 = 0.25 hours2
(0.2)2 0.25 + 0.62
0.01 + 0.36
0.37
λ2 σ s + ρ 2
Lq =
=
=
=
= 0.4625
2(1 − ρ)
2(1 − 0.6)
2(0.4)
0.8
Wq = Lq /λ = 0.4625/0.2 = 2.3125 hours
W = Wq + Ws = Wq + E [S] = 2.3125 + 3 = 5.3125 hours.
Therefore, customers will experience an average speedup of 25 − 5.3125 = 19.6875 hours.
Please note that the information that the bank has 200,000 customers is only relevant insofar that
it tells us that the bank has a large customer pool and that a exponential arrival process is a proper
model, hence we can use the Pollaczek-Khinchin formula. The exact number of customers is not
16
important, only that the customer pool is large. The other piece of information needed to apply
the Pollaczek-Khinchin formula is that there is only one server.
It is possible to take a shortcut in this problem by realizing that the new office equipment just
subtracts a constant from the service time and therefore cannot alter the variance of the service
time. So we don’t really need to recompute σ 2 when we do the last part of the problem. A common
error on this problem is to miscalculate E [S 2 ]: for example, the formula (0.5 · 22 + 0.5 · 32 ) + 22 does
not yield the correct value of E [S 2 ] in part (b). If you use this incorrect value, you get a negative
variance, which is impossible.
Scripting sales calls
Letting A stand for the event of being willing to buy deal A and event B stand for the being willing
to buy deal B, we compute the following probabilities from the table:
P{A} = P{A ∩ B} + P A ∩ B = 0.02 + 0.01 = 0.03
P{B} = P{A ∩ B} + P A ∩ B = 0.02 + 0.03 = 0.05
P A∩B
0.01
1
=
P A|B =
=
≈ 0.010526
1 − 0.05
95
P B
P B∩A
3
0.03
=
≈ 0.030928
=
P B|A =
1 − 0.03
97
P A
We may then develop the tree shown on the next page. It is best to offer deal B first, and the
expected profit per call is $15.50.
17
0.03
Buys A
$ 350.00
$ 350.00 $ 350.00
0.030928
Buys B
Offer A first
0
0.5
Allows an offer of B
14.625
0
$ 275.00
$ 275.00 $ 275.00
8.505155
0.97
Refuses A
0.969072
Refuses B
0
0
0
0
4.252577
0.5
Hangs up
2
0
15.5
0
0
0.05
Buys B
$ 275.00
$ 275.00 $ 275.00
0.010526
Buys A
Offer B first
0
0.5
Allows and offer of A
15.5
0
0.95
Refuses B
3.684211
$ 350.00
$ 350.00 $ 350.00
0.989474
Rufuses A
0
0
0
0
1.842105
0.5
Hangs up
0
Tree for the “scripting sales calls” question
18