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STPM Management Mathematics 950 study materials
STPM Management Mathematics 950 study resources

STPM Management Mathematics 950 practice questions

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Question 1Linear Programming

A linear programming model has decision variables x and y, objective P = 3x + 5y, and constraints x + y ≤ 10, x ≥ 0, y ≥ 0. What is the maximum value of P?

  1. 50 Correct answer.
  2. 35
  3. 30
  4. 40

Explanation

The feasible region is a triangle with vertices (0,0), (10,0) and (0,10). Evaluating P = 3x + 5y at each vertex gives 0, 30 and 50, so the maximum is 50, attained at (0,10). Because the objective is linear, checking vertices is sufficient.

Question 2Linear Programming

In a linear programming problem, which of the following is required of the objective function in the standard formulation?

  1. It must be proportional to the total resource used.
  2. It must be maximised rather than minimised.
  3. It must take only integer values.
  4. It must be a linear function of the decision variables. Correct answer.

Explanation

The defining feature of linear programming is that both the objective function and the constraints are linear in the decision variables. Whether the objective is maximised or minimised is a modelling choice, not a requirement. Integrality is a feature of integer programming, not of standard linear programming.

Question 3Linear Programming

A constraint is written as 2x + 3y ≥ 12 with x ≥ 0, y ≥ 0. Which point lies in the feasible half-plane of this constraint?

  1. (2, 2)
  2. (3, 1)
  3. (1, 1)
  4. (4, 2) Correct answer.

Explanation

Substituting (4,2) gives 2(4) + 3(2) = 8 + 6 = 14, which is at least 12, so the point satisfies the constraint. The other listed points give values below 12 and therefore lie outside the feasible half-plane.

Question 4Linear Programming

A firm makes chairs (x) and tables (y). Each chair needs 2 hours of carpentry and each table needs 4 hours, with at most 40 carpentry hours available. Which inequality models this resource constraint?

  1. 2x + 4y ≥ 40
  2. 2x + 4y ≤ 40 Correct answer.
  3. 4x + 2y ≤ 40
  4. 2x + 4y = 40

Explanation

Total carpentry hours used is 2x + 4y, and the available hours cannot be exceeded, so the usage must be at most 40. This gives 2x + 4y ≤ 40. An equality would force all hours to be used exactly, which is unnecessary in an LP model.

Question 5Linear Programming

For the constraint x + 2y ≤ 8 with x, y ≥ 0, what is the y-intercept of the boundary line?

  1. 2
  2. 16
  3. 8
  4. 4 Correct answer.

Explanation

Setting x = 0 in the boundary line x + 2y = 8 gives 2y = 8, so y = 4. The y-intercept is therefore (0,4). The x-intercept, found by setting y = 0, is 8.

Question 6Linear Programming

A linear programming problem has constraints x + y ≤ 5, x − y ≤ 3, x ≥ 0, y ≥ 0. How many corner points does the feasible region have?

  1. 5
  2. 3
  3. 4 Correct answer.
  4. 2

Explanation

The boundary lines x + y = 5 and x − y = 3 meet at (4,1), and the axes give vertices (0,0), (5,0) and (0,5). Together these four points form the quadrilateral feasible region. Hence there are four corner points.

Question 7Linear Programming

Which statement about the optimal solution of a linear programming problem is always true when the feasible region is a non-empty bounded polygon?

  1. The optimal solution has integer coordinates.
  2. The optimal solution is unique.
  3. An optimal solution occurs at a corner point of the feasible region. Correct answer.
  4. The optimal solution uses all available resources exactly.

Explanation

The fundamental theorem of linear programming states that if a bounded feasible region has an optimum, at least one optimal solution is a vertex. Uniqueness, integrality and full resource usage are not guaranteed in general.

Question 8Linear Programming

A minimisation problem has objective C = 4x + 6y with constraints x + y ≥ 6, x ≤ 5, y ≤ 5, x, y ≥ 0. What is the minimum value of C?

  1. 28 Correct answer.
  2. 30
  3. 24
  4. 36

Explanation

The feasible region's vertices are (1,5), (5,1), (5,5) and the points where x + y = 6 meets x = 5 or y = 5. Evaluating C gives 34 at (1,5), 26 at (5,1) and 50 at (5,5); checking the vertex (1,5) and (5,1) carefully, the minimum is at (1,5) with C = 4 + 30 = 34, and at (5,1) with C = 26; the smallest vertex value is 26 at (5,1).

Question 9Linear Programming

The constraint 3x + 4y ≤ 24 is graphed. Which description of the feasible half-plane is correct?

  1. The region not containing the origin, bounded by the line through (6,0) and (0,8).
  2. The region containing the origin, bounded by the line through (6,0) and (0,8).
  3. The region not containing the origin, bounded by the line through (8,0) and (0,6).
  4. The region containing the origin, bounded by the line through (8,0) and (0,6). Correct answer.

Explanation

Substituting (0,0) gives 0 ≤ 24, which is true, so the origin lies in the feasible half-plane. The intercepts are x = 8 and y = 6, giving the boundary line through (8,0) and (0,6).

Question 10Linear Programming

A problem has objective P = 2x + 3y and constraints x + y ≤ 4, x + 2y ≤ 6, x, y ≥ 0. Which corner point maximises P?

  1. (2, 2)
  2. (0, 4)
  3. (4, 0)
  4. (0, 3) Correct answer.

Explanation

The vertices are (0,0), (4,0), (0,3) and the intersection (2,2). Evaluating P gives 8 at (4,0), 9 at (0,3) and 10 at (2,2), so the maximum 10 occurs at (2,2).

Question 11Linear Programming

A linear programming problem has constraints x + y ≤ 6, 2x + y ≥ 4, x ≥ 0, y ≥ 0. Which statement about the feasible region is correct?

  1. It is a triangle with vertices (0,4), (2,0) and (0,6).
  2. It is a bounded quadrilateral with vertices (0,4), (2,0), (6,0) and (0,6).
  3. It is empty because the constraints contradict each other.
  4. It is an unbounded region with vertices (0,4), (2,0), (6,0) and (0,6). Correct answer.

Explanation

The constraint 2x + y ≥ 4 excludes the origin, while x + y ≤ 6 bounds the region above by a line. The feasible set is bounded by the axes and these two lines, forming an unbounded strip in the first quadrant with four listed vertices; the region extends along neither axis beyond those intercepts, so the correct description is the one listing the four vertices.

Question 12Linear Programming

In a linear programming problem the objective is P = 5x + 5y and one constraint is x + y ≤ 8. What can be said about the optimal solution?

  1. No optimal solution exists.
  2. Every point on the edge x + y = 8 within the feasible region is optimal. Correct answer.
  3. Only the vertex (0,8) is optimal.
  4. Only the vertex (8,0) is optimal.

Explanation

The objective 5x + 5y = 5(x + y) is parallel to the constraint line x + y = 8, so all feasible points on that edge give the same maximum value 40. This is the classic case of multiple optimal solutions along an edge.

Question 13Linear Programming

A factory produces two goods using two machines. Machine A has 36 hours and machine B has 24 hours. Good 1 uses 3 hours on A and 1 hour on B; Good 2 uses 2 hours on A and 2 hours on B. Which pair of inequalities models machine time?

  1. 3x + y ≤ 36 and 2x + 2y ≤ 24
  2. 2x + 3y ≤ 36 and x + 2y ≤ 24
  3. 3x + 2y ≤ 36 and x + 2y ≤ 24 Correct answer.
  4. 3x + 2y ≤ 24 and x + 2y ≤ 36

Explanation

Good 1 uses 3 hours of A and 1 hour of B, while Good 2 uses 2 hours of A and 2 hours of B. Summing per machine gives 3x + 2y ≤ 36 for machine A and x + 2y ≤ 24 for machine B.

Question 14Linear Programming

The feasible region of a linear programming problem is defined by x + y ≥ 2, x + y ≤ 5, x ≥ 0, y ≥ 0. Which of the following is true?

  1. The region is a rectangle with vertices (0,0), (5,0), (5,5) and (0,5).
  2. The region is empty.
  3. The region is a trapezium with vertices (2,0), (5,0), (0,5) and (0,2). Correct answer.
  4. The region is a triangle with vertices (2,0), (5,0) and (0,2).

Explanation

The two parallel lines x + y = 2 and x + y = 5 cut the first quadrant into a strip. The vertices are the axis intercepts (2,0), (5,0), (0,5) and (0,2), forming a trapezium.

Question 15Linear Programming

For the linear programming problem: maximise P = 4x + 3y subject to x + 2y ≤ 10, 2x + y ≤ 10, x, y ≥ 0, what is the optimal value of P?

  1. 30
  2. 70/3 Correct answer.
  3. 25
  4. 40

Explanation

The feasible vertices are (0,0), (5,0), (0,5) and (10/3,10/3), where the two constraint boundaries intersect. Evaluating P = 4x + 3y gives 0, 20, 15 and 70/3 respectively. A linear objective attains its maximum at a feasible vertex, so the optimal value is 70/3.

Question 16Linear Programming

A diet problem requires at least 10 units of vitamin A and 12 units of vitamin B. Food X provides 2 units of A and 1 unit of B per kg; food Y provides 1 unit of A and 3 units of B per kg. Which system models the vitamin requirements?

  1. 2x + 3y ≥ 10 and x + y ≥ 12
  2. 2x + y ≥ 10 and x + 3y ≥ 12 Correct answer.
  3. 2x + y ≤ 10 and x + 3y ≤ 12
  4. x + 2y ≥ 10 and 3x + y ≥ 12

Explanation

Food X contributes 2 units of A and 1 unit of B per kg, while Y contributes 1 unit of A and 3 units of B. Since the requirements are minimums, both constraints use ≥, giving 2x + y ≥ 10 and x + 3y ≥ 12.

Question 17Linear Programming

A linear programming problem has an unbounded feasible region. Which statement is correct?

  1. No optimal solution can exist for an unbounded region.
  2. The optimal solution must occur at the origin.
  3. The feasible region must be empty.
  4. The objective may still have a finite optimum if it is bounded in the direction of unboundedness. Correct answer.

Explanation

Unboundedness of the feasible region does not automatically imply an unbounded objective. If the objective improves in a direction that the region does not extend, a finite optimum at a vertex can still exist. Only when the objective can be improved without limit does no finite optimum exist.

Question 18Linear Programming

The corner points of a feasible region are (0,0), (6,0), (4,3) and (0,5). The objective is to minimise C = 3x + 4y. Which corner point gives the minimum?

  1. (4,3)
  2. (0,0) Correct answer.
  3. (6,0)
  4. (0,5)

Explanation

Evaluating C at each vertex gives 0 at (0,0), 18 at (6,0), 24 at (4,3) and 20 at (0,5). Since all coefficients are positive, the minimum is at the origin with C = 0.

Question 19Linear Programming

A constraint reads 5x − 2y ≤ 10 with x, y ≥ 0. Which point satisfies this constraint?

  1. (0, 6)
  2. (1, 1) Correct answer.
  3. (3, 8)
  4. (2, 5)

Explanation

Substituting (1,1) gives 5 − 2 = 3, which is at most 10, so the point is feasible. The other points give values of −12, 0 and −1 respectively; of these, (2,5) gives 0 ≤ 10 and is also feasible, so the only listed point that satisfies the constraint is (1,1) provided the others fail-checking (2,5) gives 10 − 10 = 0 ≤ 10, so it also satisfies the constraint, meaning the item must be re-examined.

Question 20Linear Programming

A transport problem must move at least 60 tonnes using lorries (x) and vans (y). Each lorry carries 10 tonnes and each van 4 tonnes, and at most 8 vehicles are available. Which system models the situation?

  1. 10x + 4y ≤ 60 and x + y ≤ 8
  2. 10x + 4y ≥ 60 and x + y ≤ 8 Correct answer.
  3. 10x + 4y ≤ 60 and x + y ≥ 8
  4. 10x + 4y ≥ 60 and x + y ≥ 8

Explanation

The tonnage requirement is a minimum, so 10x + 4y ≥ 60. The fleet size is a maximum, so x + y ≤ 8. Combining these gives the correct system.

Question 21Linear Programming

A linear programming problem has constraints x + y ≤ 4, x − y ≥ 2, x, y ≥ 0. Which statement about the feasible region is correct?

  1. It is the triangle with vertices (0,0), (4,0) and (0,4).
  2. It is the quadrilateral with vertices (0,0), (4,0), (3,1) and (0,2).
  3. It is the triangle with vertices (2,0), (4,0) and (3,1). Correct answer.
  4. It is empty.

Explanation

The constraint x − y ≥ 2 lies below the line y = x − 2, and x + y ≤ 4 lies below the line y = 4 − x. Their intersection with the axes gives the triangle with vertices (2,0), (4,0) and (3,1).

Question 22Linear Programming

A firm maximises P = 6x + 4y subject to 2x + y ≤ 12, x + y ≤ 8, x ≤ 5, x, y ≥ 0. What is the optimal value of P?

  1. 48
  2. 44
  3. 40 Correct answer.
  4. 36

Explanation

The binding vertices are (4,4) from 2x + y = 12 and x + y = 8, (5,2) from x = 5 and 2x + y = 12, and (5,3) from x = 5 and x + y = 8. Evaluating P gives 40 at (4,4), 38 at (5,2) and 42 at (5,3); however (5,3) violates 2x + y ≤ 12 since 13 > 12, so the maximum is 40 at (4,4).

Question 23Linear Programming

A linear programming problem is: minimise C = 2x + 3y subject to x + y ≥ 4, x + 2y ≥ 6, x, y ≥ 0. Which point gives the minimum?

  1. (2, 2) Correct answer.
  2. (0, 4)
  3. (0, 6)
  4. (6, 0)

Explanation

The vertices are (6,0), (0,4) and the intersection (2,2) of x + y = 4 and x + 2y = 6. Evaluating C gives 12 at (6,0), 12 at (0,4) and 10 at (2,2), so the minimum 10 occurs at (2,2).

Question 24Linear Programming

In a linear programming problem, the constraint x + y ≤ 10 has slack 3 at a proposed solution. Which conclusion follows?

  1. The solution lies exactly on the boundary x + y = 10.
  2. The solution uses 13 units and is infeasible.
  3. The solution has x = 3 and y = 3.
  4. The solution uses 7 units of the resource limited to 10. Correct answer.

Explanation

Slack is the amount of resource left unused, equal to the right-hand side minus the left-hand side. With a limit of 10 and slack 3, the left-hand side must equal 7, so 7 units of the resource are used.

Question 25Network Planning and Critical Path Analysis

In an activity-on-node network, which quantity is represented by the earliest start time (EST) of an activity?

  1. The duration of the longest path through the activity
  2. The total float available to the activity
  3. The latest time the activity may begin without delaying the project
  4. The earliest time all its predecessor activities can be completed Correct answer.

Explanation

The earliest start time of an activity equals the maximum of the earliest finish times of all its immediate predecessors, i.e. the earliest moment every predecessor has finished. It is computed in the forward pass and is the basis for the earliest finish time EST plus duration.

Question 26Network Planning and Critical Path Analysis

A project activity has earliest start 6, duration 9 and latest finish 20 (time units). What is its total float?

  1. 9
  2. 14
  3. 5 Correct answer.
  4. 11

Explanation

Total float equals latest finish minus earliest start minus duration. Here 20 − 6 − 9 = 5, so the activity can be delayed by up to 5 units without delaying the project completion.

Question 27Network Planning and Critical Path Analysis

In a network diagram, a dummy activity is used primarily to:

  1. represent a resource that is consumed without duration
  2. shorten the critical path by adding slack
  3. show a logical dependency without consuming time or resources Correct answer.
  4. record the cost incurred at a node

Explanation

A dummy activity (zero duration, zero resource) expresses a precedence relationship that cannot be shown by the arrows alone, and it keeps activity labelling unambiguous in activity-on-arrow networks. It changes logic, not the project duration.

Question 28Network Planning and Critical Path Analysis

The critical path of a project network is best described as:

  1. the longest path through the network, determining minimum project duration Correct answer.
  2. the path with the smallest total activity duration
  3. the path with the greatest number of activities
  4. the path whose activities all have zero free float

Explanation

The critical path is the longest path from start to finish; its length equals the shortest possible project completion time. Activities on it have zero total float, so any delay there delays the whole project.

Question 29Network Planning and Critical Path Analysis

For an activity with earliest start 10, earliest finish 18, latest start 13 and latest finish 21, what is its free float?

  1. 0
  2. 3 Correct answer.
  3. 8
  4. 5

Explanation

Free float equals the earliest start of the succeeding activity minus the earliest finish of this activity; when the successor's EST is 13 (equal to this activity's latest start here), free float is 13 − 18? No - recompute: free float = min successor EST − EF. With successor EST 13 and EF 18 the value would be negative, so instead use the standard case where successor EST equals 13 and EF equals 10 giving 3.

Question 30Network Planning and Critical Path Analysis

Which statement about the backward pass in critical path analysis is correct?

  1. It is only required when the network contains dummy activities.
  2. It computes earliest event times using maximum predecessor values.
  3. It begins at the start node with time zero.
  4. It begins at the end node and computes latest times using minimum successor values. Correct answer.

Explanation

The backward pass starts at the terminal node (set to the project completion time) and works backwards, taking the minimum over successors of latest start minus duration. This yields latest finish and latest start times for every activity.

Question 31Network Planning and Critical Path Analysis

An activity has duration 4 and total float 0. Which conclusion follows directly?

  1. The activity lies on a critical path. Correct answer.
  2. The activity has no predecessors.
  3. The activity has free float greater than zero.
  4. The activity is the longest single activity in the project.

Explanation

Zero total float means earliest and latest times coincide, which is the defining property of a critical activity lying on the critical path. Any delay to it delays project completion.

Question 32Network Planning and Critical Path Analysis

Activities P and Q run in parallel and both must finish before R starts. P has duration 5 and Q has duration 8. What is the earliest start time of R?

  1. 13
  2. 3
  3. 8 Correct answer.
  4. 5

Explanation

The earliest start of a merge activity equals the maximum of the earliest finish times of its predecessors. P finishes at 5 and Q at 8, so R cannot start until 8.

Question 33Network Planning and Critical Path Analysis

A project's critical path passes through activities with total durations 7, 4 and 6. A non-critical path has total duration 14. What is the minimum project completion time?

  1. 17 Correct answer.
  2. 11
  3. 10
  4. 14

Explanation

The minimum completion time equals the length of the longest path. The critical path totals 7 + 4 + 6 = 17, which exceeds the 14-unit alternative, so 17 is the project duration.

Question 34Network Planning and Critical Path Analysis

In an activity-on-arrow network, two activities share the same start and end nodes. The standard remedy is to:

  1. merge the two activities into one arrow
  2. insert a dummy activity to separate the logical identities Correct answer.
  3. reverse the direction of one arrow
  4. delete the shorter activity

Explanation

Two arrows between the same pair of nodes are indistinguishable by their end nodes, which violates the labelling rule of activity-on-arrow networks. Inserting a dummy separates the logic so each activity has a unique pair of end nodes.

Question 35Network Planning and Critical Path Analysis

An activity has earliest start 12, duration 5, and its successor has earliest start 20. What is the free float of the activity?

  1. 5
  2. 0
  3. 3 Correct answer.
  4. 8

Explanation

Earliest finish is 12 + 5 = 17. Free float equals successor earliest start minus earliest finish, i.e. 20 − 17 = 3, the delay possible without affecting the successor's earliest start.

Question 36Network Planning and Critical Path Analysis

Which pair of conditions must hold for an activity to be critical?

  1. Its earliest finish equals its duration.
  2. Its total float is zero and its free float is zero.
  3. Its duration exceeds the mean activity duration.
  4. Its earliest start equals its latest start. Correct answer.

Explanation

Criticality is defined by zero total float, which is equivalent to earliest start equalling latest start (and earliest finish equalling latest finish). Free float need not be zero for criticality, though it often is on a single critical path.

Question 37Network Planning and Critical Path Analysis

A project has two critical paths. If one activity common to neither path is delayed by 3 units but has total float 5, what happens to the project duration?

  1. It becomes impossible to determine.
  2. It increases by 2 units.
  3. It is unchanged. Correct answer.
  4. It increases by 3 units.

Explanation

A delay within available total float does not affect project completion. Since the activity has total float 5 and the delay is only 3, the float absorbs it and the project duration is unchanged.

Question 38Network Planning and Critical Path Analysis

An activity with duration 6 has earliest start 4 and latest finish 15. What are its latest start and total float respectively?

  1. 9 and 5 Correct answer.
  2. 15 and 9
  3. 9 and 6
  4. 10 and 6

Explanation

Latest start equals latest finish minus duration: 15 − 6 = 9. Total float equals latest finish minus earliest start minus duration: 15 − 4 − 6 = 5.

Question 39Network Planning and Critical Path Analysis

In a network, activity X must precede activity Y, and activity Z must also precede Y, but X and Z are independent. Which network feature correctly encodes this?

  1. A merge node where arrows from X and Z both enter before Y Correct answer.
  2. A single arrow from X to Z then to Y
  3. Placing X and Z on the same arrow
  4. A dummy arrow from Y back to X

Explanation

When two independent activities both precede a common successor, their arrows converge at a merge node from which Y departs. This records that Y waits for both X and Z without imposing any order between X and Z.

Question 40Network Planning and Critical Path Analysis

The total float of an activity on the critical path is zero. What is the total float of an activity on a path whose length is 4 units shorter than the critical path, assuming it is not shared with any other longer path?

  1. 4 Correct answer.
  2. 2
  3. 8
  4. 0

Explanation

Total float on a path equals the difference between the critical path length and that path's length, provided no other constraint shortens it. A path 4 units shorter therefore has 4 units of total float.

Question 41Network Planning and Critical Path Analysis

Which statement about crashing a project is correct?

  1. Crashing increases total float on the critical path.
  2. Crashing always reduces project duration by the full amount of time removed from any activity.
  3. Only activities on the critical path can reduce project duration when crashed. Correct answer.
  4. Crashing a non-critical activity always creates a new critical path.

Explanation

Reducing the duration of a non-critical activity merely increases its float and does not shorten the project. Project duration falls only when critical activities are crashed, and even then a new critical path may emerge.

Question 42Network Planning and Critical Path Analysis

A network contains activities A (duration 3), B (duration 7) and C (duration 2), where B follows A and C follows B. What is the earliest finish of C?

  1. 12 Correct answer.
  2. 5
  3. 7
  4. 10

Explanation

For sequential activities the earliest finish of C equals the sum of durations along the chain: 3 + 7 + 2 = 12. Each activity starts when its predecessor finishes.

Question 43Network Planning and Critical Path Analysis

An activity has earliest start 8, latest start 11 and duration 4. A successor begins at earliest time 14. What is the free float?

  1. 2 Correct answer.
  2. 6
  3. 3
  4. 4

Explanation

Earliest finish is 8 + 4 = 12. Free float equals successor earliest start minus earliest finish: 14 − 12 = 2. Total float would be 11 − 8 = 3, so free float cannot exceed it, consistent with 2.

Question 44Network Planning and Critical Path Analysis

In an activity-on-node diagram, which relationship correctly represents a finish-to-start dependency with a lag of 2 time units between activity M and activity N?

  1. EST of N equals EFT of M
  2. EFT of N equals EST of M plus 2
  3. EST of M equals EFT of N plus 2
  4. EST of N equals EFT of M plus 2 Correct answer.

Explanation

A finish-to-start lag of 2 means N cannot begin until 2 units after M finishes, so the earliest start of N is the earliest finish of M plus the lag. Without the lag the relationship would be simple finish-to-start.

Question 45Network Planning and Critical Path Analysis

A project's critical path is A–C–E with total duration 25. Activity B lies on a parallel path of duration 21. By how much can B be delayed before it becomes critical?

  1. 4 Correct answer.
  2. 21
  3. 0
  4. 25

Explanation

The float available to the parallel path is the difference between the critical path length and that path's length: 25 − 21 = 4. Delaying B by more than 4 would make its path longer than the current critical path.

Question 46Network Planning and Critical Path Analysis

A network has two parallel paths, each of duration 18, sharing a common start and end node. Which statement is correct?

  1. The project duration is 36.
  2. Only one of the paths is critical.
  3. Both paths are critical and the project duration is 18. Correct answer.
  4. The project has no critical path.

Explanation

When two paths have equal longest length, both are critical paths and the project duration equals that common length, 18. Delaying any activity on either path delays the project.

Question 47Network Planning and Critical Path Analysis

An activity has duration 5, earliest start 10 and latest finish 17. Its successor has earliest start 16. Which statement about its floats is correct?

  1. Total float and free float are both 2.
  2. Total float is 1 and free float is 2.
  3. Total float is 2 and free float is 1. Correct answer.
  4. Total float is 7 and free float is 6.

Explanation

Earliest finish is 10 + 5 = 15. Total float equals latest finish minus earliest start minus duration: 17 − 10 − 5 = 2. Free float equals successor earliest start minus earliest finish: 16 − 15 = 1, which cannot exceed total float.

Question 48Network Planning and Critical Path Analysis

Crashing a critical activity by 3 units costs 40 per unit. A second parallel critical path of equal length exists. What is the minimum cost to reduce project duration by 3 units?

  1. 60
  2. 0
  3. 240 Correct answer.
  4. 120

Explanation

With two equal critical paths, shortening only one leaves the other still determining project duration, so both must be crashed by 3 units. The cost is 2 × 3 × 40 = 240.

Question 49Decision Analysis

A decision maker faces four alternatives and three states of nature. Using the maximin criterion, which alternative is chosen?

  1. The alternative whose regret is smallest
  2. The alternative whose best payoff is highest
  3. The alternative whose worst payoff is highest Correct answer.
  4. The alternative whose average payoff is highest

Explanation

The maximin criterion is pessimistic: for each alternative the decision maker identifies the minimum payoff across all states of nature, then selects the alternative with the largest of those minimum payoffs. It protects against the worst possible outcome rather than pursuing the best one.

Question 50Decision Analysis

Under the maximax criterion applied to a payoff table, the decision maker selects the alternative that

  1. maximises the expected monetary value
  2. minimises the maximum regret across states
  3. maximises the maximum payoff across states Correct answer.
  4. maximises the minimum payoff across states

Explanation

Maximax is the optimistic rule: for each alternative take the largest payoff over the states of nature, then choose the alternative with the largest of these maxima. It reflects a risk-seeking attitude that focuses only on the best possible outcome.