6.2 Alternate Day Working, Negative Work & Wage Distribution Based on Work Done

The Alternate Day Working Framework

In alternate-day working problems, workers do not operate simultaneously; instead, they take turns in a prescribed sequential rotation (e.g., Worker $A$ on Day 1, Worker $B$ on Day 2, Worker $A$ on Day 3, and so on).

The Multi-Step Cycle Algorithm

1. Establish Total Work Units ($W$): Compute $W = \text{LCM}(t_A, t_B, \dots)$.
2. Determine Individual Daily Efficiencies: Compute $E_A = W/t_A, E_B = W/t_B$, etc.
3. Define the Base Cycle Duration & Net Cycle Output ($W_{\text{cycle}}$):
   • For two workers ($A$ and $B$), the elementary cycle length is 2 days.
   • Work delivered in one complete 2-day cycle: $W_{\text{cycle}} = E_A + E_B$.
   • For three workers ($A, B, C$), the cycle length is 3 days, with $W_{\text{cycle}} = E_A + E_B + E_C$.
4. Calculate Maximum Completed Integer Cycles ($k$):

$$k = \left\lfloor \frac{W}{W_{\text{cycle}}} \right\rfloor$$

• Total days elapsed during complete cycles $= k \times (\text{cycle length})$.
• Work completed during these cycles $= k \times W_{\text{cycle}}$.

1. Calculate Remaining Work ($R$):

$$R = W - (k \times W_{\text{cycle}})$$

1. Step Through Remaining Units Sequentially:
   • Who begins the next cycle? If $A$ starts, evaluate whether $R \le E_A$.
   • If $R \le E_A$, worker $A$ completes the remaining work in a fractional day: $\Delta t = \frac{R}{E_A}$ days.
   • If $R > E_A$, worker $A$ works 1 full day, leaving $(R - E_A)$ units. Worker $B$ then works on the subsequent day to complete the balance in $\frac{R - E_A}{E_B}$ days.

+-----------------------------------------------------------------------------+
|                     ALTERNATE WORKING CYCLE ALGORITHM                       |
+-----------------------------------------------------------------------------+
| Total Work W = LCM(t_A, t_B)                                                |
| Net Output per 2-Day Cycle: W_cycle = E_A + E_B                             |
|                               │                                             |
|                               ▼                                             |
| Compute Full Cycles: k = floor(W / W_cycle)                                 |
| Days Elapsed = k * 2  |  Work Done = k * W_cycle                            |
|                               │                                             |
|                               ▼                                             |
| Remaining Work R = W - (k * W_cycle)                                        |
|                               │                                             |
|          ┌────────────────────┴────────────────────┐                        |
|          ▼ (If R <= E_A)                           ▼ (If R > E_A)           |
| A finishes in R / E_A days             A works 1 full day (R' = R - E_A)    |
| Total Time = (2k) + (R / E_A)          B finishes in R' / E_B days          |
|                                        Total Time = (2k + 1) + (R' / E_B)   |
+-----------------------------------------------------------------------------+

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Negative Work (Builders vs. Destroyers & The Peak Threshold Trap)

Negative work arises when one entity adds to progress (e.g., track construction, embankment filling, masonry wall erection) while another entity actively subtracts from it (e.g., soil erosion, water seepage, brick demolition).

Let Builder $A$ complete work in $x$ days ($E_A = +W/x$) and Destroyer $B$ undo the entire work in $y$ days ($E_B = W/y$ is the positive magnitude of removal). Assume the builder cannot finish in its first turn and $x<y$, so complete two-day cycles make positive progress. The signed destruction contribution is $-E_B$, giving net output $E_A-E_B$.

The Net Cycle Rate

In an alternating schedule where Builder $A$ works on Day 1 and Destroyer $B$ acts on Day 2:

$$\text{Net Progress per 2-Day Cycle} = E_A - E_B > 0$$

The Rigorous Peak Threshold Protocol

1. Reserve the Builder's Daily Output:

$$\text{Threshold Target} = W - E_A$$

1. Compute Integer Cycles Required to Reach the Threshold: Find the smallest integer number of cycles $k$ such that:

$$k \times (E_A - E_B) \ge W - E_A$$

1. Evaluate Cumulative Work at the End of Cycle $k$:

$$W_k = k \times (E_A - E_B)$$

Days elapsed $= 2k$. 4. Complete the Final Quantum on Day $(2k + 1)$: The remaining work is $R = W - W_k$. Because $W_k \ge W - E_A$, we are guaranteed that $R \le E_A$.

$$\text{Fractional Day Required by Builder} = \frac{R}{E_A}$$ $$\text{Total Days} = 2k + \frac{R}{E_A}$$

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Leaving and Joining Mid-Project

Real-world railway contracts frequently experience mid-project labour fluctuations. These scenarios divide into two mathematical structures:

+-----------------------------------------------------------------------------+
|                        MID-PROJECT WORK TURNOVER                            |
+-----------------------------------------------------------------------------+
| FORWARD DEPARTURE (Leaves x days AFTER START):                              |
| Work Done = (E_A + E_B) * x                                                 |
| Balance Work = W - [(E_A + E_B) * x]                                        |
| Remaining Days = Balance Work / E_B                                         |
+-----------------------------------------------------------------------------+
| BACKWARD DEPARTURE (Leaves x days BEFORE COMPLETION):                       |
| Method 1: Total Days T -> (T - x) * E_A + T * E_B = W                       |
| Method 2: VIRTUAL WORK ADDITION SHORTCUT                                    |
| Credit A with 'phantom' output: W_virtual = W + (x * E_A)                   |
| All workers now work full duration T: T = W_virtual / (E_A + E_B)           |
+-----------------------------------------------------------------------------+

The Virtual Work Addition Shortcut

When a worker leaves $x$ days before completion, the standard algebraic approach sets total time as $T$. A faster mental shortcut is the Virtual Work Method:

1. If worker $A$ had not left early, they would have produced an additional $x \times E_A$ units of work.
2. Add this virtual work to the actual target: $W_{\text{virtual}} = W + (x \cdot E_A)$.
3. Now, pretend that all workers remained on site for the entire project duration $T$.
4. Solve directly in one step:

$$T = \frac{W + (x \cdot E_A)}{E_{\text{total}}}$$

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Wage Distribution Principles Based on Work Done

In the exercises here, the contract explicitly pays in proportion to work delivered. This is a stipulated arithmetic sharing rule. It is not a statement about statutory wages, time-based pay or actual railway employment terms.

The Work-Share Wage Formula

If workers $A, B,$ and $C$ work for durations $t_A, t_B,$ and $t_C$ with daily efficiencies $E_A, E_B,$ and $E_C$:

$$\text{Work Delivered by Individual } i = W_i = E_i \times t_i$$ $$\text{Ratio of Wages} = (E_A \cdot t_A) : (E_B \cdot t_B) : (E_C \cdot t_C)$$ $$\text{Individual Wage Share} = \left( \frac{W_i}{W_{\text{total}}} \right) \times \text{Total Contract Amount}$$

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Contractor Estimates & Deadline Remediation

A fictional project may ask how many equally productive workers are needed to meet a fixed deadline. Assume unchanged hours, productivity and work specification; round a noninteger required workforce upward when only whole workers can be assigned.

Workforce Reinforcement Formula

Suppose a contractor agrees to finish a project in $D$ days with an initial workforce of $M_1$ workers. After $d_1$ days, an inspection reveals that only a fraction $W_1$ of the total work is complete:

• Remaining calendar days: $d_2 = D - d_1$.
• Remaining work: $W_2 = W_{\text{total}} - W_1$.
• Required total workforce $M_2$ to finish precisely on time:

$$\frac{M_1 \cdot d_1}{W_1} = \frac{M_2 \cdot d_2}{W_2} \implies M_2 = \frac{M_1 \cdot d_1 \cdot W_2}{W_1 \cdot d_2}$$ $$\text{Additional Workers Required } (\Delta M) = M_2 - M_1$$

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Worked Examples

Worked Example 1: Alternate Day Track Relaying with Fractional Completion

Problem: Gang A can relay a standard station crossover in 12 days, while Gang B can relay the same crossover in 18 days. Due to single-line traffic constraints, they work on alternate days, with Gang A working on Day 1, Gang B on Day 2, Gang A on Day 3, and so on. In exactly how many days will the crossover relaying be completed?

• Step 1: Compute Total Work Units and Efficiencies:

$$\text{Total Work } W = \text{LCM}(12, 18) = 36 \text{ units}$$ $$E_A = \frac{36}{12} = 3 \text{ units/day}, \quad E_B = \frac{36}{18} = 2 \text{ units/day}$$

• Step 2: Determine 2-Day Cycle Output:

$$\text{Work per 2-day cycle } = E_A + E_B = 3 + 2 = 5 \text{ units}$$

• Step 3: Calculate Complete Integer Cycles:

$$k = \left\lfloor \frac{36}{5} \right\rfloor = 7 \text{ cycles}$$

• Elapsed time for 7 complete cycles $= 7 \times 2 = 14 \text{ days}$.
• Cumulative work completed $= 7 \times 5 = 35 \text{ units}$.
• Step 4: Evaluate Remaining Work:

$$R = 36 - 35 = 1 \text{ unit}$$

• Step 5: Determine Fractional Time for Gang A on Day 15: It is Gang A's turn to work. Gang A delivers 3 units per full day.

$$\Delta t = \frac{R}{E_A} = \frac{1 \text{ unit}}{3 \text{ units/day}} = \frac{1}{3} \text{ day}$$

• Conclusion: Total time taken is $14 + \frac{1}{3} = 14\frac{1}{3} \text{ days}$ (one third of a working day beyond 14 days; converting to hours requires the stated working-day length).

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Worked Example 2: Negative Work (Embankment Construction vs. Monsoon Erosion)

Problem: A railway civil construction gang can construct a protective earth embankment along a flood-prone line in 20 days. However, heavy tidal backflow and monsoon erosion can wash away an entire completed embankment in 30 days. If the construction gang works on Day 1, tidal erosion acts on Day 2, and they alternate continuously, in how many days will the embankment be completed for the first time?

• Step 1: Establish Units and Daily Rates:

$$\text{Total Work } W = \text{LCM}(20, 30) = 60 \text{ units}$$ $$E_{\text{gang}} = +\frac{60}{20} = +3 \text{ units/day}, \quad E_{\text{erosion}} = -\frac{60}{30} = -2 \text{ units/day}$$

• Step 2: Determine Net Progress per 2-Day Cycle:

$$\text{Net Cycle Progress} = (+3) + (-2) = +1 \text{ unit every 2 days}$$

• Step 3: Apply the Peak Threshold Rule: Reserve the builder gang's full daily progress from the total target:

$$\text{Threshold Target} = W - E_{\text{gang}} = 60 - 3 = 57 \text{ units}$$

• Step 4: Compute Cycles Required to Reach Threshold:

$$\text{Cycles Required } k = \frac{57 \text{ units}}{1 \text{ unit/cycle}} = 57 \text{ cycles}$$

• Elapsed days for 57 complete cycles $= 57 \times 2 = 114 \text{ days}$.
• Cumulative embankment constructed $= 57 \times 1 = 57 \text{ units}$.
• Step 5: Final Advance on Day 115: On Day 115, the construction gang takes its turn:

$$\text{Output delivered on Day 115} = +3 \text{ units}$$ $$\text{Cumulative Work} = 57 + 3 = 60 \text{ units (100\% COMPLETE!)}$$

Because the target of 60 units is reached at the end of Day 115, the embankment is finished and handed over. Erosion does not act.

• Conclusion: The embankment is completed in exactly 115 days.

(Note: Falling into the trap of $(60 / 1) \times 2 = 120 \text{ days}$ overestimates the time by 5 full days!)

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Worked Example 3: Wage Distribution with Early Departures

Problem: Three railway signalling technicians—A, B, and C—are contracted to overhaul an interlocking relay panel for a lump-sum contract of ₹12,000. Working alone, A can finish in 12 days, B in 15 days, and C in 20 days. All three begin work together. After 3 days, technician A leaves the site. Technicians B and C continue together, but technician B leaves 1 day before the work is finally completed. Technician C finishes the remaining balance alone. Assuming payment proportional to work delivered, calculate each technician’s share.

• Step 1: Determine Total Work Units and Individual Efficiencies:

$$\text{Total Work } W = \text{LCM}(12, 15, 20) = 60 \text{ units}$$ $$E_A = \frac{60}{12} = 5 \text{ units/day}, \quad E_B = \frac{60}{15} = 4 \text{ units/day}, \quad E_C = \frac{60}{20} = 3 \text{ units/day}$$

• Step 2: Track Progress Across Phases:
  • Phase 1 (First 3 days - A, B, and C together):

$$\text{Work done} = 3 \times (E_A + E_B + E_C) = 3 \times (5 + 4 + 3) = 3 \times 12 = 36 \text{ units}$$

Remaining work $= 60 - 36 = 24 \text{ units}$. Technician A departs.

• Phase 3 (Final 1 day - C works alone): Technician B leaves 1 day before completion, meaning C works the last day alone:

$$\text{Work done by C} = 1 \times E_C = 1 \times 3 = 3 \text{ units}$$

• Phase 2 (Intermediate days - B and C together):

$$\text{Work completed in Phase 2} = 24 - 3 = 21 \text{ units}$$ $$\text{Duration of Phase 2} = \frac{21}{E_B + E_C} = \frac{21}{4 + 3} = \frac{21}{7} = 3 \text{ days}$$

• Step 3: Total Days and Work Delivered by Each Technician:
  • Technician A: Worked 3 days in Phase 1.

$$W_A = 3 \times 5 = 15 \text{ units}$$

• Technician B: Worked in Phase 1 (3 days) + Phase 2 (3 days) $= 6 \text{ days}$.

$$W_B = 6 \times 4 = 24 \text{ units}$$

• Technician C: Worked in Phase 1 (3 days) + Phase 2 (3 days) + Phase 3 (1 day) $= 7 \text{ days}$.

$$W_C = 7 \times 3 = 21 \text{ units}$$

(Check: $15 + 24 + 21 = 60 \text{ units}$.)

• Step 4: Distribute Contract Wages Proportional to Work Delivered:

$$\text{Wage per unit of work} = \frac{\text{INR } 12,000}{60 \text{ units}} = \text{INR } 200 \text{ per unit}$$ $$\text{Share of A} = 15 \times \text{INR } 200 = \text{INR } 3,000$$ $$\text{Share of B} = 24 \times \text{INR } 200 = \text{INR } 4,800$$ $$\text{Share of C} = 21 \times \text{INR } 200 = \text{INR } 4,200$$

(Check: $\text{INR } 3,000 + \text{INR } 4,800 + \text{INR } 4,200 = \text{INR } 12,000$.)

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Calculation Traps & Tactical Shortcuts

1. The Over-Cycle Trap in Alternating Work: In problems where remaining work $R$ is smaller than the next worker's single-day capacity, candidates frequently award an entire integer day rather than calculating the fraction $\frac{R}{E}$. If $R = 2$ units and the worker does $5$ units/day, the added time is strictly $+0.4$ days (not $+1$ day).
2. The Builder-Destroyer Fallacy: Dividing total capacity directly by net cycle yield in negative work problems treats destruction as inevitable even after the target is met. Always deduct the builder's positive stride before computing cycle multiples.
3. Nominal Day Wage Splitting: When allocating contract funds, dividing money by total days present (e.g., $3 : 6 : 7$) instead of work units produced ($15 : 24 : 21$) completely ignores varying skill levels and produces incorrect answers.

The peak-threshold formula assumes constant rates and positive net progress per complete cycle. If the builder can finish within its first turn, stop then; no complete cycle is needed. If the builder cannot finish in one turn and every complete cycle loses all its progress, the target is never reached under the stated repeating schedule.