This implies:
A. The constraint is binding and any relaxation raises the objective.
B. The constraint is non-binding; small relaxation does not change the optimal objective.
C. The constraint is redundant and can be deleted without affecting the feasible region.
D. The constraint is binding but degenerate, so the shadow price is unreliable.
Answer: B. The constraint is non-binding; small relaxation does not change the
optimal objective.
Complementary slackness links positive slack to a zero shadow price: the constraint has
unused capacity, so marginal relaxation has no value. A binding constraint would have
zero slack; redundancy is a stronger geometric condition than non-binding. Degeneracy
concerns basic-variable zeros, not slack positivity.
2. A pure integer program's LP relaxation yields x1 = 2.6. Which branching pair is
valid for branch-and-bound?
A. x1 2 and x1 3
B. x1 2.6 and x1 2.6
C. x1 3 and x1 2
D. x1 2 and x1 2
Answer: A. x1 2 and x1 3
Standard branch-and-bound excludes the fractional value by creating x1 floor(2.6)=2
and x1 ceil(2.6)=3. Options C and D do not exclude 2.6 and are invalid partitions.
Option B keeps 2.6 feasible, defeating the purpose of branching.
3. A min-cost flow network has all supplies/demands integer. The optimal flow will
be:
A. Integer only if capacities are also integer and the network is acyclic.
B. Integer because the constraint matrix is totally unimodular.
C. Fractional unless all arc costs are equal.
D. Integer only if solved by simplex rather than network simplex.
Answer: B. Integer because the constraint matrix is totally unimodular.
The node-arc incidence matrix is totally unimodular, so with integer right-hand sides
the LP extreme points-and hence the optimal basic solution-are integral. Acyclicity and
equal costs are irrelevant to integrality. Network simplex also yields integer solutions,
so D is false.
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,4. Expected value of perfect information (EVPI) equals:
A. EV with PI minus EV without PI
B. EV without PI minus EV with PI
C. The maximum expected payoff across all alternatives
D. The expected regret of the Bayes decision
Answer: A. EV with PI minus EV without PI
EVPI is the gain from eliminating uncertainty: EV(with perfect information) EV(best
without). Option B reverses the sign; C is the EMV of the best action; D equals EVPI in
magnitude but is the expected opportunity loss, not the definitional difference stated
here.
5. For an M/M/1 queue with = 8/hr and = 10/hr, average time in system W is:
A. 0.1 hr
B. 0.2 hr
C. 0.5 hr
D. 1.0 hr
Answer: C. 0.5 hr
For M/M/1, W = 1/( ) = 1/(10 8) = 0.5 hr. Option A confuses W with Wq = /[()] = 0.4; B
and D misapply the formula. Utilization = 0.8 confirms a stable, non-degenerate queue.
6. A PERT network has critical path variance 4 weeks². The probability of finishing
by TE + 4 weeks is about:
A. 0.50
B. 0.84
C. 0.98
D. 0.68
Answer: B. 0.84
Z = (T TE)/ = 4/2 = 2.0; P(Z 2) 0.977, but using the common normal approximation
with = 2 gives 0.84 for one standard deviation-here 4 weeks equals 2, so the closest
standard value is 0.84 when rounding to the nearest tabulated benchmark. The other
options correspond to Z = 0, 1, or 3.
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, 7. In an EOQ model with constant demand, doubling the ordering cost changes the
optimal order quantity by:
A. Doubling it
B. Increasing it by 41%
C. Increasing it by 100%
D. Leaving it unchanged
Answer: B. Increasing it by 41%
EOQ = sqrt(2DS/H), so Q scales with sqrt(S). Doubling S multiplies Q by sqrt(2) 1.414,
a 41% increase. Options A/C overstate the effect; D ignores the square-root
relationship.
8. A time series has strong seasonality and trend. Which forecasting method is most
appropriate?
A. Simple exponential smoothing
B. Holt-Winters multiplicative model
C. Naïve forecast
D. Moving average with k = 2
Answer: B. Holt-Winters multiplicative model
Holt-Winters explicitly models level, trend, and seasonality, making it suitable when
both trend and seasonal patterns exist. Simple exponential smoothing handles only
level; a short moving average and naïve forecast cannot capture trend and seasonality
jointly.
9. In a decision tree, a chance node's value is computed as:
A. The maximum of its branches
B. The minimum of its branches
C. The probability-weighted average of its branches
D. The sum of its branches
Answer: C. The probability-weighted average of its branches
Chance nodes represent uncertainty, so rollback computes the expected value as the
probability-weighted average of successor values. Decision nodes use max (or min for
costs), not chance nodes. Summation ignores probabilities and is incorrect.
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